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

Optimal. Leaf size=307 \[ -\frac{6 b^2 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{5 c^2}+\frac{1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}+\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{12 i b d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}+\frac{1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{5 a b d^3 x}{2 c}+\frac{3 b^2 d^3 \log \left (c^2 x^2+1\right )}{2 c^2}-\frac{13 i b^2 d^3 \tan ^{-1}(c x)}{10 c^2}-\frac{1}{30} i b^2 c d^3 x^3+\frac{13 i b^2 d^3 x}{10 c}-\frac{5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac{1}{4} b^2 d^3 x^2 \]

[Out]

(-5*a*b*d^3*x)/(2*c) + (((13*I)/10)*b^2*d^3*x)/c - (b^2*d^3*x^2)/4 - (I/30)*b^2*c*d^3*x^3 - (((13*I)/10)*b^2*d
^3*ArcTan[c*x])/c^2 - (5*b^2*d^3*x*ArcTan[c*x])/(2*c) - ((6*I)/5)*b*d^3*x^2*(a + b*ArcTan[c*x]) + (b*c*d^3*x^3
*(a + b*ArcTan[c*x]))/2 + (I/10)*b*c^2*d^3*x^4*(a + b*ArcTan[c*x]) + (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x])^2)
/(4*c^2) - (d^3*(1 + I*c*x)^5*(a + b*ArcTan[c*x])^2)/(5*c^2) - (((12*I)/5)*b*d^3*(a + b*ArcTan[c*x])*Log[2/(1
- I*c*x)])/c^2 + (3*b^2*d^3*Log[1 + c^2*x^2])/(2*c^2) - (6*b^2*d^3*PolyLog[2, 1 - 2/(1 - I*c*x)])/(5*c^2)

________________________________________________________________________________________

Rubi [A]  time = 0.613652, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 38, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used = {4876, 4864, 4846, 260, 4852, 321, 203, 266, 43, 1586, 4854, 2402, 2315, 302} \[ -\frac{6 b^2 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{5 c^2}+\frac{1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}+\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{12 i b d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}+\frac{1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{5 a b d^3 x}{2 c}+\frac{3 b^2 d^3 \log \left (c^2 x^2+1\right )}{2 c^2}-\frac{13 i b^2 d^3 \tan ^{-1}(c x)}{10 c^2}-\frac{1}{30} i b^2 c d^3 x^3+\frac{13 i b^2 d^3 x}{10 c}-\frac{5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac{1}{4} b^2 d^3 x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*b*d^3*x)/(2*c) + (((13*I)/10)*b^2*d^3*x)/c - (b^2*d^3*x^2)/4 - (I/30)*b^2*c*d^3*x^3 - (((13*I)/10)*b^2*d
^3*ArcTan[c*x])/c^2 - (5*b^2*d^3*x*ArcTan[c*x])/(2*c) - ((6*I)/5)*b*d^3*x^2*(a + b*ArcTan[c*x]) + (b*c*d^3*x^3
*(a + b*ArcTan[c*x]))/2 + (I/10)*b*c^2*d^3*x^4*(a + b*ArcTan[c*x]) + (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x])^2)
/(4*c^2) - (d^3*(1 + I*c*x)^5*(a + b*ArcTan[c*x])^2)/(5*c^2) - (((12*I)/5)*b*d^3*(a + b*ArcTan[c*x])*Log[2/(1
- I*c*x)])/c^2 + (3*b^2*d^3*Log[1 + c^2*x^2])/(2*c^2) - (6*b^2*d^3*PolyLog[2, 1 - 2/(1 - I*c*x)])/(5*c^2)

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 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 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 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps

