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

Optimal. Leaf size=416 \[ -i c^3 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2+3 i b c^2 d^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )-3 i b c^2 d^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {7}{2} c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2-2 i b c^2 d^3 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+6 i b c^2 d^3 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-6 c^2 d^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+3 b^2 c^2 d^3 \text {Li}_2\left (\frac {2}{1-i c x}-1\right )+b^2 c^2 d^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )+\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )-\frac {1}{2} b^2 c^2 d^3 \log \left (c^2 x^2+1\right )+b^2 c^2 d^3 \log (x) \]

[Out]

-b*c*d^3*(a+b*arctan(c*x))/x+7/2*c^2*d^3*(a+b*arctan(c*x))^2-1/2*d^3*(a+b*arctan(c*x))^2/x^2-3*I*b*c^2*d^3*(a+
b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))+3*I*b*c^2*d^3*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))+6*c^2*d^3*(a
+b*arctan(c*x))^2*arctanh(-1+2/(1+I*c*x))+b^2*c^2*d^3*ln(x)-I*c^3*d^3*x*(a+b*arctan(c*x))^2-1/2*b^2*c^2*d^3*ln
(c^2*x^2+1)+6*I*b*c^2*d^3*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))+3*b^2*c^2*d^3*polylog(2,-1+2/(1-I*c*x))+b^2*c^2*
d^3*polylog(2,1-2/(1+I*c*x))-3*I*c*d^3*(a+b*arctan(c*x))^2/x-2*I*b*c^2*d^3*(a+b*arctan(c*x))*ln(2/(1+I*c*x))+3
/2*b^2*c^2*d^3*polylog(3,1-2/(1+I*c*x))-3/2*b^2*c^2*d^3*polylog(3,-1+2/(1+I*c*x))

