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

Optimal. Leaf size=403 \[ -\frac {3 i b c^2 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (-c x+i)}-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^2 (-c x+i)}-\frac {3 c^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {4 i b c^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac {6 c^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d^2 x}-\frac {2 b^2 c^2 \text {Li}_2\left (\frac {2}{1-i c x}-1\right )}{d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d^2}-\frac {b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d^2}+\frac {i b^2 c^2}{2 d^2 (-c x+i)}+\frac {b^2 c^2 \log (x)}{d^2}-\frac {i b^2 c^2 \tan ^{-1}(c x)}{2 d^2} \]

[Out]

-3*I*b*c^2*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))/d^2+2*I*c*(a+b*arctan(c*x))^2/d^2/x-b*c*(a+b*arctan(c*x
))/d^2/x-b*c^2*(a+b*arctan(c*x))/d^2/(I-c*x)-2*c^2*(a+b*arctan(c*x))^2/d^2-1/2*(a+b*arctan(c*x))^2/d^2/x^2-I*c
^2*(a+b*arctan(c*x))^2/d^2/(I-c*x)+1/2*I*b^2*c^2/d^2/(I-c*x)+6*c^2*(a+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))
/d^2+b^2*c^2*ln(x)/d^2-3*c^2*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/d^2-1/2*b^2*c^2*ln(c^2*x^2+1)/d^2-1/2*I*b^2*c
^2*arctan(c*x)/d^2-2*b^2*c^2*polylog(2,-1+2/(1-I*c*x))/d^2-4*I*b*c^2*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))/d^2-3
/2*b^2*c^2*polylog(3,-1+2/(1+I*c*x))/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.93, antiderivative size = 403, normalized size of antiderivative = 1.00, number of steps used = 31, number of rules used = 21, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.840, Rules used = {4876, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 2447, 4850, 4988, 4994, 6610, 4864, 4862, 627, 44, 203, 4854} \[ -\frac {3 i b c^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac {2 b^2 c^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {3 b^2 c^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d^2}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (-c x+i)}-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^2 (-c x+i)}-\frac {3 c^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {4 i b c^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^2}-\frac {6 c^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d^2 x}-\frac {b^2 c^2 \log \left (c^2 x^2+1\right )}{2 d^2}+\frac {i b^2 c^2}{2 d^2 (-c x+i)}+\frac {b^2 c^2 \log (x)}{d^2}-\frac {i b^2 c^2 \tan ^{-1}(c x)}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 4918

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f^2), 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] && LtQ[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 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^3 (d+i c d x)^2} \, dx &=\int \left (\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x^3}-\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^3 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (-i+c x)^2}+\frac {3 c^3 \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^3} \, dx}{d^2}-\frac {(2 i c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx}{d^2}-\frac {\left (3 c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx}{d^2}-\frac {\left (i c^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{d^2}+\frac {\left (3 c^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{-i+c x} \, dx}{d^2}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (i-c x)}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}+\frac {(b c) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx}{d^2}-\frac {\left (4 i b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx}{d^2}-\frac {\left (2 i b c^3\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^2}+\frac {\left (6 b c^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}{d^2}+\frac {\left (12 b c^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}{d^2}\\ &=-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (i-c x)}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {3 i b c^2 \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)}{x^2} \, dx}{d^2}+\frac {\left (4 b c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx}{d^2}-\frac {\left (b c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^2}-\frac {\left (6 b c^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}{d^2}+\frac {\left (6 b c^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}{d^2}+\frac {\left (3 i b^2 c^3\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 c \left (a+b \tan ^{-1}(c x)\right )}{d^2 x}-\frac {b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^2 (i-c x)}-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (i-c x)}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {4 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {3 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 d^2}+\frac {\left (b^2 c^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d^2}-\frac {\left (3 i b^2 c^3\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 (3 i b^2 c^3\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 (4 i b^2 c^3\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d^2}-\frac {\left (b^2 c^3\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^2}\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d^2 x}-\frac {b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^2 (i-c x)}-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (i-c x)}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {4 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {3 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^2}+\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (b^2 c^3\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{d^2}\\ &=-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d^2 x}-\frac {b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^2 (i-c x)}-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (i-c x)}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {4 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {3 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^2}+\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (b^2 c^3\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 {\left (b^2 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d^2}\\ &=\frac {i b^2 c^2}{2 d^2 (i-c x)}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d^2 x}-\frac {b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^2 (i-c x)}-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (i-c x)}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}+\frac {b^2 c^2 \log (x)}{d^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d^2}-\frac {4 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {3 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^2}-\frac {\left (i b^2 c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^2}\\ &=\frac {i b^2 c^2}{2 d^2 (i-c x)}-\frac {i b^2 c^2 \tan ^{-1}(c x)}{2 d^2}-\frac {b c \left (a+b \tan ^{-1}(c x)\right )}{d^2 x}-\frac {b c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^2 (i-c x)}-\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tan ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 i c \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )^2}{d^2 (i-c x)}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{d^2}+\frac {b^2 c^2 \log (x)}{d^2}-\frac {3 c^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {b^2 c^2 \log \left (1+c^2 x^2\right )}{2 d^2}-\frac {4 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )}{d^2}-\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d^2}-\frac {3 i b c^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d^2}\\ \end {align*}

