3.91 \(\int \frac {(d+i c d x)^3 (a+b \tan ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=429 \[ -b c^3 d^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )+b c^3 d^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {11}{6} i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {20}{3} b c^3 d^3 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 i c^3 d^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}+\frac {10}{3} i b^2 c^3 d^3 \text {Li}_2\left (\frac {2}{1-i c x}-1\right )+\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )-\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )+3 i b^2 c^3 d^3 \log (x)-\frac {1}{3} b^2 c^3 d^3 \tan ^{-1}(c x)-\frac {b^2 c^2 d^3}{3 x}-\frac {3}{2} i b^2 c^3 d^3 \log \left (c^2 x^2+1\right ) \]

[Out]

-1/3*b^2*c^2*d^3/x-1/3*b^2*c^3*d^3*arctan(c*x)-1/3*b*c*d^3*(a+b*arctan(c*x))/x^2-3/2*I*c*d^3*(a+b*arctan(c*x))
^2/x^2+10/3*I*b^2*c^3*d^3*polylog(2,-1+2/(1-I*c*x))-1/3*d^3*(a+b*arctan(c*x))^2/x^3-1/2*I*b^2*c^3*d^3*polylog(
3,-1+2/(1+I*c*x))+3*c^2*d^3*(a+b*arctan(c*x))^2/x-3/2*I*b^2*c^3*d^3*ln(c^2*x^2+1)+2*I*c^3*d^3*(a+b*arctan(c*x)
)^2*arctanh(-1+2/(1+I*c*x))+3*I*b^2*c^3*d^3*ln(x)-20/3*b*c^3*d^3*(a+b*arctan(c*x))*ln(2-2/(1-I*c*x))-3*I*b*c^2
*d^3*(a+b*arctan(c*x))/x-b*c^3*d^3*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))+b*c^3*d^3*(a+b*arctan(c*x))*poly
log(2,-1+2/(1+I*c*x))+11/6*I*c^3*d^3*(a+b*arctan(c*x))^2+1/2*I*b^2*c^3*d^3*polylog(3,1-2/(1+I*c*x))

________________________________________________________________________________________

Rubi [A]  time = 0.89, antiderivative size = 429, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 17, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {4876, 4852, 4918, 325, 203, 4924, 4868, 2447, 266, 36, 29, 31, 4884, 4850, 4988, 4994, 6610} \[ -b c^3 d^3 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+b c^3 d^3 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {10}{3} i b^2 c^3 d^3 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+\frac {1}{2} i b^2 c^3 d^3 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c^3 d^3 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {11}{6} i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {20}{3} b c^3 d^3 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 i c^3 d^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {3}{2} i b^2 c^3 d^3 \log \left (c^2 x^2+1\right )-\frac {b^2 c^2 d^3}{3 x}+3 i b^2 c^3 d^3 \log (x)-\frac {1}{3} b^2 c^3 d^3 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^3*(a + b*ArcTan[c*x])^2)/x^4,x]

[Out]

-(b^2*c^2*d^3)/(3*x) - (b^2*c^3*d^3*ArcTan[c*x])/3 - (b*c*d^3*(a + b*ArcTan[c*x]))/(3*x^2) - ((3*I)*b*c^2*d^3*
(a + b*ArcTan[c*x]))/x + ((11*I)/6)*c^3*d^3*(a + b*ArcTan[c*x])^2 - (d^3*(a + b*ArcTan[c*x])^2)/(3*x^3) - (((3
*I)/2)*c*d^3*(a + b*ArcTan[c*x])^2)/x^2 + (3*c^2*d^3*(a + b*ArcTan[c*x])^2)/x - (2*I)*c^3*d^3*(a + b*ArcTan[c*
x])^2*ArcTanh[1 - 2/(1 + I*c*x)] + (3*I)*b^2*c^3*d^3*Log[x] - ((3*I)/2)*b^2*c^3*d^3*Log[1 + c^2*x^2] - (20*b*c
^3*d^3*(a + b*ArcTan[c*x])*Log[2 - 2/(1 - I*c*x)])/3 + ((10*I)/3)*b^2*c^3*d^3*PolyLog[2, -1 + 2/(1 - I*c*x)] -
b*c^3*d^3*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + b*c^3*d^3*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2
/(1 + I*c*x)] + (I/2)*b^2*c^3*d^3*PolyLog[3, 1 - 2/(1 + I*c*x)] - (I/2)*b^2*c^3*d^3*PolyLog[3, -1 + 2/(1 + I*c
*x)]

