3.1261 \(\int \frac{x^3 (a+b \tan ^{-1}(c x))^2}{d+e x^2} \, dx\)

Optimal. Leaf size=590 \[ -\frac{i b d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e^2}+\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}+\frac{d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e^2}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}-\frac{a b x}{c e}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^2 e}-\frac{b^2 x \tan ^{-1}(c x)}{c e} \]

[Out]

-((a*b*x)/(c*e)) - (b^2*x*ArcTan[c*x])/(c*e) + (a + b*ArcTan[c*x])^2/(2*c^2*e) + (x^2*(a + b*ArcTan[c*x])^2)/(
2*e) + (d*(a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/e^2 - (d*(a + b*ArcTan[c*x])^2*Log[(2*c*(Sqrt[-d] - Sqrt[e
]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*e^2) - (d*(a + b*ArcTan[c*x])^2*Log[(2*c*(Sqrt[-d] + Sqrt[e]
*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(2*e^2) + (b^2*Log[1 + c^2*x^2])/(2*c^2*e) - (I*b*d*(a + b*ArcTa
n[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e^2 + ((I/2)*b*d*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sq
rt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/e^2 + ((I/2)*b*d*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(S
qrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/e^2 + (b^2*d*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*
e^2) - (b^2*d*PolyLog[3, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(4*e^2) - (
b^2*d*PolyLog[3, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(4*e^2)

________________________________________________________________________________________

Rubi [A]  time = 0.49888, antiderivative size = 590, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4916, 4852, 4846, 260, 4884, 4980, 4858} \[ -\frac{i b d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e^2}+\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{4 e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{4 e^2}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}+\frac{d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}-i \sqrt{e}\right )}\right )}{2 e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{(1-i c x) \left (c \sqrt{-d}+i \sqrt{e}\right )}\right )}{2 e^2}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}-\frac{a b x}{c e}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^2 e}-\frac{b^2 x \tan ^{-1}(c x)}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*x)/(c*e)) - (b^2*x*ArcTan[c*x])/(c*e) + (a + b*ArcTan[c*x])^2/(2*c^2*e) + (x^2*(a + b*ArcTan[c*x])^2)/(
2*e) + (d*(a + b*ArcTan[c*x])^2*Log[2/(1 - I*c*x)])/e^2 - (d*(a + b*ArcTan[c*x])^2*Log[(2*c*(Sqrt[-d] - Sqrt[e
]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(2*e^2) - (d*(a + b*ArcTan[c*x])^2*Log[(2*c*(Sqrt[-d] + Sqrt[e]
*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(2*e^2) + (b^2*Log[1 + c^2*x^2])/(2*c^2*e) - (I*b*d*(a + b*ArcTa
n[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e^2 + ((I/2)*b*d*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sq
rt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/e^2 + ((I/2)*b*d*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(S
qrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/e^2 + (b^2*d*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*
e^2) - (b^2*d*PolyLog[3, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*x))/((c*Sqrt[-d] - I*Sqrt[e])*(1 - I*c*x))])/(4*e^2) - (
b^2*d*PolyLog[3, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*x))/((c*Sqrt[-d] + I*Sqrt[e])*(1 - I*c*x))])/(4*e^2)

Rule 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTan[c*x])^p)
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[c*x])^(p +
1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rule 4858

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^2*Log[2/
(1 - I*c*x)])/e, x] + (Simp[((a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] + Sim
p[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 - I*c*x)])/e, x] - Simp[(I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 -
(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] - Simp[(b^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/(2*e), x] + Simp
[(b^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && Ne
Q[c^2*d^2 + e^2, 0]

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{d+e x^2} \, dx &=\frac{\int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{e}-\frac{d \int \frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{d+e x^2} \, dx}{e}\\ &=\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}-\frac{(b c) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{e}-\frac{d \int \left (-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 \sqrt{e} \left (\sqrt{-d}-\sqrt{e} x\right )}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 \sqrt{e} \left (\sqrt{-d}+\sqrt{e} x\right )}\right ) \, dx}{e}\\ &=\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{d \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{\sqrt{-d}-\sqrt{e} x} \, dx}{2 e^{3/2}}-\frac{d \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{\sqrt{-d}+\sqrt{e} x} \, dx}{2 e^{3/2}}-\frac{b \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c e}+\frac{b \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c e}\\ &=-\frac{a b x}{c e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{4 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{4 e^2}-\frac{b^2 \int \tan ^{-1}(c x) \, dx}{c e}\\ &=-\frac{a b x}{c e}-\frac{b^2 x \tan ^{-1}(c x)}{c e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{4 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{4 e^2}+\frac{b^2 \int \frac{x}{1+c^2 x^2} \, dx}{e}\\ &=-\frac{a b x}{c e}-\frac{b^2 x \tan ^{-1}(c x)}{c e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 c^2 e}-\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}+\frac{i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1-i c x}\right )}{2 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} x\right )}{\left (c \sqrt{-d}-i \sqrt{e}\right ) (1-i c x)}\right )}{4 e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} x\right )}{\left (c \sqrt{-d}+i \sqrt{e}\right ) (1-i c x)}\right )}{4 e^2}\\ \end{align*}

