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

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

[Out]

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

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Rubi [A]  time = 0.60, 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 = {5205, 12, 1831, 275, 203, 292, 31, 634, 617, 204, 628, 295, 618} \[ \frac {(d+e x)^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b d \log \left (c^{2/3} x^2+1\right )}{2 \sqrt [3]{c}}-\frac {b d \log \left (c^{4/3} x^4-c^{2/3} x^2+1\right )}{4 \sqrt [3]{c}}+\frac {\sqrt {3} b d \tan ^{-1}\left (\frac {1-2 c^{2/3} x^2}{\sqrt {3}}\right )}{2 \sqrt [3]{c}}-\frac {\sqrt {3} b e \log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}+\frac {\sqrt {3} b e \log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b e \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}+\frac {b e \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b e \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt {3}\right )}{4 c^{2/3}}-\frac {b d^2 \tan ^{-1}\left (c x^3\right )}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e*ArcTan[c^(1/3)*x])/(2*c^(2/3)) - (b*d^2*ArcTan[c*x^3])/(2*e) + ((d + e*x)^2*(a + b*ArcTan[c*x^3]))/(2*e)
+ (b*e*ArcTan[Sqrt[3] - 2*c^(1/3)*x])/(4*c^(2/3)) - (b*e*ArcTan[Sqrt[3] + 2*c^(1/3)*x])/(4*c^(2/3)) + (Sqrt[3
]*b*d*ArcTan[(1 - 2*c^(2/3)*x^2)/Sqrt[3]])/(2*c^(1/3)) + (b*d*Log[1 + c^(2/3)*x^2])/(2*c^(1/3)) - (Sqrt[3]*b*e
*Log[1 - Sqrt[3]*c^(1/3)*x + c^(2/3)*x^2])/(8*c^(2/3)) + (Sqrt[3]*b*e*Log[1 + Sqrt[3]*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 203

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

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 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 295

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 - 1)*m*Pi)/n] - s*Cos[((2*k - 1)*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(
(2*k - 1)*Pi)/n]*x + s^2*x^2), x] + Int[(r*Cos[((2*k - 1)*m*Pi)/n] + s*Cos[((2*k - 1)*(m + 1)*Pi)/n]*x)/(r^2 +
2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x]; (2*(-1)^(m/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] && PosQ[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 5205

Int[((a_.) + ArcTan[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcTan[
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 \tan ^{-1}\left (c x^3\right )\right ) \, dx &=\frac {(d+e x)^2 \left (a+b \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 \tan ^{-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 {1}{2} \sqrt {3} \sqrt [3]{c} x}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{2 \sqrt [3]{c}}-\frac {(b e) \int \frac {-\frac {1}{2}-\frac {1}{2} \sqrt {3} \sqrt [3]{c} x}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{2 \sqrt [3]{c}}\\ &=-\frac {b e \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tan ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-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 {\left (\sqrt {3} b e\right ) \int \frac {-\sqrt {3} \sqrt [3]{c}+2 c^{2/3} x}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{2/3}}+\frac {\left (\sqrt {3} b e\right ) \int \frac {\sqrt {3} \sqrt [3]{c}+2 c^{2/3} x}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 c^{2/3}}-\frac {(b e) \int \frac {1}{1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 \sqrt [3]{c}}-\frac {(b e) \int \frac {1}{1+\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2} \, dx}{8 \sqrt [3]{c}}\\ &=-\frac {b e \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tan ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-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 e \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}+\frac {\sqrt {3} b e \log \left (1+\sqrt {3} \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 {(b e) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1-\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 \sqrt {3} c^{2/3}}+\frac {(b e) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1+\frac {2 \sqrt [3]{c} x}{\sqrt {3}}\right )}{4 \sqrt {3} c^{2/3}}\\ &=-\frac {b e \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tan ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b e \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b e \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{c} x\right )}{4 c^{2/3}}+\frac {b d \log \left (1+c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac {\sqrt {3} b e \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}+\frac {\sqrt {3} b e \log \left (1+\sqrt {3} \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 {b e \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}-\frac {b d^2 \tan ^{-1}\left (c x^3\right )}{2 e}+\frac {(d+e x)^2 \left (a+b \tan ^{-1}\left (c x^3\right )\right )}{2 e}+\frac {b e \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b e \tan ^{-1}\left (\sqrt {3}+2 \sqrt [3]{c} x\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 d \log \left (1+c^{2/3} x^2\right )}{2 \sqrt [3]{c}}-\frac {\sqrt {3} b e \log \left (1-\sqrt {3} \sqrt [3]{c} x+c^{2/3} x^2\right )}{8 c^{2/3}}+\frac {\sqrt {3} b e \log \left (1+\sqrt {3} \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 = 310, normalized size = 1.09 \[ a d x+\frac {1}{2} a e x^2-\frac {b d \left (-2 \log \left (c^{2/3} x^2+1\right )+\log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )+\log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )-2 \sqrt {3} \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )-2 \sqrt {3} \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt {3}\right )\right )}{4 \sqrt [3]{c}}-\frac {\sqrt {3} b e \log \left (c^{2/3} x^2-\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}+\frac {\sqrt {3} b e \log \left (c^{2/3} x^2+\sqrt {3} \sqrt [3]{c} x+1\right )}{8 c^{2/3}}-\frac {b e \tan ^{-1}\left (\sqrt [3]{c} x\right )}{2 c^{2/3}}+\frac {b e \tan ^{-1}\left (\sqrt {3}-2 \sqrt [3]{c} x\right )}{4 c^{2/3}}-\frac {b e \tan ^{-1}\left (2 \sqrt [3]{c} x+\sqrt {3}\right )}{4 c^{2/3}}+b d x \tan ^{-1}\left (c x^3\right )+\frac {1}{2} b e x^2 \tan ^{-1}\left (c x^3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> no explicit roots found