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Mathematica [A]  time = 3.14, size = 491, normalized size = 1.22 \[ \frac {12 a^2 c^2 \log \left (c^2 x^2+1\right )+\frac {8 i a^2 c^2}{c x-i}-24 a^2 c^2 \log (x)+24 i a^2 c^2 \tan ^{-1}(c x)+\frac {16 i a^2 c}{x}-\frac {4 a^2}{x^2}+4 i a b c^2 \left (-8 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+2 \tan ^{-1}(c x) \left (\frac {i}{c^2 x^2}+\frac {4}{c x}+6 i \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+\sin \left (2 \tan ^{-1}(c x)\right )+i \cos \left (2 \tan ^{-1}(c x)\right )+i\right )+6 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+\frac {2 i}{c x}+12 \tan ^{-1}(c x)^2-i \sin \left (2 \tan ^{-1}(c x)\right )+\cos \left (2 \tan ^{-1}(c x)\right )\right )-b^2 c^2 \left (-8 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+\frac {4 \tan ^{-1}(c x)^2}{c^2 x^2}+24 i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+16 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+12 \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {16 i \tan ^{-1}(c x)^2}{c x}+20 \tan ^{-1}(c x)^2+\frac {8 \tan ^{-1}(c x)}{c x}+24 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+32 i \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )-4 i \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )-4 \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )+2 i \sin \left (2 \tan ^{-1}(c x)\right )+4 \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )-4 i \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )-2 \cos \left (2 \tan ^{-1}(c x)\right )-i \pi ^3\right )}{8 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-4*a^2)/x^2 + ((16*I)*a^2*c)/x + ((8*I)*a^2*c^2)/(-I + c*x) + (24*I)*a^2*c^2*ArcTan[c*x] - 24*a^2*c^2*Log[x]
+ 12*a^2*c^2*Log[1 + c^2*x^2] - b^2*c^2*((-I)*Pi^3 + (8*ArcTan[c*x])/(c*x) + 20*ArcTan[c*x]^2 + (4*ArcTan[c*x
]^2)/(c^2*x^2) - ((16*I)*ArcTan[c*x]^2)/(c*x) - 2*Cos[2*ArcTan[c*x]] - (4*I)*ArcTan[c*x]*Cos[2*ArcTan[c*x]] +
4*ArcTan[c*x]^2*Cos[2*ArcTan[c*x]] + 24*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + (32*I)*ArcTan[c*x]*Log
[1 - E^((2*I)*ArcTan[c*x])] - 8*Log[(c*x)/Sqrt[1 + c^2*x^2]] + (24*I)*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[
c*x])] + 16*PolyLog[2, E^((2*I)*ArcTan[c*x])] + 12*PolyLog[3, E^((-2*I)*ArcTan[c*x])] + (2*I)*Sin[2*ArcTan[c*x
]] - 4*ArcTan[c*x]*Sin[2*ArcTan[c*x]] - (4*I)*ArcTan[c*x]^2*Sin[2*ArcTan[c*x]]) + (4*I)*a*b*c^2*((2*I)/(c*x) +
12*ArcTan[c*x]^2 + Cos[2*ArcTan[c*x]] - 8*Log[(c*x)/Sqrt[1 + c^2*x^2]] + 6*PolyLog[2, E^((2*I)*ArcTan[c*x])]
- I*Sin[2*ArcTan[c*x]] + 2*ArcTan[c*x]*(I + I/(c^2*x^2) + 4/(c*x) + I*Cos[2*ArcTan[c*x]] + (6*I)*Log[1 - E^((2
*I)*ArcTan[c*x])] + Sin[2*ArcTan[c*x]])))/(8*d^2)

