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

Optimal. Leaf size=300 \[ -i b d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+a b c d^2 x+2 i c d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{5}{2} d^2 \left (a+b \tan ^{-1}(c x)\right )^2+4 i b d^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} b^2 d^2 \log \left (c^2 x^2+1\right )+b^2 c d^2 x \tan ^{-1}(c x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.57812, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 19, 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, 4850, 4988, 4884, 4994, 6610, 4852, 4916, 260} \[ -i b d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d^2 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+a b c d^2 x+2 i c d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{5}{2} d^2 \left (a+b \tan ^{-1}(c x)\right )^2+4 i b d^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} b^2 d^2 \log \left (c^2 x^2+1\right )+b^2 c d^2 x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*b*c*d^2*x + b^2*c*d^2*x*ArcTan[c*x] - (5*d^2*(a + b*ArcTan[c*x])^2)/2 + (2*I)*c*d^2*x*(a + b*ArcTan[c*x])^2
- (c^2*d^2*x^2*(a + b*ArcTan[c*x])^2)/2 + 2*d^2*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (4*I)*b*d^2
*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] - (b^2*d^2*Log[1 + c^2*x^2])/2 - 2*b^2*d^2*PolyLog[2, 1 - 2/(1 + I*c*x
)] - I*b*d^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*d^2*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2
/(1 + I*c*x)] - (b^2*d^2*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/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 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 4850

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

Rule 4988

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

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 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]

Rubi steps

\begin{align*} \int \frac{(d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx &=\int \left (2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-c^2 d^2 x \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^2 \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+\left (2 i c d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx-\left (c^2 d^2\right ) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=2 i c d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-\left (4 b c d^2\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (4 i b c^2 d^2\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\left (b c^3 d^2\right ) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 i c d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+\left (4 i b c d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx+\left (b c d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx-\left (b c d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx+\left (2 b c d^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-\left (2 b c d^2\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=a b c d^2 x-\frac{5}{2} d^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 i c d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )+\left (i b^2 c d^2\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (4 i b^2 c d^2\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (b^2 c d^2\right ) \int \tan ^{-1}(c x) \, dx\\ &=a b c d^2 x+b^2 c d^2 x \tan ^{-1}(c x)-\frac{5}{2} d^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 i c d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )-\left (4 b^2 d^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )-\left (b^2 c^2 d^2\right ) \int \frac{x}{1+c^2 x^2} \, dx\\ &=a b c d^2 x+b^2 c d^2 x \tan ^{-1}(c x)-\frac{5}{2} d^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 i c d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \log \left (1+c^2 x^2\right )-2 b^2 d^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )-i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d^2 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.651142, size = 360, normalized size = 1.2 \[ \frac{1}{2} d^2 \left (2 i a b (\text{PolyLog}(2,-i c x)-\text{PolyLog}(2,i c x))+4 b^2 \left (\text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\tan ^{-1}(c x) \left ((1+i c x) \tan ^{-1}(c x)+2 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )\right )+2 b^2 \left (i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+i \tan ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+\frac{2}{3} i \tan ^{-1}(c x)^3+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )+a^2 \left (-c^2\right ) x^2+4 i a^2 c x+2 a^2 \log (c x)-2 a b \left (\left (c^2 x^2+1\right ) \tan ^{-1}(c x)-c x\right )+4 i a b \left (2 c x \tan ^{-1}(c x)-\log \left (c^2 x^2+1\right )\right )-b^2 \log \left (c^2 x^2+1\right )-b^2 \left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2+2 b^2 c x \tan ^{-1}(c x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^2*((4*I)*a^2*c*x - a^2*c^2*x^2 + 2*b^2*c*x*ArcTan[c*x] - b^2*(1 + c^2*x^2)*ArcTan[c*x]^2 - 2*a*b*(-(c*x) +
(1 + c^2*x^2)*ArcTan[c*x]) + 2*a^2*Log[c*x] + (4*I)*a*b*(2*c*x*ArcTan[c*x] - Log[1 + c^2*x^2]) - b^2*Log[1 + c
^2*x^2] + 4*b^2*(ArcTan[c*x]*((1 + I*c*x)*ArcTan[c*x] + (2*I)*Log[1 + E^((2*I)*ArcTan[c*x])]) + PolyLog[2, -E^
((2*I)*ArcTan[c*x])]) + (2*I)*a*b*(PolyLog[2, (-I)*c*x] - PolyLog[2, I*c*x]) + 2*b^2*((-I/24)*Pi^3 + ((2*I)/3)
*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])]
+ I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + PolyLo
g[3, E^((-2*I)*ArcTan[c*x])]/2 - PolyLog[3, -E^((2*I)*ArcTan[c*x])]/2)))/2

