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

Optimal. Leaf size=263 \[ -\frac {3 i b^2 c \text {Li}_2\left (\frac {2}{1-i c x}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {3 i b^2 c \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 d}+\frac {3 b c \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac {i c \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{d}+\frac {3 b^3 c \text {Li}_3\left (\frac {2}{1-i c x}-1\right )}{2 d}-\frac {3 b^3 c \text {Li}_4\left (\frac {2}{i c x+1}-1\right )}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.60, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4870, 4852, 4924, 4868, 4884, 4992, 6610, 4994, 4998} \[ -\frac {3 i b^2 c \text {PolyLog}\left (2,-1+\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac {3 i b^2 c \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 d}+\frac {3 b c \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac {3 b^3 c \text {PolyLog}\left (3,-1+\frac {2}{1-i c x}\right )}{2 d}-\frac {3 b^3 c \text {PolyLog}\left (4,-1+\frac {2}{1+i c x}\right )}{4 d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \log \left (2-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac {i c \log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*c*(a + b*ArcTan[c*x])^3)/d - (a + b*ArcTan[c*x])^3/(d*x) + (3*b*c*(a + b*ArcTan[c*x])^2*Log[2 - 2/(1 - I
*c*x)])/d - (I*c*(a + b*ArcTan[c*x])^3*Log[2 - 2/(1 + I*c*x)])/d - ((3*I)*b^2*c*(a + b*ArcTan[c*x])*PolyLog[2,
-1 + 2/(1 - I*c*x)])/d + (3*b*c*(a + b*ArcTan[c*x])^2*PolyLog[2, -1 + 2/(1 + I*c*x)])/(2*d) + (3*b^3*c*PolyLo
g[3, -1 + 2/(1 - I*c*x)])/(2*d) - (((3*I)/2)*b^2*c*(a + b*ArcTan[c*x])*PolyLog[3, -1 + 2/(1 + I*c*x)])/d - (3*
b^3*c*PolyLog[4, -1 + 2/(1 + I*c*x)])/(4*d)

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 4870

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[1/d, I
nt[(f*x)^m*(a + b*ArcTan[c*x])^p, x], x] - Dist[e/(d*f), Int[((f*x)^(m + 1)*(a + b*ArcTan[c*x])^p)/(d + e*x),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0] && LtQ[m, -1]

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 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 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[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 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 4998

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a
+ b*ArcTan[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] - Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[k
+ 1, u])/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[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 {\left (a+b \tan ^{-1}(c x)\right )^3}{x^2 (d+i c d x)} \, dx &=-\left ((i c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{x (d+i c d x)} \, dx\right )+\frac {\int \frac {\left (a+b \tan ^{-1}(c x)\right )^3}{x^2} \, dx}{d}\\ &=-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {(3 b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac {\left (3 i b c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {(3 i b c) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x (i+c x)} \, dx}{d}-\frac {\left (3 b^2 c^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {\left (6 b^2 c^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}{d}+\frac {\left (3 i b^3 c^2\right ) \int \frac {\text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 b^3 c \text {Li}_4\left (-1+\frac {2}{1+i c x}\right )}{4 d}+\frac {\left (3 i b^3 c^2\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-i c x}\right )}{d}-\frac {i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac {2}{1+i c x}\right )}{d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-i c x}\right )}{d}+\frac {3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {3 b^3 c \text {Li}_3\left (-1+\frac {2}{1-i c x}\right )}{2 d}-\frac {3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )}{2 d}-\frac {3 b^3 c \text {Li}_4\left (-1+\frac {2}{1+i c x}\right )}{4 d}\\ \end {align*}

