3.32 \(\int (d+e x)^2 (a+b \tanh ^{-1}(c x^3)) \, dx\)

Optimal. Leaf size=336 \[ \frac {(d+e x)^3 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{3 e}+\frac {b d^2 \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {\sqrt {3} b d^2 \tan ^{-1}\left (\frac {2 c^{2/3} x^2+1}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}-\frac {b d^2 \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{4 \sqrt [3]{c}}+\frac {b d e \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{4 c^{2/3}}-\frac {b d e \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{4 c^{2/3}}-\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2 c^{2/3}}-\frac {b d e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{c^{2/3}}+\frac {b \left (c d^3+e^3\right ) \log \left (1-c x^3\right )}{6 c e}-\frac {b \left (c d^3-e^3\right ) \log \left (c x^3+1\right )}{6 c e} \]

[Out]

-b*d*e*arctanh(c^(1/3)*x)/c^(2/3)+1/3*(e*x+d)^3*(a+b*arctanh(c*x^3))/e+1/2*b*d^2*ln(1-c^(2/3)*x^2)/c^(1/3)+1/4
*b*d*e*ln(1-c^(1/3)*x+c^(2/3)*x^2)/c^(2/3)-1/4*b*d*e*ln(1+c^(1/3)*x+c^(2/3)*x^2)/c^(2/3)+1/6*b*(c*d^3+e^3)*ln(
-c*x^3+1)/c/e-1/6*b*(c*d^3-e^3)*ln(c*x^3+1)/c/e-1/4*b*d^2*ln(1+c^(2/3)*x^2+c^(4/3)*x^4)/c^(1/3)+1/2*b*d*e*arct
an(-1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(1/2)/c^(2/3)+1/2*b*d*e*arctan(1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(
1/2)/c^(2/3)+1/2*b*d^2*arctan(1/3*(1+2*c^(2/3)*x^2)*3^(1/2))*3^(1/2)/c^(1/3)

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Rubi [A]  time = 0.51, antiderivative size = 332, normalized size of antiderivative = 0.99, number of steps used = 25, number of rules used = 14, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {6742, 6091, 275, 292, 31, 634, 617, 204, 628, 6097, 296, 618, 206, 260} \[ \frac {a (d+e x)^3}{3 e}+\frac {b d^2 \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac {b d^2 \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{4 \sqrt [3]{c}}+\frac {\sqrt {3} b d^2 \tan ^{-1}\left (\frac {2 c^{2/3} x^2+1}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}+\frac {b d e \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{4 c^{2/3}}-\frac {b d e \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{4 c^{2/3}}-\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{2 c^{2/3}}-\frac {b d e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{c^{2/3}}+\frac {b e^2 \log \left (1-c^2 x^6\right )}{6 c}+b d^2 x \tanh ^{-1}\left (c x^3\right )+b d e x^2 \tanh ^{-1}\left (c x^3\right )+\frac {1}{3} b e^2 x^3 \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*(a + b*ArcTanh[c*x^3]),x]

[Out]

(a*(d + e*x)^3)/(3*e) - (Sqrt[3]*b*d*e*ArcTan[1/Sqrt[3] - (2*c^(1/3)*x)/Sqrt[3]])/(2*c^(2/3)) + (Sqrt[3]*b*d*e
*ArcTan[1/Sqrt[3] + (2*c^(1/3)*x)/Sqrt[3]])/(2*c^(2/3)) + (Sqrt[3]*b*d^2*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/
(2*c^(1/3)) - (b*d*e*ArcTanh[c^(1/3)*x])/c^(2/3) + b*d^2*x*ArcTanh[c*x^3] + b*d*e*x^2*ArcTanh[c*x^3] + (b*e^2*
x^3*ArcTanh[c*x^3])/3 + (b*d^2*Log[1 - c^(2/3)*x^2])/(2*c^(1/3)) + (b*d*e*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(4
*c^(2/3)) - (b*d*e*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(4*c^(2/3)) - (b*d^2*Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4])/
(4*c^(1/3)) + (b*e^2*Log[1 - c^2*x^6])/(6*c)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 6091

Int[ArcTanh[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTanh[c*x^n], x] - Dist[c*n, Int[x^n/(1 - c^2*x^(2*n)), x]
, x] /; FreeQ[{c, n}, x]