________________________________________________________________________________________

Rubi [A]  time = 0.75, antiderivative size = 416, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 20, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 2447, 4850, 4988, 4994, 6610} \[ 3 i b c^2 d^3 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-3 i b c^2 d^3 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+3 b^2 c^2 d^3 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+b^2 c^2 d^3 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+\frac {3}{2} b^2 c^2 d^3 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {7}{2} c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2-i c^3 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-2 i b c^2 d^3 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+6 i b c^2 d^3 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-6 c^2 d^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{2} b^2 c^2 d^3 \log \left (c^2 x^2+1\right )+b^2 c^2 d^3 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*d^3*(a + b*ArcTan[c*x]))/x) + (7*c^2*d^3*(a + b*ArcTan[c*x])^2)/2 - (d^3*(a + b*ArcTan[c*x])^2)/(2*x^2)
- ((3*I)*c*d^3*(a + b*ArcTan[c*x])^2)/x - I*c^3*d^3*x*(a + b*ArcTan[c*x])^2 - 6*c^2*d^3*(a + b*ArcTan[c*x])^2
*ArcTanh[1 - 2/(1 + I*c*x)] + b^2*c^2*d^3*Log[x] - (2*I)*b*c^2*d^3*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] - (b
^2*c^2*d^3*Log[1 + c^2*x^2])/2 + (6*I)*b*c^2*d^3*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)] + 3*b^2*c^2*d^3*Po
lyLog[2, -1 + 2/(1 - I*c*x)] + b^2*c^2*d^3*PolyLog[2, 1 - 2/(1 + I*c*x)] + (3*I)*b*c^2*d^3*(a + b*ArcTan[c*x])
*PolyLog[2, 1 - 2/(1 + I*c*x)] - (3*I)*b*c^2*d^3*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] + (3*b^2*c
^2*d^3*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 - (3*b^2*c^2*d^3*PolyLog[3, -1 + 2/(1 + I*c*x)])/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 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 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 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 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 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 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 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 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 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 {(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx &=\int \left (-i c^3 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}+\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^2}-\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d^3 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx+\left (3 i c d^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx-\left (3 c^2 d^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx-\left (i c^3 d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{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 )^2-6 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\left (b c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx+\left (6 i b c^2 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx+\left (12 b c^3 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 (2 i b c^4 d^3\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=4 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}{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 )^2-6 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\left (b c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (6 b c^2 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx-\left (2 i b c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx-\left (b c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx-\left (6 b c^3 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 b c^3 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\\ &=-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {7}{2} 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}{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 )^2-6 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-2 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )+6 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\left (b^2 c^2 d^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx+\left (2 i b^2 c^3 d^3\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (3 i b^2 c^3 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 i b^2 c^3 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 i b^2 c^3 d^3\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {7}{2} 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}{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 )^2-6 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-2 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )+6 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+3 b^2 c^2 d^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-3 i b c^2 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} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} \left (b^2 c^2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )+\left (2 b^2 c^2 d^3\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )\\ &=-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {7}{2} 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}{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 )^2-6 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-2 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )+6 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+3 b^2 c^2 d^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+b^2 c^2 d^3 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-3 i b c^2 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} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} \left (b^2 c^2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 c^4 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {7}{2} 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}{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 )^2-6 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+b^2 c^2 d^3 \log (x)-2 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 c^2 d^3 \log \left (1+c^2 x^2\right )+6 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+3 b^2 c^2 d^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )+b^2 c^2 d^3 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-3 i b c^2 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} b^2 c^2 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 c^2 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 1.15, size = 500, normalized size = 1.20 \[ \frac {1}{2} d^3 \left (-2 i a^2 c^3 x-6 a^2 c^2 \log (x)-\frac {6 i a^2 c}{x}-\frac {a^2}{x^2}-6 i a b c^2 (\text {Li}_2(-i c x)-\text {Li}_2(i c x))-2 i a b c^2 \left (2 c x \tan ^{-1}(c x)-\log \left (c^2 x^2+1\right )\right )-\frac {6 i a b c \left (c x \left (\log \left (c^2 x^2+1\right )-2 \log (c x)\right )+2 \tan ^{-1}(c x)\right )}{x}-\frac {2 a b \left (\tan ^{-1}(c x)+c x \left (c x \tan ^{-1}(c x)+1\right )\right )}{x^2}-2 i b^2 c^2 \left (\tan ^{-1}(c x) \left ((c x-i) \tan ^{-1}(c x)+2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-i \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )\right )+6 b^2 c^2 \left (-i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )-i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )-\frac {2}{3} i \tan ^{-1}(c x)^3-\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+\tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+\frac {i \pi ^3}{24}\right )-\frac {b^2 \left (-2 c^2 x^2 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2+2 c x \tan ^{-1}(c x)\right )}{x^2}+\frac {6 b^2 c \left (c x \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+\tan ^{-1}(c x) \left ((c x-i) \tan ^{-1}(c x)+2 i c x \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )\right )}{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(-(a^2/x^2) - ((6*I)*a^2*c)/x - (2*I)*a^2*c^3*x - (2*a*b*(ArcTan[c*x] + c*x*(1 + c*x*ArcTan[c*x])))/x^2 -
6*a^2*c^2*Log[x] - (b^2*(2*c*x*ArcTan[c*x] + (1 + c^2*x^2)*ArcTan[c*x]^2 - 2*c^2*x^2*Log[(c*x)/Sqrt[1 + c^2*x
^2]]))/x^2 - (2*I)*a*b*c^2*(2*c*x*ArcTan[c*x] - Log[1 + c^2*x^2]) - ((6*I)*a*b*c*(2*ArcTan[c*x] + c*x*(-2*Log[
c*x] + Log[1 + c^2*x^2])))/x - (2*I)*b^2*c^2*(ArcTan[c*x]*((-I + c*x)*ArcTan[c*x] + 2*Log[1 + E^((2*I)*ArcTan[
c*x])]) - I*PolyLog[2, -E^((2*I)*ArcTan[c*x])]) + (6*b^2*c*(ArcTan[c*x]*((-I + c*x)*ArcTan[c*x] + (2*I)*c*x*Lo
g[1 - E^((2*I)*ArcTan[c*x])]) + c*x*PolyLog[2, E^((2*I)*ArcTan[c*x])]))/x - (6*I)*a*b*c^2*(PolyLog[2, (-I)*c*x
] - PolyLog[2, I*c*x]) + 6*b^2*c^2*((I/24)*Pi^3 - ((2*I)/3)*ArcTan[c*x]^3 - ArcTan[c*x]^2*Log[1 - E^((-2*I)*Ar
cTan[c*x])] + ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])]
- I*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2 + PolyLog[3, -E^((2*
I)*ArcTan[c*x])]/2)))/2