\begin{align*} \int x (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (\frac{i (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{i (d+i c d x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{c d}\right ) \, dx\\ &=\frac{i \int (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c}-\frac{i \int (d+i c d x)^4 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}+\frac{(2 b) \int \left (-15 d^5 \left (a+b \tan ^{-1}(c x)\right )-11 i c d^5 x \left (a+b \tan ^{-1}(c x)\right )+5 c^2 d^5 x^2 \left (a+b \tan ^{-1}(c x)\right )+i c^3 d^5 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{16 i \left (i d^5-c d^5 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}\right ) \, dx}{5 c d^2}-\frac{b \int \left (-7 d^4 \left (a+b \tan ^{-1}(c x)\right )-4 i c d^4 x \left (a+b \tan ^{-1}(c x)\right )+c^2 d^4 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{8 i \left (i d^4-c d^4 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}\right ) \, dx}{2 c d}\\ &=\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac{(32 i b) \int \frac{\left (i d^5-c d^5 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c d^2}+\frac{(4 i b) \int \frac{\left (i d^4-c d^4 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c d}+\left (2 i b d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx-\frac{1}{5} \left (22 i b d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac{\left (7 b d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}-\frac{\left (6 b d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}-\frac{1}{2} \left (b c d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\left (2 b c d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac{1}{5} \left (2 i b c^2 d^3\right ) \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=-\frac{5 a b d^3 x}{2 c}-\frac{6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac{(32 i b) \int \frac{a+b \tan ^{-1}(c x)}{-\frac{i}{d^5}-\frac{c x}{d^5}} \, dx}{5 c d^2}+\frac{(4 i b) \int \frac{a+b \tan ^{-1}(c x)}{-\frac{i}{d^4}-\frac{c x}{d^4}} \, dx}{c d}+\frac{\left (7 b^2 d^3\right ) \int \tan ^{-1}(c x) \, dx}{2 c}-\frac{\left (6 b^2 d^3\right ) \int \tan ^{-1}(c x) \, dx}{c}-\left (i b^2 c d^3\right ) \int \frac{x^2}{1+c^2 x^2} \, dx+\frac{1}{5} \left (11 i b^2 c d^3\right ) \int \frac{x^2}{1+c^2 x^2} \, dx+\frac{1}{6} \left (b^2 c^2 d^3\right ) \int \frac{x^3}{1+c^2 x^2} \, dx-\frac{1}{3} \left (2 b^2 c^2 d^3\right ) \int \frac{x^3}{1+c^2 x^2} \, dx-\frac{1}{10} \left (i b^2 c^3 d^3\right ) \int \frac{x^4}{1+c^2 x^2} \, dx\\ &=-\frac{5 a b d^3 x}{2 c}+\frac{6 i b^2 d^3 x}{5 c}-\frac{5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac{6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac{12 i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{5 c^2}-\frac{1}{2} \left (7 b^2 d^3\right ) \int \frac{x}{1+c^2 x^2} \, dx+\left (6 b^2 d^3\right ) \int \frac{x}{1+c^2 x^2} \, dx+\frac{\left (i b^2 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{c}-\frac{\left (11 i b^2 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{5 c}-\frac{\left (4 i b^2 d^3\right ) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{c}+\frac{\left (32 i b^2 d^3\right ) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{5 c}+\frac{1}{12} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )-\frac{1}{3} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )-\frac{1}{10} \left (i b^2 c^3 d^3\right ) \int \left (-\frac{1}{c^4}+\frac{x^2}{c^2}+\frac{1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{5 a b d^3 x}{2 c}+\frac{13 i b^2 d^3 x}{10 c}-\frac{1}{30} i b^2 c d^3 x^3-\frac{6 i b^2 d^3 \tan ^{-1}(c x)}{5 c^2}-\frac{5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac{6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac{12 i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{5 c^2}+\frac{5 b^2 d^3 \log \left (1+c^2 x^2\right )}{4 c^2}+\frac{\left (4 b^2 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )}{c^2}-\frac{\left (32 b^2 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )}{5 c^2}-\frac{\left (i b^2 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx}{10 c}+\frac{1}{12} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )-\frac{1}{3} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{5 a b d^3 x}{2 c}+\frac{13 i b^2 d^3 x}{10 c}-\frac{1}{4} b^2 d^3 x^2-\frac{1}{30} i b^2 c d^3 x^3-\frac{13 i b^2 d^3 \tan ^{-1}(c x)}{10 c^2}-\frac{5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac{6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac{d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac{12 i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{5 c^2}+\frac{3 b^2 d^3 \log \left (1+c^2 x^2\right )}{2 c^2}-\frac{6 b^2 d^3 \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{5 c^2}\\ \end{align*}

