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

Optimal. Leaf size=391 \[ \frac {3 b c \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {9 i b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)}+\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-c x+i)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (-c x+i)^2}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}+\frac {2 b c \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {3 i c \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {6 i c \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {i b^2 c \text {Li}_2\left (\frac {2}{1-i c x}-1\right )}{d^3}-\frac {3 i b^2 c \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d^3}-\frac {19 b^2 c}{16 d^3 (-c x+i)}-\frac {i b^2 c}{16 d^3 (-c x+i)^2}+\frac {19 b^2 c \tan ^{-1}(c x)}{16 d^3} \]

[Out]

-1/16*I*b^2*c/d^3/(I-c*x)^2-19/16*b^2*c/d^3/(I-c*x)+19/16*b^2*c*arctan(c*x)/d^3+1/4*b*c*(a+b*arctan(c*x))/d^3/
(I-c*x)^2-9/4*I*b*c*(a+b*arctan(c*x))/d^3/(I-c*x)+1/8*I*c*(a+b*arctan(c*x))^2/d^3-(a+b*arctan(c*x))^2/d^3/x+1/
2*I*c*(a+b*arctan(c*x))^2/d^3/(I-c*x)^2+2*c*(a+b*arctan(c*x))^2/d^3/(I-c*x)+6*I*c*(a+b*arctan(c*x))^2*arctanh(
-1+2/(1+I*c*x))/d^3-3*I*c*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/d^3+2*b*c*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))/d^
3-I*b^2*c*polylog(2,-1+2/(1-I*c*x))/d^3+3*b*c*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))/d^3-3/2*I*b^2*c*poly
log(3,-1+2/(1+I*c*x))/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.98, antiderivative size = 391, normalized size of antiderivative = 1.00, number of steps used = 36, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {4876, 4852, 4924, 4868, 2447, 4850, 4988, 4884, 4994, 6610, 4864, 4862, 627, 44, 203, 4854} \[ \frac {3 b c \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {i b^2 c \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d^3}-\frac {3 i b^2 c \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {9 i b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)}+\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-c x+i)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (-c x+i)^2}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}+\frac {2 b c \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {3 i c \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {6 i c \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {19 b^2 c}{16 d^3 (-c x+i)}-\frac {i b^2 c}{16 d^3 (-c x+i)^2}+\frac {19 b^2 c \tan ^{-1}(c x)}{16 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/16)*b^2*c)/(d^3*(I - c*x)^2) - (19*b^2*c)/(16*d^3*(I - c*x)) + (19*b^2*c*ArcTan[c*x])/(16*d^3) + (b*c*(a
+ b*ArcTan[c*x]))/(4*d^3*(I - c*x)^2) - (((9*I)/4)*b*c*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) + ((I/8)*c*(a + b*
ArcTan[c*x])^2)/d^3 - (a + b*ArcTan[c*x])^2/(d^3*x) + ((I/2)*c*(a + b*ArcTan[c*x])^2)/(d^3*(I - c*x)^2) + (2*c
*(a + b*ArcTan[c*x])^2)/(d^3*(I - c*x)) - ((6*I)*c*(a + b*ArcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)])/d^3 - ((3
*I)*c*(a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/d^3 + (2*b*c*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/d^3 -
(I*b^2*c*PolyLog[2, -1 + 2/(1 - I*c*x)])/d^3 + (3*b*c*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d^3
- (((3*I)/2)*b^2*c*PolyLog[3, -1 + 2/(1 + I*c*x)])/d^3