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Mathematica [A]  time = 1.49, size = 436, normalized size = 1.66 \[ -\frac {-i a^3 c \log \left (c^2 x^2+1\right )+2 i a^3 c \log (x)+2 a^3 c \tan ^{-1}(c x)+\frac {2 a^3}{x}+3 a^2 b c \left (2 \left (-\log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )+\tan ^{-1}(c x)^2+\tan ^{-1}(c x) \left (\frac {1}{c x}+i \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )\right )+\text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )\right )+6 i a b^2 c \left (i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+\text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {i \tan ^{-1}(c x)^2}{c x}+\tan ^{-1}(c x)^2+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+2 i \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )-\frac {i \pi ^3}{24}\right )+2 i b^3 c \left (\frac {3}{2} i \left (\tan ^{-1}(c x)+2 i\right ) \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+\frac {3}{2} \left (\tan ^{-1}(c x)+i\right ) \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {3}{4} i \text {Li}_4\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {i \tan ^{-1}(c x)^3}{c x}-\tan ^{-1}(c x)^3+\tan ^{-1}(c x)^3 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+3 i \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\frac {i \pi ^4}{64}+\frac {\pi ^3}{8}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*((2*a^3)/x + 2*a^3*c*ArcTan[c*x] + (2*I)*a^3*c*Log[x] - I*a^3*c*Log[1 + c^2*x^2] + 3*a^2*b*c*(2*(ArcTan[c
*x]^2 + ArcTan[c*x]*(1/(c*x) + I*Log[1 - E^((2*I)*ArcTan[c*x])]) - Log[(c*x)/Sqrt[1 + c^2*x^2]]) + PolyLog[2,
E^((2*I)*ArcTan[c*x])]) + (6*I)*a*b^2*c*((-1/24*I)*Pi^3 + ArcTan[c*x]^2 - (I*ArcTan[c*x]^2)/(c*x) + ArcTan[c*x
]^2*Log[1 - E^((-2*I)*ArcTan[c*x])] + (2*I)*ArcTan[c*x]*Log[1 - E^((2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog
[2, E^((-2*I)*ArcTan[c*x])] + PolyLog[2, E^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2) + (2*I
)*b^3*c*(Pi^3/8 - (I/64)*Pi^4 - ArcTan[c*x]^3 - (I*ArcTan[c*x]^3)/(c*x) + (3*I)*ArcTan[c*x]^2*Log[1 - E^((-2*I
)*ArcTan[c*x])] + ArcTan[c*x]^3*Log[1 - E^((-2*I)*ArcTan[c*x])] + ((3*I)/2)*ArcTan[c*x]*(2*I + ArcTan[c*x])*Po
lyLog[2, E^((-2*I)*ArcTan[c*x])] + (3*(I + ArcTan[c*x])*PolyLog[3, E^((-2*I)*ArcTan[c*x])])/2 - ((3*I)/4)*Poly
Log[4, E^((-2*I)*ArcTan[c*x])]))/d

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{3} \log \left (-\frac {c x + i}{c x - i}\right )^{3} - 6 i \, a b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 12 \, a^{2} b \log \left (-\frac {c x + i}{c x - i}\right ) + 8 i \, a^{3}}{8 \, c d x^{3} - 8 i \, d x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^3*log(-(c*x + I)/(c*x - I))^3 - 6*I*a*b^2*log(-(c*x + I)/(c*x - I))^2 - 12*a^2*b*log(-(c*x + I)/(
c*x - I)) + 8*I*a^3)/(8*c*d*x^3 - 8*I*d*x^2), 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((a+b*arctan(c*x))^3/x^2/(d+I*c*d*x),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 1.32, size = 11233, normalized size = 42.71 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \frac {a^{3}}{c x^{3} - i x^{2}}\, dx + \int \frac {b^{3} \operatorname {atan}^{3}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx + \int \frac {3 a b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx + \int \frac {3 a^{2} b \operatorname {atan}{\left (c x \right )}}{c x^{3} - i x^{2}}\, dx\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))**3/x**2/(d+I*c*d*x),x)

[Out]

-I*(Integral(a**3/(c*x**3 - I*x**2), x) + Integral(b**3*atan(c*x)**3/(c*x**3 - I*x**2), x) + Integral(3*a*b**2
*atan(c*x)**2/(c*x**3 - I*x**2), x) + Integral(3*a**2*b*atan(c*x)/(c*x**3 - I*x**2), x))/d

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