3.1.88 \(\int \frac {(d+i c d x)^3 (a+b \text {ArcTan}(c x))^2}{x} \, dx\) [88]

Optimal. Leaf size=385 \[ 3 a b c d^3 x-\frac {1}{3} i b^2 c d^3 x+\frac {1}{3} i b^2 d^3 \text {ArcTan}(c x)+3 b^2 c d^3 x \text {ArcTan}(c x)+\frac {1}{3} i b c^2 d^3 x^2 (a+b \text {ArcTan}(c x))-\frac {29}{6} d^3 (a+b \text {ArcTan}(c x))^2+3 i c d^3 x (a+b \text {ArcTan}(c x))^2-\frac {3}{2} c^2 d^3 x^2 (a+b \text {ArcTan}(c x))^2-\frac {1}{3} i c^3 d^3 x^3 (a+b \text {ArcTan}(c x))^2+2 d^3 (a+b \text {ArcTan}(c x))^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\frac {20}{3} i b d^3 (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 d^3 \log \left (1+c^2 x^2\right )-\frac {10}{3} b^2 d^3 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-i b d^3 (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )+i b d^3 (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d^3 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^3 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \]

[Out]

3*a*b*c*d^3*x+1/3*I*b*c^2*d^3*x^2*(a+b*arctan(c*x))+I*b*d^3*(a+b*arctan(c*x))*polylog(2,-1+2/(1+I*c*x))+3*b^2*
c*d^3*x*arctan(c*x)-1/3*I*c^3*d^3*x^3*(a+b*arctan(c*x))^2-29/6*d^3*(a+b*arctan(c*x))^2-I*b*d^3*(a+b*arctan(c*x
))*polylog(2,1-2/(1+I*c*x))-3/2*c^2*d^3*x^2*(a+b*arctan(c*x))^2-1/3*I*b^2*c*d^3*x-2*d^3*(a+b*arctan(c*x))^2*ar
ctanh(-1+2/(1+I*c*x))+1/3*I*b^2*d^3*arctan(c*x)-3/2*b^2*d^3*ln(c^2*x^2+1)-10/3*b^2*d^3*polylog(2,1-2/(1+I*c*x)
)+3*I*c*d^3*x*(a+b*arctan(c*x))^2+20/3*I*b*d^3*(a+b*arctan(c*x))*ln(2/(1+I*c*x))-1/2*b^2*d^3*polylog(3,1-2/(1+
I*c*x))+1/2*b^2*d^3*polylog(3,-1+2/(1+I*c*x))

________________________________________________________________________________________

Rubi [A]
time = 0.55, antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {4996, 4930, 5040, 4964, 2449, 2352, 4942, 5108, 5004, 5114, 6745, 4946, 5036, 266, 327, 209} \begin {gather*} -\frac {1}{3} i c^3 d^3 x^3 (a+b \text {ArcTan}(c x))^2-\frac {3}{2} c^2 d^3 x^2 (a+b \text {ArcTan}(c x))^2+\frac {1}{3} i b c^2 d^3 x^2 (a+b \text {ArcTan}(c x))-i b d^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))+i b d^3 \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) (a+b \text {ArcTan}(c x))+3 i c d^3 x (a+b \text {ArcTan}(c x))^2-\frac {29}{6} d^3 (a+b \text {ArcTan}(c x))^2+\frac {20}{3} i b d^3 \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))+2 d^3 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2+3 a b c d^3 x+\frac {1}{3} i b^2 d^3 \text {ArcTan}(c x)+3 b^2 c d^3 x \text {ArcTan}(c x)-\frac {3}{2} b^2 d^3 \log \left (c^2 x^2+1\right )-\frac {10}{3} b^2 d^3 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )-\frac {1}{2} b^2 d^3 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d^3 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )-\frac {1}{3} i b^2 c d^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3*a*b*c*d^3*x - (I/3)*b^2*c*d^3*x + (I/3)*b^2*d^3*ArcTan[c*x] + 3*b^2*c*d^3*x*ArcTan[c*x] + (I/3)*b*c^2*d^3*x^
2*(a + b*ArcTan[c*x]) - (29*d^3*(a + b*ArcTan[c*x])^2)/6 + (3*I)*c*d^3*x*(a + b*ArcTan[c*x])^2 - (3*c^2*d^3*x^
2*(a + b*ArcTan[c*x])^2)/2 - (I/3)*c^3*d^3*x^3*(a + b*ArcTan[c*x])^2 + 2*d^3*(a + b*ArcTan[c*x])^2*ArcTanh[1 -
2/(1 + I*c*x)] + ((20*I)/3)*b*d^3*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)] - (3*b^2*d^3*Log[1 + c^2*x^2])/2 - (
10*b^2*d^3*PolyLog[2, 1 - 2/(1 + I*c*x)])/3 - I*b*d^3*(a + b*ArcTan[c*x])*PolyLog[2, 1 - 2/(1 + I*c*x)] + I*b*
d^3*(a + b*ArcTan[c*x])*PolyLog[2, -1 + 2/(1 + I*c*x)] - (b^2*d^3*PolyLog[3, 1 - 2/(1 + I*c*x)])/2 + (b^2*d^3*
PolyLog[3, -1 + 2/(1 + I*c*x)])/2

