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

Optimal. Leaf size=139 \[ \frac {3 i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c d}-\frac {3 b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}+\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c d}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{i c x+1}\right )}{4 c d} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4854, 4884, 4994, 4998, 6610} \[ \frac {3 i b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c d}-\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d}+\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{1+i c x}\right )}{4 c d}+\frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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}{d+i c d x} \, dx &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {(3 i 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}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {\left (3 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}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c d}-\frac {\left (3 i b^3\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 \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {3 b \left (a+b \tan ^{-1}(c x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 i b^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_3\left (1-\frac {2}{1+i c x}\right )}{2 c d}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1+i c x}\right )}{4 c d}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 133, normalized size = 0.96 \[ \frac {i \left (4 \log \left (\frac {2 d}{d+i c d x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3+3 i b \left (2 \text {Li}_2\left (\frac {c x+i}{c x-i}\right ) \left (a+b \tan ^{-1}(c x)\right )^2-b \left (2 i \text {Li}_3\left (\frac {c x+i}{c x-i}\right ) \left (a+b \tan ^{-1}(c x)\right )+b \text {Li}_4\left (\frac {c x+i}{c x-i}\right )\right )\right )\right )}{4 c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{3} \log \left (-\frac {c x + i}{c x - i}\right )^{3} - 6 i \, a b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - 12 \, a^{2} b \log \left (-\frac {c x + i}{c x - i}\right ) + 8 i \, a^{3}}{8 \, c d x - 8 i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 0.32, size = 2044, normalized size = 14.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2/c*a*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+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+
I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-3/2/c*a*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+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+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+3/2/c*a*b^2/d*Pi*csgn((1+I*c*x
)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/2/c*b^3/d*Pi*csgn
((1+I*c*x)^2/(c^2*x^2+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))*csgn((1+I*c*x)^2/(c^
2*x^2+1))*arctan(c*x)^3-3/2/c*a*b^2/d*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+
I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2-I/c*b^3/d*ln(1+I*c*x)*arctan(c*x)^3+3/2*I/c*b^3/d*arctan(c*x)*polylog
(3,-(1+I*c*x)^2/(c^2*x^2+1))+3/2*I/c*a*b^2/d*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+3/c*a*b^2/d*arctan(c*x)*polyl
og(2,-(1+I*c*x)^2/(c^2*x^2+1))+3/2/c*a^2*b/d*dilog(1/2*I*c*x+1/2)+3/4/c*a^2*b/d*ln(1+I*c*x)^2+3/2/c*b^3/d*arct
an(c*x)^2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-1/2*I/c*a^3/d*ln(c^2*x^2+1)+3/2/c*a*b^2/d*Pi*csgn(I/((1+I*c*x)^2
/(c^2*x^2+1)+1))*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*arcta
n(c*x)^2-3/4/c*b^3/d*polylog(4,-(1+I*c*x)^2/(c^2*x^2+1))+1/c*a^3/d*arctan(c*x)+2/c*a*b^2/d*arctan(c*x)^3+1/2/c
*b^3/d*Pi*arctan(c*x)^3+1/2/c*b^3/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+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+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^3-1/2/c*b^3/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+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+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^3-3*I/c*a*b^
2/d*ln(1+I*c*x)*arctan(c*x)^2+3*I/c*a*b^2/d*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-3*I/c*a^2*b/d*ln(1+I
*c*x)*arctan(c*x)-1/2/c*b^3/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+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)^3-3/2/c*a*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*
arctan(c*x)^2+1/2/c*b^3/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn((1+I*c*x)^2/(c^2
*x^2+1))*arctan(c*x)^3+3/2/c*a*b^2/d*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c
*x)^2-3/2/c*a*b^2/d*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+3/2/c*a*b^2
/d*Pi*arctan(c*x)^2+1/2/c*b^3/d*arctan(c*x)^4+I/c*b^3/d*arctan(c*x)^3*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-1/2/c*b^
3/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^3+1/2/c*b^3/d*Pi*csgn(I*(1+I*c*
x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^3-1/2/c*b^3/d*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((
1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^3-3/2/c*a^2*b/d*ln(1/2-1/2*I*c*x)*ln(1+I*c*x)+3/2/c*a^2*b/d*ln(1/2-1/
2*I*c*x)*ln(1/2*I*c*x+1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {i \, a^{3} \log \left (i \, c d x + d\right )}{c d} + \frac {16 \, b^{3} \arctan \left (c x\right )^{4} - b^{3} \log \left (c^{2} x^{2} + 1\right )^{4} + {\left (b^{3} c {\left (\frac {4 \, \log \left (c^{2} d x^{2} + d\right ) \log \left (c^{2} x^{2} + 1\right )^{3}}{c^{2} d} + \frac {\frac {4 \, {\left (\log \left (c^{2} x^{2} + 1\right )^{3} + 3 \, \log \left (c^{2} x^{2} + 1\right )^{2} \log \relax (d)\right )} \log \left (c^{2} x^{2} + 1\right )}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )^{4} + 4 \, \log \left (c^{2} x^{2} + 1\right )^{3} \log \relax (d)}{c^{2}}}{d} - \frac {6 \, {\left (\log \left (c^{2} x^{2} + 1\right )^{2} + 2 \, \log \left (c^{2} x^{2} + 1\right ) \log \relax (d)\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{c^{2} d}\right )} + \frac {16 \, b^{3} \arctan \left (c x\right )^{4}}{c d} + \frac {128 \, a b^{2} \arctan \left (c x\right )^{3}}{c d} + \frac {192 \, a^{2} b \arctan \left (c x\right )^{2}}{c d}\right )} c d - 4 i \, c d \int \frac {32 \, {\left (b^{3} c x \arctan \left (c x\right )^{3} + 3 \, a b^{2} c x \arctan \left (c x\right )^{2} + 3 \, a^{2} b c x \arctan \left (c x\right )\right )}}{c^{2} d x^{2} + d}\,{d x}}{128 \, c d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a^3*log(I*c*d*x + d)/(c*d) + 1/128*(16*b^3*arctan(c*x)^4 + 16*I*b^3*arctan(c*x)^3*log(c^2*x^2 + 1) + 4*I*b^
3*arctan(c*x)*log(c^2*x^2 + 1)^3 - b^3*log(c^2*x^2 + 1)^4 + 16*(b^3*arctan(c*x)^4/(c*d) + 8*b^3*c*integrate(1/
16*x*log(c^2*x^2 + 1)^3/(c^2*d*x^2 + d), x) + 8*a*b^2*arctan(c*x)^3/(c*d) + 12*a^2*b*arctan(c*x)^2/(c*d))*c*d
- 128*I*c*d*integrate(1/32*(40*b^3*c*x*arctan(c*x)^3 + 6*b^3*c*x*arctan(c*x)*log(c^2*x^2 + 1)^2 + 96*a*b^2*c*x
*arctan(c*x)^2 + 96*a^2*b*c*x*arctan(c*x) + 12*b^3*arctan(c*x)^2*log(c^2*x^2 + 1) + b^3*log(c^2*x^2 + 1)^3)/(c
^2*d*x^2 + d), x))/(c*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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