Mathematica [A]  time = 1.28737, size = 325, normalized size = 1.06 \[ \frac{d^3 \left (-72 b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-12 i a^2 c^5 x^5-45 a^2 c^4 x^4+60 i a^2 c^3 x^3+30 a^2 c^2 x^2+6 i a b c^4 x^4+30 a b c^3 x^3-72 i a b c^2 x^2+72 i a b \log \left (c^2 x^2+1\right )+6 b \tan ^{-1}(c x) \left (a \left (-4 i c^5 x^5-15 c^4 x^4+20 i c^3 x^3+10 c^2 x^2+25\right )+b \left (i c^4 x^4+5 c^3 x^3-12 i c^2 x^2-25 c x-13 i\right )-24 i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-150 a b c x-18 i a b-2 i b^2 c^3 x^3-15 b^2 c^2 x^2+90 b^2 \log \left (c^2 x^2+1\right )+78 i b^2 c x+3 b^2 (1-4 i c x) (c x-i)^4 \tan ^{-1}(c x)^2-15 b^2\right )}{60 c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*((-18*I)*a*b - 15*b^2 - 150*a*b*c*x + (78*I)*b^2*c*x + 30*a^2*c^2*x^2 - (72*I)*a*b*c^2*x^2 - 15*b^2*c^2*x
^2 + (60*I)*a^2*c^3*x^3 + 30*a*b*c^3*x^3 - (2*I)*b^2*c^3*x^3 - 45*a^2*c^4*x^4 + (6*I)*a*b*c^4*x^4 - (12*I)*a^2
*c^5*x^5 + 3*b^2*(1 - (4*I)*c*x)*(-I + c*x)^4*ArcTan[c*x]^2 + 6*b*ArcTan[c*x]*(b*(-13*I - 25*c*x - (12*I)*c^2*
x^2 + 5*c^3*x^3 + I*c^4*x^4) + a*(25 + 10*c^2*x^2 + (20*I)*c^3*x^3 - 15*c^4*x^4 - (4*I)*c^5*x^5) - (24*I)*b*Lo
g[1 + E^((2*I)*ArcTan[c*x])]) + (72*I)*a*b*Log[1 + c^2*x^2] + 90*b^2*Log[1 + c^2*x^2] - 72*b^2*PolyLog[2, -E^(
(2*I)*ArcTan[c*x])]))/(60*c^2)

