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

Optimal. Leaf size=356 \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d}-\frac{4 b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{i b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^2 d}-\frac{a b x}{c^3 d}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}-\frac{5 \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d}+\frac{8 i b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^4 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^4 d}-\frac{i b^2 x}{3 c^3 d}-\frac{b^2 x \tan ^{-1}(c x)}{c^3 d}+\frac{i b^2 \tan ^{-1}(c x)}{3 c^4 d} \]

[Out]

-((a*b*x)/(c^3*d)) - ((I/3)*b^2*x)/(c^3*d) + ((I/3)*b^2*ArcTan[c*x])/(c^4*d) - (b^2*x*ArcTan[c*x])/(c^3*d) + (
(I/3)*b*x^2*(a + b*ArcTan[c*x]))/(c^2*d) - (5*(a + b*ArcTan[c*x])^2)/(6*c^4*d) + (I*x*(a + b*ArcTan[c*x])^2)/(
c^3*d) + (x^2*(a + b*ArcTan[c*x])^2)/(2*c^2*d) - ((I/3)*x^3*(a + b*ArcTan[c*x])^2)/(c*d) + (((8*I)/3)*b*(a + b
*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^4*d) + ((a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^4*d) + (b^2*Log[1 +
c^2*x^2])/(2*c^4*d) - (4*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(3*c^4*d) + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1
- 2/(1 + I*c*x)])/(c^4*d) + (b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*c^4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.823159, antiderivative size = 356, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 14, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.56, Rules used = {4866, 4852, 4916, 321, 203, 4920, 4854, 2402, 2315, 4846, 260, 4884, 4994, 6610} \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d}-\frac{4 b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^4 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac{i b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^2 d}-\frac{a b x}{c^3 d}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}-\frac{5 \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^4 d}+\frac{8 i b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^4 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 c^4 d}-\frac{i b^2 x}{3 c^3 d}-\frac{b^2 x \tan ^{-1}(c x)}{c^3 d}+\frac{i b^2 \tan ^{-1}(c x)}{3 c^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*b*x)/(c^3*d)) - ((I/3)*b^2*x)/(c^3*d) + ((I/3)*b^2*ArcTan[c*x])/(c^4*d) - (b^2*x*ArcTan[c*x])/(c^3*d) + (
(I/3)*b*x^2*(a + b*ArcTan[c*x]))/(c^2*d) - (5*(a + b*ArcTan[c*x])^2)/(6*c^4*d) + (I*x*(a + b*ArcTan[c*x])^2)/(
c^3*d) + (x^2*(a + b*ArcTan[c*x])^2)/(2*c^2*d) - ((I/3)*x^3*(a + b*ArcTan[c*x])^2)/(c*d) + (((8*I)/3)*b*(a + b
*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^4*d) + ((a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/(c^4*d) + (b^2*Log[1 +
c^2*x^2])/(2*c^4*d) - (4*b^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/(3*c^4*d) + (I*b*(a + b*ArcTan[c*x])*PolyLog[2, 1
- 2/(1 + I*c*x)])/(c^4*d) + (b^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/(2*c^4*d)

