3.1.91 \(\int \frac {(d+i c d x)^3 (a+b \text {ArcTan}(c x))^2}{x^4} \, dx\) [91]

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

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

________________________________________________________________________________________

Rubi [A]
time = 0.64, 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 = {4996, 4946, 5038, 331, 209, 5044, 4988, 2497, 272, 36, 29, 31, 5004, 4942, 5108, 5114, 6745} \begin {gather*} -b c^3 d^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))+b c^3 d^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))+\frac {11}{6} i c^3 d^3 (a+b \text {ArcTan}(c x))^2-\frac {20}{3} b c^3 d^3 \log \left (2-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))-2 i c^3 d^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2+\frac {3 c^2 d^3 (a+b \text {ArcTan}(c x))^2}{x}-\frac {3 i b c^2 d^3 (a+b \text {ArcTan}(c x))}{x}-\frac {d^3 (a+b \text {ArcTan}(c x))^2}{3 x^3}-\frac {3 i c d^3 (a+b \text {ArcTan}(c x))^2}{2 x^2}-\frac {b c d^3 (a+b \text {ArcTan}(c x))}{3 x^2}-\frac {1}{3} b^2 c^3 d^3 \text {ArcTan}(c x)+\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 {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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b^2*c^2*d^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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 272

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 331

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 2497

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 4942

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 4988

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 4996

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 5004

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 5038

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 5044

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 5108

Int[(ArcTanh[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[L
og[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 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 6745

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 ) \text {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 ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (3 i b^2 c^5 d^3\right ) \text {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 = 0.43, size = 595, normalized size = 1.39 \begin {gather*} \frac {d^3 \left (-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 \text {ArcTan}(c x)-72 i a b c x \text {ArcTan}(c x)-8 b^2 c x \text {ArcTan}(c x)+144 a b c^2 x^2 \text {ArcTan}(c x)-72 i b^2 c^2 x^2 \text {ArcTan}(c x)-72 i a b c^3 x^3 \text {ArcTan}(c x)-8 b^2 c^3 x^3 \text {ArcTan}(c x)-8 b^2 \text {ArcTan}(c x)^2-36 i b^2 c x \text {ArcTan}(c x)^2+72 b^2 c^2 x^2 \text {ArcTan}(c x)^2+44 i b^2 c^3 x^3 \text {ArcTan}(c x)^2+16 b^2 c^3 x^3 \text {ArcTan}(c x)^3-24 i b^2 c^3 x^3 \text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )-160 b^2 c^3 x^3 \text {ArcTan}(c x) \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )+24 i b^2 c^3 x^3 \text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )-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 \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )+80 a b c^3 x^3 \log \left (1+c^2 x^2\right )+24 b^2 c^3 x^3 \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )+24 b^2 c^3 x^3 \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )+80 i b^2 c^3 x^3 \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )+24 a b c^3 x^3 \text {PolyLog}(2,-i c x)-24 a b c^3 x^3 \text {PolyLog}(2,i c x)-12 i b^2 c^3 x^3 \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )+12 i b^2 c^3 x^3 \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )\right )}{24 x^3} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 10.27, size = 1726, normalized size = 4.02

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1726\)
default	\(\text {Expression too large to display}\)	\(1726\)

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,method=_RETURNVERBOSE)

[Out]