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 4 i \, a b \log \left (-\frac {c x + i}{c x - i}\right ) - 4 \, a^{2}}{4 \, {\left (c^{2} d^{2} x^{5} - 2 i \, c d^{2} x^{4} - d^{2} x^{3}\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/4*(b^2*log(-(c*x + I)/(c*x - I))^2 - 4*I*a*b*log(-(c*x + I)/(c*x - I)) - 4*a^2)/(c^2*d^2*x^5 - 2*I*
c*d^2*x^4 - d^2*x^3), 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^3/(d+I*c*d*x)^2,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 7.14, size = 2378, normalized size = 5.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*b^2/d^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)+3*I*c^2*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+3/2*I*c^2*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)^2-3/2*I*c^2*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)^2-3/2*I*c^2*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+3*I*c^
2*a*b/d^2*ln(c*x)*ln(1-I*c*x)-3*I*c^2*a*b/d^2*ln(c*x)*ln(1+I*c*x)+2*I*c^2*a*b/d^2*arctan(c*x)/(c*x-I)-3*I*c^2*
a*b/d^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))-a*b/d^2*arctan(c*x)/x^2-3*c^2*b^2/d^2*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c
^2*x^2+1))+c^2*b^2/d^2*arctan(c*x)/(2*c*x-2*I)-3*c^2*b^2/d^2*arctan(c*x)^2*ln(c*x)+3*c^2*b^2/d^2*arctan(c*x)^2
*ln((1+I*c*x)^2/(c^2*x^2+1)-1)-3*c^2*b^2/d^2*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-3*c^2*b^2/d^2*arc
tan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/4*c^3*b^2/d^2/(c*x-I)*x+I*c^2*a^2/d^2/(c*x-I)+c^2*a*b/d^2/(c*x-
I)+3*c^2*b^2/d^2*ln(c*x-I)*arctan(c*x)^2+2*I*c*a^2/d^2/x-1/4*I*c^2*b^2/d^2/(c*x-I)+2*I*c^2*b^2/d^2*arctan(c*x)
^3+3*I*c^2*a^2/d^2*arctan(c*x)+3/2*I*c^2*b^2/d^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))*arctan(c*x)^2-3/2*I*c^2*b^2/d^2*arctan(c*x)^
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))+c*b^2/d^2*arctan(c*x)/x*(c^2*x^2+1)^(1/2)+I*c^2*b^2/d^2*arctan(c*x)^2/(c*x-I)-6
*c^2*a*b/d^2*arctan(c*x)*ln(c*x)+6*c^2*a*b/d^2*arctan(c*x)*ln(c*x-I)+3*I*c^2*a*b/d^2*dilog(1-I*c*x)+3/2*I*c^2*
a*b/d^2*ln(c*x-I)^2+2*I*c*b^2/d^2*arctan(c*x)^2/x-4*I*c^2*a*b/d^2*ln(c*x)-3*I*c^2*a*b/d^2*dilog(-1/2*I*(I+c*x)
)-3*I*c^2*a*b/d^2*dilog(1+I*c*x)+6*I*c^2*b^2/d^2*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*c^2*b
^2/d^2*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))-4*I*c^2*b^2/d^2*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+
1)^(1/2))+2*I*c^2*a*b/d^2*ln(c^2*x^2+1)-9/2*I*c^2*b^2/d^2*Pi*arctan(c*x)^2-c*a*b/d^2/x+3/2*I*c^2*b^2/d^2*Pi*cs
gn(((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*c*a*b/d^2*arctan(c*x)/x-I*c^3*
b^2/d^2*arctan(c*x)/(2*c*x-2*I)*x-1/2*b^2/d^2*arctan(c*x)^2/x^2-2*c^2*b^2/d^2*arctan(c*x)^2-6*c^2*b^2/d^2*poly
log(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-6*c^2*b^2/d^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))-3*c^2*a^2/d^2*ln(c*x)
+3/2*c^2*a^2/d^2*ln(c^2*x^2+1)+c^2*b^2/d^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+4*c^2*b^2/d^2*dilog((1+I*c*x)/(c^
2*x^2+1)^(1/2))-4*c^2*b^2/d^2*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*I*c^2*b^2/d^2*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)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2+3/2*I*c^2*b^2/d^2*Pi*c
sgn(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*I*c^2*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+3/2*I*c^2*b^2/d^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*arctan(c*x)^2-3/2*I*c^2*b^2/d^2*arctan(c*x)^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))+3/2*I*c^2*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*a^2/d^2/x^2

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {a^{2}}{c^{2} x^{5} - 2 i c x^{4} - x^{3}}\, dx + \int \frac {b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c^{2} x^{5} - 2 i c x^{4} - x^{3}}\, dx + \int \frac {2 a b \operatorname {atan}{\left (c x \right )}}{c^{2} x^{5} - 2 i c x^{4} - x^{3}}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(a**2/(c**2*x**5 - 2*I*c*x**4 - x**3), x) + Integral(b**2*atan(c*x)**2/(c**2*x**5 - 2*I*c*x**4 - x**
3), x) + Integral(2*a*b*atan(c*x)/(c**2*x**5 - 2*I*c*x**4 - x**3), x))/d**2

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