________________________________________________________________________________________

Maple [C]  time = 1.217, size = 1542, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d^2*a^2*ln(c*x)+2*d^2*b^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+4*d^2*b^2*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/
2))+4*d^2*b^2*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+d^2*b^2*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+2*d^2*b^2*polylog(3
,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*d^2*b^2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+a*b*c*d^2*x+b^2*c*d^2*x*arctan(
c*x)-1/2*I*d^2*b^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2
*x^2+1)+1))^2*arctan(c*x)^2-1/2*I*d^2*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*c
sgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+4*I*d^2*a*b*arctan(c*x)*c*x+1/2*I
*d^2*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/(
(1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-1/2*I*d^2*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I*((1+I*c*
x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2*I*d^2*b^2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1
)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2+1/2*I*d^2*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1))*csgn(I/
((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+3/2
*d^2*b^2*arctan(c*x)^2+I*d^2*a*b*ln(c*x)*ln(1+I*c*x)-I*d^2*a*b*ln(c*x)*ln(1-I*c*x)-d^2*a*b*arctan(c*x)*c^2*x^2
+2*I*d^2*b^2*arctan(c*x)^2*c*x-1/2*I*d^2*b^2*Pi*csgn(((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^
2*arctan(c*x)^2+1/2*I*d^2*b^2*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)
^2-1/2*d^2*a^2*c^2*x^2-d^2*b^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+d^2*b^2*arctan(c*x)^2*ln(1+(1+I*c*x
)/(c^2*x^2+1)^(1/2))+d^2*b^2*arctan(c*x)^2*ln(c*x)+d^2*b^2*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-I*d
^2*b^2*arctan(c*x)-d^2*a*b*arctan(c*x)-1/2*d^2*b^2*arctan(c*x)^2*c^2*x^2+2*I*d^2*a^2*c*x+I*d^2*b^2*arctan(c*x)
*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+1/2*I*d^2*b^2*Pi*arctan(c*x)^2-2*I*d^2*b^2*arctan(c*x)*polylog(2,-(1+I*c*
x)/(c^2*x^2+1)^(1/2))-2*I*d^2*b^2*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+4*I*d^2*b^2*arctan(c*x)*l
n(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+4*I*d^2*b^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+I*d^2*a*b*dilog
(1+I*c*x)+2*d^2*a*b*arctan(c*x)*ln(c*x)-I*d^2*a*b*dilog(1-I*c*x)-2*I*d^2*a*b*ln(c^2*x^2+1)

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

[Out]

