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

Optimal. Leaf size=383 \[ \frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (-c x+i)}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (-c x+i)^2}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (-c x+i)}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (-c x+i)^2}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac{2 i b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}+\frac{3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3}+\frac{21 i b^2}{16 c^4 d^3 (-c x+i)}+\frac{b^2}{16 c^4 d^3 (-c x+i)^2}-\frac{21 i b^2 \tan ^{-1}(c x)}{16 c^4 d^3} \]

[Out]

b^2/(16*c^4*d^3*(I - c*x)^2) + (((21*I)/16)*b^2)/(c^4*d^3*(I - c*x)) - (((21*I)/16)*b^2*ArcTan[c*x])/(c^4*d^3)
+ ((I/4)*b*(a + b*ArcTan[c*x]))/(c^4*d^3*(I - c*x)^2) - (11*b*(a + b*ArcTan[c*x]))/(4*c^4*d^3*(I - c*x)) + (3
*(a + b*ArcTan[c*x])^2)/(8*c^4*d^3) + (I*x*(a + b*ArcTan[c*x])^2)/(c^3*d^3) - (a + b*ArcTan[c*x])^2/(2*c^4*d^3
*(I - c*x)^2) - ((3*I)*(a + b*ArcTan[c*x])^2)/(c^4*d^3*(I - c*x)) + ((2*I)*b*(a + b*ArcTan[c*x])*Log[2/(1 + I*
c*x)])/(c^4*d^3) + (3*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^4*d^3) - (b^2*PolyLog[2, 1 - 2/(1 + I*c*x)]
)/(c^4*d^3) + ((3*I)*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^4*d^3) + (3*b^2*PolyLog[3, 1 - 2/
(1 + I*c*x)])/(2*c^4*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.67463, antiderivative size = 383, normalized size of antiderivative = 1., number of steps used = 31, number of rules used = 14, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.56, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4864, 4862, 627, 44, 203, 4884, 4994, 6610} \[ \frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (-c x+i)}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (-c x+i)^2}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (-c x+i)}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (-c x+i)^2}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac{2 i b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}+\frac{3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3}+\frac{21 i b^2}{16 c^4 d^3 (-c x+i)}+\frac{b^2}{16 c^4 d^3 (-c x+i)^2}-\frac{21 i b^2 \tan ^{-1}(c x)}{16 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^2/(16*c^4*d^3*(I - c*x)^2) + (((21*I)/16)*b^2)/(c^4*d^3*(I - c*x)) - (((21*I)/16)*b^2*ArcTan[c*x])/(c^4*d^3)
+ ((I/4)*b*(a + b*ArcTan[c*x]))/(c^4*d^3*(I - c*x)^2) - (11*b*(a + b*ArcTan[c*x]))/(4*c^4*d^3*(I - c*x)) + (3
*(a + b*ArcTan[c*x])^2)/(8*c^4*d^3) + (I*x*(a + b*ArcTan[c*x])^2)/(c^3*d^3) - (a + b*ArcTan[c*x])^2/(2*c^4*d^3
*(I - c*x)^2) - ((3*I)*(a + b*ArcTan[c*x])^2)/(c^4*d^3*(I - c*x)) + ((2*I)*b*(a + b*ArcTan[c*x])*Log[2/(1 + I*
c*x)])/(c^4*d^3) + (3*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^4*d^3) - (b^2*PolyLog[2, 1 - 2/(1 + I*c*x)]
)/(c^4*d^3) + ((3*I)*b*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^4*d^3) + (3*b^2*PolyLog[3, 1 - 2/
(1 + I*c*x)])/(2*c^4*d^3)

Rule 4876

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

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 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 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 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 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 4864