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int (d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\int \left (a (d+e x)^2+b (d+e x)^2 \tanh ^{-1}\left (c x^3\right )\right ) \, dx\\ &=\frac {a (d+e x)^3}{3 e}+b \int (d+e x)^2 \tanh ^{-1}\left (c x^3\right ) \, dx\\ &=\frac {a (d+e x)^3}{3 e}+b \int \left (d^2 \tanh ^{-1}\left (c x^3\right )+2 d e x \tanh ^{-1}\left (c x^3\right )+e^2 x^2 \tanh ^{-1}\left (c x^3\right )\right ) \, dx\\ &=\frac {a (d+e x)^3}{3 e}+\left (b d^2\right ) \int \tanh ^{-1}\left (c x^3\right ) \, dx+(2 b d e) \int x \tanh ^{-1}\left (c x^3\right ) \, dx+\left (b e^2\right ) \int x^2 \tanh ^{-1}\left (c x^3\right ) \, dx\\ &=\frac {a (d+e x)^3}{3 e}+b d^2 x \tanh ^{-1}\left (c x^3\right )+b d e x^2 \tanh ^{-1}\left (c x^3\right )+\frac {1}{3} b e^2 x^3 \tanh ^{-1}\left (c x^3\right )-\left (3 b c d^2\right ) \int \frac {x^3}{1-c^2 x^6} \, dx-(3 b c d e) \int \frac {x^4}{1-c^2 x^6} \, dx-\left (b c e^2\right ) \int \frac {x^5}{1-c^2 x^6} \, dx\\ &=\frac {a (d+e x)^3}{3 e}+b d^2 x \tanh ^{-1}\left (c x^3\right )+b d e x^2 \tanh ^{-1}\left (c x^3\right )+\frac {1}{3} b e^2 x^3 \tanh ^{-1}\left (c x^3\right )+\frac {b e^2 \log \left (1-c^2 x^6\right )}{6 c}-\frac {1}{2} \left (3 b c d^2\right ) \operatorname {Subst}\left (\int \frac {x}{1-c^2 x^3} \, dx,x,x^2\right )-\frac {(b d e) \int \frac {1}{1-c^{2/3} x^2} \, dx}{\sqrt [3]{c}}-\frac {(b d e) \int \frac {-\frac {1}{2}-\frac {\sqrt [3]{c} x}{2}}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{\sqrt [3]{c}}-\frac {(b d e) \int \frac {-\frac {1}{2}+\frac {\sqrt [3]{c} x}{2}}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{\sqrt [3]{c}}\\ &=\frac {a (d+e x)^3}{3 e}-\frac {b d e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{c^{2/3}}+b d^2 x \tanh ^{-1}\left (c x^3\right )+b d e x^2 \tanh ^{-1}\left (c x^3\right )+\frac {1}{3} b e^2 x^3 \tanh ^{-1}\left (c x^3\right )+\frac {b e^2 \log \left (1-c^2 x^6\right )}{6 c}-\frac {1}{2} \left (b \sqrt [3]{c} d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^{2/3} x} \, dx,x,x^2\right )+\frac {1}{2} \left (b \sqrt [3]{c} d^2\right ) \operatorname {Subst}\left (\int \frac {1-c^{2/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )+\frac {(b d e) \int \frac {-\sqrt [3]{c}+2 c^{2/3} x}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{4 c^{2/3}}-\frac {(b d e) \int \frac {\sqrt [3]{c}+2 c^{2/3} x}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{4 c^{2/3}}+\frac {(3 b d e) \int \frac {1}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{4 \sqrt [3]{c}}+\frac {(3 b d e) \int \frac {1}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{4 \sqrt [3]{c}}\\ &=\frac {a (d+e x)^3}{3 e}-\frac {b d e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{c^{2/3}}+b d^2 x \tanh ^{-1}\left (c x^3\right )+b d e x^2 \tanh ^{-1}\left (c x^3\right )+\frac {1}{3} b e^2 x^3 \tanh ^{-1}\left (c x^3\right )+\frac {b d^2 \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {b d e \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{4 c^{2/3}}-\frac {b d e \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{4 c^{2/3}}+\frac {b e^2 \log \left (1-c^2 x^6\right )}{6 c}-\frac {\left (b d^2\right ) \operatorname {Subst}\left (\int \frac {c^{2/3}+2 c^{4/3} x}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )}{4 \sqrt [3]{c}}+\frac {1}{4} \left (3 b \sqrt [3]{c} d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^{2/3} x+c^{4/3} x^2} \, dx,x,x^2\right )+\frac {(3 b d e) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {(3 b d e) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{c} x\right )}{2 c^{2/3}}\\ &=\frac {a (d+e x)^3}{3 e}-\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{2 c^{2/3}}-\frac {b d e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{c^{2/3}}+b d^2 x \tanh ^{-1}\left (c x^3\right )+b d e x^2 \tanh ^{-1}\left (c x^3\right )+\frac {1}{3} b e^2 x^3 \tanh ^{-1}\left (c x^3\right )+\frac {b d^2 \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {b d e \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{4 c^{2/3}}-\frac {b d e \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{4 c^{2/3}}-\frac {b d^2 \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{4 \sqrt [3]{c}}+\frac {b e^2 \log \left (1-c^2 x^6\right )}{6 c}-\frac {\left (3 b d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 c^{2/3} x^2\right )}{2 \sqrt [3]{c}}\\ &=\frac {a (d+e x)^3}{3 e}-\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {\sqrt {3} b d e \tan ^{-1}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{2 c^{2/3}}+\frac {\sqrt {3} b d^2 \tan ^{-1}\left (\frac {1+2 c^{2/3} x^2}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}-\frac {b d e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{c^{2/3}}+b d^2 x \tanh ^{-1}\left (c x^3\right )+b d e x^2 \tanh ^{-1}\left (c x^3\right )+\frac {1}{3} b e^2 x^3 \tanh ^{-1}\left (c x^3\right )+\frac {b d^2 \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {b d e \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{4 c^{2/3}}-\frac {b d e \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{4 c^{2/3}}-\frac {b d^2 \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{4 \sqrt [3]{c}}+\frac {b e^2 \log \left (1-c^2 x^6\right )}{6 c}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 299, normalized size = 0.89 \[ \frac {12 a c d^2 x+12 a c d e x^2+4 a c e^2 x^3-3 b \sqrt [3]{c} d \left (\sqrt [3]{c} d-e\right ) \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )-3 b \sqrt [3]{c} d \left (\sqrt [3]{c} d+e\right ) \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )+2 b e^2 \log \left (1-c^2 x^6\right )+4 b c x \tanh ^{-1}\left (c x^3\right ) \left (3 d^2+3 d e x+e^2 x^2\right )+6 b \sqrt [3]{c} d \left (\sqrt [3]{c} d+e\right ) \log \left (1-\sqrt [3]{c} x\right )+6 b \sqrt [3]{c} d \left (\sqrt [3]{c} d-e\right ) \log \left (\sqrt [3]{c} x+1\right )+6 \sqrt {3} b \sqrt [3]{c} d \left (\sqrt [3]{c} d+e\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x-1}{\sqrt {3}}\right )-6 \sqrt {3} b \sqrt [3]{c} d \left (\sqrt [3]{c} d-e\right ) \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x+1}{\sqrt {3}}\right )}{12 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2*(a + b*ArcTanh[c*x^3]),x]

