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

Optimal. Leaf size=277 \[ \frac {3 b^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d}-\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac {3 i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}+\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {3 i b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d}-\frac {3 i b^3 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c^2 d}+\frac {3 i b^3 \text {Li}_4\left (1-\frac {2}{i c x+1}\right )}{4 c^2 d} \]

[Out]

(a+b*arctan(c*x))^3/c^2/d-I*x*(a+b*arctan(c*x))^3/c/d-3*I*b*(a+b*arctan(c*x))^2*ln(2/(1+I*c*x))/c^2/d-(a+b*arc
tan(c*x))^3*ln(2/(1+I*c*x))/c^2/d+3*b^2*(a+b*arctan(c*x))*polylog(2,1-2/(1+I*c*x))/c^2/d-3/2*I*b*(a+b*arctan(c
*x))^2*polylog(2,1-2/(1+I*c*x))/c^2/d-3/2*I*b^3*polylog(3,1-2/(1+I*c*x))/c^2/d-3/2*b^2*(a+b*arctan(c*x))*polyl
og(3,1-2/(1+I*c*x))/c^2/d+3/4*I*b^3*polylog(4,1-2/(1+I*c*x))/c^2/d

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Rubi [A]  time = 0.50, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4866, 4846, 4920, 4854, 4884, 4994, 6610, 4998} \[ \frac {3 b^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d}-\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac {3 i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac {3 i b^3 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c^2 d}+\frac {3 i b^3 \text {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c^2 d}+\frac {\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {3 i b \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac {i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4846

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

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4866

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f)/e, 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] && GtQ[m, 0]

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 4920

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

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Mathematica [A]  time = 0.80, size = 393, normalized size = 1.42 \[ -\frac {i \left (2 i a^3 \log \left (c^2 x^2+1\right )+4 a^3 c x-4 a^3 \tan ^{-1}(c x)-6 a^2 b \log \left (c^2 x^2+1\right )-12 a^2 b \tan ^{-1}(c x)^2+12 a^2 b c x \tan ^{-1}(c x)-12 i a^2 b \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-6 b \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right ) \left (2 b (a+i b) \tan ^{-1}(c x)+a (a+2 i b)+b^2 \tan ^{-1}(c x)^2\right )+6 b^2 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right ) \left (-i a-i b \tan ^{-1}(c x)+b\right )-8 a b^2 \tan ^{-1}(c x)^3-12 i a b^2 \tan ^{-1}(c x)^2+12 a b^2 c x \tan ^{-1}(c x)^2-12 i a b^2 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+24 a b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+3 b^3 \text {Li}_4\left (-e^{2 i \tan ^{-1}(c x)}\right )-2 b^3 \tan ^{-1}(c x)^4-4 i b^3 \tan ^{-1}(c x)^3+4 b^3 c x \tan ^{-1}(c x)^3-4 i b^3 \tan ^{-1}(c x)^3 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+12 b^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{4 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/4*I)*(4*a^3*c*x - 4*a^3*ArcTan[c*x] + 12*a^2*b*c*x*ArcTan[c*x] - 12*a^2*b*ArcTan[c*x]^2 - (12*I)*a*b^2*Ar
cTan[c*x]^2 + 12*a*b^2*c*x*ArcTan[c*x]^2 - 8*a*b^2*ArcTan[c*x]^3 - (4*I)*b^3*ArcTan[c*x]^3 + 4*b^3*c*x*ArcTan[
c*x]^3 - 2*b^3*ArcTan[c*x]^4 - (12*I)*a^2*b*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 24*a*b^2*ArcTan[c*x]*
Log[1 + E^((2*I)*ArcTan[c*x])] - (12*I)*a*b^2*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + 12*b^3*ArcTan[c*x
]^2*Log[1 + E^((2*I)*ArcTan[c*x])] - (4*I)*b^3*ArcTan[c*x]^3*Log[1 + E^((2*I)*ArcTan[c*x])] + (2*I)*a^3*Log[1
+ c^2*x^2] - 6*a^2*b*Log[1 + c^2*x^2] - 6*b*(a*(a + (2*I)*b) + 2*(a + I*b)*b*ArcTan[c*x] + b^2*ArcTan[c*x]^2)*
PolyLog[2, -E^((2*I)*ArcTan[c*x])] + 6*b^2*((-I)*a + b - I*b*ArcTan[c*x])*PolyLog[3, -E^((2*I)*ArcTan[c*x])] +
3*b^3*PolyLog[4, -E^((2*I)*ArcTan[c*x])]))/(c^2*d)

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fricas [F]  time = 1.33, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{3} x \log \left (-\frac {c x + i}{c x - i}\right )^{3} - 6 i \, a b^{2} x \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 12 \, a^{2} b x \log \left (-\frac {c x + i}{c x - i}\right ) + 8 i \, a^{3} x}{8 \, c d x - 8 i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 0.98, size = 5478, normalized size = 19.78 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctan(c*x))^3/(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(x*(a+b*arctan(c*x))^3/(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 \[ \int \frac {x\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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