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

Rule 4862

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

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 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 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 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^3 \left (a+b \tan ^{-1}(c x)\right )^2}{(d+i c d x)^3} \, dx &=\int \left (\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (-i+c x)^3}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (-i+c x)^2}-\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (-i+c x)}\right ) \, dx\\ &=\frac{i \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c^3 d^3}-\frac{(3 i) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{c^3 d^3}+\frac{\int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^3} \, dx}{c^3 d^3}-\frac{3 \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{-i+c x} \, dx}{c^3 d^3}\\ &=\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (i-c x)^2}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (i-c x)}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}-\frac{(6 i b) \int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}+\frac{b \int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^3}+\frac{a+b \tan ^{-1}(c x)}{4 (-i+c x)^2}-\frac{a+b \tan ^{-1}(c x)}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}-\frac{(6 b) \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^3 d^3}-\frac{(2 i b) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (i-c x)^2}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (i-c x)}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}-\frac{(i b) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{2 c^3 d^3}+\frac{(2 i b) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c^3 d^3}+\frac{b \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{4 c^3 d^3}-\frac{b \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{4 c^3 d^3}-\frac{(3 b) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^3 d^3}+\frac{(3 b) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^3 d^3}-\frac{\left (3 i b^2\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d^3}\\ &=\frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)^2}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (i-c x)^2}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (i-c x)}+\frac{2 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}-\frac{\left (i b^2\right ) \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{4 c^3 d^3}-\frac{\left (2 i b^2\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d^3}+\frac{b^2 \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{4 c^3 d^3}-\frac{\left (3 b^2\right ) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^3 d^3}\\ &=\frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)^2}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (i-c x)^2}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (i-c x)}+\frac{2 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c^4 d^3}-\frac{\left (i b^2\right ) \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{4 c^3 d^3}+\frac{b^2 \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{4 c^3 d^3}-\frac{\left (3 b^2\right ) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{c^3 d^3}\\ &=\frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)^2}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (i-c x)^2}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (i-c x)}+\frac{2 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}-\frac{\left (i b^2\right ) \int \left (-\frac{i}{2 (-i+c x)^3}+\frac{1}{4 (-i+c x)^2}-\frac{1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{4 c^3 d^3}+\frac{b^2 \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{4 c^3 d^3}-\frac{\left (3 b^2\right ) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}\\ &=\frac{b^2}{16 c^4 d^3 (i-c x)^2}+\frac{21 i b^2}{16 c^4 d^3 (i-c x)}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)^2}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (i-c x)^2}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (i-c x)}+\frac{2 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}+\frac{\left (i b^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{16 c^3 d^3}+\frac{\left (i b^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{8 c^3 d^3}-\frac{\left (3 i b^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{2 c^3 d^3}\\ &=\frac{b^2}{16 c^4 d^3 (i-c x)^2}+\frac{21 i b^2}{16 c^4 d^3 (i-c x)}-\frac{21 i b^2 \tan ^{-1}(c x)}{16 c^4 d^3}+\frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)^2}-\frac{11 b \left (a+b \tan ^{-1}(c x)\right )}{4 c^4 d^3 (i-c x)}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^4 d^3}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 c^4 d^3 (i-c x)^2}-\frac{3 i \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d^3 (i-c x)}+\frac{2 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d^3}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^4 d^3}+\frac{3 b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d^3}\\ \end{align*}

