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

Optimal. Leaf size=402 \[ 3 b c d^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-3 b c d^3 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-i b^2 c d^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )-3 i b^2 c d^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )-\frac{3}{2} i b^2 c d^3 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{3}{2} i b^2 c d^3 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-\frac{1}{2} i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+i a b c^2 d^3 x-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{9}{2} i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-6 b c d^3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 b c d^3 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+6 i c d^3 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} i b^2 c d^3 \log \left (c^2 x^2+1\right )+i b^2 c^2 d^3 x \tan ^{-1}(c x) \]

[Out]

I*a*b*c^2*d^3*x + I*b^2*c^2*d^3*x*ArcTan[c*x] - ((9*I)/2)*c*d^3*(a + b*ArcTan[c*x])^2 - (d^3*(a + b*ArcTan[c*x
])^2)/x - 3*c^2*d^3*x*(a + b*ArcTan[c*x])^2 - (I/2)*c^3*d^3*x^2*(a + b*ArcTan[c*x])^2 + (6*I)*c*d^3*(a + b*Arc
Tan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] - 6*b*c*d^3*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] - (I/2)*b^2*c*d^3*Lo
g[1 + c^2*x^2] + 2*b*c*d^3*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)] - I*b^2*c*d^3*PolyLog[2, -1 + 2/(1 - I*c
*x)] - (3*I)*b^2*c*d^3*PolyLog[2, 1 - 2/(1 + I*c*x)] + 3*b*c*d^3*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c
*x)] - 3*b*c*d^3*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - ((3*I)/2)*b^2*c*d^3*PolyLog[3, 1 - 2/(1
+ I*c*x)] + ((3*I)/2)*b^2*c*d^3*PolyLog[3, -1 + 2/(1 + I*c*x)]

________________________________________________________________________________________

Rubi [A]  time = 0.736219, antiderivative size = 402, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 17, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.68, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4852, 4924, 4868, 2447, 4850, 4988, 4884, 4994, 6610, 4916, 260} \[ 3 b c d^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-3 b c d^3 \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-i b^2 c d^3 \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right )-3 i b^2 c d^3 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )-\frac{3}{2} i b^2 c d^3 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{3}{2} i b^2 c d^3 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-\frac{1}{2} i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+i a b c^2 d^3 x-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{9}{2} i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-6 b c d^3 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 b c d^3 \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+6 i c d^3 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} i b^2 c d^3 \log \left (c^2 x^2+1\right )+i b^2 c^2 d^3 x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

I*a*b*c^2*d^3*x + I*b^2*c^2*d^3*x*ArcTan[c*x] - ((9*I)/2)*c*d^3*(a + b*ArcTan[c*x])^2 - (d^3*(a + b*ArcTan[c*x
])^2)/x - 3*c^2*d^3*x*(a + b*ArcTan[c*x])^2 - (I/2)*c^3*d^3*x^2*(a + b*ArcTan[c*x])^2 + (6*I)*c*d^3*(a + b*Arc
Tan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)] - 6*b*c*d^3*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] - (I/2)*b^2*c*d^3*Lo
g[1 + c^2*x^2] + 2*b*c*d^3*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)] - I*b^2*c*d^3*PolyLog[2, -1 + 2/(1 - I*c
*x)] - (3*I)*b^2*c*d^3*PolyLog[2, 1 - 2/(1 + I*c*x)] + 3*b*c*d^3*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c
*x)] - 3*b*c*d^3*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - ((3*I)/2)*b^2*c*d^3*PolyLog[3, 1 - 2/(1
+ I*c*x)] + ((3*I)/2)*b^2*c*d^3*PolyLog[3, -1 + 2/(1 + I*c*x)]

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 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 4924