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giac [A]  time = 0.99, size = 257, normalized size = 0.90 \[ \frac {1}{2} \, b x^{2} \arctan \left (c x^{3}\right ) e + b d x \arctan \left (c x^{3}\right ) + \frac {1}{2} \, a x^{2} e + a d x + \frac {b d {\left | c \right |}^{\frac {2}{3}} \log \left (x^{2} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{2 \, c} - \frac {b c \arctan \left (x {\left | c \right |}^{\frac {1}{3}}\right ) e}{2 \, {\left | c \right |}^{\frac {5}{3}}} + \frac {{\left (2 \, \sqrt {3} b c d {\left | c \right |}^{\frac {2}{3}} - b c {\left | c \right |}^{\frac {1}{3}} e\right )} \arctan \left ({\left (2 \, x + \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{4 \, c^{2}} - \frac {{\left (2 \, \sqrt {3} b c d {\left | c \right |}^{\frac {2}{3}} + b c {\left | c \right |}^{\frac {1}{3}} e\right )} \arctan \left ({\left (2 \, x - \frac {\sqrt {3}}{{\left | c \right |}^{\frac {1}{3}}}\right )} {\left | c \right |}^{\frac {1}{3}}\right )}{4 \, c^{2}} + \frac {{\left (\sqrt {3} b c {\left | c \right |}^{\frac {1}{3}} e - 2 \, b c d {\left | c \right |}^{\frac {2}{3}}\right )} \log \left (x^{2} + \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{8 \, c^{2}} - \frac {{\left (\sqrt {3} b c {\left | c \right |}^{\frac {1}{3}} e + 2 \, b c d {\left | c \right |}^{\frac {2}{3}}\right )} \log \left (x^{2} - \frac {\sqrt {3} x}{{\left | c \right |}^{\frac {1}{3}}} + \frac {1}{{\left | c \right |}^{\frac {2}{3}}}\right )}{8 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*x^2*arctan(c*x^3)*e + b*d*x*arctan(c*x^3) + 1/2*a*x^2*e + a*d*x + 1/2*b*d*abs(c)^(2/3)*log(x^2 + 1/abs(c
)^(2/3))/c - 1/2*b*c*arctan(x*abs(c)^(1/3))*e/abs(c)^(5/3) + 1/4*(2*sqrt(3)*b*c*d*abs(c)^(2/3) - b*c*abs(c)^(1
/3)*e)*arctan((2*x + sqrt(3)/abs(c)^(1/3))*abs(c)^(1/3))/c^2 - 1/4*(2*sqrt(3)*b*c*d*abs(c)^(2/3) + b*c*abs(c)^
(1/3)*e)*arctan((2*x - sqrt(3)/abs(c)^(1/3))*abs(c)^(1/3))/c^2 + 1/8*(sqrt(3)*b*c*abs(c)^(1/3)*e - 2*b*c*d*abs
(c)^(2/3))*log(x^2 + sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/c^2 - 1/8*(sqrt(3)*b*c*abs(c)^(1/3)*e + 2*b*c*d*
abs(c)^(2/3))*log(x^2 - sqrt(3)*x/abs(c)^(1/3) + 1/abs(c)^(2/3))/c^2