Mathematica [A]  time = 1.52805, size = 507, normalized size = 1.32 \[ \frac{4 i a b \left (-48 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-16 \log \left (c^2 x^2+1\right )-96 \tan ^{-1}(c x)^2-20 i \sin \left (2 \tan ^{-1}(c x)\right )+i \sin \left (4 \tan ^{-1}(c x)\right )+20 \cos \left (2 \tan ^{-1}(c x)\right )-\cos \left (4 \tan ^{-1}(c x)\right )+4 \tan ^{-1}(c x) \left (8 c x-24 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+10 \sin \left (2 \tan ^{-1}(c x)\right )-\sin \left (4 \tan ^{-1}(c x)\right )+10 i \cos \left (2 \tan ^{-1}(c x)\right )-i \cos \left (4 \tan ^{-1}(c x)\right )\right )\right )+i b^2 \left (-64 \left (3 \tan ^{-1}(c x)+i\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-96 i \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )-128 \tan ^{-1}(c x)^3+64 c x \tan ^{-1}(c x)^2-64 i \tan ^{-1}(c x)^2-192 i \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+128 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+80 \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )-8 \tan ^{-1}(c x)^2 \sin \left (4 \tan ^{-1}(c x)\right )-80 i \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )+4 i \tan ^{-1}(c x) \sin \left (4 \tan ^{-1}(c x)\right )-40 \sin \left (2 \tan ^{-1}(c x)\right )+\sin \left (4 \tan ^{-1}(c x)\right )+80 i \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )-8 i \tan ^{-1}(c x)^2 \cos \left (4 \tan ^{-1}(c x)\right )+80 \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )-4 \tan ^{-1}(c x) \cos \left (4 \tan ^{-1}(c x)\right )-40 i \cos \left (2 \tan ^{-1}(c x)\right )+i \cos \left (4 \tan ^{-1}(c x)\right )\right )-96 a^2 \log \left (c^2 x^2+1\right )+64 i a^2 c x+\frac{192 i a^2}{c x-i}-\frac{32 a^2}{(c x-i)^2}-192 i a^2 \tan ^{-1}(c x)}{64 c^4 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((64*I)*a^2*c*x - (32*a^2)/(-I + c*x)^2 + ((192*I)*a^2)/(-I + c*x) - (192*I)*a^2*ArcTan[c*x] - 96*a^2*Log[1 +
c^2*x^2] + (4*I)*a*b*(-96*ArcTan[c*x]^2 + 20*Cos[2*ArcTan[c*x]] - Cos[4*ArcTan[c*x]] - 16*Log[1 + c^2*x^2] - 4
8*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (20*I)*Sin[2*ArcTan[c*x]] + 4*ArcTan[c*x]*(8*c*x + (10*I)*Cos[2*ArcTan[
c*x]] - I*Cos[4*ArcTan[c*x]] - (24*I)*Log[1 + E^((2*I)*ArcTan[c*x])] + 10*Sin[2*ArcTan[c*x]] - Sin[4*ArcTan[c*
x]]) + I*Sin[4*ArcTan[c*x]]) + I*b^2*((-64*I)*ArcTan[c*x]^2 + 64*c*x*ArcTan[c*x]^2 - 128*ArcTan[c*x]^3 - (40*I
)*Cos[2*ArcTan[c*x]] + 80*ArcTan[c*x]*Cos[2*ArcTan[c*x]] + (80*I)*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + I*Cos[4*A
rcTan[c*x]] - 4*ArcTan[c*x]*Cos[4*ArcTan[c*x]] - (8*I)*ArcTan[c*x]^2*Cos[4*ArcTan[c*x]] + 128*ArcTan[c*x]*Log[
1 + E^((2*I)*ArcTan[c*x])] - (192*I)*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - 64*(I + 3*ArcTan[c*x])*Pol
yLog[2, -E^((2*I)*ArcTan[c*x])] - (96*I)*PolyLog[3, -E^((2*I)*ArcTan[c*x])] - 40*Sin[2*ArcTan[c*x]] - (80*I)*A
rcTan[c*x]*Sin[2*ArcTan[c*x]] + 80*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]] + Sin[4*ArcTan[c*x]] + (4*I)*ArcTan[c*x]*S
in[4*ArcTan[c*x]] - 8*ArcTan[c*x]^2*Sin[4*ArcTan[c*x]]))/(64*c^4*d^3)

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Maple [C]  time = 0.625, size = 5012, normalized size = 13.1 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{-i \, b^{2} x^{3} \log \left (-\frac{c x + i}{c x - i}\right )^{2} - 4 \, a b x^{3} \log \left (-\frac{c x + i}{c x - i}\right ) + 4 i \, a^{2} x^{3}}{4 \, c^{3} d^{3} x^{3} - 12 i \, c^{2} d^{3} x^{2} - 12 \, c d^{3} x + 4 i \, d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-I*b^2*x^3*log(-(c*x + I)/(c*x - I))^2 - 4*a*b*x^3*log(-(c*x + I)/(c*x - I)) + 4*I*a^2*x^3)/(4*c^3*d
^3*x^3 - 12*I*c^2*d^3*x^2 - 12*c*d^3*x + 4*I*d^3), 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/(d+I*c*d*x)**3,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}}{{\left (i \, c d x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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