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 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 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 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 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 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 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 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 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 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 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 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 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 {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2 (d+i c d x)^3} \, dx &=\int \left (\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x^2}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-i+c x)^3}+\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-i+c x)^2}+\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-i+c x)}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx}{d^3}-\frac {(3 i c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx}{d^3}-\frac {\left (i c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^3} \, dx}{d^3}+\frac {\left (3 i c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{-i+c x} \, dx}{d^3}+\frac {\left (2 c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{d^3}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (i-c x)}-\frac {6 i c \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {(2 b c) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d^3}-\frac {\left (i b c^2\right ) \int \left (-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^3}+\frac {a+b \tan ^{-1}(c x)}{4 (-i+c x)^2}-\frac {a+b \tan ^{-1}(c x)}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{d^3}+\frac {\left (6 i b c^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}{d^3}+\frac {\left (12 i b c^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}{d^3}+\frac {\left (4 b c^2\right ) \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^3}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (i-c x)}-\frac {6 i c \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {(2 i b c) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{d^3}-\frac {\left (i b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{4 d^3}+\frac {\left (i b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{4 d^3}-\frac {\left (2 i b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^3}+\frac {\left (2 i b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{d^3}-\frac {\left (6 i b c^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}{d^3}+\frac {\left (6 i b c^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}{d^3}-\frac {\left (b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{2 d^3}-\frac {\left (3 b^2 c^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}\\ &=\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}-\frac {9 i b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (i-c x)}-\frac {6 i c \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^3}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i b^2 c \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 d^3}-\frac {\left (i b^2 c^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{4 d^3}-\frac {\left (2 i b^2 c^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^3}-\frac {\left (b^2 c^2\right ) \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{4 d^3}-\frac {\left (2 b^2 c^2\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d^3}+\frac {\left (3 b^2 c^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}-\frac {\left (3 b^2 c^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}\\ &=\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}-\frac {9 i b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (i-c x)}-\frac {6 i c \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b^2 c \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^3}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i b^2 c \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {\left (i b^2 c^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{4 d^3}-\frac {\left (2 i b^2 c^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{d^3}-\frac {\left (b^2 c^2\right ) \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{4 d^3}\\ &=\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}-\frac {9 i b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (i-c x)}-\frac {6 i c \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b^2 c \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^3}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i b^2 c \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {\left (i b^2 c^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{4 d^3}-\frac {\left (2 i b^2 c^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{d^3}-\frac {\left (b^2 c^2\right ) \int \left (-\frac {i}{2 (-i+c x)^3}+\frac {1}{4 (-i+c x)^2}-\frac {1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{4 d^3}\\ &=-\frac {i b^2 c}{16 d^3 (i-c x)^2}-\frac {19 b^2 c}{16 d^3 (i-c x)}+\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}-\frac {9 i