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

Optimal. Leaf size=285 \[ \frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b d \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {\sqrt {3} b d \tan ^{-1}\left (\frac {2 c^{2/3} x^2+1}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}-\frac {b d \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{4 \sqrt [3]{c}}+\frac {b e \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b e \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{4 c^{2/3}}-\frac {b e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tanh ^{-1}\left (c x^3\right )}{2 e} \]

[Out]

-1/2*b*e*arctanh(c^(1/3)*x)/c^(2/3)-1/2*b*d^2*arctanh(c*x^3)/e+1/2*(e*x+d)^2*(a+b*arctanh(c*x^3))/e+1/2*b*d*ln
(1-c^(2/3)*x^2)/c^(1/3)+1/8*b*e*ln(1-c^(1/3)*x+c^(2/3)*x^2)/c^(2/3)-1/8*b*e*ln(1+c^(1/3)*x+c^(2/3)*x^2)/c^(2/3
)-1/4*b*d*ln(1+c^(2/3)*x^2+c^(4/3)*x^4)/c^(1/3)+1/4*b*e*arctan(-1/3*3^(1/2)+2/3*c^(1/3)*x*3^(1/2))*3^(1/2)/c^(
2/3)+1/4*b*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*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 = 0.45, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 13, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {6273, 12, 1831, 275, 206, 292, 31, 634, 617, 204, 628, 296, 618} \[ \frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b d \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac {b d \log \left (c^{4/3} x^4+c^{2/3} x^2+1\right )}{4 \sqrt [3]{c}}+\frac {\sqrt {3} b d \tan ^{-1}\left (\frac {2 c^{2/3} x^2+1}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}+\frac {b e \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b e \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{4 c^{2/3}}-\frac {b e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tanh ^{-1}\left (c x^3\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[3]*b*e*ArcTan[1/Sqrt[3] - (2*c^(1/3)*x)/Sqrt[3]])/(4*c^(2/3)) + (Sqrt[3]*b*e*ArcTan[1/Sqrt[3] + (2*c^(1
/3)*x)/Sqrt[3]])/(4*c^(2/3)) + (Sqrt[3]*b*d*ArcTan[(1 + 2*c^(2/3)*x^2)/Sqrt[3]])/(2*c^(1/3)) - (b*e*ArcTanh[c^
(1/3)*x])/(2*c^(2/3)) - (b*d^2*ArcTanh[c*x^3])/(2*e) + ((d + e*x)^2*(a + b*ArcTanh[c*x^3]))/(2*e) + (b*d*Log[1
- c^(2/3)*x^2])/(2*c^(1/3)) + (b*e*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3)) - (b*e*Log[1 + c^(1/3)*x + c
^(2/3)*x^2])/(8*c^(2/3)) - (b*d*Log[1 + c^(2/3)*x^2 + c^(4/3)*x^4])/(4*c^(1/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 1831

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[((c*x)^(m + ii)*(Coeff[Pq,
x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2)))/(c^ii*(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; Fr
eeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n

Rule 6273

Int[((a_.) + ArcTanh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/(1 - u^2), x],
x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m
+ 1), u, x] && FalseQ[PowerVariableExpn[u, m + 1, x]]

