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

Optimal. Leaf size=410 \[ \frac {3 i b^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac {3 i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d}+\frac {3 i b^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}+\frac {3 b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c^3 d}+\frac {3 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}-\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c^3 d}+\frac {3 i b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c^3 d}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c^3 d}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{i c x+1}\right )}{4 c^3 d} \]

[Out]

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

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Rubi [A]  time = 0.86, antiderivative size = 410, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {4866, 4852, 4916, 4846, 4920, 4854, 2402, 2315, 4884, 4994, 6610, 4998} \[ \frac {3 i b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}-\frac {3 i b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d}+\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}-\frac {3 b^3 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c^3 d}+\frac {3 i b^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d}+\frac {3 i b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c^3 d}+\frac {3 b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}-\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c^3 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(a + b*ArcTan[c*x])^2)/(2*c^3*d) + (((3*I)/2)*b*x*(a + b*ArcTan[c*x])^2)/(c^2*d) + ((I/2)*(a + b*ArcTan[
c*x])^3)/(c^3*d) + (x*(a + b*ArcTan[c*x])^3)/(c^2*d) - ((I/2)*x^2*(a + b*ArcTan[c*x])^3)/(c*d) + ((3*I)*b^2*(a
+ b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^3*d) + (3*b*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^3*d) - (I*(a
+ b*ArcTan[c*x])^3*Log[2/(1 + I*c*x)])/(c^3*d) - (3*b^3*PolyLog[2, 1 - 2/(1 + I*c*x)])/(2*c^3*d) + ((3*I)*b^2
*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^3*d) + (3*b*(a + b*ArcTan[c*x])^2*PolyLog[2, 1 - 2/(1 +
I*c*x)])/(2*c^3*d) + (3*b^3*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*c^3*d) - (((3*I)/2)*b^2*(a + b*ArcTan[c*x])*Pol
yLog[3, 1 - 2/(1 + I*c*x)])/(c^3*d) - (3*b^3*PolyLog[4, 1 - 2/(1 + I*c*x)])/(4*c^3*d)

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 4866

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f)/e, Int[((f*x)^(m - 1)*(a + b*ArcTan[c*x])^p)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0] && GtQ[m, 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 4920

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u
])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - c*x))^2, 0]