[Out]

(12*a*c*d^2*x + 12*a*c*d*e*x^2 + 4*a*c*e^2*x^3 + 6*Sqrt[3]*b*c^(1/3)*d*(c^(1/3)*d + e)*ArcTan[(-1 + 2*c^(1/3)*
x)/Sqrt[3]] - 6*Sqrt[3]*b*c^(1/3)*d*(c^(1/3)*d - e)*ArcTan[(1 + 2*c^(1/3)*x)/Sqrt[3]] + 4*b*c*x*(3*d^2 + 3*d*e
*x + e^2*x^2)*ArcTanh[c*x^3] + 6*b*c^(1/3)*d*(c^(1/3)*d + e)*Log[1 - c^(1/3)*x] + 6*b*c^(1/3)*d*(c^(1/3)*d - e
)*Log[1 + c^(1/3)*x] - 3*b*c^(1/3)*d*(c^(1/3)*d - e)*Log[1 - c^(1/3)*x + c^(2/3)*x^2] - 3*b*c^(1/3)*d*(c^(1/3)
*d + e)*Log[1 + c^(1/3)*x + c^(2/3)*x^2] + 2*b*e^2*Log[1 - c^2*x^6])/(12*c)

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fricas [C]  time = 2.84, size = 9282, normalized size = 27.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(a+b*arctanh(c*x^3)),x, algorithm="fricas")

[Out]

1/24*(8*a*c*e^2*x^3 + 24*a*c*d*e*x^2 + 24*a*c*d^2*x - 2*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^
2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*
d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3
*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*c*log(15*b^2*c*d^3*e^2 + b^2*e^5 +
1/4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3
)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*
(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^
(1/3)*(I*sqrt(3) + 1))^2*c^2*e - 1/2*(3*b*c^2*d^3 - 2*b*c*e^3)*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)
*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (
27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(
9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1)) + 9*(b^2*c^2*d^5 + b^2*c*d^2
*e^3)*x) - 2*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*
d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/
2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*
b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*c*log(15*b^2*c*d^3*e^2 - b^2*e^5 - 1/4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*
e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2
/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c
^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2*e - 1/2*(3*b*c
^2*d^3 + 2*b*c*e^3)*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 -
27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/
c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6
+ e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1)) - 9*(b^2*c^2*d^5 - b^2*c*d^2*e^3)*x) + (6*b*e^2 + (2*(1/2)^(2/3)*(b^2*e
^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^
3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^
3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*c
+ 3*sqrt(1/3)*c*sqrt((144*b^2*c*d^3*e - 4*b^2*e^4 - 4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)
*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^
6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e
- e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*b*c*e^2 - (2*(1/2)^(2/3)*(b^2*e^4/c^2
+ (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e -
e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^
3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2)/c
^2))*log(-15*b^2*c*d^3*e^2 + b^2*e^5 + 1/4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3
) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^
3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3
*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2*e + 1/2*(3*b*c^2*d^3 + 2*b*c*e^3)*(2*(1/2)
^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^
2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c
^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqr
t(3) + 1)) - 18*(b^2*c^2*d^5 - b^2*c*d^2*e^3)*x + 3/4*sqrt(1/3)*(6*b*c^2*d^3 - 2*b*c*e^3 - (2*(1/2)^(2/3)*(b^2
*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*
d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*
d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*
c^2*e)*sqrt((144*b^2*c*d^3*e - 4*b^2*e^4 - 4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt
(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*
b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b
^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*b*c*e^2 - (2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d
^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*
e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^
3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2)/c^2)) + (6
*b*e^2 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3
- e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(
1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/
c^3)^(1/3)*(I*sqrt(3) + 1))*c - 3*sqrt(1/3)*c*sqrt((144*b^2*c*d^3*e - 4*b^2*e^4 - 4*(2*(1/2)^(2/3)*(b^2*e^4/c^
2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e -
e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e
^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*b*c*e^2
- (2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)
*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(
2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(
1/3)*(I*sqrt(3) + 1))^2*c^2)/c^2))*log(-15*b^2*c*d^3*e^2 + b^2*e^5 + 1/4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^
3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e
^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3
/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2*e + 1/2*(3*b
*c^2*d^3 + 2*b*c*e^3)*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3