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 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 {(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^4} \, dx &=\int \left (\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^4}+\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^3}-\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^2}-\frac {i c^3 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^4} \, dx+\left (3 i c d^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^3} \, dx-\left (3 c^2 d^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x^2} \, dx-\left (i c^3 d^3\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\frac {1}{3} \left (2 b c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3 \left (1+c^2 x^2\right )} \, dx+\left (3 i b c^2 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )} \, dx-\left (6 b c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx+\left (4 i b c^4 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\\ &=3 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}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\frac {1}{3} \left (2 b c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx+\left (3 i b c^2 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (6 i b c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx-\frac {1}{3} \left (2 b c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x \left (1+c^2 x^2\right )} \, dx-\left (2 i b c^4 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 (2 i b c^4 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-\left (3 i b c^4 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {11}{6} 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}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-6 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )-b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{3} \left (b^2 c^2 d^3\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac {1}{3} \left (2 i b c^3 d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{x (i+c x)} \, dx+\left (3 i b^2 c^3 d^3\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx+\left (b^2 c^4 d^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (b^2 c^4 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 b^2 c^4 d^3\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b^2 c^2 d^3}{3 x}-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {11}{6} 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}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\frac {20}{3} b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+3 i b^2 c^3 d^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )-b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} \left (3 i b^2 c^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac {1}{3} \left (b^2 c^4 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx+\frac {1}{3} \left (2 b^2 c^4 d^3\right ) \int \frac {\log \left (2-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b^2 c^2 d^3}{3 x}-\frac {1}{3} b^2 c^3 d^3 \tan ^{-1}(c x)-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {11}{6} 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}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\frac {20}{3} b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+\frac {10}{3} i b^2 c^3 d^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )-b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} \left (3 i b^2 c^3 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (3 i b^2 c^5 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b^2 c^2 d^3}{3 x}-\frac {1}{3} b^2 c^3 d^3 \tan ^{-1}(c x)-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac {3 i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {11}{6} 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}{3 x^3}-\frac {3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}+\frac {3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-2 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+3 i b^2 c^3 d^3 \log (x)-\frac {3}{2} i b^2 c^3 d^3 \log \left (1+c^2 x^2\right )-\frac {20}{3} b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-i c x}\right )+\frac {10}{3} i b^2 c^3 d^3 \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )-b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )-\frac {1}{2} i b^2 c^3 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.80, size = 595, normalized size = 1.39 \[ \frac {d^3 \left (-24 i a^2 c^3 x^3 \log (x)+72 a^2 c^2 x^2-36 i a^2 c x-8 a^2+24 a b c^3 x^3 \text {Li}_2(-i c x)-24 a b c^3 x^3 \text {Li}_2(i c x)-160 a b c^3 x^3 \log (c x)-72 i a b c^3 x^3 \tan ^{-1}(c x)-72 i a b c^2 x^2+144 a b c^2 x^2 \tan ^{-1}(c x)+80 a b c^3 x^3 \log \left (c^2 x^2+1\right )-8 a b c x-72 i a b c x \tan ^{-1}(c x)-16 a b \tan ^{-1}(c x)+24 b^2 c^3 x^3 \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+24 b^2 c^3 x^3 \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+80 i b^2 c^3 x^3 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-12 i b^2 c^3 x^3 \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )+12 i b^2 c^3 x^3 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )-\pi ^3 b^2 c^3 x^3+16 b^2 c^3 x^3 \tan ^{-1}(c x)^3+44 i b^2 c^3 x^3 \tan ^{-1}(c x)^2-8 b^2 c^3 x^3 \tan ^{-1}(c x)-24 i b^2 c^3 x^3 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-160 b^2 c^3 x^3 \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+24 i b^2 c^3 x^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-8 b^2 c^2 x^2+72 b^2 c^2 x^2 \tan ^{-1}(c x)^2-72 i b^2 c^2 x^2 \tan ^{-1}(c x)+72 i b^2 c^3 x^3 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )-36 i b^2 c x \tan ^{-1}(c x)^2-8 b^2 c x \tan ^{-1}(c x)-8 b^2 \tan ^{-1}(c x)^2\right )}{24 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x])^2)/x^4,x]

[Out]