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 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 327

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

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 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 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 5036

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d*(f^2/e), 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] && GtQ[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 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} \, dx &=\int \left (3 i c 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 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^3 \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+\left (3 i c d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx-\left (3 c^2 d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx-\left (i c^3 d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=3 i c d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-\left (4 b c 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 (6 i b c^2 d^3\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\left (3 b c^3 d^3\right ) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\frac {1}{3} \left (2 i b c^4 d^3\right ) \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2+3 i c d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+\left (6 i b c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx+\left (2 b c 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 b c 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 b c d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx-\left (3 b c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx+\frac {1}{3} \left (2 i b c^2 d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx-\frac {1}{3} \left (2 i b c^2 d^3\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=3 a b c d^3 x+\frac {1}{3} i b c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {29}{6} d^3 \left (a+b \tan ^{-1}(c x)\right )^2+3 i c d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+2 d^3 \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+6 i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b 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 (2 i b c d^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx+\left (i b^2 c d^3\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 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 d^3\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (3 b^2 c d^3\right ) \int \tan ^{-1}(c x) \, dx-\frac {1}{3} \left (i b^2 c^3 d^3\right ) \int \frac {x^2}{1+c^2 x^2} \, dx\\ &=3 a b c d^3 x-\frac {1}{3} i b^2 c d^3 x+3 b^2 c d^3 x \tan ^{-1}(c x)+\frac {1}{3} i b c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {29}{6} d^3 \left (a+b \tan ^{-1}(c x)\right )^2+3 i c d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+2 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} i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b 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} b^2 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )-\left (6 b^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 {1}{3} \left (i b^2 c d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx-\frac {1}{3} \left (2 i b^2 c d^3\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (3 b^2 c^2 d^3\right ) \int \frac {x}{1+c^2 x^2} \, dx\\ &=3 a b c d^3 x-\frac {1}{3} i b^2 c d^3 x+\frac {1}{3} i b^2 d^3 \tan ^{-1}(c x)+3 b^2 c d^3 x \tan ^{-1}(c x)+\frac {1}{3} i b c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {29}{6} d^3 \left (a+b \tan ^{-1}(c x)\right )^2+3 i c d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+2 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} i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 