b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (i-c x)}-\frac {6 i c \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b^2 c \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^3}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i b^2 c \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}+\frac {\left (b^2 c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{16 d^3}+\frac {\left (b^2 c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{8 d^3}+\frac {\left (b^2 c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{d^3}\\ &=-\frac {i b^2 c}{16 d^3 (i-c x)^2}-\frac {19 b^2 c}{16 d^3 (i-c x)}+\frac {19 b^2 c \tan ^{-1}(c x)}{16 d^3}+\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}-\frac {9 i b c \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (i-c x)}-\frac {6 i c \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {2 b c \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^3}-\frac {i b^2 c \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^3}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i b^2 c \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 3.85, size = 549, normalized size = 1.40 \[ -\frac {-96 i a^2 c \log \left (c^2 x^2+1\right )+\frac {128 a^2 c}{c x-i}-\frac {32 i a^2 c}{(c x-i)^2}+192 i a^2 c \log (x)+192 a^2 c \tan ^{-1}(c x)+\frac {64 a^2}{x}+\frac {4 a b \left (c x \left (-32 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )-20 i \sin \left (2 \tan ^{-1}(c x)\right )-i \sin \left (4 \tan ^{-1}(c x)\right )+20 \cos \left (2 \tan ^{-1}(c x)\right )+\cos \left (4 \tan ^{-1}(c x)\right )\right )+48 c x \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+96 c x \tan ^{-1}(c x)^2+4 \tan ^{-1}(c x) \left (24 i c x \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+10 c x \sin \left (2 \tan ^{-1}(c x)\right )+c x \sin \left (4 \tan ^{-1}(c x)\right )+10 i c x \cos \left (2 \tan ^{-1}(c x)\right )+i c x \cos \left (4 \tan ^{-1}(c x)\right )+8\right )\right )}{x}-i b^2 c \left (-192 i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )-64 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-96 \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )+\frac {64 i \tan ^{-1}(c x)^2}{c x}-64 \tan ^{-1}(c x)^2-192 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-128 i \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+80 i \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )+8 i \tan ^{-1}(c x)^2 \sin \left (4 \tan ^{-1}(c x)\right )+80 \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )+4 \tan ^{-1}(c x) \sin \left (4 \tan ^{-1}(c x)\right )-40 i \sin \left (2 \tan ^{-1}(c x)\right )-i \sin \left (4 \tan ^{-1}(c x)\right )-80 \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )-8 \tan ^{-1}(c x)^2 \cos \left (4 \tan ^{-1}(c x)\right )+80 i \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )+4 i \tan ^{-1}(c x) \cos \left (4 \tan ^{-1}(c x)\right )+40 \cos \left (2 \tan ^{-1}(c x)\right )+\cos \left (4 \tan ^{-1}(c x)\right )+8 i \pi ^3\right )}{64 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/64*((64*a^2)/x - ((32*I)*a^2*c)/(-I + c*x)^2 + (128*a^2*c)/(-I + c*x) + 192*a^2*c*ArcTan[c*x] + (192*I)*a^2
*c*Log[x] - (96*I)*a^2*c*Log[1 + c^2*x^2] - I*b^2*c*((8*I)*Pi^3 - 64*ArcTan[c*x]^2 + ((64*I)*ArcTan[c*x]^2)/(c
*x) + 40*Cos[2*ArcTan[c*x]] + (80*I)*ArcTan[c*x]*Cos[2*ArcTan[c*x]] - 80*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + Co
s[4*ArcTan[c*x]] + (4*I)*ArcTan[c*x]*Cos[4*ArcTan[c*x]] - 8*ArcTan[c*x]^2*Cos[4*ArcTan[c*x]] - 192*ArcTan[c*x]
^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - (128*I)*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] - (192*I)*ArcTan[c*x]*
PolyLog[2, E^((-2*I)*ArcTan[c*x])] - 64*PolyLog[2, E^((2*I)*ArcTan[c*x])] - 96*PolyLog[3, E^((-2*I)*ArcTan[c*x
])] - (40*I)*Sin[2*ArcTan[c*x]] + 80*ArcTan[c*x]*Sin[2*ArcTan[c*x]] + (80*I)*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]]
- I*Sin[4*ArcTan[c*x]] + 4*ArcTan[c*x]*Sin[4*ArcTan[c*x]] + (8*I)*ArcTan[c*x]^2*Sin[4*ArcTan[c*x]]) + (4*a*b*(
96*c*x*ArcTan[c*x]^2 + 48*c*x*PolyLog[2, E^((2*I)*ArcTan[c*x])] + c*x*(20*Cos[2*ArcTan[c*x]] + Cos[4*ArcTan[c*
x]] - 32*Log[(c*x)/Sqrt[1 + c^2*x^2]] - (20*I)*Sin[2*ArcTan[c*x]] - I*Sin[4*ArcTan[c*x]]) + 4*ArcTan[c*x]*(8 +
(10*I)*c*x*Cos[2*ArcTan[c*x]] + I*c*x*Cos[4*ArcTan[c*x]] + (24*I)*c*x*Log[1 - E^((2*I)*ArcTan[c*x])] + 10*c*x
*Sin[2*ArcTan[c*x]] + c*x*Sin[4*ArcTan[c*x]])))/x)/d^3

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fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-i \, b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 4 \, a b \log \left (-\frac {c x + i}{c x - i}\right ) + 4 i \, a^{2}}{4 \, c^{3} d^{3} x^{5} - 12 i \, c^{2} d^{3} x^{4} - 12 \, c d^{3} x^{3} + 4 i \, d^{3} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-I*b^2*log(-(c*x + I)/(c*x - I))^2 - 4*a*b*log(-(c*x + I)/(c*x - I)) + 4*I*a^2)/(4*c^3*d^3*x^5 - 12*
I*c^2*d^3*x^4 - 12*c*d^3*x^3 + 4*I*d^3*x^2), x)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 1.62, size = 9659, normalized size = 24.70 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2/(x^2*(d + c*d*x*1i)^3),x)

[Out]

int((a + b*atan(c*x))^2/(x^2*(d + c*d*x*1i)^3), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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