c^3*(1/2*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))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2-I*a^2*d^3*ln(c*x)+3*I*d^3*b^2*ln((1+I*c*x)/(c^2*x^
2+1)^(1/2)-1)+3*I*d^3*b^2*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*d^3*b^2*polylog(3,(1+I*c*x)/(c^2*x^2+1)^(1/2))
+20/3*I*d^3*b^2*dilog(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*I*d^3*b^2*polylog(3,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/2*I
*d^3*b^2*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+11/6*I*d^3*b^2*arctan(c*x)^2-20/3*I*d^3*b^2*dilog((1+I*c*x)/(c^2*
x^2+1)^(1/2))+10/3*a*b*ln(c^2*x^2+1)*d^3+8/3*d^3*b^2*arctan(c*x)-2*I*d^3*a*b*arctan(c*x)*ln(c*x)-2/3*d^3*a*b*a
rctan(c*x)/c^3/x^3-3*I*d^3*a*b/c/x-3/2*I*d^3*b^2*arctan(c*x)^2/c^2/x^2-3*I*d^3*b^2*arctan(c*x)/c/x-1/2*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*arct
an(c*x)^2-1/2*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))^2*csgn(I/((1+I*c*x)^2
/(c^2*x^2+1)+1))*arctan(c*x)^2-1/2*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+1/2*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))*a
rctan(c*x)^2+6*d^3*a*b*arctan(c*x)/c/x+3*a^2*d^3/c/x+d^3*a*b*dilog(1+I*c*x)-d^3*a*b*dilog(1-I*c*x)-20/3*d^3*a*
b*ln(c*x)-20/3*d^3*b^2*arctan(c*x)*ln(1+(1+I*c*x)/(c^2*x^2+1)^(1/2))-2*d^3*b^2*arctan(c*x)*polylog(2,(1+I*c*x)
/(c^2*x^2+1)^(1/2))-2*d^3*b^2*arctan(c*x)*polylog(2,-(1+I*c*x)/(c^2*x^2+1)^(1/2))+d^3*b^2*arctan(c*x)*polylog(
2,-(1+I*c*x)^2/(c^2*x^2+1))+1/2*d^3*b^2*Pi*arctan(c*x)^2-3*I*d^3*a*b*arctan(c*x)/c^2/x^2-1/3*a^2*d^3/c^3/x^3-1
/3*d^3*b^2*arctan(c*x)/c^2/x^2-1/3*d^3*b^2*arctan(c*x)^2/c^3/x^3-3*I*d^3*a*b*arctan(c*x)+I*d^3*b^2*arctan(c*x)
^2*ln((1+I*c*x)^2/(c^2*x^2+1)-1)+1/3*I*d^3*b^2/(1+I*c*x+(c^2*x^2+1)^(1/2))*(c^2*x^2+1)^(1/2)-I*d^3*b^2*arctan(
c*x)^2*ln(c*x)-1/3*I*d^3*b^2/(I*c*x-(c^2*x^2+1)^(1/2)+1)*(c^2*x^2+1)^(1/2)-I*d^3*b^2*arctan(c*x)^2*ln(1+(1+I*c
*x)/(c^2*x^2+1)^(1/2))-I*d^3*b^2*arctan(c*x)^2*ln(1-(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/2*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+1/2*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+1/2*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*d^3*b^2*arctan(c*x)^2/c/x+d^3*a*b*ln(c*x)*ln(1+I*c*x)-d^3*a*b*ln(c*x)*ln(1-I
*c*x)-3/2*I*a^2*d^3/c^2/x^2-1/3*d^3*a*b/c^2/x^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

