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

Optimal. Leaf size=590 \[ \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 e}-\frac {i b d \text {Li}_2\left (1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac {i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{2 e^2}+\frac {i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{2 e^2}+\frac {d \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{e^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}-i \sqrt {e}\right )}\right )}{2 e^2}-\frac {d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} x\right )}{(1-i c x) \left (c \sqrt {-d}+i \sqrt {e}\right )}\right )}{2 e^2}+\frac {x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}-\frac {a b x}{c e}+\frac {b^2 \log \left (c^2 x^2+1\right )}{2 c^2 e}+\frac {b^2 d \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 e^2}-\frac {b^2 d \text {Li}_3\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} x\right )}{\left (c \sqrt {-d}-i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^2}-\frac {b^2 d \text {Li}_3\left (1-\frac {2 c \left (\sqrt {e} x+\sqrt {-d}\right )}{\left (\sqrt {-d} c+i \sqrt {e}\right ) (1-i c x)}\right )}{4 e^2}-\frac {b^2 x \tan ^{-1}(c x)}{c e} \]

[Out]

-a*b*x/c/e-b^2*x*arctan(c*x)/c/e+1/2*(a+b*arctan(c*x))^2/c^2/e+1/2*x^2*(a+b*arctan(c*x))^2/e+d*(a+b*arctan(c*x
))^2*ln(2/(1-I*c*x))/e^2+1/2*b^2*ln(c^2*x^2+1)/c^2/e-1/2*d*(a+b*arctan(c*x))^2*ln(2*c*((-d)^(1/2)-x*e^(1/2))/(
1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/e^2-1/2*d*(a+b*arctan(c*x))^2*ln(2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-
d)^(1/2)+I*e^(1/2)))/e^2-I*b*d*(a+b*arctan(c*x))*polylog(2,1-2/(1-I*c*x))/e^2+1/2*I*b*d*(a+b*arctan(c*x))*poly
log(2,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/e^2+1/2*I*b*d*(a+b*arctan(c*x))*polylog
(2,1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/e^2+1/2*b^2*d*polylog(3,1-2/(1-I*c*x))/e^2
-1/4*b^2*d*polylog(3,1-2*c*((-d)^(1/2)-x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)-I*e^(1/2)))/e^2-1/4*b^2*d*polylog(3,
1-2*c*((-d)^(1/2)+x*e^(1/2))/(1-I*c*x)/(c*(-d)^(1/2)+I*e^(1/2)))/e^2

________________________________________________________________________________________

Rubi [A]  time = 0.50, antiderivative size = 590, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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 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 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])

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*}

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Mathematica [B]  time = 10.89, size = 1569, 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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{3} \arctan \left (c x\right )^{2} + 2 \, a b x^{3} \arctan \left (c x\right ) + a^{2} x^{3}}{e x^{2} + d}, x\right ) \]

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

[Out]

Timed out

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maple [F]  time = 56.82, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a +b \arctan \left (c x \right )\right )^{2}}{e \,x^{2}+d}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} {\left (\frac {x^{2}}{e} - \frac {d \log \left (e x^{2} + d\right )}{e^{2}}\right )} + \int \frac {b^{2} x^{3} \arctan \left (c x\right )^{2} + 2 \, a b x^{3} \arctan \left (c x\right )}{e x^{2} + d}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{e\,x^2+d} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{d + e x^{2}}\, dx \]

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]

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

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