- 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^
2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^
6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1)) - 18*(b^2*c^2*d^5 - b^2*c*d^2*e^3)*x - 3/4*sqrt(1/3)*(6*b*c^2*d^3 - 2
*b*c*e^3 - (2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^
3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)
^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^
3/c^3)^(1/3)*(I*sqrt(3) + 1))*c^2*e)*sqrt((144*b^2*c*d^3*e - 4*b^2*e^4 - 4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 + (9*c*
d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3
*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d
^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*b*c*e^2 - (2*(1/
2)^(2/3)*(b^2*e^4/c^2 + (9*c*d^3*e - e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/
c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6
/c^3 - 27*(c*d^3 - e^3)*b^3*d^3/c^2 + 3*(9*c*d^3*e - e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*s
qrt(3) + 1))^2*c^2)/c^2)) + (6*b*e^2 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) +
1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^
3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2
/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*c + 3*sqrt(1/3)*c*sqrt(-(144*b^2*c*d^3*e + 4*b^2*e^4
+ 4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^
3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)
*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)
^(1/3)*(I*sqrt(3) + 1))*b*c*e^2 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2
*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1
/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3
+ (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2)/c^2))*log(-15*b^2*c*d^3*e^2 - b^2*e^5 - 1/4*(2*(1/
2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/
c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6
/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*s
qrt(3) + 1))^2*c^2*e + 1/2*(3*b*c^2*d^3 - 2*b*c*e^3)*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*
(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6
+ e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e
+ e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1)) + 18*(b^2*c^2*d^5 + b^2*c*d^2*e^3)*x +
3/4*sqrt(1/3)*(6*b*c^2*d^3 + 2*b*c*e^3 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3)
+ 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3
/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*
e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*c^2*e)*sqrt(-(144*b^2*c*d^3*e + 4*b^2*e^4 + 4*(2*
(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d
^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*
e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(
I*sqrt(3) + 1))*b*c*e^2 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6
/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*
b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^
2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2)/c^2)) + (6*b*e^2 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*
e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2
/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c
^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*c - 3*sqrt(1/3)*c*sq
rt(-(144*b^2*c*d^3*e + 4*b^2*e^4 + 4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)
/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)
^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c
^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*b*c*e^2 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e
^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3
+ (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 -
3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2)/c^2))*log(-15*b^2*
c*d^3*e^2 - b^2*e^5 - 1/4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6
/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*
b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^
2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2*e + 1/2*(3*b*c^2*d^3 - 2*b*c*e^3)*(2*(1/2)^(2/3)*(b^2*e^4/c
^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e
+ e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 +
e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1)) + 18*(
b^2*c^2*d^5 + b^2*c*d^2*e^3)*x - 3/4*sqrt(1/3)*(6*b*c^2*d^3 + 2*b*c*e^3 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d
^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*
e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^
3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*c^2*e)*sqrt(-(144
*b^2*c*d^3*e + 4*b^2*e^4 + 4*(2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/c^2)*(-I*sqrt(3) + 1)/(2*b^3*
e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3) -
2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27
*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))*b*c*e^2 + (2*(1/2)^(2/3)*(b^2*e^4/c^2 - (9*c*d^3*e + e^4)*b^2/
c^2)*(-I*sqrt(3) + 1)/(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d^3*e + e^4)*b^3*e^2/c^3 + (27*c^
2*d^6 + e^6)*b^3/c^3)^(1/3) - 2*b*e^2/c + (1/2)^(1/3)*(2*b^3*e^6/c^3 + 27*(c*d^3 + e^3)*b^3*d^3/c^2 - 3*(9*c*d
^3*e + e^4)*b^3*e^2/c^3 + (27*c^2*d^6 + e^6)*b^3/c^3)^(1/3)*(I*sqrt(3) + 1))^2*c^2)/c^2)) + 4*(b*c*e^2*x^3 + 3
*b*c*d*e*x^2 + 3*b*c*d^2*x)*log(-(c*x^3 + 1)/(c*x^3 - 1)))/c