-I*a^2*c^3*d^3*log(x) + 3*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*a*b*c^2*d^3 - 3*I*((c*arctan(c*x
) + 1/x)*c + arctan(c*x)/x^2)*a*b*c*d^3 + 1/3*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)
/x^3)*a*b*d^3 + 3*a^2*c^2*d^3/x - 3/2*I*a^2*c*d^3/x^2 - 1/3*a^2*d^3/x^3 - 1/96*(24*(3*b^2*c^3*d^3*arctan(c*x)^
3 + 48*b^2*c^5*d^3*integrate(1/48*x^5*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^6 + x^4), x) + 36*b^2*c^4*d^3*integr
ate(1/48*x^4*log(c^2*x^2 + 1)^2/(c^2*x^6 + x^4), x) - 144*b^2*c^4*d^3*integrate(1/48*x^4*log(c^2*x^2 + 1)/(c^2
*x^6 + x^4), x) - 96*b^2*c^3*d^3*integrate(1/48*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^6 + x^4), x) + 432*b^2
*c^3*d^3*integrate(1/48*x^3*arctan(c*x)/(c^2*x^6 + x^4), x) + 288*b^2*c^2*d^3*integrate(1/48*x^2*arctan(c*x)^2
/(c^2*x^6 + x^4), x) + 24*b^2*c^2*d^3*integrate(1/48*x^2*log(c^2*x^2 + 1)^2/(c^2*x^6 + x^4), x) + 88*b^2*c^2*d
^3*integrate(1/48*x^2*log(c^2*x^2 + 1)/(c^2*x^6 + x^4), x) - 144*b^2*c*d^3*integrate(1/48*x*arctan(c*x)*log(c^
2*x^2 + 1)/(c^2*x^6 + x^4), x) - 32*b^2*c*d^3*integrate(1/48*x*arctan(c*x)/(c^2*x^6 + x^4), x) - 144*b^2*d^3*i
ntegrate(1/48*arctan(c*x)^2/(c^2*x^6 + x^4), x) - 12*b^2*d^3*integrate(1/48*log(c^2*x^2 + 1)^2/(c^2*x^6 + x^4)
, x))*x^3 + I*(3456*b^2*c^5*d^3*integrate(1/48*x^5*arctan(c*x)^2/(c^2*x^6 + x^4), x) + 9216*a*b*c^5*d^3*integr
ate(1/48*x^5*arctan(c*x)/(c^2*x^6 + x^4), x) + b^2*c^3*d^3*log(c^2*x^2 + 1)^3 + 72*b^2*c^3*d^3*arctan(c*x)^2 -
3456*b^2*c^4*d^3*integrate(1/48*x^4*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^6 + x^4), x) - 6912*b^2*c^3*d^3*integ
rate(1/48*x^3*arctan(c*x)^2/(c^2*x^6 + x^4), x) - 576*b^2*c^3*d^3*integrate(1/48*x^3*log(c^2*x^2 + 1)^2/(c^2*x
^6 + x^4), x) + 9216*a*b*c^3*d^3*integrate(1/48*x^3*arctan(c*x)/(c^2*x^6 + x^4), x) + 5184*b^2*c^3*d^3*integra
te(1/48*x^3*log(c^2*x^2 + 1)/(c^2*x^6 + x^4), x) - 2304*b^2*c^2*d^3*integrate(1/48*x^2*arctan(c*x)*log(c^2*x^2
+ 1)/(c^2*x^6 + x^4), x) - 4224*b^2*c^2*d^3*integrate(1/48*x^2*arctan(c*x)/(c^2*x^6 + x^4), x) - 10368*b^2*c*
d^3*integrate(1/48*x*arctan(c*x)^2/(c^2*x^6 + x^4), x) - 864*b^2*c*d^3*integrate(1/48*x*log(c^2*x^2 + 1)^2/(c^
2*x^6 + x^4), x) - 384*b^2*c*d^3*integrate(1/48*x*log(c^2*x^2 + 1)/(c^2*x^6 + x^4), x) + 1152*b^2*d^3*integrat
e(1/48*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^6 + x^4), x))*x^3 - 4*(18*b^2*c^2*d^3*x^2 - 9*I*b^2*c*d^3*x - 2*b^2
*d^3)*arctan(c*x)^2 - 4*(18*I*b^2*c^2*d^3*x^2 + 9*b^2*c*d^3*x - 2*I*b^2*d^3)*arctan(c*x)*log(c^2*x^2 + 1) + (1
8*b^2*c^2*d^3*x^2 - 9*I*b^2*c*d^3*x - 2*b^2*d^3)*log(c^2*x^2 + 1)^2)/x^3

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 - 3*I*a*b*c^2*
d^3*x^2 - 3*a*b*c*d^3*x + I*a*b*d^3)*log(-(c*x + I)/(c*x - I)))/x^4, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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