d^3 \log \left (1+c^2 x^2\right )-3 b^2 d^3 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b 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} b^2 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{3} \left (2 b^2 d^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )\\ &=3 a b c d^3 x-\frac {1}{3} i b^2 c d^3 x+\frac {1}{3} i b^2 d^3 \tan ^{-1}(c x)+3 b^2 c d^3 x \tan ^{-1}(c x)+\frac {1}{3} i b c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {29}{6} d^3 \left (a+b \tan ^{-1}(c x)\right )^2+3 i c d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+2 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} i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-\frac {3}{2} b^2 d^3 \log \left (1+c^2 x^2\right )-\frac {10}{3} b^2 d^3 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b 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} b^2 d^3 \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d^3 \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.46, size = 465, normalized size = 1.21 \begin {gather*} -\frac {1}{24} i d^3 \left (b^2 \pi ^3-72 a^2 c x+72 i a b c x+8 b^2 c x-36 i a^2 c^2 x^2-8 a b c^2 x^2+8 a^2 c^3 x^3-72 i a b \text {ArcTan}(c x)-8 b^2 \text {ArcTan}(c x)-144 a b c x \text {ArcTan}(c x)+72 i b^2 c x \text {ArcTan}(c x)-72 i a b c^2 x^2 \text {ArcTan}(c x)-8 b^2 c^2 x^2 \text {ArcTan}(c x)+16 a b c^3 x^3 \text {ArcTan}(c x)+44 i b^2 \text {ArcTan}(c x)^2-72 b^2 c x \text {ArcTan}(c x)^2-36 i b^2 c^2 x^2 \text {ArcTan}(c x)^2+8 b^2 c^3 x^3 \text {ArcTan}(c x)^2-16 b^2 \text {ArcTan}(c x)^3+24 i b^2 \text {ArcTan}(c x)^2 \log \left (1-e^{-2 i \text {ArcTan}(c x)}\right )-160 b^2 \text {ArcTan}(c x) \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )-24 i b^2 \text {ArcTan}(c x)^2 \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )+24 i a^2 \log (c x)+80 a b \log \left (1+c^2 x^2\right )-36 i b^2 \log \left (1+c^2 x^2\right )-24 b^2 \text {ArcTan}(c x) \text {PolyLog}\left (2,e^{-2 i \text {ArcTan}(c x)}\right )-8 b^2 (-10 i+3 \text {ArcTan}(c x)) \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )-24 a b \text {PolyLog}(2,-i c x)+24 a b \text {PolyLog}(2,i c x)+12 i b^2 \text {PolyLog}\left (3,e^{-2 i \text {ArcTan}(c x)}\right )-12 i b^2 \text {PolyLog}\left (3,-e^{2 i \text {ArcTan}(c x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/24*I)*d^3*(b^2*Pi^3 - 72*a^2*c*x + (72*I)*a*b*c*x + 8*b^2*c*x - (36*I)*a^2*c^2*x^2 - 8*a*b*c^2*x^2 + 8*a^2
*c^3*x^3 - (72*I)*a*b*ArcTan[c*x] - 8*b^2*ArcTan[c*x] - 144*a*b*c*x*ArcTan[c*x] + (72*I)*b^2*c*x*ArcTan[c*x] -
(72*I)*a*b*c^2*x^2*ArcTan[c*x] - 8*b^2*c^2*x^2*ArcTan[c*x] + 16*a*b*c^3*x^3*ArcTan[c*x] + (44*I)*b^2*ArcTan[c
*x]^2 - 72*b^2*c*x*ArcTan[c*x]^2 - (36*I)*b^2*c^2*x^2*ArcTan[c*x]^2 + 8*b^2*c^3*x^3*ArcTan[c*x]^2 - 16*b^2*Arc
Tan[c*x]^3 + (24*I)*b^2*ArcTan[c*x]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] - 160*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*A
rcTan[c*x])] - (24*I)*b^2*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + (24*I)*a^2*Log[c*x] + 80*a*b*Log[1 +
c^2*x^2] - (36*I)*b^2*Log[1 + c^2*x^2] - 24*b^2*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])] - 8*b^2*(-10*I
+ 3*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - 24*a*b*PolyLog[2, (-I)*c*x] + 24*a*b*PolyLog[2, I*c*x] +
(12*I)*b^2*PolyLog[3, E^((-2*I)*ArcTan[c*x])] - (12*I)*b^2*PolyLog[3, -E^((2*I)*ArcTan[c*x])])

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 6.46, size = 1651, normalized size = 4.29

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1651\)
default	\(\text {Expression too large to display}\)	\(1651\)

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,x,method=_RETURNVERBOSE)

[Out]

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

[Out]