Rule 4998

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a
+ b*ArcTan[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] - Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[k
+ 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[u^2 - (1 -
(2*I)/(I - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{d+i c d x} \, dx &=\frac {i \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{d+i c d x} \, dx}{c}-\frac {i \int x \left (a+b \tan ^{-1}(c x)\right )^3 \, dx}{c d}\\ &=-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}-\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d+i c d x} \, dx}{c^2}+\frac {(3 i b) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{2 d}+\frac {\int \left (a+b \tan ^{-1}(c x)\right )^3 \, dx}{c^2 d}\\ &=\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {(3 i b) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{2 c^2 d}-\frac {(3 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{2 c^2 d}+\frac {(3 i b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}-\frac {(3 b) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{c d}\\ &=\frac {3 i b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {(3 b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{i-c x} \, dx}{c^2 d}-\frac {\left (3 b^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}-\frac {\left (3 i b^2\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c d}\\ &=-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {3 i b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {\left (3 i b^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c^2 d}-\frac {\left (6 b^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}+\frac {\left (3 i b^3\right ) \int \frac {\text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 c^2 d}\\ &=-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {3 i b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1+i c x}\right )}{4 c^3 d}-\frac {\left (3 i b^3\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}-\frac {\left (3 i b^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d}\\ &=-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {3 i b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1+i c x}\right )}{4 c^3 d}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^3 d}\\ &=-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d}+\frac {3 i b x \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c^3 d}+\frac {x \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c^3 d}-\frac {3 b^3 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^3 d}+\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c^3 d}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1+i c x}\right )}{4 c^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.03, size = 541, normalized size = 1.32 \[ -\frac {i \left (2 a^3 c^2 x^2-2 a^3 \log \left (c^2 x^2+1\right )+4 i a^3 c x-4 i a^3 \tan ^{-1}(c x)-6 i a^2 b \log \left (c^2 x^2+1\right )+6 a^2 b c^2 x^2 \tan ^{-1}(c x)-6 a^2 b c x-12 i a^2 b \tan ^{-1}(c x)^2+6 a^2 b \tan ^{-1}(c x)+12 i a^2 b c x \tan ^{-1}(c x)+12 a^2 b \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+6 a b^2 \log \left (c^2 x^2+1\right )+6 a b^2 c^2 x^2 \tan ^{-1}(c x)^2+6 b^2 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right ) \left (a+b \tan ^{-1}(c x)+i b\right )-8 i a b^2 \tan ^{-1}(c x)^3+18 a b^2 \tan ^{-1}(c x)^2+12 i a b^2 c x \tan ^{-1}(c x)^2-12 a b^2 c x \tan ^{-1}(c x)+12 a b^2 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+24 i a b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-6 i b \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right ) \left (a+b \tan ^{-1}(c x)+i b\right )^2+2 b^3 c^2 x^2 \tan ^{-1}(c x)^3+3 i b^3 \text {Li}_4\left (-e^{2 i \tan ^{-1}(c x)}\right )-2 i b^3 \tan ^{-1}(c x)^4+6 b^3 \tan ^{-1}(c x)^3+4 i b^3 c x \tan ^{-1}(c x)^3+6 i b^3 \tan ^{-1}(c x)^2-6 b^3 c x \tan ^{-1}(c x)^2+4 b^3 \tan ^{-1}(c x)^3 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+12 i b^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-12 b^3 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{4 c^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/4*I)*((4*I)*a^3*c*x - 6*a^2*b*c*x + 2*a^3*c^2*x^2 - (4*I)*a^3*ArcTan[c*x] + 6*a^2*b*ArcTan[c*x] + (12*I)*
a^2*b*c*x*ArcTan[c*x] - 12*a*b^2*c*x*ArcTan[c*x] + 6*a^2*b*c^2*x^2*ArcTan[c*x] - (12*I)*a^2*b*ArcTan[c*x]^2 +
18*a*b^2*ArcTan[c*x]^2 + (6*I)*b^3*ArcTan[c*x]^2 + (12*I)*a*b^2*c*x*ArcTan[c*x]^2 - 6*b^3*c*x*ArcTan[c*x]^2 +
6*a*b^2*c^2*x^2*ArcTan[c*x]^2 - (8*I)*a*b^2*ArcTan[c*x]^3 + 6*b^3*ArcTan[c*x]^3 + (4*I)*b^3*c*x*ArcTan[c*x]^3
+ 2*b^3*c^2*x^2*ArcTan[c*x]^3 - (2*I)*b^3*ArcTan[c*x]^4 + 12*a^2*b*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])]
+ (24*I)*a*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - 12*b^3*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])]
+ 12*a*b^2*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (12*I)*b^3*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x
])] + 4*b^3*ArcTan[c*x]^3*Log[1 + E^((2*I)*ArcTan[c*x])] - 2*a^3*Log[1 + c^2*x^2] - (6*I)*a^2*b*Log[1 + c^2*x^
2] + 6*a*b^2*Log[1 + c^2*x^2] - (6*I)*b*(a + I*b + b*ArcTan[c*x])^2*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 6*b^2
*(a + I*b + b*ArcTan[c*x])*PolyLog[3, -E^((2*I)*ArcTan[c*x])] + (3*I)*b^3*PolyLog[4, -E^((2*I)*ArcTan[c*x])]))
/(c^3*d)