Rubi steps

\begin {align*} \int (d+e x) \left (a+b \tanh ^{-1}\left (c x^3\right )\right ) \, dx &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}-\frac {b \int \frac {3 c x^2 (d+e x)^2}{1-c^2 x^6} \, dx}{2 e}\\ &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}-\frac {(3 b c) \int \frac {x^2 (d+e x)^2}{1-c^2 x^6} \, dx}{2 e}\\ &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}-\frac {(3 b c) \int \left (\frac {d^2 x^2}{1-c^2 x^6}+\frac {2 d e x^3}{1-c^2 x^6}+\frac {e^2 x^4}{1-c^2 x^6}\right ) \, dx}{2 e}\\ &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}-(3 b c d) \int \frac {x^3}{1-c^2 x^6} \, dx-\frac {\left (3 b c d^2\right ) \int \frac {x^2}{1-c^2 x^6} \, dx}{2 e}-\frac {1}{2} (3 b c e) \int \frac {x^4}{1-c^2 x^6} \, dx\\ &=\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}-\frac {1}{2} (3 b c d) \operatorname {Subst}\left (\int \frac {x}{1-c^2 x^3} \, dx,x,x^2\right )-\frac {\left (b c d^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,x^3\right )}{2 e}-\frac {(b e) \int \frac {1}{1-c^{2/3} x^2} \, dx}{2 \sqrt [3]{c}}-\frac {(b e) \int \frac {-\frac {1}{2}-\frac {\sqrt [3]{c} x}{2}}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{2 \sqrt [3]{c}}-\frac {(b e) \int \frac {-\frac {1}{2}+\frac {\sqrt [3]{c} x}{2}}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{2 \sqrt [3]{c}}\\ &=-\frac {b e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tanh ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}-\frac {1}{2} \left (b \sqrt [3]{c} d\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\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 e) \int \frac {-\sqrt [3]{c}+2 c^{2/3} x}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{2/3}}-\frac {(b e) \int \frac {\sqrt [3]{c}+2 c^{2/3} x}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{2/3}}+\frac {(3 b e) \int \frac {1}{1-\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 \sqrt [3]{c}}+\frac {(3 b e) \int \frac {1}{1+\sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 \sqrt [3]{c}}\\ &=-\frac {b e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tanh ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b d \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {b e \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}-\frac {b e \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}-\frac {(b d) \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\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 e) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {(3 b e) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{c} x\right )}{4 c^{2/3}}\\ &=-\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}-\frac {b e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tanh ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b d \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {b e \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}-\frac {b e \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}-\frac {b d \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{4 \sqrt [3]{c}}-\frac {(3 b d) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 c^{2/3} x^2\right )}{2 \sqrt [3]{c}}\\ &=-\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {1-2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {1+2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b d \tan ^{-1}\left (\frac {1+2 c^{2/3} x^2}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}-\frac {b e \tanh ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tanh ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tanh ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b d \log \left (1-c^{2/3} x^2\right )}{2 \sqrt [3]{c}}+\frac {b e \log \left (1-\sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}-\frac {b e \log \left (1+\sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}-\frac {b d \log \left (1+c^{2/3} x^2+c^{4/3} x^4\right )}{4 \sqrt [3]{c}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 333, normalized size = 1.17 \[ a d x+\frac {1}{2} a e x^2-\frac {b d \left (\log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )+\log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )-2 \log \left (1-\sqrt [3]{c} x\right )-2 \log \left (\sqrt [3]{c} x+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x-1}{\sqrt {3}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x+1}{\sqrt {3}}\right )\right )}{4 \sqrt [3]{c}}+\frac {b e \log \left (c^{2/3} x^2-\sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b e \log \left (c^{2/3} x^2+\sqrt [3]{c} x+1\right )}{8 c^{2/3}}+\frac {b e \log \left (1-\sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b e \log \left (\sqrt [3]{c} x+1\right )}{4 c^{2/3}}+\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x-1}{\sqrt {3}}\right )}{4 c^{2/3}}+\frac {\sqrt {3} b e \tan ^{-1}\left (\frac {2 \sqrt [3]{c} x+1}{\sqrt {3}}\right )}{4 c^{2/3}}+b d x \tanh ^{-1}\left (c x^3\right )+\frac {1}{2} b e x^2 \tanh ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*d*x + (a*e*x^2)/2 + (Sqrt[3]*b*e*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]])/(4*c^(2/3)) + (Sqrt[3]*b*e*ArcTan[(1 +
2*c^(1/3)*x)/Sqrt[3]])/(4*c^(2/3)) + b*d*x*ArcTanh[c*x^3] + (b*e*x^2*ArcTanh[c*x^3])/2 + (b*e*Log[1 - c^(1/3)*
x])/(4*c^(2/3)) - (b*e*Log[1 + c^(1/3)*x])/(4*c^(2/3)) + (b*e*Log[1 - c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3)) -
(b*e*Log[1 + c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3)) - (b*d*(-2*Sqrt[3]*ArcTan[(-1 + 2*c^(1/3)*x)/Sqrt[3]] + 2*S
qrt[3]*ArcTan[(1 + 2*c^(1/3)*x)/Sqrt[3]] - 2*Log[1 - c^(1/3)*x] - 2*Log[1 + c^(1/3)*x] + Log[1 - c^(1/3)*x + c
^(2/3)*x^2] + Log[1 + c^(1/3)*x + c^(2/3)*x^2]))/(4*c^(1/3))