-1/3*I*a^2*c^3*d^3*x^3 - 36*I*b^2*c^5*d^3*integrate(1/48*x^5*arctan(c*x)^2/(c^2*x^3 + x), x) - 12*b^2*c^5*d^3*
integrate(1/48*x^5*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) - 3*I*b^2*c^5*d^3*integrate(1/48*x^5*log(c^2
*x^2 + 1)^2/(c^2*x^3 + x), x) - 96*I*a*b*c^5*d^3*integrate(1/48*x^5*arctan(c*x)/(c^2*x^3 + x), x) - 8*b^2*c^5*
d^3*integrate(1/48*x^5*arctan(c*x)/(c^2*x^3 + x), x) - 4*I*b^2*c^5*d^3*integrate(1/48*x^5*log(c^2*x^2 + 1)/(c^
2*x^3 + x), x) - 108*b^2*c^4*d^3*integrate(1/48*x^4*arctan(c*x)^2/(c^2*x^3 + x), x) + 36*I*b^2*c^4*d^3*integra
te(1/48*x^4*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) - 9*b^2*c^4*d^3*integrate(1/48*x^4*log(c^2*x^2 + 1)
^2/(c^2*x^3 + x), x) - 288*a*b*c^4*d^3*integrate(1/48*x^4*arctan(c*x)/(c^2*x^3 + x), x) + 44*I*b^2*c^4*d^3*int
egrate(1/48*x^4*arctan(c*x)/(c^2*x^3 + x), x) - 22*b^2*c^4*d^3*integrate(1/48*x^4*log(c^2*x^2 + 1)/(c^2*x^3 +
x), x) - 3/2*a^2*c^2*d^3*x^2 + 72*I*b^2*c^3*d^3*integrate(1/48*x^3*arctan(c*x)^2/(c^2*x^3 + x), x) + 24*b^2*c^
3*d^3*integrate(1/48*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 6*I*b^2*c^3*d^3*integrate(1/48*x^3*l
og(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) - 96*I*a*b*c^3*d^3*integrate(1/48*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 108*
b^2*c^3*d^3*integrate(1/48*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 54*I*b^2*c^3*d^3*integrate(1/48*x^3*log(c^2*x^2
+ 1)/(c^2*x^3 + x), x) + 3/4*I*b^2*d^3*arctan(c*x)^3 - 72*b^2*c^2*d^3*integrate(1/48*x^2*arctan(c*x)^2/(c^2*x
^3 + x), x) + 24*I*b^2*c^2*d^3*integrate(1/48*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) - 192*a*b*c^2
*d^3*integrate(1/48*x^2*arctan(c*x)/(c^2*x^3 + x), x) - 72*I*b^2*c^2*d^3*integrate(1/48*x^2*arctan(c*x)/(c^2*x
^3 + x), x) - 1/48*b^2*d^3*log(c^2*x^2 + 1)^3 + 3*I*a^2*c*d^3*x + 36*b^2*c*d^3*integrate(1/48*x*arctan(c*x)*lo
g(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 9*I*b^2*c*d^3*integrate(1/48*x*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 3/16*
b^2*d^3*log(c^2*x^2 + 1)^2 + 3*I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d^3 + 36*b^2*d^3*integrate(1/48*ar
ctan(c*x)^2/(c^2*x^3 + x), x) - 12*I*b^2*d^3*integrate(1/48*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 3
*b^2*d^3*integrate(1/48*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 96*a*b*d^3*integrate(1/48*arctan(c*x)/(c^2*x^3
+ x), x) + a^2*d^3*log(x) + 1/24*(-2*I*b^2*c^3*d^3*x^3 - 9*b^2*c^2*d^3*x^2 + 18*I*b^2*c*d^3*x)*arctan(c*x)^2 +
1/24*(2*b^2*c^3*d^3*x^3 - 9*I*b^2*c^2*d^3*x^2 - 18*b^2*c*d^3*x)*arctan(c*x)*log(c^2*x^2 + 1) - 1/96*(-2*I*b^2
*c^3*d^3*x^3 - 9*b^2*c^2*d^3*x^2 + 18*I*b^2*c*d^3*x)*log(c^2*x^2 + 1)^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,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, 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,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,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} \,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,x)

[Out]

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

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