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maple [A]  time = 0.12, size = 314, normalized size = 1.10 \[ \frac {a \,x^{2} e}{2}+a d x +\frac {b \arctan \left (c \,x^{3}\right ) x^{2} e}{2}+b \arctan \left (c \,x^{3}\right ) d x +\frac {b c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}} d \ln \left (x^{2}+\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right )}{2}-\frac {b e \arctan \left (\frac {x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}\right )}{2 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\frac {b \,c^{3} \ln \left (x^{2}-\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right ) \left (\frac {1}{c^{2}}\right )^{\frac {5}{3}} d}{4}-\frac {b c \ln \left (x^{2}-\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right ) \sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {5}{6}} e}{8}-\frac {b \,c^{3} \left (\frac {1}{c^{2}}\right )^{\frac {5}{3}} \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\sqrt {3}\right ) \sqrt {3}\, d}{2}-\frac {b \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}-\sqrt {3}\right ) e}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\frac {b c \ln \left (x^{2}+\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right ) \sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {5}{6}} e}{8}-\frac {b c \ln \left (x^{2}+\sqrt {3}\, \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}} x +\left (\frac {1}{c^{2}}\right )^{\frac {1}{3}}\right ) \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}} d}{4}-\frac {b \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\sqrt {3}\right ) e}{4 c \left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\frac {b c \left (\frac {1}{c^{2}}\right )^{\frac {2}{3}} \arctan \left (\frac {2 x}{\left (\frac {1}{c^{2}}\right )^{\frac {1}{6}}}+\sqrt {3}\right ) \sqrt {3}\, d}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2*e+a*d*x+1/2*b*arctan(c*x^3)*x^2*e+b*arctan(c*x^3)*d*x+1/2*b*c*(1/c^2)^(2/3)*d*ln(x^2+(1/c^2)^(1/3))-
1/2*b/c*e/(1/c^2)^(1/6)*arctan(x/(1/c^2)^(1/6))-1/4*b*c^3*ln(x^2-3^(1/2)*(1/c^2)^(1/6)*x+(1/c^2)^(1/3))*(1/c^2
)^(5/3)*d-1/8*b*c*ln(x^2-3^(1/2)*(1/c^2)^(1/6)*x+(1/c^2)^(1/3))*3^(1/2)*(1/c^2)^(5/6)*e-1/2*b*c^3*(1/c^2)^(5/3
)*arctan(2*x/(1/c^2)^(1/6)-3^(1/2))*3^(1/2)*d-1/4*b/c/(1/c^2)^(1/6)*arctan(2*x/(1/c^2)^(1/6)-3^(1/2))*e+1/8*b*
c*ln(x^2+3^(1/2)*(1/c^2)^(1/6)*x+(1/c^2)^(1/3))*3^(1/2)*(1/c^2)^(5/6)*e-1/4*b*c*ln(x^2+3^(1/2)*(1/c^2)^(1/6)*x
+(1/c^2)^(1/3))*(1/c^2)^(2/3)*d-1/4*b/c/(1/c^2)^(1/6)*arctan(2*x/(1/c^2)^(1/6)+3^(1/2))*e+1/2*b*c*(1/c^2)^(2/3
)*arctan(2*x/(1/c^2)^(1/6)+3^(1/2))*3^(1/2)*d