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fricas [C]  time = 2.10, size = 3928, normalized size = 13.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*e*x^2 + a*d*x + 1/8*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*
b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*log
(2*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/
2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*b*c*d^2 + 4*b^2*d*e^2 + 1/
4*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/2
)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c*e + (8*b^2*c*d^3 + b^2*
e^3)*x) - 1/16*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1
/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1) - 2*sqrt(3/2)*s
qrt(1/2)*sqrt(-(32*b^2*d*e + (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^
3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^
2*c)/c))*log(-(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/
3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*b*c*d^2 - 2*b^2
*d*e^2 - 1/8*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3
)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c*e + 1/4*sqrt
(3/2)*sqrt(1/2)*(8*b*c*d^2 - (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^
3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*
c*e)*sqrt(-(32*b^2*d*e + (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b
^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c)
/c) + (8*b^2*c*d^3 + b^2*e^3)*x) - 1/16*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8
*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqr
t(3) + 1) + 2*sqrt(3/2)*sqrt(1/2)*sqrt(-(32*b^2*d*e + (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3
)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2
)^(1/3)*(I*sqrt(3) + 1))^2*c)/c))*log(-(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*
c*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt
(3) + 1))*b*c*d^2 - 2*b^2*d*e^2 - 1/8*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 + (8*c
*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(
3) + 1))^2*c*e - 1/4*sqrt(3/2)*sqrt(1/2)*(8*b*c*d^2 - (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3
)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2
)^(1/3)*(I*sqrt(3) + 1))*c*e)*sqrt(-(32*b^2*d*e + (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^
3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) - (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 + (8*c*d^3 - e^3)*b^3/c^2)^(1
/3)*(I*sqrt(3) + 1))^2*c)/c) + (8*b^2*c*d^3 + b^2*e^3)*x) + 1/16*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*
c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e
^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1) + 2*sqrt(3/2)*sqrt(1/2)*sqrt((32*b^2*d*e - (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(
3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 -
(8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c)/c))*log((4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d
^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)
*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*b*c*d^2 - 2*b^2*d*e^2 + 1/8*(4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^
3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*
b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c*e + 1/4*sqrt(3/2)*sqrt(1/2)*(8*b*c*d^2 - (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3
) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 -
(8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*c*e)*sqrt((32*b^2*d*e - (4*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1
)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/c^2 - (8*c*
d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c)/c) - (8*b^2*c*d^3 - b^2*e^3)*x) + 1/16*(4*(1/2)^(2/3)*b^2*d*e*
(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*
b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1) - 2*sqrt(3/2)*sqrt(1/2)*sqrt((32*b^2*d*e - (4*(1/2)^(
2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*
c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c)/c))*log((4*(1/2)^(2/3)*b^2*d*e*(-I
*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3
/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*b*c*d^2 - 2*b^2*d*e^2 + 1/8*(4*(1/2)^(2/3)*b^2*d*e*(-I*
sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^3/
c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c*e - 1/4*sqrt(3/2)*sqrt(1/2)*(8*b*c*d^2 - (4*(1/2)^(2
/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c
*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*c*e)*sqrt((32*b^2*d*e - (4*(1/2)^(2/3)*b
^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3
+ e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c)/c) - (8*b^2*c*d^3 - b^2*e^3)*x) - 1/8*(4
*(1/2)^(2/3)*b^2*d*e*(-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1
/3)*((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*log(-2*(4*(1/2)^(2/3)*b^2*d*e*(
-I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b
^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))*b*c*d^2 + 4*b^2*d*e^2 - 1/4*(4*(1/2)^(2/3)*b^2*d*e*(-
I*sqrt(3) + 1)/(((8*c*d^3 + e^3)*b^3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*c) + (1/2)^(1/3)*((8*c*d^3 + e^3)*b^
3/c^2 - (8*c*d^3 - e^3)*b^3/c^2)^(1/3)*(I*sqrt(3) + 1))^2*c*e - (8*b^2*c*d^3 - b^2*e^3)*x) + 1/4*(b*e*x^2 + 2*
b*d*x)*log(-(c*x^3 + 1)/(c*x^3 - 1))