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 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 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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 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 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 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} \, dx &=\frac{i \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d+i c d x} \, dx}{c}-\frac{i \int x^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}-\frac{\int \frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{d+i c d x} \, dx}{c^2}+\frac{(2 i b) \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 d}+\frac{\int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c^2 d}\\ &=\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}-\frac{i \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{d+i c d x} \, dx}{c^3}+\frac{i \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c^3 d}+\frac{(2 i b) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{3 c^2 d}-\frac{(2 i b) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 c^2 d}-\frac{b \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c d}\\ &=\frac{i b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{3 c^4 d}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d}+\frac{(2 i b) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{3 c^3 d}-\frac{b \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^3 d}+\frac{b \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^3 d}-\frac{(2 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}-\frac{(2 i b) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c^2 d}-\frac{\left (i b^2\right ) \int \frac{x^2}{1+c^2 x^2} \, dx}{3 c d}\\ &=-\frac{a b x}{c^3 d}-\frac{i b^2 x}{3 c^3 d}+\frac{i b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^2 d}-\frac{5 \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}+\frac{2 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d}+\frac{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}+\frac{(2 i b) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c^3 d}+\frac{\left (i b^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{3 c^3 d}-\frac{\left (2 i b^2\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{3 c^3 d}-\frac{\left (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}-\frac{b^2 \int \tan ^{-1}(c x) \, dx}{c^3 d}\\ &=-\frac{a b x}{c^3 d}-\frac{i b^2 x}{3 c^3 d}+\frac{i b^2 \tan ^{-1}(c x)}{3 c^4 d}-\frac{b^2 x \tan ^{-1}(c x)}{c^3 d}+\frac{i b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^2 d}-\frac{5 \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}+\frac{8 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d}+\frac{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}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d}-\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 )}{3 c^4 d}-\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}+\frac{b^2 \int \frac{x}{1+c^2 x^2} \, dx}{c^2 d}\\ &=-\frac{a b x}{c^3 d}-\frac{i b^2 x}{3 c^3 d}+\frac{i b^2 \tan ^{-1}(c x)}{3 c^4 d}-\frac{b^2 x \tan ^{-1}(c x)}{c^3 d}+\frac{i b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^2 d}-\frac{5 \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}+\frac{8 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{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}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d}-\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}\\ &=-\frac{a b x}{c^3 d}-\frac{i b^2 x}{3 c^3 d}+\frac{i b^2 \tan ^{-1}(c x)}{3 c^4 d}-\frac{b^2 x \tan ^{-1}(c x)}{c^3 d}+\frac{i b x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c^2 d}-\frac{5 \left (a+b \tan ^{-1}(c x)\right )^2}{6 c^4 d}+\frac{i x \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d}+\frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{i x^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c d}+\frac{8 i b \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^4 d}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 c^4 d}-\frac{4 b^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{3 c^4 d}+\frac{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}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^4 d}\\ \end{align*}

Mathematica [A]  time = 1.01068, size = 421, normalized size = 1.18 \[ -\frac{i a b \left (3 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-8 \log \left (\frac{1}{\sqrt{c^2 x^2+1}}\right )+\left (c^2 x^2+1\right ) \left (2 c x \tan ^{-1}(c x)+3 i \tan ^{-1}(c x)-1\right )-3 i c x+6 \tan ^{-1}(c x)^2-8 c x \tan ^{-1}(c x)+6 i \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{3 c^4 d}-\frac{i b^2 \left (\left (6 \tan ^{-1}(c x)+8 i\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+3 i \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )-6 i \log \left (\frac{1}{\sqrt{c^2 x^2+1}}\right )+2 c x \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2+3 i \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2-2 \left (c^2 x^2+1\right ) \tan ^{-1}(c x)+2 c x+4 \tan ^{-1}(c x)^3-8 c x \tan ^{-1}(c x)^2+8 i \tan ^{-1}(c x)^2-6 i c x \tan ^{-1}(c x)+6 i \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-16 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{6 c^4 d}+\frac{a^2 x^2}{2 c^2 d}-\frac{a^2 \log \left (c^2 x^2+1\right )}{2 c^4 d}+\frac{i a^2 x}{c^3 d}-\frac{i a^2 \tan ^{-1}(c x)}{c^4 d}-\frac{i a^2 x^3}{3 c d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*a^2*x)/(c^3*d) + (a^2*x^2)/(2*c^2*d) - ((I/3)*a^2*x^3)/(c*d) - (I*a^2*ArcTan[c*x])/(c^4*d) - (a^2*Log[1 + c
^2*x^2])/(2*c^4*d) - ((I/3)*a*b*((-3*I)*c*x - 8*c*x*ArcTan[c*x] + 6*ArcTan[c*x]^2 + (1 + c^2*x^2)*(-1 + (3*I)*
ArcTan[c*x] + 2*c*x*ArcTan[c*x]) + (6*I)*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - 8*Log[1/Sqrt[1 + c^2*x^2
]] + 3*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/(c^4*d) - ((I/6)*b^2*(2*c*x - (6*I)*c*x*ArcTan[c*x] - 2*(1 + c^2*x
^2)*ArcTan[c*x] + (8*I)*ArcTan[c*x]^2 - 8*c*x*ArcTan[c*x]^2 + (3*I)*(1 + c^2*x^2)*ArcTan[c*x]^2 + 2*c*x*(1 + c
^2*x^2)*ArcTan[c*x]^2 + 4*ArcTan[c*x]^3 - 16*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + (6*I)*ArcTan[c*x]^2*
Log[1 + E^((2*I)*ArcTan[c*x])] - (6*I)*Log[1/Sqrt[1 + c^2*x^2]] + (8*I + 6*ArcTan[c*x])*PolyLog[2, -E^((2*I)*A
rcTan[c*x])] + (3*I)*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/(c^4*d)

