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

Optimal. Leaf size=337 \[ i b c^2 d^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )-i b c^2 d^2 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {3}{2} c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2+4 i b c^2 d^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 c^2 d^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 b^2 c^2 d^2 \text {Li}_2\left (\frac {2}{1-i c x}-1\right )+\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )-\frac {1}{2} b^2 c^2 d^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )-\frac {1}{2} b^2 c^2 d^2 \log \left (c^2 x^2+1\right )+b^2 c^2 d^2 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.65, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4876, 4852, 4918, 266, 36, 29, 31, 4884, 4924, 4868, 2447, 4850, 4988, 4994, 6610} \[ i b c^2 d^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-i b c^2 d^2 \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 b^2 c^2 d^2 \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right )+\frac {1}{2} b^2 c^2 d^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 c^2 d^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )+\frac {3}{2} c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )^2+4 i b c^2 d^2 \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-2 c^2 d^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 x^2}-\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x}-\frac {b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{2} b^2 c^2 d^2 \log \left (c^2 x^2+1\right )+b^2 c^2 d^2 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.90, size = 388, normalized size = 1.15 \[ -\frac {d^2 \left (2 a^2 c^2 x^2 \log (x)+4 i a^2 c x+a^2+2 i a b c^2 x^2 (\text {Li}_2(-i c x)-\text {Li}_2(i c x))+4 i a b c x \left (c x \left (\log \left (c^2 x^2+1\right )-2 \log (c x)\right )+2 \tan ^{-1}(c x)\right )+2 a b \left (\tan ^{-1}(c x)+c x \left (c x \tan ^{-1}(c x)+1\right )\right )+\frac {1}{12} b^2 c^2 x^2 \left (24 i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+24 i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+12 \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-12 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+16 i \tan ^{-1}(c x)^3+24 \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-24 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-i \pi ^3\right )+b^2 \left (-2 c^2 x^2 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+\left (c^2 x^2+1\right ) \tan ^{-1}(c x)^2+2 c x \tan ^{-1}(c x)\right )+4 i b^2 c x \left (i c x \left (\tan ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )\right )+\tan ^{-1}(c x)^2-2 c x \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(d^2*(a^2 + (4*I)*a^2*c*x + 2*a*b*(ArcTan[c*x] + c*x*(1 + c*x*ArcTan[c*x])) + 2*a^2*c^2*x^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]]) + (4*I)*a*b*c*x*(2*
ArcTan[c*x] + c*x*(-2*Log[c*x] + Log[1 + c^2*x^2])) + (4*I)*b^2*c*x*(ArcTan[c*x]^2 - 2*c*x*ArcTan[c*x]*Log[1 -
E^((2*I)*ArcTan[c*x])] + I*c*x*(ArcTan[c*x]^2 + PolyLog[2, E^((2*I)*ArcTan[c*x])])) + (2*I)*a*b*c^2*x^2*(Poly
Log[2, (-I)*c*x] - PolyLog[2, I*c*x]) + (b^2*c^2*x^2*((-I)*Pi^3 + (16*I)*ArcTan[c*x]^3 + 24*ArcTan[c*x]^2*Log[
1 - E^((-2*I)*ArcTan[c*x])] - 24*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2,
E^((-2*I)*ArcTan[c*x])] + (24*I)*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 12*PolyLog[3, E^((-2*I)*ArcT
an[c*x])] - 12*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/12))/x^2

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {4 \, a^{2} c^{2} d^{2} x^{2} - 8 i \, a^{2} c d^{2} x - 4 \, a^{2} d^{2} - {\left (b^{2} c^{2} d^{2} x^{2} - 2 i \, b^{2} c d^{2} x - b^{2} d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - {\left (-4 i \, a b c^{2} d^{2} x^{2} - 8 \, a b c d^{2} x + 4 i \, a b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-1/4*(4*a^2*c^2*d^2*x^2 - 8*I*a^2*c*d^2*x - 4*a^2*d^2 - (b^2*c^2*d^2*x^2 - 2*I*b^2*c*d^2*x - b^2*d^2)
*log(-(c*x + I)/(c*x - I))^2 - (-4*I*a*b*c^2*d^2*x^2 - 8*a*b*c*d^2*x + 4*I*a*b*d^2)*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)^2*(a+b*arctan(c*x))^2/x^3,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 6.41, size = 1647, normalized size = 4.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d^2*a^2/x^2-I*c^2*d^2*a*b*dilog(1+I*c*x)+4*I*c^2*d^2*a*b*ln(c*x)+4*I*c^2*d^2*b^2*arctan(c*x)*ln(1+(1+I*c*
x)/(c^2*x^2+1)^(1/2))-2*I*c*d^2*b^2*arctan(c*x)^2/x-I*c^2*d^2*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+
1))+2*I*c^2*d^2*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+2*I*c^2*d^2*b^2*arctan(c*x)*polylog(2,
(1+I*c*x)/(c^2*x^2+1)^(1/2))-I*c^2*d^2*b^2*arctan(c*x)-2*I*c*d^2*a^2/x+I*c^2*d^2*a*b*ln(c*x)*ln(1-I*c*x)-2*I*c
^2*d^2*a*b*ln(c^2*x^2+1)-2*c^2*d^2*a*b*arctan(c*x)*ln(c*x)+I*c^2*d^2*a*b*dilog(1-I*c*x)-1/2*I*c^2*d^2*b^2*arct
an(c*x)^2*Pi-2*c^2*d^2*b^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))+3/2*c^2*d^2*b^2*arctan(c*x)^2-4*c^2*d^2*b^2*
dilog((1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*c^2*d^2*b^2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+c^2*d^2*b^2*ln(1+(1+I*c
*x)/(c^2*x^2+1)^(1/2))-1/2*d^2*b^2/x^2*arctan(c*x)^2-c^2*d^2*a^2*ln(c*x)+c^2*d^2*b^2*ln((1+I*c*x)/(c^2*x^2+1)^
(1/2)-1)+4*c^2*d^2*b^2*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*c^2*d^2*b^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/
2))-c*d^2*a*b/x-d^2*a*b*arctan(c*x)/x^2-c^2*d^2*b^2*arctan(c*x)^2*ln(c*x)-c^2*d^2*b^2*arctan(c*x)^2*ln(1+(1+I*
c*x)/(c^2*x^2+1)^(1/2))-c^2*d^2*b^2*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))+c^2*d^2*b^2*arctan(c*x)^2*
ln((1+I*c*x)^2/(c^2*x^2+1)-1)-c^2*d^2*a*b*arctan(c*x)-c*d^2*b^2*arctan(c*x)/x-1/2*I*c^2*d^2*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))+1/2*I*c^2*d^2*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*I*c^2*d^2*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*I*c^2*d^2*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))+1/2*I*c^2*d^2*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))*c
sgn(((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-I*c^2*d^2*a*b*ln(c*x)*ln(1+I*c*x)
+1/2*I*c^2*d^2*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*I*c^2*
d^2*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))^3-1/2*I*c^2*d^2*b^2*P
i*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)^2-4*I*c*d^2*a*b*arctan(c*x)/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((d+I*c*d*x)^2*(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 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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