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Maple [B]  time = 0.1, size = 656, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*c*d^3*b^2*arctan(c*x)^2*x^3-1/5*I*c^3*d^3*b^2*arctan(c*x)^2*x^5+1/10*I*c^2*d^3*b^2*arctan(c*x)*x^4+6/5*I/c^2
*d^3*b^2*ln(c^2*x^2+1)*arctan(c*x)+1/10*I*c^2*d^3*a*b*x^4+6/5*I/c^2*d^3*a*b*ln(c^2*x^2+1)-3/2*c^2*d^3*a*b*arct
an(c*x)*x^4-3/4*c^2*d^3*a^2*x^4+1/2*d^3*b^2*arctan(c*x)^2*x^2-3/10/c^2*d^3*b^2*ln(c*x+I)^2+3/10/c^2*d^3*b^2*ln
(c*x-I)^2+5/4/c^2*d^3*b^2*arctan(c*x)^2-3/5/c^2*d^3*b^2*dilog(1/2*I*(c*x-I))+3/5/c^2*d^3*b^2*dilog(-1/2*I*(c*x
+I))-3/5/c^2*d^3*b^2*ln(c*x-I)*ln(c^2*x^2+1)+3/5/c^2*d^3*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))+3/5/c^2*d^3*b^2*ln(c
*x+I)*ln(c^2*x^2+1)-3/5/c^2*d^3*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))-1/5*I*c^3*d^3*a^2*x^5-6/5*I*d^3*b^2*arctan(c*x
)*x^2-6/5*I*d^3*a*b*x^2+1/2*c*d^3*b^2*arctan(c*x)*x^3-3/4*c^2*d^3*b^2*arctan(c*x)^2*x^4+d^3*a*b*arctan(c*x)*x^
2+5/2/c^2*d^3*a*b*arctan(c*x)+I*c*d^3*a^2*x^3+1/2*c*d^3*a*b*x^3-2/5*I*c^3*d^3*a*b*arctan(c*x)*x^5+2*I*c*d^3*a*
b*arctan(c*x)*x^3+1/2*d^3*a^2*x^2-1/4*b^2*d^3*x^2+13/10*I*b^2*d^3*x/c-5/2*a*b*d^3*x/c-5/2*b^2*d^3*x*arctan(c*x
)/c-1/30*I*b^2*c*d^3*x^3-13/10*I*b^2*d^3*arctan(c*x)/c^2+3/2*b^2*d^3*ln(c^2*x^2+1)/c^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*I*a^2*c^3*d^3*x^5 - 3/4*a^2*c^2*d^3*x^4 - 1/10*I*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^
2*x^2 + 1)/c^6))*a*b*c^3*d^3 + I*a^2*c*d^3*x^3 + 1/2*b^2*d^3*x^2*arctan(c*x)^2 - 1/2*(3*x^4*arctan(c*x) - c*((
c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*a*b*c^2*d^3 + I*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^
4))*a*b*c*d^3 + 1/2*a^2*d^3*x^2 + (x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a*b*d^3 - 1/2*(2*c*(x/c^2 -
arctan(c*x)/c^3)*arctan(c*x) + (arctan(c*x)^2 - log(c^2*x^2 + 1))/c^2)*b^2*d^3 - 1/320*(16*I*b^2*c^3*d^3*x^5 +
60*b^2*c^2*d^3*x^4 - 80*I*b^2*c*d^3*x^3)*arctan(c*x)^2 + 1/320*(16*b^2*c^3*d^3*x^5 - 60*I*b^2*c^2*d^3*x^4 - 8
0*b^2*c*d^3*x^3)*arctan(c*x)*log(c^2*x^2 + 1) - 1/320*(-4*I*b^2*c^3*d^3*x^5 - 15*b^2*c^2*d^3*x^4 + 20*I*b^2*c*
d^3*x^3)*log(c^2*x^2 + 1)^2 - I*integrate(1/80*(60*(b^2*c^5*d^3*x^6 - 2*b^2*c^3*d^3*x^4 - 3*b^2*c*d^3*x^2)*arc
tan(c*x)^2 + 5*(b^2*c^5*d^3*x^6 - 2*b^2*c^3*d^3*x^4 - 3*b^2*c*d^3*x^2)*log(c^2*x^2 + 1)^2 - 2*(19*b^2*c^4*d^3*
x^5 - 20*b^2*c^2*d^3*x^3)*arctan(c*x) + (4*b^2*c^5*d^3*x^6 - 35*b^2*c^3*d^3*x^4 - 60*(b^2*c^4*d^3*x^5 + b^2*c^
2*d^3*x^3)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x) - integrate(1/80*(180*(b^2*c^4*d^3*x^5 + b^2*c^2*d
^3*x^3)*arctan(c*x)^2 + 15*(b^2*c^4*d^3*x^5 + b^2*c^2*d^3*x^3)*log(c^2*x^2 + 1)^2 + 2*(4*b^2*c^5*d^3*x^6 - 35*
b^2*c^3*d^3*x^4)*arctan(c*x) + (19*b^2*c^4*d^3*x^5 - 20*b^2*c^2*d^3*x^3 + 20*(b^2*c^5*d^3*x^6 - 2*b^2*c^3*d^3*
x^4 - 3*b^2*c*d^3*x^2)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{80} \,{\left (4 i \, b^{2} c^{3} d^{3} x^{5} + 15 \, b^{2} c^{2} d^{3} x^{4} - 20 i \, b^{2} c d^{3} x^{3} - 10 \, b^{2} d^{3} x^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} +{\rm integral}\left (\frac{-20 i \, a^{2} c^{5} d^{3} x^{6} - 60 \, a^{2} c^{4} d^{3} x^{5} + 40 i \, a^{2} c^{3} d^{3} x^{4} - 40 \, a^{2} c^{2} d^{3} x^{3} + 60 i \, a^{2} c d^{3} x^{2} + 20 \, a^{2} d^{3} x +{\left (20 \, a b c^{5} d^{3} x^{6} +{\left (-60 i \, a b - 4 \, b^{2}\right )} c^{4} d^{3} x^{5} - 5 \,{\left (8 \, a b - 3 i \, b^{2}\right )} c^{3} d^{3} x^{4} +{\left (-40 i \, a b + 20 \, b^{2}\right )} c^{2} d^{3} x^{3} - 10 \,{\left (6 \, a b + i \, b^{2}\right )} c d^{3} x^{2} + 20 i \, a b d^{3} x\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{20 \,{\left (c^{2} x^{2} + 1\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(4*I*b^2*c^3*d^3*x^5 + 15*b^2*c^2*d^3*x^4 - 20*I*b^2*c*d^3*x^3 - 10*b^2*d^3*x^2)*log(-(c*x + I)/(c*x - I)
)^2 + integral(1/20*(-20*I*a^2*c^5*d^3*x^6 - 60*a^2*c^4*d^3*x^5 + 40*I*a^2*c^3*d^3*x^4 - 40*a^2*c^2*d^3*x^3 +
60*I*a^2*c*d^3*x^2 + 20*a^2*d^3*x + (20*a*b*c^5*d^3*x^6 + (-60*I*a*b - 4*b^2)*c^4*d^3*x^5 - 5*(8*a*b - 3*I*b^2
)*c^3*d^3*x^4 + (-40*I*a*b + 20*b^2)*c^2*d^3*x^3 - 10*(6*a*b + I*b^2)*c*d^3*x^2 + 20*I*a*b*d^3*x)*log(-(c*x +
I)/(c*x - I)))/(c^2*x^2 + 1), x)

________________________________________________________________________________________

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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