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maxima [A]  time = 0.41, size = 232, normalized size = 0.81 \[ \frac {1}{2} \, a e x^{2} - \frac {1}{4} \, {\left (c {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, c^{\frac {4}{3}} x^{2} - c^{\frac {2}{3}}\right )}}{3 \, c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}} + \frac {\log \left (c^{\frac {4}{3}} x^{4} - c^{\frac {2}{3}} x^{2} + 1\right )}{c^{\frac {4}{3}}} - \frac {2 \, \log \left (\frac {c^{\frac {2}{3}} x^{2} + 1}{c^{\frac {2}{3}}}\right )}{c^{\frac {4}{3}}}\right )} - 4 \, x \arctan \left (c x^{3}\right )\right )} b d + \frac {1}{8} \, {\left (4 \, x^{2} \arctan \left (c x^{3}\right ) + c {\left (\frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} + \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {\sqrt {3} \log \left (c^{\frac {2}{3}} x^{2} - \sqrt {3} c^{\frac {1}{3}} x + 1\right )}{c^{\frac {5}{3}}} - \frac {4 \, \arctan \left (c^{\frac {1}{3}} x\right )}{c^{\frac {5}{3}}} - \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x + \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}} - \frac {2 \, \arctan \left (\frac {2 \, c^{\frac {2}{3}} x - \sqrt {3} c^{\frac {1}{3}}}{c^{\frac {1}{3}}}\right )}{c^{\frac {5}{3}}}\right )}\right )} b e + a d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(a+b*arctan(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*arctan(c*x^3))*b*d + 1/8*(4*x^2
*arctan(c*x^3) + c*(sqrt(3)*log(c^(2/3)*x^2 + sqrt(3)*c^(1/3)*x + 1)/c^(5/3) - sqrt(3)*log(c^(2/3)*x^2 - sqrt(
3)*c^(1/3)*x + 1)/c^(5/3) - 4*arctan(c^(1/3)*x)/c^(5/3) - 2*arctan((2*c^(2/3)*x + sqrt(3)*c^(1/3))/c^(1/3))/c^
(5/3) - 2*arctan((2*c^(2/3)*x - sqrt(3)*c^(1/3))/c^(1/3))/c^(5/3)))*b*e + a*d*x

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mupad [B]  time = 0.44, size = 485, normalized size = 1.70 \[ \mathrm {atan}\left (c\,x^3\right )\,\left (\frac {b\,e\,x^2}{2}+b\,d\,x\right )+\left (\sum _{k=1}^6\ln \left (-\mathrm {root}\left (4096\,a^6\,c^4-1024\,a^3\,b^3\,c^3\,d^3+576\,a^2\,b^4\,c^2\,d^2\,e^2-48\,a\,b^5\,c\,d\,e^4+64\,b^6\,c^2\,d^6+b^6\,e^6,a,k\right )\,\left (\mathrm {root}\left (4096\,a^6\,c^4-1024\,a^3\,b^3\,c^3\,d^3+576\,a^2\,b^4\,c^2\,d^2\,e^2-48\,a\,b^5\,c\,d\,e^4+64\,b^6\,c^2\,d^6+b^6\,e^6,a,k\right )\,\left (\mathrm {root}\left (4096\,a^6\,c^4-1024\,a^3\,b^3\,c^3\,d^3+576\,a^2\,b^4\,c^2\,d^2\,e^2-48\,a\,b^5\,c\,d\,e^4+64\,b^6\,c^2\,d^6+b^6\,e^6,a,k\right )\,\left (-486\,b^2\,c^{10}\,e^2\,x+1944\,b^2\,c^{10}\,d\,e+\mathrm {root}\left (4096\,a^6\,c^4-1024\,a^3\,b^3\,c^3\,d^3+576\,a^2\,b^4\,c^2\,d^2\,e^2-48\,a\,b^5\,c\,d\,e^4+64\,b^6\,c^2\,d^6+b^6\,e^6,a,k\right )\,b\,c^{11}\,d\,x\,3888\right )-\frac {243\,b^3\,c^9\,e^3}{2}\right )-486\,b^4\,c^{10}\,d^4\,x\right )-\frac {243\,b^5\,c^9\,d^4\,e}{2}-\frac {243\,b^5\,c^9\,d^3\,e^2\,x}{4}\right )\,\mathrm {root}\left (4096\,a^6\,c^4-1024\,a^3\,b^3\,c^3\,d^3+576\,a^2\,b^4\,c^2\,d^2\,e^2-48\,a\,b^5\,c\,d\,e^4+64\,b^6\,c^2\,d^6+b^6\,e^6,a,k\right )\right )+a\,d\,x+\frac {a\,e\,x^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x^3))*(d + e*x),x)