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giac [A]  time = 4.76, size = 383, normalized size = 1.14 \[ -\frac {\sqrt {3} {\left (b c d^{2} {\left | c \right |}^{\frac {2}{3}} - b c d {\left | c \right |}^{\frac {1}{3}} e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{2 \, c^{2}} + \frac {\sqrt {3} {\left (b c d^{2} {\left | c \right |}^{\frac {2}{3}} + b c d {\left | c \right |}^{\frac {1}{3}} e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{2 \, c^{2}} + \frac {b c x^{3} e^{2} \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 3 \, b c d x^{2} e \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 2 \, a c x^{3} e^{2} + 6 \, a c d x^{2} e + 3 \, b c d^{2} x \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + 6 \, a c d^{2} x + b e^{2} \log \left (c^{2} x^{6} - 1\right )}{6 \, c} - \frac {{\left (b c d^{2} {\left | c \right |}^{\frac {2}{3}} + b c d {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left (x^{2} + \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{4 \, c^{2}} - \frac {{\left (b c d^{2} {\left | c \right |}^{\frac {2}{3}} - b c d {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left (x^{2} - \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{4 \, c^{2}} + \frac {{\left (b c d^{2} {\left | c \right |}^{\frac {2}{3}} - b c d {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left ({\left | x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{2 \, c^{2}} + \frac {{\left (b c d^{2} {\left | c \right |}^{\frac {2}{3}} + b c d {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left ({\left | x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{2 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(a+b*arctanh(c*x^3)),x, algorithm="giac")

[Out]