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-4 i \, a^{2} c^{3} d^{3} x^{3} - 12 \, a^{2} c^{2} d^{3} x^{2} + 12 i \, a^{2} c d^{3} x + 4 \, a^{2} d^{3} + {\left (i \, b^{2} c^{3} d^{3} x^{3} + 3 \, b^{2} c^{2} d^{3} x^{2} - 3 i \, b^{2} c d^{3} x - b^{2} d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + {\left (4 \, a b c^{3} d^{3} x^{3} - 12 i \, a b c^{2} d^{3} x^{2} - 12 \, a b c d^{3} x + 4 i \, a b d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, x^{3}}, x\right ) \]

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

[Out]

Timed out

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maple [C]  time = 6.75, size = 1840, normalized size = 4.42 \[ \text {result too large to display} \]

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^3,x)

[Out]

-d^3*a*b*arctan(c*x)/x^2+3/2*I*c^2*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)^2-2*I*c^3*d^3*a*b*arctan(c*x)*x-6*I*c*d^3*a*b*arctan(c*x)/x-1/2*d^3*a^2/x^2+3/2*c^2*d^3*b^2*polylo
g(3,-(1+I*c*x)^2/(c^2*x^2+1))-6*c^2*d^3*b^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*c^2*d^3*b^2*dilog(1-I*(1
+I*c*x)/(c^2*x^2+1)^(1/2))-6*c^2*d^3*b^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+c^2*d^3*b^2*ln(1+(1+I*c*x)/(c^
2*x^2+1)^(1/2))-6*c^2*d^3*b^2*dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))+6*c^2*d^3*b^2*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(
1/2))+c^2*d^3*b^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)-1/2*d^3*b^2*arctan(c*x)^2/x^2+3/2*c^2*d^3*b^2*arctan(c*x)^
2-c*d^3*a*b/x-3*c^2*d^3*b^2*arctan(c*x)^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-3*c^2*d^3*b^2*arctan(c*x)^2*ln(1-(
1+I*c*x)/(c^2*x^2+1)^(1/2))+3*c^2*d^3*b^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)-3*c^2*d^3*b^2*arctan(c*x
)^2*ln(c*x)-c^2*d^3*a*b*arctan(c*x)-I*c^3*d^3*a^2*x-3*I*c*d^3*a^2/x-3/2*I*c^2*d^3*b^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))-3/2*I*c^2*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*I*c^2*d^3*a*b*ln(c*x)*ln(1+I*c*x)+3*I*c^2*d^3*a*b*ln(c*x)*ln(1-I*c*x)-3/2*I*c^2*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*c^2*d^3*a^2*ln(c*x)-2*c^2*d^3*b^2*dil
og(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+c*d^3*b^2*arctan(c*x)/x*(c^2*x^2+1)^(1/2)-6*c^2*d^3*a*b*arctan(c*x)*ln(c*x
)-I*c^3*d^3*b^2*arctan(c*x)^2*x-3*I*c*d^3*b^2*arctan(c*x)^2/x-3*I*c^2*d^3*a*b*dilog(1+I*c*x)+3*I*c^2*d^3*a*b*d
ilog(1-I*c*x)+6*I*c^2*d^3*a*b*ln(c*x)-2*I*c^2*d^3*b^2*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*c^2*
d^3*b^2*arctan(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-3*I*c^2*d^3*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^
2*x^2+1))+6*I*c^2*d^3*b^2*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*c^2*d^3*b^2*arctan(c*x)*polylog(2,
-(1+I*c*x)/(c^2*x^2+1)^(1/2))+6*I*c^2*d^3*b^2*arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*c^2*d^3*a
*b*ln(c^2*x^2+1)-3/2*I*c^2*d^3*b^2*arctan(c*x)^2*Pi+3/2*I*c^2*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*I
*c^2*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*I*c^2*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*I*c^2*d^3*b^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))

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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