[Out]

atan(c*x^3)*(b*d*x + (b*e*x^2)/2) + symsum(log(- root(4096*a^6*c^4 - 1024*a^3*b^3*c^3*d^3 + 576*a^2*b^4*c^2*d^
2*e^2 - 48*a*b^5*c*d*e^4 + 64*b^6*c^2*d^6 + b^6*e^6, a, k)*(root(4096*a^6*c^4 - 1024*a^3*b^3*c^3*d^3 + 576*a^2
*b^4*c^2*d^2*e^2 - 48*a*b^5*c*d*e^4 + 64*b^6*c^2*d^6 + b^6*e^6, a, k)*(root(4096*a^6*c^4 - 1024*a^3*b^3*c^3*d^
3 + 576*a^2*b^4*c^2*d^2*e^2 - 48*a*b^5*c*d*e^4 + 64*b^6*c^2*d^6 + b^6*e^6, a, k)*(1944*b^2*c^10*d*e - 486*b^2*
c^10*e^2*x + 3888*root(4096*a^6*c^4 - 1024*a^3*b^3*c^3*d^3 + 576*a^2*b^4*c^2*d^2*e^2 - 48*a*b^5*c*d*e^4 + 64*b
^6*c^2*d^6 + b^6*e^6, a, 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^4*e)/2 -
(243*b^5*c^9*d^3*e^2*x)/4)*root(4096*a^6*c^4 - 1024*a^3*b^3*c^3*d^3 + 576*a^2*b^4*c^2*d^2*e^2 - 48*a*b^5*c*d*
e^4 + 64*b^6*c^2*d^6 + b^6*e^6, a, k), k, 1, 6) + a*d*x + (a*e*x^2)/2