-1/2*sqrt(3)*(b*c*d^2*abs(c)^(2/3) - b*c*d*abs(c)^(1/3)*e)*arctan(1/3*sqrt(3)*(2*x + 1/abs(c)^(1/3))*abs(c)^(1
/3))/c^2 + 1/2*sqrt(3)*(b*c*d^2*abs(c)^(2/3) + b*c*d*abs(c)^(1/3)*e)*arctan(1/3*sqrt(3)*(2*x - 1/abs(c)^(1/3))
*abs(c)^(1/3))/c^2 + 1/6*(b*c*x^3*e^2*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 3*b*c*d*x^2*e*log(-(c*x^3 + 1)/(c*x^3 -
1)) + 2*a*c*x^3*e^2 + 6*a*c*d*x^2*e + 3*b*c*d^2*x*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 6*a*c*d^2*x + b*e^2*log(c^2*
x^6 - 1))/c - 1/4*(b*c*d^2*abs(c)^(2/3) + b*c*d*abs(c)^(1/3)*e)*log(x^2 + x/abs(c)^(1/3) + 1/abs(c)^(2/3))/c^2
- 1/4*(b*c*d^2*abs(c)^(2/3) - b*c*d*abs(c)^(1/3)*e)*log(x^2 - x/abs(c)^(1/3) + 1/abs(c)^(2/3))/c^2 + 1/2*(b*c
*d^2*abs(c)^(2/3) - b*c*d*abs(c)^(1/3)*e)*log(abs(x + 1/abs(c)^(1/3)))/c^2 + 1/2*(b*c*d^2*abs(c)^(2/3) + b*c*d
*abs(c)^(1/3)*e)*log(abs(x - 1/abs(c)^(1/3)))/c^2

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maple [A]  time = 0.04, size = 500, normalized size = 1.49 \[ \frac {a \,x^{3} e^{2}}{3}+a d e \,x^{2}+a x \,d^{2}+\frac {a \,d^{3}}{3 e}+\frac {b \,e^{2} \arctanh \left (c \,x^{3}\right ) x^{3}}{3}+b e \arctanh \left (c \,x^{3}\right ) x^{2} d +b \arctanh \left (c \,x^{3}\right ) x \,d^{2}+\frac {b \arctanh \left (c \,x^{3}\right ) d^{3}}{3 e}+\frac {b \,d^{2} \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \,d^{2} \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \,d^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b e d \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b e d \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b e d \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b \ln \left (c \,x^{3}-1\right ) d^{3}}{6 e}+\frac {b \,e^{2} \ln \left (c \,x^{3}-1\right )}{6 c}+\frac {b \,d^{2} \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b \,d^{2} \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}+\frac {b \,d^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {2}{3}}}-\frac {b e d \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b e d \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b e d \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{2 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b \ln \left (c \,x^{3}+1\right ) d^{3}}{6 e}+\frac {b \,e^{2} \ln \left (c \,x^{3}+1\right )}{6 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*(a+b*arctanh(c*x^3)),x)

[Out]

1/3*a*x^3*e^2+a*d*e*x^2+a*x*d^2+1/3*a/e*d^3+1/3*b*e^2*arctanh(c*x^3)*x^3+b*e*arctanh(c*x^3)*x^2*d+b*arctanh(c*
x^3)*x*d^2+1/3*b/e*arctanh(c*x^3)*d^3+1/2*b/c*d^2/(1/c)^(2/3)*ln(x-(1/c)^(1/3))-1/4*b/c*d^2/(1/c)^(2/3)*ln(x^2
+(1/c)^(1/3)*x+(1/c)^(2/3))-1/2*b/c*d^2/(1/c)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))+1/2*b*e/c*
d/(1/c)^(1/3)*ln(x-(1/c)^(1/3))-1/4*b*e/c*d/(1/c)^(1/3)*ln(x^2+(1/c)^(1/3)*x+(1/c)^(2/3))+1/2*b*e/c*d*3^(1/2)/
(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))+1/6*b/e*ln(c*x^3-1)*d^3+1/6*b*e^2/c*ln(c*x^3-1)+1/2*b/c*d^
2/(1/c)^(2/3)*ln(x+(1/c)^(1/3))-1/4*b/c*d^2/(1/c)^(2/3)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))+1/2*b/c*d^2/(1/c)^(2
/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x-1))-1/2*b*e/c*d/(1/c)^(1/3)*ln(x+(1/c)^(1/3))+1/4*b*e/c*d/(1/c
)^(1/3)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))+1/2*b*e/c*d*3^(1/2)/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x-
1))-1/6*b/e*ln(c*x^3+1)*d^3+1/6*b*e^2/c*ln(c*x^3+1)