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

Rule 4868

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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 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)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx &=\int \left (-3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^2}+\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-i c^3 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^3 \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx+\left (3 i c d^3\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx-\left (3 c^2 d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx-\left (i c^3 d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+6 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+\left (2 b c d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx-\left (12 i b c^2 d^3\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 (6 b c^3 d^3\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\left (i b c^4 d^3\right ) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-4 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+6 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+\left (2 i b c d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx+\left (i b c^2 d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx-\left (i b c^2 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx+\left (6 i b c^2 d^3\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 (6 i b c^2 d^3\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-\left (6 b c^2 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx\\ &=i a b c^2 d^3 x-\frac{9}{2} i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+6 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-6 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )+2 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )+3 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )-3 b c d^3 \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^2 d^3\right ) \int \tan ^{-1}(c x) \, dx-\left (2 b^2 c^2 d^3\right ) \int \frac{\log \left (2-\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx-\left (3 b^2 c^2 d^3\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (3 b^2 c^2 d^3\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (6 b^2 c^2 d^3\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=i a b c^2 d^3 x+i b^2 c^2 d^3 x \tan ^{-1}(c x)-\frac{9}{2} i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+6 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-6 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )+2 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )-i b^2 c d^3 \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )+3 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )-3 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{3}{2} i b^2 c d^3 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{3}{2} i b^2 c d^3 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )-\left (6 i b^2 c d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )-\left (i b^2 c^3 d^3\right ) \int \frac{x}{1+c^2 x^2} \, dx\\ &=i a b c^2 d^3 x+i b^2 c^2 d^3 x \tan ^{-1}(c x)-\frac{9}{2} i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+6 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-6 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )-\frac{1}{2} i b^2 c d^3 \log \left (1+c^2 x^2\right )+2 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )-i b^2 c d^3 \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )-3 i b^2 c d^3 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+3 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )-3 b c d^3 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{3}{2} i b^2 c d^3 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{3}{2} i b^2 c d^3 \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.592865, size = 512, normalized size = 1.27 \[ \frac{d^3 \left (-24 a b c x \text{PolyLog}(2,-i c x)+24 a b c x \text{PolyLog}(2,i c x)-24 b^2 c x \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )-24 b^2 c x \left (\tan ^{-1}(c x)-i\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-8 i b^2 c x \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+12 i b^2 c x \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-12 i b^2 c x \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )-4 i a^2 c^3 x^3-24 a^2 c^2 x^2+24 i a^2 c x \log (x)-8 a^2+8 i a b c^2 x^2+16 a b c x \log \left (c^2 x^2+1\right )-8 i a b c^3 x^3 \tan ^{-1}(c x)-48 a b c^2 x^2 \tan ^{-1}(c x)+16 a b c x \log (c x)-8 i a b c x \tan ^{-1}(c x)-16 a b \tan ^{-1}(c x)-4 i b^2 c x \log \left (c^2 x^2+1\right )-4 i b^2 c^3 x^3 \tan ^{-1}(c x)^2-24 b^2 c^2 x^2 \tan ^{-1}(c x)^2+8 i b^2 c^2 x^2 \tan ^{-1}(c x)+\pi ^3 b^2 c x-16 b^2 c x \tan ^{-1}(c x)^3+12 i b^2 c x \tan ^{-1}(c x)^2-8 b^2 \tan ^{-1}(c x)^2+24 i b^2 c x \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+16 b^2 c x \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )-24 i b^2 c x \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-48 b^2 c x \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{8 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(-8*a^2 + b^2*c*Pi^3*x - 24*a^2*c^2*x^2 + (8*I)*a*b*c^2*x^2 - (4*I)*a^2*c^3*x^3 - 16*a*b*ArcTan[c*x] - (8
*I)*a*b*c*x*ArcTan[c*x] - 48*a*b*c^2*x^2*ArcTan[c*x] + (8*I)*b^2*c^2*x^2*ArcTan[c*x] - (8*I)*a*b*c^3*x^3*ArcTa
n[c*x] - 8*b^2*ArcTan[c*x]^2 + (12*I)*b^2*c*x*ArcTan[c*x]^2 - 24*b^2*c^2*x^2*ArcTan[c*x]^2 - (4*I)*b^2*c^3*x^3
*ArcTan[c*x]^2 - 16*b^2*c*x*ArcTan[c*x]^3 + (24*I)*b^2*c*x*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + 16*
b^2*c*x*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] - 48*b^2*c*x*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] - (
24*I)*b^2*c*x*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*a^2*c*x*Log[x] + 16*a*b*c*x*Log[c*x] + 16*
a*b*c*x*Log[1 + c^2*x^2] - (4*I)*b^2*c*x*Log[1 + c^2*x^2] - 24*b^2*c*x*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan
[c*x])] - 24*b^2*c*x*(-I + ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - (8*I)*b^2*c*x*PolyLog[2, E^((2*I)
*ArcTan[c*x])] - 24*a*b*c*x*PolyLog[2, (-I)*c*x] + 24*a*b*c*x*PolyLog[2, I*c*x] + (12*I)*b^2*c*x*PolyLog[3, E^
((-2*I)*ArcTan[c*x])] - (12*I)*b^2*c*x*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/(8*x)

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Maple [C]  time = 4.881, size = 1739, normalized size = 4.3 \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)^3*(a+b*arctan(c*x))^2/x^2,x)

[Out]

