3.1.90 \(\int \frac {(d+i c d x)^3 (a+b \text {ArcTan}(c x))^2}{x^3} \, dx\) [90]

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

[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-I*c^3*d^3*x*(a+b*
arctan(c*x))^2-3*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)-3*I*c*d^3*(a+b*arctan(c*x))^2/x-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))-2*I*b*c^2*d^3*(a+b*arctan(c*x))*ln(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
/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 = 0.54, 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 = {4996, 4930, 5040, 4964, 2449, 2352, 4946, 5038, 272, 36, 29, 31, 5004, 5044, 4988, 2497, 4942, 5108, 5114, 6745} \begin {gather*} -i c^3 d^3 x (a+b \text {ArcTan}(c x))^2+3 i b c^2 d^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))-3 i b c^2 d^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))+\frac {7}{2} c^2 d^3 (a+b \text {ArcTan}(c x))^2-2 i b c^2 d^3 \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))+6 i b c^2 d^3 \log \left (2-\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))-6 c^2 d^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2-\frac {d^3 (a+b \text {ArcTan}(c x))^2}{2 x^2}-\frac {3 i c d^3 (a+b \text {ArcTan}(c x))^2}{x}-\frac {b c d^3 (a+b \text {ArcTan}(c x))}{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) \end {gather*}

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 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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 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 4930

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

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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 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 5040

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*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 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^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 ) \text {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 ) \text {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 ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 c^4 d^3\right ) \text {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 = 0.69, size = 500, normalized size = 1.20 \begin {gather*} \frac {1}{2} d^3 \left (-\frac {a^2}{x^2}-\frac {6 i a^2 c}{x}-2 i a^2 c^3 x-\frac {2 a b (\text {ArcTan}(c x)+c x (1+c x \text {ArcTan}(c x)))}{x^2}-6 a^2 c^2 \log (x)-\frac {b^2 \left (2 c x \text {ArcTan}(c x)+\left (1+c^2 x^2\right ) \text {ArcTan}(c x)^2-2 c^2 x^2 \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )\right )}{x^2}-2 i a b c^2 \left (2 c x \text {ArcTan}(c x)-\log \left (1+c^2 x^2\right )\right )-\frac {6 i a b c \left (2 \text {ArcTan}(c x)+c x \left (-2 \log (c x)+\log \left (1+c^2 x^2\right )\right )\right )}{x}-2 i b^2 c^2 \left (\text {ArcTan}(c x) \left ((-i+c x) \text {ArcTan}(c x)+2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-i \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )+\frac {6 b^2 c \left (\text {ArcTan}(c x) \left ((-i+c x) \text {ArcTan}(c x)+2 i c x \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )\right )+c x \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )\right )}{x}-6 i a b c^2 (\text {PolyLog}(2,-i c x)-\text {PolyLog}(2,i c x))+6 b^2 c^2 \left (\frac {i \pi ^3}{24}-\frac {2}{3} i \text {ArcTan}(c x)^3-\text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )+\text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )-i \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )-i \text {ArcTan}(c x) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 7.02, size = 1746, normalized size = 4.20

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1746\)
default	\(\text {Expression too large to display}\)	\(1746\)

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,method=_RETURNVERBOSE)

[Out]

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

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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^3,x, algorithm="maxima")

[Out]

-I*a^2*c^3*d^3*x - I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*c^2*d^3 - 3*a^2*c^2*d^3*log(x) - 3*I*(c*(log(c
^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*a*b*c*d^3 - ((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*a*b*d^3 - 3
*I*a^2*c*d^3/x - 1/2*a^2*d^3/x^2 - 1/32*(16*I*(24*b^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)^2/(c^2*x^5 + x^3)
, x) + 2*b^2*c^5*d^3*integrate(1/16*x^5*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x) + 8*b^2*c^5*d^3*integrate(1/16*
x^5*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) - b^2*c^2*d^3*arctan(c*x)^3 - 24*b^2*c^4*d^3*integrate(1/16*x^4*arcta
n(c*x)*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) - 16*b^2*c^4*d^3*integrate(1/16*x^4*arctan(c*x)/(c^2*x^5 + x^3), x
) - 4*b^2*c^3*d^3*integrate(1/16*x^3*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x) + 24*b^2*c^3*d^3*integrate(1/16*x^
3*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) - 16*b^2*c^2*d^3*integrate(1/16*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x
^5 + x^3), x) - 56*b^2*c^2*d^3*integrate(1/16*x^2*arctan(c*x)/(c^2*x^5 + x^3), x) - 72*b^2*c*d^3*integrate(1/1
6*x*arctan(c*x)^2/(c^2*x^5 + x^3), x) - 6*b^2*c*d^3*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x) -
4*b^2*c*d^3*integrate(1/16*x*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) + 8*b^2*d^3*integrate(1/16*arctan(c*x)*log(c
^2*x^2 + 1)/(c^2*x^5 + x^3), x))*x^2 + (128*b^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x
^5 + x^3), x) + 256*b^2*c^5*d^3*integrate(1/16*x^5*arctan(c*x)/(c^2*x^5 + x^3), x) + 1152*b^2*c^4*d^3*integrat
e(1/16*x^4*arctan(c*x)^2/(c^2*x^5 + x^3), x) + 3072*a*b*c^4*d^3*integrate(1/16*x^4*arctan(c*x)/(c^2*x^5 + x^3)
, x) + b^2*c^2*d^3*log(c^2*x^2 + 1)^3 + 24*b^2*c^2*d^3*arctan(c*x)^2 - 256*b^2*c^3*d^3*integrate(1/16*x^3*arct
an(c*x)*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) + 2*b^2*c^2*d^3*log(c^2*x^2 + 1)^2 + 768*b^2*c^2*d^3*integrate(1/
16*x^2*arctan(c*x)^2/(c^2*x^5 + x^3), x) + 64*b^2*c^2*d^3*integrate(1/16*x^2*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3
), x) + 3072*a*b*c^2*d^3*integrate(1/16*x^2*arctan(c*x)/(c^2*x^5 + x^3), x) + 448*b^2*c^2*d^3*integrate(1/16*x
^2*log(c^2*x^2 + 1)/(c^2*x^5 + x^3), x) - 384*b^2*c*d^3*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^5
+ x^3), x) - 128*b^2*c*d^3*integrate(1/16*x*arctan(c*x)/(c^2*x^5 + x^3), x) - 384*b^2*d^3*integrate(1/16*arct
an(c*x)^2/(c^2*x^5 + x^3), x) - 32*b^2*d^3*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^5 + x^3), x))*x^2 - 4*(-2*
I*b^2*c^3*d^3*x^3 - 6*I*b^2*c*d^3*x - b^2*d^3)*arctan(c*x)^2 - 4*(2*b^2*c^3*d^3*x^3 + 6*b^2*c*d^3*x - I*b^2*d^
3)*arctan(c*x)*log(c^2*x^2 + 1) + (-2*I*b^2*c^3*d^3*x^3 - 6*I*b^2*c*d^3*x - b^2*d^3)*log(c^2*x^2 + 1)^2)/x^2

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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^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 - 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^3, 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**3,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^3,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^3} \,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^3,x)

[Out]

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

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