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maxima [A]  time = 0.42, size = 295, normalized size = 0.88 \[ \frac {1}{3} \, a e^{2} x^{3} + a d e x^{2} + \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}} - \frac {\log \left (c^{\frac {4}{3}} x^{4} + c^{\frac {2}{3}} x^{2} + 1\right )}{c^{\frac {4}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {2}{3}} x^{2} - 1}{c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}}\right )} + 4 \, x \operatorname {artanh}\left (c x^{3}\right )\right )} b d^{2} + \frac {1}{4} \, {\left (4 \, x^{2} \operatorname {artanh}\left (c x^{3}\right ) + c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x - c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} - \frac {\log \left (c^{\frac {2}{3}} x^{2} + c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} + \frac {\log \left (c^{\frac {2}{3}} x^{2} - c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x + 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x - 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}}\right )}\right )} b d e + a d^{2} x + \frac {{\left (2 \, c x^{3} \operatorname {artanh}\left (c x^{3}\right ) + \log \left (-c^{2} x^{6} + 1\right )\right )} b e^{2}}{6 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(a+b*arctanh(c*x^3)),x, algorithm="maxima")

[Out]

1/3*a*e^2*x^3 + a*d*e*x^2 + 1/4*(c*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(4/3)*x^2 + c^(2/3))/c^(2/3))/c^(4/3) -
log(c^(4/3)*x^4 + c^(2/3)*x^2 + 1)/c^(4/3) + 2*log((c^(2/3)*x^2 - 1)/c^(2/3))/c^(4/3)) + 4*x*arctanh(c*x^3))*b
*d^2 + 1/4*(4*x^2*arctanh(c*x^3) + c*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(2/3)*x + c^(1/3))/c^(1/3))/c^(5/3) +
2*sqrt(3)*arctan(1/3*sqrt(3)*(2*c^(2/3)*x - c^(1/3))/c^(1/3))/c^(5/3) - log(c^(2/3)*x^2 + c^(1/3)*x + 1)/c^(5/
3) + log(c^(2/3)*x^2 - c^(1/3)*x + 1)/c^(5/3) - 2*log((c^(1/3)*x + 1)/c^(1/3))/c^(5/3) + 2*log((c^(1/3)*x - 1)
/c^(1/3))/c^(5/3)))*b*d*e + a*d^2*x + 1/6*(2*c*x^3*arctanh(c*x^3) + log(-c^2*x^6 + 1))*b*e^2/c

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mupad [B]  time = 1.82, size = 1081, normalized size = 3.22 \[ \left (\sum _{k=1}^3\ln \left (x\,\left (162\,b^5\,c^9\,d^8\,e^2+6\,b^5\,c^7\,d^2\,e^8\right )+\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2-162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,\left (x\,\left (486\,b^4\,c^{10}\,d^8-90\,b^4\,c^8\,d^2\,e^6\right )+\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2-162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,\left (\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2-162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,\left (3888\,b^2\,c^{10}\,d^3\,e-\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2-162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,b\,c^{11}\,d^2\,x\,3888+648\,b^2\,c^{10}\,d^2\,e^2\,x\right )-972\,b^3\,c^9\,d^3\,e^3+324\,b^3\,c^9\,d^2\,e^4\,x\right )\right )+243\,b^5\,c^9\,d^9\,e+9\,b^5\,c^7\,d^3\,e^7\right )\,\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2-162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\right )+\left (\sum _{k=1}^3\ln \left (x\,\left (162\,b^5\,c^9\,d^8\,e^2+6\,b^5\,c^7\,d^2\,e^8\right )+\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2+162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,\left (x\,\left (486\,b^4\,c^{10}\,d^8-90\,b^4\,c^8\,d^2\,e^6\right )+\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2+162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,\left (\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2+162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,\left (3888\,b^2\,c^{10}\,d^3\,e-\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2+162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\,b\,c^{11}\,d^2\,x\,3888+648\,b^2\,c^{10}\,d^2\,e^2\,x\right )-972\,b^3\,c^9\,d^3\,e^3+324\,b^3\,c^9\,d^2\,e^4\,x\right )\right )+243\,b^5\,c^9\,d^9\,e+9\,b^5\,c^7\,d^3\,e^7\right )\,\mathrm {root}\left (216\,c^3\,z^3-108\,b\,c^2\,e^2\,z^2+162\,b^2\,c^2\,d^3\,e\,z+18\,b^2\,c\,e^4\,z-27\,b^3\,c^2\,d^6-b^3\,e^6,z,k\right )\right )+\ln \left (c\,x^3+1\right )\,\left (\frac {b\,d^2\,x}{2}+\frac {b\,d\,e\,x^2}{2}+\frac {b\,e^2\,x^3}{6}\right )-\ln \left (1-c\,x^3\right )\,\left (\frac {b\,d^2\,x}{2}+\frac {b\,d\,e\,x^2}{2}+\frac {b\,e^2\,x^3}{6}\right )+\frac {a\,e^2\,x^3}{3}+a\,d^2\,x+a\,d\,e\,x^2 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))*(d + e*x)^2,x)