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giac [A]  time = 2.18, size = 304, normalized size = 1.07 \[ \frac {1}{4} \, b x^{2} e \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + \frac {1}{2} \, a x^{2} e + \frac {1}{2} \, b d x \log \left (-\frac {c x^{3} + 1}{c x^{3} - 1}\right ) + a d x - \frac {\sqrt {3} {\left (2 \, b c d {\left | c \right |}^{\frac {2}{3}} - b c {\left | c \right |}^{\frac {1}{3}} e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{4 \, c^{2}} + \frac {\sqrt {3} {\left (2 \, b c d {\left | c \right |}^{\frac {2}{3}} + b c {\left | c \right |}^{\frac {1}{3}} e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{4 \, c^{2}} - \frac {{\left (2 \, b c d {\left | c \right |}^{\frac {2}{3}} + b c {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left (x^{2} + \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{8 \, c^{2}} - \frac {{\left (2 \, b c d {\left | c \right |}^{\frac {2}{3}} - b c {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left (x^{2} - \frac {x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{8 \, c^{2}} + \frac {{\left (2 \, b c d {\left | c \right |}^{\frac {2}{3}} - b c {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left ({\left | x + \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{4 \, c^{2}} + \frac {{\left (2 \, b c d {\left | c \right |}^{\frac {2}{3}} + b c {\left | c \right |}^{\frac {1}{3}} e\right )} \log \left ({\left | x - \frac {1}{{\left | c \right |}^{\frac {1}{3}}} \right |}\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^2*e*log(-(c*x^3 + 1)/(c*x^3 - 1)) + 1/2*a*x^2*e + 1/2*b*d*x*log(-(c*x^3 + 1)/(c*x^3 - 1)) + a*d*x - 1/
4*sqrt(3)*(2*b*c*d*abs(c)^(2/3) - b*c*abs(c)^(1/3)*e)*arctan(1/3*sqrt(3)*(2*x + 1/abs(c)^(1/3))*abs(c)^(1/3))/
c^2 + 1/4*sqrt(3)*(2*b*c*d*abs(c)^(2/3) + b*c*abs(c)^(1/3)*e)*arctan(1/3*sqrt(3)*(2*x - 1/abs(c)^(1/3))*abs(c)
^(1/3))/c^2 - 1/8*(2*b*c*d*abs(c)^(2/3) + b*c*abs(c)^(1/3)*e)*log(x^2 + x/abs(c)^(1/3) + 1/abs(c)^(2/3))/c^2 -
1/8*(2*b*c*d*abs(c)^(2/3) - b*c*abs(c)^(1/3)*e)*log(x^2 - x/abs(c)^(1/3) + 1/abs(c)^(2/3))/c^2 + 1/4*(2*b*c*d
*abs(c)^(2/3) - b*c*abs(c)^(1/3)*e)*log(abs(x + 1/abs(c)^(1/3)))/c^2 + 1/4*(2*b*c*d*abs(c)^(2/3) + b*c*abs(c)^
(1/3)*e)*log(abs(x - 1/abs(c)^(1/3)))/c^2