(d^3*(-8*a^2 - (36*I)*a^2*c*x - 8*a*b*c*x + 72*a^2*c^2*x^2 - (72*I)*a*b*c^2*x^2 - 8*b^2*c^2*x^2 - b^2*c^3*Pi^3
*x^3 - 16*a*b*ArcTan[c*x] - (72*I)*a*b*c*x*ArcTan[c*x] - 8*b^2*c*x*ArcTan[c*x] + 144*a*b*c^2*x^2*ArcTan[c*x] -
(72*I)*b^2*c^2*x^2*ArcTan[c*x] - (72*I)*a*b*c^3*x^3*ArcTan[c*x] - 8*b^2*c^3*x^3*ArcTan[c*x] - 8*b^2*ArcTan[c*
x]^2 - (36*I)*b^2*c*x*ArcTan[c*x]^2 + 72*b^2*c^2*x^2*ArcTan[c*x]^2 + (44*I)*b^2*c^3*x^3*ArcTan[c*x]^2 + 16*b^2
*c^3*x^3*ArcTan[c*x]^3 - (24*I)*b^2*c^3*x^3*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - 160*b^2*c^3*x^3*Ar
cTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + (24*I)*b^2*c^3*x^3*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - (
24*I)*a^2*c^3*x^3*Log[x] - 160*a*b*c^3*x^3*Log[c*x] + (72*I)*b^2*c^3*x^3*Log[(c*x)/Sqrt[1 + c^2*x^2]] + 80*a*b
*c^3*x^3*Log[1 + c^2*x^2] + 24*b^2*c^3*x^3*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] + 24*b^2*c^3*x^3*Arc
Tan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (80*I)*b^2*c^3*x^3*PolyLog[2, E^((2*I)*ArcTan[c*x])] + 24*a*b*c^
3*x^3*PolyLog[2, (-I)*c*x] - 24*a*b*c^3*x^3*PolyLog[2, I*c*x] - (12*I)*b^2*c^3*x^3*PolyLog[3, E^((-2*I)*ArcTan
[c*x])] + (12*I)*b^2*c^3*x^3*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/(24*x^3)

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

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^4,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^4, 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^4,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 8.40, size = 1814, normalized size = 4.23 \[ \text {result too large to display} \]

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^4,x)

[Out]

3*c^2*d^3*a^2/x-1/3*d^3*b^2*arctan(c*x)^2/x^3+6*c^2*d^3*a*b*arctan(c*x)/x+c^3*d^3*a*b*ln(c*x)*ln(1+I*c*x)-c^3*
d^3*a*b*ln(c*x)*ln(1-I*c*x)+I*c^3*d^3*b^2*arctan(c*x)^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+1/2*c^3*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)^2-1/2*c^3*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+1/2*c^3*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*I*c^2*d^3*a*b/x-3/2*I*c*d^3*b^2*arctan(c*x)^2/x^2-3*I*c^2*d
^3*b^2*arctan(c*x)/x-I*c^3*d^3*b^2*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-I*c^3*d^3*b^2*arctan(c*x)^2
*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/3*I*c^3*d^3*b^2/(I*c*x-(c^2*x^2+1)^(1/2)+1)*(c^2*x^2+1)^(1/2)-I*c^3*d^3*b
^2*arctan(c*x)^2*ln(c*x)+8/3*b^2*c^3*d^3*arctan(c*x)-1/3*d^3*a^2/x^3-1/2*c^3*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)^2-1/2*c^3*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*arcta
n(c*x)^2+1/2*c^3*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))*arctan(c*x)^2-1/2*c^3*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)^2-2
*I*c^3*d^3*a*b*arctan(c*x)*ln(c*x)-3*I*c*d^3*a*b*arctan(c*x)/x^2+1/3*I*c^3*d^3*b^2/(I*c*x+(c^2*x^2+1)^(1/2)+1)
*(c^2*x^2+1)^(1/2)-3*I*c^3*d^3*a*b*arctan(c*x)+1/2*c^3*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))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)-1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+1/2*
I*c^3*d^3*b^2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-20/3*I*c^3*d^3*b^2*dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*c^
3*d^3*b^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*I*c^3*d^3*b^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))+3*I*c^3*d^
3*b^2*ln((1+I*c*x)/(c^2*x^2+1)^(1/2)-1)-2*I*c^3*d^3*b^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+20/3*I*c^3*d^3*
b^2*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-3/2*I*c*d^3*a^2/x^2+11/6*I*c^3*d^3*b^2*arctan(c*x)^2-I*c^3*d^3*a^2*ln
(c*x)-c^3*d^3*a*b*dilog(1-I*c*x)-20/3*c^3*d^3*b^2*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*c^3*d^3*b^2*
arctan(c*x)*polylog(2,(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*c^3*d^3*b^2*Pi*arctan(c*x)^2+3*c^2*d^3*b^2*arctan(c*x)^
2/x-1/3*c*d^3*b^2*arctan(c*x)/x^2-20/3*c^3*d^3*a*b*ln(c*x)+10/3*c^3*d^3*a*b*ln(c^2*x^2+1)+c^3*d^3*a*b*dilog(1+
I*c*x)-1/3*c*d^3*a*b/x^2-2/3*d^3*a*b*arctan(c*x)/x^3-2*c^3*d^3*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1
)^(1/2))+c^3*d^3*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+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^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3}{x^4} \,d x \]

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^4,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i)^3)/x^4, 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**4,x)

[Out]

Timed out

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