[Out]

symsum(log(x*(6*b^5*c^7*d^2*e^8 + 162*b^5*c^9*d^8*e^2) + root(216*c^3*z^3 - 108*b*c^2*e^2*z^2 - 162*b^2*c^2*d^
3*e*z + 18*b^2*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k)*(x*(486*b^4*c^10*d^8 - 90*b^4*c^8*d^2*e^6) + root(216
*c^3*z^3 - 108*b*c^2*e^2*z^2 - 162*b^2*c^2*d^3*e*z + 18*b^2*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k)*(root(21
6*c^3*z^3 - 108*b*c^2*e^2*z^2 - 162*b^2*c^2*d^3*e*z + 18*b^2*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k)*(3888*b
^2*c^10*d^3*e - 3888*root(216*c^3*z^3 - 108*b*c^2*e^2*z^2 - 162*b^2*c^2*d^3*e*z + 18*b^2*c*e^4*z - 27*b^3*c^2*
d^6 - b^3*e^6, z, k)*b*c^11*d^2*x + 648*b^2*c^10*d^2*e^2*x) - 972*b^3*c^9*d^3*e^3 + 324*b^3*c^9*d^2*e^4*x)) +
243*b^5*c^9*d^9*e + 9*b^5*c^7*d^3*e^7)*root(216*c^3*z^3 - 108*b*c^2*e^2*z^2 - 162*b^2*c^2*d^3*e*z + 18*b^2*c*e
^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k), k, 1, 3) + symsum(log(x*(6*b^5*c^7*d^2*e^8 + 162*b^5*c^9*d^8*e^2) + ro
ot(216*c^3*z^3 - 108*b*c^2*e^2*z^2 + 162*b^2*c^2*d^3*e*z + 18*b^2*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k)*(x
*(486*b^4*c^10*d^8 - 90*b^4*c^8*d^2*e^6) + root(216*c^3*z^3 - 108*b*c^2*e^2*z^2 + 162*b^2*c^2*d^3*e*z + 18*b^2
*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k)*(root(216*c^3*z^3 - 108*b*c^2*e^2*z^2 + 162*b^2*c^2*d^3*e*z + 18*b^
2*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k)*(3888*b^2*c^10*d^3*e - 3888*root(216*c^3*z^3 - 108*b*c^2*e^2*z^2 +
162*b^2*c^2*d^3*e*z + 18*b^2*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k)*b*c^11*d^2*x + 648*b^2*c^10*d^2*e^2*x)
- 972*b^3*c^9*d^3*e^3 + 324*b^3*c^9*d^2*e^4*x)) + 243*b^5*c^9*d^9*e + 9*b^5*c^7*d^3*e^7)*root(216*c^3*z^3 - 1
08*b*c^2*e^2*z^2 + 162*b^2*c^2*d^3*e*z + 18*b^2*c*e^4*z - 27*b^3*c^2*d^6 - b^3*e^6, z, k), k, 1, 3) + log(c*x^
3 + 1)*((b*e^2*x^3)/6 + (b*d^2*x)/2 + (b*d*e*x^2)/2) - log(1 - c*x^3)*((b*e^2*x^3)/6 + (b*d^2*x)/2 + (b*d*e*x^
2)/2) + (a*e^2*x^3)/3 + a*d^2*x + a*d*e*x^2

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*(a+b*atanh(c*x**3)),x)

[Out]

Timed out

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