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{3} x^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{3} - 6 i \, a b^{2} x^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 12 \, a^{2} b x^{2} \log \left (-\frac {c x + i}{c x - i}\right ) + 8 i \, a^{3} x^{2}}{8 \, c d x - 8 i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^3*x^2*log(-(c*x + I)/(c*x - I))^3 - 6*I*a*b^2*x^2*log(-(c*x + I)/(c*x - I))^2 - 12*a^2*b*x^2*log(
-(c*x + I)/(c*x - I)) + 8*I*a^3*x^2)/(8*c*d*x - 8*I*d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 12.33, size = 1725, normalized size = 4.21 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*b^3/d*arctan(c*x)^3*x-1/2*I/c*a^3/d*x^2+3/2/c^3*b*a^2/d*ln(c*x-I)*ln(-1/2*I*(I+c*x))-3/c^3*a*b^2/d*arcta
n(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+6/c^3*a*b^2/d*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+6/c^3
*a*b^2/d*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*I/c^2*b*a^2/d*x-9/4*I/c^3*b*a^2/d*arctan(c*x)-1/2
*I/c*b^3/d*arctan(c*x)^3*x^2+3/2*I/c^2*b^3/d*arctan(c*x)^2*x-3/2*I/c^3*a*b^2/d*polylog(3,-(1+I*c*x)^2/(c^2*x^2
+1))-6*I/c^3*a*b^2/d*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-6*I/c^3*a*b^2/d*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1
/2))+3*I/c^3*a*b^2/d*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+3/8*I/c^3*b*a^2/d*arctan(1/6*c^3*x^3+7/6*c*x)-3/8*I/c^3*b*a
^2/d*arctan(1/2*c*x)-I/c^3*b^3/d*arctan(c*x)^3*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-3/2*I/c^3*b^3/d*arctan(c*x)*polyl
og(3,-(1+I*c*x)^2/(c^2*x^2+1))+3*I/c^3*b^3/d*arctan(c*x)*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-3*I/c^3*b^3/d*arctan(c*
x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+3/4*I/c^3*b*a^2/d*arctan(1/2*c*x-1/2*I)-9/2*I/c^3*a*b^2/d*arctan(c*x)^2
+3/c^3*a*b^2/d*Pi*arctan(c*x)^2+3/c^2*a*b^2/d*arctan(c*x)^2*x+3/c^2*b*a^2/d*arctan(c*x)*x+1/2*I/c^3*a^3/d*ln(c
^2*x^2+1)-3/2*I/c^3*b^3/d*arctan(c*x)^3+3/2/c^3*b*a^2/d+1/c^2*a^3/d*x+3/2/c^3*b^3/d*arctan(c*x)^2+3/2/c^3*b^3/
d*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+3/2/c^3*b^3/d*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-1/c^3*a^3/d*arctan(c*x
)+3/4/c^3*b^3/d*polylog(4,-(1+I*c*x)^2/(c^2*x^2+1))-1/2/c^3*b^3/d*arctan(c*x)^4-2/c^3*a*b^2/d*arctan(c*x)^3+3/
c^3*b^3/d*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-3/2/c^3*b^3/d*arctan(c*x)^2*polylog(2,-(1+I*c*x)^2/(c^2*
x^2+1))+3/c^3*a*b^2/d*arctan(c*x)+3/2/c^3*b*a^2/d*dilog(-1/2*I*(I+c*x))-3/4/c^3*b*a^2/d*ln(c*x-I)^2-3/16/c^3*b
*a^2/d*ln(c^4*x^4+10*c^2*x^2+9)-9/8/c^3*b*a^2/d*ln(c^2*x^2+1)-3/2/c^3*a*b^2/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+
1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-3
/2/c^3*a*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2-3/c^3*a*b^2/d*Pi*c
sgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+3*I/c^3*b*a^2/d*arctan(c*x)*ln(c*x-I)
-3*I/c^3*a*b^2/d*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))+3*I/c^3*a*b^2/d*arctan(c*x)^2*ln(c*x-I)-3/2*I/c
*b*a^2/d*arctan(c*x)*x^2-3/2*I/c*a*b^2/d*arctan(c*x)^2*x^2+3*I/c^2*a*b^2/d*arctan(c*x)*x+3/2/c^3*a*b^2/d*Pi*cs
gn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-3/
2/c^3*a*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arc
tan(c*x)^2

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*(a + b*atan(c*x))^3)/(d + c*d*x*1i),x)

[Out]

int((x^2*(a + b*atan(c*x))^3)/(d + c*d*x*1i), x)

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

[Out]

Timed out

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