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maple [A]  time = 0.03, size = 362, normalized size = 1.27 \[ \frac {a \,x^{2} e}{2}+a d x +\frac {b \arctanh \left (c \,x^{3}\right ) x^{2} e}{2}+b \arctanh \left (c \,x^{3}\right ) d x +\frac {b d \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 \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 \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 \ln \left (x -\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}-\frac {b e \ln \left (x^{2}+\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{8 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b e \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}+1\right )}{3}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b d \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 \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 \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 \ln \left (x +\left (\frac {1}{c}\right )^{\frac {1}{3}}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b e \ln \left (x^{2}-\left (\frac {1}{c}\right )^{\frac {1}{3}} x +\left (\frac {1}{c}\right )^{\frac {2}{3}}\right )}{8 c \left (\frac {1}{c}\right )^{\frac {1}{3}}}+\frac {b e \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {1}{c}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{4 c \left (\frac {1}{c}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2*e+a*d*x+1/2*b*arctanh(c*x^3)*x^2*e+b*arctanh(c*x^3)*d*x+1/2*b*d/c/(1/c)^(2/3)*ln(x-(1/c)^(1/3))-1/4*
b*d/c/(1/c)^(2/3)*ln(x^2+(1/c)^(1/3)*x+(1/c)^(2/3))-1/2*b*d/c/(1/c)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c)^
(1/3)*x+1))+1/4*b*e/c/(1/c)^(1/3)*ln(x-(1/c)^(1/3))-1/8*b*e/c/(1/c)^(1/3)*ln(x^2+(1/c)^(1/3)*x+(1/c)^(2/3))+1/
4*b*e*3^(1/2)/c/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x+1))+1/2*b*d/c/(1/c)^(2/3)*ln(x+(1/c)^(1/3))-1/
4*b*d/c/(1/c)^(2/3)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))+1/2*b*d/c/(1/c)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c
)^(1/3)*x-1))-1/4*b*e/c/(1/c)^(1/3)*ln(x+(1/c)^(1/3))+1/8*b*e/c/(1/c)^(1/3)*ln(x^2-(1/c)^(1/3)*x+(1/c)^(2/3))+
1/4*b*e*3^(1/2)/c/(1/c)^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c)^(1/3)*x-1))

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maxima [A]  time = 0.42, size = 248, normalized size = 0.87 \[ \frac {1}{2} \, a e x^{2} + \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} + c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}} - \frac {\log \left (c^{\frac {4}{3}} x^{4} + c^{\frac {2}{3}} x^{2} + 1\right )}{c^{\frac {4}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {2}{3}} x^{2} - 1}{c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}}\right )} + 4 \, x \operatorname {artanh}\left (c x^{3}\right )\right )} b d + \frac {1}{8} \, {\left (4 \, x^{2} \operatorname {artanh}\left (c x^{3}\right ) + c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x + c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} + \frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {2}{3}} x - c^{\frac {1}{3}}\right )}}{3 \, c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} - \frac {\log \left (c^{\frac {2}{3}} x^{2} + c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} + \frac {\log \left (c^{\frac {2}{3}} x^{2} - c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x + 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} + \frac {2 \, \log \left (\frac {c^{\frac {1}{3}} x - 1}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}}\right )}\right )} b e + a d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*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 + 1/8*(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*e + a*d*x