Mathematica [B]  time = 9.59908, size = 1567, normalized size = 2.66 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^2*e*x^2 - 2*a^2*d*Log[d + e*x^2] + 4*a*b*(-((e*x)/c) - I*d*ArcTan[c*x]^2 + ArcTan[c*x]*(e*(c^(-2) + x^2)
+ 2*d*Log[1 + E^((2*I)*ArcTan[c*x])]) - I*d*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (2*d*(-(c^2*d) + e)*((-I)*Arc
Tan[c*x]^2 + (2*I)*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[(c*e*x)/Sqrt[c^2*d*e]] + (-ArcSin[Sqrt[(c^2*d)/(c^
2*d - e)]] + ArcTan[c*x])*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] + (ArcSin
[Sqrt[(c^2*d)/(c^2*d - e)]] + ArcTan[c*x])*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^((2*I)*ArcT
an[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] - (I/2)*(PolyLog[2, -(((c^2*d + e - 2*Sqrt[c^2*d*e
])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e))] + PolyLog[2, -(((c^2*d + e + 2*Sqrt[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(
c^2*d - e))])))/(2*c^2*d - 2*e)) + (b^2*(-4*c*e*x*ArcTan[c*x] + 2*e*ArcTan[c*x]^2 + 2*c^2*e*x^2*ArcTan[c*x]^2
+ 4*c^2*d*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - 2*c^2*d*ArcTan[c*x]^2*Log[1 + ((c*Sqrt[d] - Sqrt[e])*
E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] + Sqrt[e])] - 2*c^2*d*ArcTan[c*x]^2*Log[1 + ((c*Sqrt[d] + Sqrt[e])*E^((2*I)*
ArcTan[c*x]))/(c*Sqrt[d] - Sqrt[e])] + 2*c^2*d*ArcTan[c*x]^2*Log[1 + ((c^2*d + e - 2*Sqrt[c^2*d*e])*E^((2*I)*A
rcTan[c*x]))/(c^2*d - e)] + 4*c^2*d*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[1 + ((c^2*d + e + 2*Sqrt
[c^2*d*e])*E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] - 2*c^2*d*ArcTan[c*x]^2*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*
E^((2*I)*ArcTan[c*x]))/(c^2*d - e)] - 4*c^2*d*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[(-2*Sqrt[c^2*d
*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^((2*I)*ArcTan[c*x])) + c^2*d*(1 + E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] -
4*c^2*d*ArcTan[c*x]^2*Log[(-2*Sqrt[c^2*d*e]*E^((2*I)*ArcTan[c*x]) + e*(-1 + E^((2*I)*ArcTan[c*x])) + c^2*d*(1
+ E^((2*I)*ArcTan[c*x])))/(c^2*d - e)] + 4*c^2*d*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[((2*I)*c^2
*d - (2*I)*Sqrt[c^2*d*e] + 2*c*(-e + Sqrt[c^2*d*e])*x)/((c^2*d - e)*(I + c*x))] + 2*c^2*d*ArcTan[c*x]^2*Log[((
2*I)*c^2*d - (2*I)*Sqrt[c^2*d*e] + 2*c*(-e + Sqrt[c^2*d*e])*x)/((c^2*d - e)*(I + c*x))] + 2*e*Log[1 + c^2*x^2]
- 4*c^2*d*ArcSin[Sqrt[(c^2*d)/(c^2*d - e)]]*ArcTan[c*x]*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*(Cos[2*ArcTan[
c*x]] + I*Sin[2*ArcTan[c*x]]))/(c^2*d - e)] + 2*c^2*d*ArcTan[c*x]^2*Log[1 + ((c^2*d + e + 2*Sqrt[c^2*d*e])*(Co
s[2*ArcTan[c*x]] + I*Sin[2*ArcTan[c*x]]))/(c^2*d - e)] - (4*I)*c^2*d*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c
*x])] + (2*I)*c^2*d*ArcTan[c*x]*PolyLog[2, ((-(c*Sqrt[d]) + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] + Sqrt[
e])] + (2*I)*c^2*d*ArcTan[c*x]*PolyLog[2, -(((c*Sqrt[d] + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] - Sqrt[e]
))] + 2*c^2*d*PolyLog[3, -E^((2*I)*ArcTan[c*x])] - c^2*d*PolyLog[3, ((-(c*Sqrt[d]) + Sqrt[e])*E^((2*I)*ArcTan[
c*x]))/(c*Sqrt[d] + Sqrt[e])] - c^2*d*PolyLog[3, -(((c*Sqrt[d] + Sqrt[e])*E^((2*I)*ArcTan[c*x]))/(c*Sqrt[d] -
Sqrt[e]))]))/c^2)/(4*e^2)

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Maple [F]  time = 18.961, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3} \left ( a+b\arctan \left ( cx \right ) \right ) ^{2}}{e{x}^{2}+d}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{2}{\left (\frac{x^{2}}{e} - \frac{d \log \left (e x^{2} + d\right )}{e^{2}}\right )} + \int \frac{b^{2} x^{3} \arctan \left (c x\right )^{2} + 2 \, a b x^{3} \arctan \left (c x\right )}{e x^{2} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(x^2/e - d*log(e*x^2 + d)/e^2) + integrate((b^2*x^3*arctan(c*x)^2 + 2*a*b*x^3*arctan(c*x))/(e*x^2 + d)
, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} x^{3} \arctan \left (c x\right )^{2} + 2 \, a b x^{3} \arctan \left (c x\right ) + a^{2} x^{3}}{e x^{2} + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^3*arctan(c*x)^2 + 2*a*b*x^3*arctan(c*x) + a^2*x^3)/(e*x^2 + d), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{3}}{e x^{2} + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c*x) + a)^2*x^3/(e*x^2 + d), x)