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sympy [A]  time = 46.47, size = 3068, normalized size = 10.76 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*(d*x + e*x**2/2), Eq(c, 0)), ((a - oo*I*b)*(d*x + e*x**2/2), Eq(c, -I/x**3)), ((a + oo*I*b)*(d*x
+ e*x**2/2), Eq(c, I/x**3)), ((a - b*atan((-sqrt(3)/2 - I/2)**(-3)))*(d*x + e*x**2/2), Eq(c, -1/(x**3*(-sqrt(3
)/2 - I/2)**3))), ((a - b*atan((-sqrt(3)/2 + I/2)**(-3)))*(d*x + e*x**2/2), Eq(c, -1/(x**3*(-sqrt(3)/2 + I/2)*
*3))), ((a - b*atan((sqrt(3)/2 - I/2)**(-3)))*(d*x + e*x**2/2), Eq(c, -1/(x**3*(sqrt(3)/2 - I/2)**3))), ((a -
b*atan((sqrt(3)/2 + I/2)**(-3)))*(d*x + e*x**2/2), Eq(c, -1/(x**3*(sqrt(3)/2 + I/2)**3))), (8*(-1)**(1/6)*a*c*
*6*d*x**7*(c**(-2))**(13/6)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6))
+ 4*(-1)**(1/6)*a*c**6*e*x**8*(c**(-2))**(13/6)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c*
*4*(c**(-2))**(13/6)) + 8*(-1)**(1/6)*a*c**4*d*x*(c**(-2))**(13/6)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6)
+ 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 4*(-1)**(1/6)*a*c**4*e*x**2*(c**(-2))**(13/6)/(8*(-1)**(1/6)*c**6*x*
*6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 8*(-1)**(1/6)*b*c**6*d*x**7*(c**(-2))**(13/6)*a
tan(c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 8*(-1)**(1/3)
*b*c**6*d*x**6*(c**(-2))**(7/3)*atan(c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(
c**(-2))**(13/6)) + 4*(-1)**(1/6)*b*c**6*e*x**8*(c**(-2))**(13/6)*atan(c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-
2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 8*(-1)**(5/6)*b*c**5*d*x**6*(c**(-2))**(11/6)*log(x - (-
1)**(1/6)*(c**(-2))**(1/6))/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6))
- 6*(-1)**(5/6)*b*c**5*d*x**6*(c**(-2))**(11/6)*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)
*(c**(-2))**(1/3))/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 2*(-1)
**(5/6)*b*c**5*d*x**6*(c**(-2))**(11/6)*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2)
)**(1/3))/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 4*(-1)**(5/6)*s
qrt(3)*b*c**5*d*x**6*(c**(-2))**(11/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)/3)/(8*(-1)*
*(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) - 4*(-1)**(5/6)*sqrt(3)*b*c**5*d*x*
*6*(c**(-2))**(11/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(8*(-1)**(1/6)*c**6*x**6*(
c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 8*(-1)**(5/6)*b*c**5*d*x**6*(c**(-2))**(11/6)*log(2
)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 8*(-1)**(1/6)*b*c**4*d*
x*(c**(-2))**(13/6)*atan(c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(1
3/6)) + 8*(-1)**(1/3)*b*c**4*d*(c**(-2))**(7/3)*atan(c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-
1)**(1/6)*c**4*(c**(-2))**(13/6)) - 4*I*b*c**4*e*x**6*(c**(-2))**(3/2)*atan(c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(
c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 4*(-1)**(1/6)*b*c**4*e*x**2*(c**(-2))**(13/6)*atan(
c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 8*(-1)**(5/6)*b*c
**3*d*(c**(-2))**(11/6)*log(x - (-1)**(1/6)*(c**(-2))**(1/6))/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(
-1)**(1/6)*c**4*(c**(-2))**(13/6)) - 6*(-1)**(5/6)*b*c**3*d*(c**(-2))**(11/6)*log(4*x**2 - 4*(-1)**(1/6)*x*(c*
*(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**
4*(c**(-2))**(13/6)) + 2*(-1)**(5/6)*b*c**3*d*(c**(-2))**(11/6)*log(4*x**2 + 4*(-1)**(1/6)*x*(c**(-2))**(1/6)
+ 4*(-1)**(1/3)*(c**(-2))**(1/3))/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(
13/6)) + 4*(-1)**(5/6)*sqrt(3)*b*c**3*d*(c**(-2))**(11/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) -
sqrt(3)/3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) - 4*(-1)**(5/6)*
sqrt(3)*b*c**3*d*(c**(-2))**(11/6)*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(8*(-1)**(1/
6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 8*(-1)**(5/6)*b*c**3*d*(c**(-2))**(11
/6)*log(2)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) - 4*I*b*c**2*e*(
c**(-2))**(3/2)*atan(c*x**3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)
) - 3*b*c*e*x**6*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(8*(-1)**(1/6
)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 3*b*c*e*x**6*log(4*x**2 + 4*(-1)**(1/6
)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1
/6)*c**4*(c**(-2))**(13/6)) + 2*sqrt(3)*b*c*e*x**6*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) - sqrt(3)
/3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**4*(c**(-2))**(13/6)) + 2*sqrt(3)*b*c*e*x**6*
atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(8*(-1)**(1/6)*c**6*x**6*(c**(-2))**(13/6) + 8*
(-1)**(1/6)*c**4*(c**(-2))**(13/6)) - 3*b*e*log(4*x**2 - 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**
(-2))**(1/3))/(8*(-1)**(1/6)*c**7*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**5*(c**(-2))**(13/6)) + 3*b*e*log(4
*x**2 + 4*(-1)**(1/6)*x*(c**(-2))**(1/6) + 4*(-1)**(1/3)*(c**(-2))**(1/3))/(8*(-1)**(1/6)*c**7*x**6*(c**(-2))*
*(13/6) + 8*(-1)**(1/6)*c**5*(c**(-2))**(13/6)) + 2*sqrt(3)*b*e*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/
6)) - sqrt(3)/3)/(8*(-1)**(1/6)*c**7*x**6*(c**(-2))**(13/6) + 8*(-1)**(1/6)*c**5*(c**(-2))**(13/6)) + 2*sqrt(3
)*b*e*atan(2*(-1)**(5/6)*sqrt(3)*x/(3*(c**(-2))**(1/6)) + sqrt(3)/3)/(8*(-1)**(1/6)*c**7*x**6*(c**(-2))**(13/6
) + 8*(-1)**(1/6)*c**5*(c**(-2))**(13/6)), True))

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