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

Optimal. Leaf size=299 \[ \frac {i b \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {5 b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-c x+i)}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (-c x+i)^2}-\frac {5 \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}+\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}+\frac {b^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d^3}-\frac {11 i b^2}{16 d^3 (-c x+i)}+\frac {b^2}{16 d^3 (-c x+i)^2}+\frac {11 i b^2 \tan ^{-1}(c x)}{16 d^3} \]

[Out]

1/16*b^2/d^3/(I-c*x)^2-11/16*I*b^2/d^3/(I-c*x)+11/16*I*b^2*arctan(c*x)/d^3+1/4*I*b*(a+b*arctan(c*x))/d^3/(I-c*
x)^2+5/4*b*(a+b*arctan(c*x))/d^3/(I-c*x)-5/8*(a+b*arctan(c*x))^2/d^3-1/2*(a+b*arctan(c*x))^2/d^3/(I-c*x)^2+I*(
a+b*arctan(c*x))^2/d^3/(I-c*x)-2*(a+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))/d^3+(a+b*arctan(c*x))^2*ln(2/(1+I
*c*x))/d^3+I*b*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))/d^3+1/2*b^2*polylog(3,-1+2/(1+I*c*x))/d^3

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Rubi [A]  time = 0.79, antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 32, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {4876, 4850, 4988, 4884, 4994, 6610, 4864, 4862, 627, 44, 203, 4854} \[ \frac {i b \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}+\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^3}+\frac {5 b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (-c x+i)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-c x+i)}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (-c x+i)^2}-\frac {5 \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}+\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^3}-\frac {11 i b^2}{16 d^3 (-c x+i)}+\frac {b^2}{16 d^3 (-c x+i)^2}+\frac {11 i b^2 \tan ^{-1}(c x)}{16 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^2/(16*d^3*(I - c*x)^2) - (((11*I)/16)*b^2)/(d^3*(I - c*x)) + (((11*I)/16)*b^2*ArcTan[c*x])/d^3 + ((I/4)*b*(a
+ b*ArcTan[c*x]))/(d^3*(I - c*x)^2) + (5*b*(a + b*ArcTan[c*x]))/(4*d^3*(I - c*x)) - (5*(a + b*ArcTan[c*x])^2)
/(8*d^3) - (a + b*ArcTan[c*x])^2/(2*d^3*(I - c*x)^2) + (I*(a + b*ArcTan[c*x])^2)/(d^3*(I - c*x)) + (2*(a + b*A
rcTan[c*x])^2*ArcTanh[1 - 2/(1 + I*c*x)])/d^3 + ((a + b*ArcTan[c*x])^2*Log[2/(1 + I*c*x)])/d^3 + (I*b*(a + b*A
rcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)])/d^3 + (b^2*PolyLog[3, -1 + 2/(1 + I*c*x)])/(2*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 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 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 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 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 (d+i c d x)^3} \, dx &=\int \left (\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^3 x}+\frac {c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-i+c x)^3}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (-i+c x)^2}-\frac {c \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} \, dx}{d^3}+\frac {(i c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{d^3}+\frac {c \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^3} \, dx}{d^3}-\frac {c \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{-i+c x} \, dx}{d^3}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (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^3}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\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^3}+\frac {(b c) \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 {(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^3}-\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^3}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (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^3}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {(i b c) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{2 d^3}+\frac {(b c) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{4 d^3}-\frac {(b c) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{4 d^3}+\frac {(b c) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^3}-\frac {(b c) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{d^3}+\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^3}-\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^3}-\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^3}\\ &=\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}+\frac {5 b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}-\frac {5 \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (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^3}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 d^3}-\frac {\left (i b^2 c\right ) \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{4 d^3}+\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^3}-\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^3}+\frac {\left (b^2 c\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{4 d^3}+\frac {\left (b^2 c\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^3}\\ &=\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}+\frac {5 b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}-\frac {5 \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (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^3}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {\left (i b^2 c\right ) \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{4 d^3}+\frac {\left (b^2 c\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{4 d^3}+\frac {\left (b^2 c\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{d^3}\\ &=\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}+\frac {5 b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}-\frac {5 \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (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^3}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}-\frac {\left (i b^2 c\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 {\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}{4 d^3}+\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^3}\\ &=\frac {b^2}{16 d^3 (i-c x)^2}-\frac {11 i b^2}{16 d^3 (i-c x)}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}+\frac {5 b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}-\frac {5 \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (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^3}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}+\frac {\left (i b^2 c\right ) \int \frac {1}{1+c^2 x^2} \, dx}{16 d^3}+\frac {\left (i b^2 c\right ) \int \frac {1}{1+c^2 x^2} \, dx}{8 d^3}+\frac {\left (i b^2 c\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^3}\\ &=\frac {b^2}{16 d^3 (i-c x)^2}-\frac {11 i b^2}{16 d^3 (i-c x)}+\frac {11 i b^2 \tan ^{-1}(c x)}{16 d^3}+\frac {i b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)^2}+\frac {5 b \left (a+b \tan ^{-1}(c x)\right )}{4 d^3 (i-c x)}-\frac {5 \left (a+b \tan ^{-1}(c x)\right )^2}{8 d^3}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^3 (i-c x)^2}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{d^3 (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^3}+\frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {i b \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^3}+\frac {b^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]  time = 1.70, size = 435, normalized size = 1.45 \[ \frac {-96 a^2 \log \left (c^2 x^2+1\right )-\frac {192 i a^2}{c x-i}-\frac {96 a^2}{(c x-i)^2}+192 a^2 \log (c x)-192 i a^2 \tan ^{-1}(c x)+12 i a b \left (-16 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-32 \tan ^{-1}(c x)^2+12 i \sin \left (2 \tan ^{-1}(c x)\right )+i \sin \left (4 \tan ^{-1}(c x)\right )-12 \cos \left (2 \tan ^{-1}(c x)\right )-\cos \left (4 \tan ^{-1}(c x)\right )-4 i \tan ^{-1}(c x) \left (8 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )-6 i \sin \left (2 \tan ^{-1}(c x)\right )-i \sin \left (4 \tan ^{-1}(c x)\right )+6 \cos \left (2 \tan ^{-1}(c x)\right )+\cos \left (4 \tan ^{-1}(c x)\right )\right )\right )+b^2 \left (192 i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+96 \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )+192 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-144 i \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )-24 i \tan ^{-1}(c x)^2 \sin \left (4 \tan ^{-1}(c x)\right )-144 \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )-12 \tan ^{-1}(c x) \sin \left (4 \tan ^{-1}(c x)\right )+72 i \sin \left (2 \tan ^{-1}(c x)\right )+3 i \sin \left (4 \tan ^{-1}(c x)\right )+144 \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )+24 \tan ^{-1}(c x)^2 \cos \left (4 \tan ^{-1}(c x)\right )-144 i \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )-12 i \tan ^{-1}(c x) \cos \left (4 \tan ^{-1}(c x)\right )-72 \cos \left (2 \tan ^{-1}(c x)\right )-3 \cos \left (4 \tan ^{-1}(c x)\right )-8 i \pi ^3\right )}{192 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-96*a^2)/(-I + c*x)^2 - ((192*I)*a^2)/(-I + c*x) - (192*I)*a^2*ArcTan[c*x] + 192*a^2*Log[c*x] - 96*a^2*Log[1
+ c^2*x^2] + (12*I)*a*b*(-32*ArcTan[c*x]^2 - 12*Cos[2*ArcTan[c*x]] - Cos[4*ArcTan[c*x]] - 16*PolyLog[2, E^((2
*I)*ArcTan[c*x])] + (12*I)*Sin[2*ArcTan[c*x]] - (4*I)*ArcTan[c*x]*(6*Cos[2*ArcTan[c*x]] + Cos[4*ArcTan[c*x]] +
8*Log[1 - E^((2*I)*ArcTan[c*x])] - (6*I)*Sin[2*ArcTan[c*x]] - I*Sin[4*ArcTan[c*x]]) + I*Sin[4*ArcTan[c*x]]) +
b^2*((-8*I)*Pi^3 - 72*Cos[2*ArcTan[c*x]] - (144*I)*ArcTan[c*x]*Cos[2*ArcTan[c*x]] + 144*ArcTan[c*x]^2*Cos[2*A
rcTan[c*x]] - 3*Cos[4*ArcTan[c*x]] - (12*I)*ArcTan[c*x]*Cos[4*ArcTan[c*x]] + 24*ArcTan[c*x]^2*Cos[4*ArcTan[c*x
]] + 192*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + (192*I)*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])
] + 96*PolyLog[3, E^((-2*I)*ArcTan[c*x])] + (72*I)*Sin[2*ArcTan[c*x]] - 144*ArcTan[c*x]*Sin[2*ArcTan[c*x]] - (
144*I)*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]] + (3*I)*Sin[4*ArcTan[c*x]] - 12*ArcTan[c*x]*Sin[4*ArcTan[c*x]] - (24*I
)*ArcTan[c*x]^2*Sin[4*ArcTan[c*x]]))/(192*d^3)