-3/2*c*d^3*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*d^3*b^2*arctan(c*x)^2*c^2*x-3/2*c*d^3*b^2*Pi*arcta
n(c*x)^2+I*c*d^3*b^2*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-6*c*d^3*b^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))
-6*c*d^3*b^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*c*d^3*b^2*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*
x^2+1)^(1/2))-3*c*d^3*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+6*c*d^3*b^2*arctan(c*x)*polylog(2,-(
1+I*c*x)/(c^2*x^2+1)^(1/2))+2*c*d^3*b^2*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*c*d^3*a*b*ln(c^2*x^2+1
)+2*c*d^3*a*b*ln(c*x)-3*c*d^3*a*b*dilog(1+I*c*x)+3*c*d^3*a*b*dilog(1-I*c*x)-2*d^3*a*b*arctan(c*x)/x-1/2*I*d^3*
a^2*c^3*x^2-3/2*I*c*d^3*b^2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+6*I*c*d^3*b^2*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^
(1/2))+3*I*c*d^3*a^2*ln(c*x)+6*I*c*d^3*b^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*c*d^3*b^2*dilog(1+(1+I*
c*x)/(c^2*x^2+1)^(1/2))+2*I*c*d^3*b^2*dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*I*c*d^3*b^2*arctan(c*x)^2+I*a*b*c
^2*d^3*x+I*b^2*c^2*d^3*x*arctan(c*x)-d^3*a^2/x+6*I*c*d^3*b^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*c*d^3*
b^2*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*c*d^3*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)^2-3/2*c*d^3*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)^2-3/2*c*d^3*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-
3*c*d^3*a*b*ln(c*x)*ln(1+I*c*x)+3*c*d^3*a*b*ln(c*x)*ln(1-I*c*x)-6*d^3*a*b*arctan(c*x)*c^2*x-I*c*d^3*a*b*arctan
(c*x)+3*I*c*d^3*b^2*arctan(c*x)^2*ln(c*x)-3*I*c*d^3*b^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+3*I*c*d^3*
b^2*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*I*c*d^3*b^2*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/
2))-1/2*I*d^3*b^2*arctan(c*x)^2*c^3*x^2-3*c^2*x*a^2*d^3-d^3*b^2*arctan(c*x)^2/x+c*d^3*b^2*arctan(c*x)+3/2*c*d^
3*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+3/2*c*d^3*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))^2*arctan(c*x)^2-3/2*c*d^3*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
+3/2*c*d^3*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+6*I*c*d^3*a*b*arctan(c*x)*ln(c*x)-I*d^3*a*b*arctan(c*x)*c^3*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((d+I*c*d*x)^3*(a+b*arctan(c*x))^2/x^2,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{-4 i \, a^{2} c^{3} d^{3} x^{3} - 12 \, a^{2} c^{2} d^{3} x^{2} + 12 i \, a^{2} c d^{3} x + 4 \, a^{2} d^{3} +{\left (i \, b^{2} c^{3} d^{3} x^{3} + 3 \, b^{2} c^{2} d^{3} x^{2} - 3 i \, b^{2} c d^{3} x - b^{2} d^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} +{\left (4 \, a b c^{3} d^{3} x^{3} - 12 i \, a b c^{2} d^{3} x^{2} - 12 \, a b c d^{3} x + 4 i \, a b d^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{4 \, x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/4*(-4*I*a^2*c^3*d^3*x^3 - 12*a^2*c^2*d^3*x^2 + 12*I*a^2*c*d^3*x + 4*a^2*d^3 + (I*b^2*c^3*d^3*x^3 +
3*b^2*c^2*d^3*x^2 - 3*I*b^2*c*d^3*x - b^2*d^3)*log(-(c*x + I)/(c*x - I))^2 + (4*a*b*c^3*d^3*x^3 - 12*I*a*b*c^2
*d^3*x^2 - 12*a*b*c*d^3*x + 4*I*a*b*d^3)*log(-(c*x + I)/(c*x - I)))/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{3} \left (\int - 3 a^{2} c^{2}\, dx + \int \frac{a^{2}}{x^{2}}\, dx + \int - 3 b^{2} c^{2} \operatorname{atan}^{2}{\left (c x \right )}\, dx + \int \frac{b^{2} \operatorname{atan}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{3 i a^{2} c}{x}\, dx + \int - i a^{2} c^{3} x\, dx + \int - 6 a b c^{2} \operatorname{atan}{\left (c x \right )}\, dx + \int \frac{2 a b \operatorname{atan}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{3 i b^{2} c \operatorname{atan}^{2}{\left (c x \right )}}{x}\, dx + \int - i b^{2} c^{3} x \operatorname{atan}^{2}{\left (c x \right )}\, dx + \int \frac{6 i a b c \operatorname{atan}{\left (c x \right )}}{x}\, dx + \int - 2 i a b c^{3} 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)**3*(a+b*atan(c*x))**2/x**2,x)

[Out]

d**3*(Integral(-3*a**2*c**2, x) + Integral(a**2/x**2, x) + Integral(-3*b**2*c**2*atan(c*x)**2, x) + Integral(b
**2*atan(c*x)**2/x**2, x) + Integral(3*I*a**2*c/x, x) + Integral(-I*a**2*c**3*x, x) + Integral(-6*a*b*c**2*ata
n(c*x), x) + Integral(2*a*b*atan(c*x)/x**2, x) + Integral(3*I*b**2*c*atan(c*x)**2/x, x) + Integral(-I*b**2*c**
3*x*atan(c*x)**2, x) + Integral(6*I*a*b*c*atan(c*x)/x, x) + Integral(-2*I*a*b*c**3*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 )}^{3}{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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