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mupad [B]  time = 1.03, size = 621, normalized size = 2.18 \[ \left (\sum _{k=1}^3\ln \left (-\mathrm {root}\left (64\,c^2\,z^3+24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3+b^3\,e^3,z,k\right )\,\left (\mathrm {root}\left (64\,c^2\,z^3+24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3+b^3\,e^3,z,k\right )\,\left (\mathrm {root}\left (64\,c^2\,z^3+24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3+b^3\,e^3,z,k\right )\,\left (486\,b^2\,c^{10}\,e^2\,x-1944\,b^2\,c^{10}\,d\,e+\mathrm {root}\left (64\,c^2\,z^3+24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3+b^3\,e^3,z,k\right )\,b\,c^{11}\,d\,x\,3888\right )-\frac {243\,b^3\,c^9\,e^3}{2}\right )-486\,b^4\,c^{10}\,d^4\,x\right )+\frac {243\,b^5\,c^9\,d^4\,e}{2}+\frac {243\,b^5\,c^9\,d^3\,e^2\,x}{4}\right )\,\mathrm {root}\left (64\,c^2\,z^3+24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3+b^3\,e^3,z,k\right )\right )+\left (\sum _{k=1}^3\ln \left (-\mathrm {root}\left (64\,c^2\,z^3-24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3-b^3\,e^3,z,k\right )\,\left (\mathrm {root}\left (64\,c^2\,z^3-24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3-b^3\,e^3,z,k\right )\,\left (\mathrm {root}\left (64\,c^2\,z^3-24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3-b^3\,e^3,z,k\right )\,\left (486\,b^2\,c^{10}\,e^2\,x-1944\,b^2\,c^{10}\,d\,e+\mathrm {root}\left (64\,c^2\,z^3-24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3-b^3\,e^3,z,k\right )\,b\,c^{11}\,d\,x\,3888\right )-\frac {243\,b^3\,c^9\,e^3}{2}\right )-486\,b^4\,c^{10}\,d^4\,x\right )+\frac {243\,b^5\,c^9\,d^4\,e}{2}+\frac {243\,b^5\,c^9\,d^3\,e^2\,x}{4}\right )\,\mathrm {root}\left (64\,c^2\,z^3-24\,b^2\,c\,d\,e\,z-8\,b^3\,c\,d^3-b^3\,e^3,z,k\right )\right )+\ln \left (c\,x^3+1\right )\,\left (\frac {b\,e\,x^2}{4}+\frac {b\,d\,x}{2}\right )-\ln \left (1-c\,x^3\right )\,\left (\frac {b\,e\,x^2}{4}+\frac {b\,d\,x}{2}\right )+a\,d\,x+\frac {a\,e\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^3))*(d + e*x),x)

[Out]

symsum(log((243*b^5*c^9*d^4*e)/2 - root(64*c^2*z^3 + 24*b^2*c*d*e*z - 8*b^3*c*d^3 + b^3*e^3, z, k)*(root(64*c^
2*z^3 + 24*b^2*c*d*e*z - 8*b^3*c*d^3 + b^3*e^3, z, k)*(root(64*c^2*z^3 + 24*b^2*c*d*e*z - 8*b^3*c*d^3 + b^3*e^
3, z, k)*(486*b^2*c^10*e^2*x - 1944*b^2*c^10*d*e + 3888*root(64*c^2*z^3 + 24*b^2*c*d*e*z - 8*b^3*c*d^3 + b^3*e
^3, z, k)*b*c^11*d*x) - (243*b^3*c^9*e^3)/2) - 486*b^4*c^10*d^4*x) + (243*b^5*c^9*d^3*e^2*x)/4)*root(64*c^2*z^
3 + 24*b^2*c*d*e*z - 8*b^3*c*d^3 + b^3*e^3, z, k), k, 1, 3) + symsum(log((243*b^5*c^9*d^4*e)/2 - root(64*c^2*z
^3 - 24*b^2*c*d*e*z - 8*b^3*c*d^3 - b^3*e^3, z, k)*(root(64*c^2*z^3 - 24*b^2*c*d*e*z - 8*b^3*c*d^3 - b^3*e^3,
z, k)*(root(64*c^2*z^3 - 24*b^2*c*d*e*z - 8*b^3*c*d^3 - b^3*e^3, z, k)*(486*b^2*c^10*e^2*x - 1944*b^2*c^10*d*e
+ 3888*root(64*c^2*z^3 - 24*b^2*c*d*e*z - 8*b^3*c*d^3 - b^3*e^3, z, k)*b*c^11*d*x) - (243*b^3*c^9*e^3)/2) - 4
86*b^4*c^10*d^4*x) + (243*b^5*c^9*d^3*e^2*x)/4)*root(64*c^2*z^3 - 24*b^2*c*d*e*z - 8*b^3*c*d^3 - b^3*e^3, z, k
), k, 1, 3) + log(c*x^3 + 1)*((b*d*x)/2 + (b*e*x^2)/4) - log(1 - c*x^3)*((b*d*x)/2 + (b*e*x^2)/4) + a*d*x + (a
*e*x^2)/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)*(a+b*atanh(c*x**3)),x)

[Out]

Timed out

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