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Maple [C]  time = 2.46, size = 1331, normalized size = 3.7 \begin{align*} \text{result 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),x)

[Out]

-1/2*I/c^4*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+1/2*I/c^4*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)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-1/2*I/c^4*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*arctan(c*x)^2+1/2/c^2*b^2/d*arctan(c*x)^2*x^
2+I/c^3*a^2/d*x+5/12/c^4*a*b/d*arctan(1/2*c*x)-5/12/c^4*a*b/d*arctan(1/6*c^3*x^3+7/6*c*x)-5/6/c^4*a*b/d*arctan
(1/2*c*x-1/2*I)+11/6/c^4*a*b/d*arctan(c*x)+1/c^4*b^2/d*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-1/c^4*b^2
/d*arctan(c*x)^2*ln(c*x-I)-1/3*I/c*a^2/d*x^3-I/c^4*a^2/d*arctan(c*x)-2/3*I/c^4*b^2/d*arctan(c*x)^3+4/3*I/c^4*b
^2/d*arctan(c*x)+4/3*I/c^4*a*b/d-a*b*x/d/c^3-b^2*x*arctan(c*x)/d/c^3-1/3*I*b^2*x/d/c^3-I/c^4*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))^2*arctan(c*x)^2+I/c^4*a*b/d*ln(c*x-I)*ln(-1/2*I*(c*x+I))+2*I
/c^3*a*b/d*arctan(c*x)*x+1/3/c^4*b^2/d-2/3*I/c*a*b/d*arctan(c*x)*x^3-1/2*I/c^4*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+1/2/c^2*a^2/d*x^2-1/2/c^4*a^2/d*ln(c^2*x^2+1)+11/6/c^4*b^2
/d*arctan(c*x)^2-1/c^4*b^2/d*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/2/c^4*b^2/d*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+8
/3/c^4*b^2/d*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+8/3/c^4*b^2/d*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I/c^4
*b^2/d*Pi*arctan(c*x)^2+I/c^4*a*b/d*dilog(-1/2*I*(c*x+I))-2/c^4*a*b/d*arctan(c*x)*ln(c*x-I)+I/c^3*b^2/d*arctan
(c*x)^2*x+1/c^2*a*b/d*arctan(c*x)*x^2-11/12*I/c^4*a*b/d*ln(c^2*x^2+1)-1/2*I/c^4*a*b/d*ln(c*x-I)^2-5/24*I/c^4*a
*b/d*ln(c^4*x^4+10*c^2*x^2+9)+8/3*I/c^4*b^2/d*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-I/c^4*b^2/d*arct
an(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+8/3*I/c^4*b^2/d*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/
3*I/c^2*a*b/d*x^2-1/3*I/c*b^2/d*arctan(c*x)^2*x^3+1/3*I/c^2*b^2/d*arctan(c*x)*x^2

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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),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 d x - 4 i \, 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/(d+I*c*d*x),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*d*x
- 4*I*d), x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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),x)

[Out]

Exception raised: AttributeError

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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}}{i \, c d x + 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/(d+I*c*d*x),x, algorithm="giac")

[Out]

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