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-i \, b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 4 \, a b \log \left (-\frac {c x + i}{c x - i}\right ) + 4 i \, a^{2}}{4 \, c^{3} d^{3} x^{4} - 12 i \, c^{2} d^{3} x^{3} - 12 \, c d^{3} x^{2} + 4 i \, d^{3} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^2/x/(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^4 - 12*
I*c^2*d^3*x^3 - 12*c*d^3*x^2 + 4*I*d^3*x), 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/(d+I*c*d*x)^3,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.60, size = 2151, normalized size = 7.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2/d^3*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+b^2/d^3*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+b
^2/d^3*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+b^2/d^3*arctan(c*x)^2*ln(c*x)+3/8*I*b^2/d^3/(c*x-I)-2/3
*I*b^2/d^3*arctan(c*x)^3-I*a^2/d^3/(c*x-I)-I*a^2/d^3*arctan(c*x)-5/4*a*b/d^3*arctan(c*x)-5/4*a*b/d^3/(c*x-I)-3
*b^2/d^3*arctan(c*x)/(4*c*x-4*I)+b^2/d^3*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-1/2*I*b^2/d^3*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*b^2/d^3*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*b^2/d^3*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+1/2*I*b^2/d^3*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-I*b^2/d^3*Pi*arctan(c*x)^2*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^
2+I*a*b/d^3*ln(c*x-I)*ln(-1/2*I*(I+c*x))-I*a*b/d^3*ln(c*x)*ln(-I*(I+c*x))-I*a*b/d^3*ln(-I*c*x)*ln(-I*(-c*x+I))
+1/8*b^2/d^3*arctan(c*x)/(c*x-I)^2*c*x-1/32*I*b^2/d^3/(c*x-I)^2*c*x+I*a*b/d^3*ln(c*x)*ln(-I*(-c*x+I))-2*I*a*b/
d^3*arctan(c*x)/(c*x-I)-1/2*b^2/d^3*arctan(c*x)^2/(c*x-I)^2-b^2/d^3*ln(c*x-I)*arctan(c*x)^2+1/2*I*b^2/d^3*arct
an(c*x)^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2-1
/64*b^2/d^3/(c*x-I)^2*c^2*x^2+2*b^2/d^3*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*b^2/d^3*polylog(3,-(1+I*c*x)/
(c^2*x^2+1)^(1/2))+1/16*I*b^2/d^3*arctan(c*x)/(c*x-I)^2-2*a*b/d^3*arctan(c*x)*ln(c*x-I)-a*b/d^3*arctan(c*x)/(c
*x-I)^2-I*b^2/d^3*arctan(c*x)^2/(c*x-I)+1/4*I*a*b/d^3/(c*x-I)^2+3/8*b^2/d^3/(c*x-I)*c*x+I*a*b/d^3*dilog(-1/2*I
*(I+c*x))-I*a*b/d^3*dilog(-I*(I+c*x))-I*a*b/d^3*dilog(-I*c*x)-1/2*I*a*b/d^3*ln(c*x-I)^2-2*I*b^2/d^3*arctan(c*x
)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*b^2/d^3*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*I
*b^2/d^3*arctan(c*x)^2*Pi+2*a*b/d^3*arctan(c*x)*ln(c*x)-1/16*I*b^2/d^3*arctan(c*x)/(c*x-I)^2*c^2*x^2-1/2*I*b^2
/d^3*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^3*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*I*b^2/d^3*arctan(c*x)/(4*c*x-4*I)*c*x-1/2*I*b
^2/d^3*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^3*arctan(c*x)^2*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+
I*c*x)^2/(c^2*x^2+1)+1))^2+1/2*I*b^2/d^3*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))*cs
gn(((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^3*arctan(c*x)^2*Pi*csgn(
I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^
2+1)+1))+1/2*I*b^2/d^3*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+1/64*b^2/d^3/(c*x-I)^2-5/8*b^2/d^3*arctan(c*
x)^2-1/2*a^2/d^3*ln(c^2*x^2+1)-1/2*a^2/d^3/(c*x-I)^2+a^2/d^3*ln(c*x)

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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/(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 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{x\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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