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

Optimal. Leaf size=221 \[ \frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac{b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d^2}+\frac{b \left (a+b \tan ^{-1}(c x)\right )}{d^2 (-c x+i)}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (-c x+i)}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac{i b^2}{2 d^2 (-c x+i)}+\frac{i b^2 \tan ^{-1}(c x)}{2 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.616117, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {4876, 4850, 4988, 4884, 4994, 6610, 4864, 4862, 627, 44, 203, 4854} \[ \frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}+\frac{b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d^2}+\frac{b \left (a+b \tan ^{-1}(c x)\right )}{d^2 (-c x+i)}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (-c x+i)}-\frac{\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac{i b^2}{2 d^2 (-c x+i)}+\frac{i b^2 \tan ^{-1}(c x)}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.948125, size = 299, normalized size = 1.35 \[ \frac{-12 a b \left (2 i \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+4 i \tan ^{-1}(c x)^2+\sin \left (2 \tan ^{-1}(c x)\right )+i \cos \left (2 \tan ^{-1}(c x)\right )-2 \tan ^{-1}(c x) \left (2 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )-i \sin \left (2 \tan ^{-1}(c x)\right )+\cos \left (2 \tan ^{-1}(c x)\right )\right )\right )+b^2 \left (24 i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+12 \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )+24 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-12 i \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )-12 \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )+6 i \sin \left (2 \tan ^{-1}(c x)\right )+12 \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )-12 i \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )-6 \cos \left (2 \tan ^{-1}(c x)\right )-i \pi ^3\right )-12 a^2 \log \left (c^2 x^2+1\right )-\frac{24 i a^2}{c x-i}+24 a^2 \log (c x)-24 i a^2 \tan ^{-1}(c x)}{24 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(((-24*I)*a^2)/(-I + c*x) - (24*I)*a^2*ArcTan[c*x] + 24*a^2*Log[c*x] - 12*a^2*Log[1 + c^2*x^2] - 12*a*b*((4*I)
*ArcTan[c*x]^2 + I*Cos[2*ArcTan[c*x]] + (2*I)*PolyLog[2, E^((2*I)*ArcTan[c*x])] - 2*ArcTan[c*x]*(Cos[2*ArcTan[
c*x]] + 2*Log[1 - E^((2*I)*ArcTan[c*x])] - I*Sin[2*ArcTan[c*x]]) + Sin[2*ArcTan[c*x]]) + b^2*((-I)*Pi^3 - 6*Co
s[2*ArcTan[c*x]] - (12*I)*ArcTan[c*x]*Cos[2*ArcTan[c*x]] + 12*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + 24*ArcTan[c*x
]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + 12*PolyLog[3, E^
((-2*I)*ArcTan[c*x])] + (6*I)*Sin[2*ArcTan[c*x]] - 12*ArcTan[c*x]*Sin[2*ArcTan[c*x]] - (12*I)*ArcTan[c*x]^2*Si
n[2*ArcTan[c*x]]))/(24*d^2)

________________________________________________________________________________________

Maple [C]  time = 0.416, size = 1921, normalized size = 8.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*b^2/d^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)^2-1/2*I*b^2/d^2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((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((1+I*c*x)^2/(c^2*x^2+1))*arctan(c*x)^2+a^2/d^2*
ln(c*x)+2*b^2/d^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*b^2/d^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2
*a^2/d^2*ln(c^2*x^2+1)-1/2*I*b^2/d^2*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-I*b^2/d^2*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*a*b/d^2*ln(-1/2*I
*(c*x+I))*ln(c*x-I)+1/2*I*b^2/d^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*I*b^2/d^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-I*a^2/d^
2/(c*x-I)+b^2/d^2*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2/d^2*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^
2+1)^(1/2))-b^2/d^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+b^2/d^2*arctan(c*x)^2*ln(c*x)+1/4*I*b^2/d^2/(c
*x-I)+1/2*I*b^2/d^2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I/((1+I*c*x)^2/(c^2*x^
2+1)+1))*arctan(c*x)^2-1/2*I*b^2/d^2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn((1+I*
c*x)^2/(c^2*x^2+1))*arctan(c*x)^2+1/2*I*b^2/d^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*b^2/d^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*
b^2/d^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)^2-1/2*I*b^2/d^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*b^2/d^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)^2-2*I*a*b/d^2*arctan(c*x)/(c*x-I)-I*a*b/d^2*ln(c*x)*ln(1-I*c*
x)+I*a*b/d^2*ln(c*x)*ln(1+I*c*x)+2*I*b^2/d^2*arctan(c*x)/(4*c*x-4*I)*c*x-1/2*b^2/d^2*arctan(c*x)^2-I*a^2/d^2*a
rctan(c*x)-2*b^2/d^2*arctan(c*x)/(4*c*x-4*I)-2/3*I*b^2/d^2*arctan(c*x)^3-a*b/d^2/(c*x-I)-a*b/d^2*arctan(c*x)-b
^2/d^2*arctan(c*x)^2*ln(c*x-I)+b^2/d^2*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-2*I*b^2/d^2*arctan(c*x)*p
olylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*b^2/d^2*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*I*b^
2/d^2*Pi*arctan(c*x)^2-I*a*b/d^2*dilog(1-I*c*x)-2*a*b/d^2*arctan(c*x)*ln(c*x-I)+1/4*b^2/d^2/(c*x-I)*c*x+2*a*b/
d^2*arctan(c*x)*ln(c*x)+I*a*b/d^2*dilog(1+I*c*x)-I*b^2/d^2*arctan(c*x)^2/(c*x-I)+I*a*b/d^2*dilog(-1/2*I*(c*x+I
))-1/2*I*a*b/d^2*ln(c*x-I)^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((a+b*arctan(c*x))^2/x/(d+I*c*d*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{b^{2} \log \left (-\frac{c x + i}{c x - i}\right )^{2} - 4 i \, a b \log \left (-\frac{c x + i}{c x - i}\right ) - 4 \, a^{2}}{4 \, c^{2} d^{2} x^{3} - 8 i \, c d^{2} x^{2} - 4 \, d^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*log(-(c*x + I)/(c*x - I))^2 - 4*I*a*b*log(-(c*x + I)/(c*x - I)) - 4*a^2)/(4*c^2*d^2*x^3 - 8*I*c*
d^2*x^2 - 4*d^2*x), 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((a+b*atan(c*x))**2/x/(d+I*c*d*x)**2,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}}{{\left (i \, c d x + d\right )}^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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