-12*b^2*c^4*d^2*integrate(1/16*x^4*arctan(c*x)^2/(c^2*x^3 + x), x) + 2*I*b^2*c^4*d^2*integrate(1/8*x^4*arctan(
c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) - b^2*c^4*d^2*integrate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x)
+ 2*I*b^2*c^4*d^2*integrate(1/8*x^4*arctan(c*x)/(c^2*x^3 + x), x) - 32*a*b*c^4*d^2*integrate(1/16*x^4*arctan(c
*x)/(c^2*x^3 + x), x) - 2*b^2*c^4*d^2*integrate(1/16*x^4*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) - 1/2*a^2*c^2*d^2*
x^2 + 12*I*b^2*c^3*d^2*integrate(1/8*x^3*arctan(c*x)^2/(c^2*x^3 + x), x) + 8*b^2*c^3*d^2*integrate(1/16*x^3*ar
ctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + I*b^2*c^3*d^2*integrate(1/8*x^3*log(c^2*x^2 + 1)^2/(c^2*x^3 + x
), x) + 20*b^2*c^3*d^2*integrate(1/16*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 5*I*b^2*c^3*d^2*integrate(1/8*x^3*lo
g(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 1/2*I*b^2*d^2*arctan(c*x)^3 - 8*I*b^2*c^2*d^2*integrate(1/8*x^2*arctan(c*x)
/(c^2*x^3 + x), x) + 2*I*a^2*c*d^2*x + 8*b^2*c*d^2*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x)
, x) + I*b^2*c*d^2*integrate(1/8*x*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 1/8*b^2*d^2*log(c^2*x^2 + 1)^2 + 2*I
*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d^2 + 12*b^2*d^2*integrate(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) -
2*I*b^2*d^2*integrate(1/8*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + b^2*d^2*integrate(1/16*log(c^2*x^2
+ 1)^2/(c^2*x^3 + x), x) + 32*a*b*d^2*integrate(1/16*arctan(c*x)/(c^2*x^3 + x), x) + a^2*d^2*log(x) - 1/8*(b^2
*c^2*d^2*x^2 - 4*I*b^2*c*d^2*x)*arctan(c*x)^2 - 1/32*(4*I*b^2*c^2*d^2*x^2 + 16*b^2*c*d^2*x)*arctan(c*x)*log(c^
2*x^2 + 1) + 1/32*(b^2*c^2*d^2*x^2 - 4*I*b^2*c*d^2*x)*log(c^2*x^2 + 1)^2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{4 \, a^{2} c^{2} d^{2} x^{2} - 8 i \, a^{2} c d^{2} x - 4 \, a^{2} d^{2} -{\left (b^{2} c^{2} d^{2} x^{2} - 2 i \, b^{2} c d^{2} x - b^{2} d^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} -{\left (-4 i \, a b c^{2} d^{2} x^{2} - 8 \, a b c d^{2} x + 4 i \, a b d^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{4 \, x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-1/4*(4*a^2*c^2*d^2*x^2 - 8*I*a^2*c*d^2*x - 4*a^2*d^2 - (b^2*c^2*d^2*x^2 - 2*I*b^2*c*d^2*x - b^2*d^2)
*log(-(c*x + I)/(c*x - I))^2 - (-4*I*a*b*c^2*d^2*x^2 - 8*a*b*c*d^2*x + 4*I*a*b*d^2)*log(-(c*x + I)/(c*x - I)))
/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int \frac{a^{2}}{x}\, dx + \int 2 i a^{2} c\, dx + \int - a^{2} c^{2} x\, dx + \int \frac{b^{2} \operatorname{atan}^{2}{\left (c x \right )}}{x}\, dx + \int 2 i b^{2} c \operatorname{atan}^{2}{\left (c x \right )}\, dx + \int \frac{2 a b \operatorname{atan}{\left (c x \right )}}{x}\, dx + \int - b^{2} c^{2} x \operatorname{atan}^{2}{\left (c x \right )}\, dx + \int 4 i a b c \operatorname{atan}{\left (c x \right )}\, dx + \int - 2 a b c^{2} x \operatorname{atan}{\left (c x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**2*(Integral(a**2/x, x) + Integral(2*I*a**2*c, x) + Integral(-a**2*c**2*x, x) + Integral(b**2*atan(c*x)**2/x
, x) + Integral(2*I*b**2*c*atan(c*x)**2, x) + Integral(2*a*b*atan(c*x)/x, x) + Integral(-b**2*c**2*x*atan(c*x)
**2, x) + Integral(4*I*a*b*c*atan(c*x), x) + Integral(-2*a*b*c**2*x*atan(c*x), x))

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

[Out]

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