3.18 \(\int \frac{(a+b \tan ^{-1}(c x))^3}{d+e x} \, dx\)
Optimal. Leaf size=320 \[ \frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e}-\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 e}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e}+\frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{3 i b^3 \text{PolyLog}\left (4,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{4 e}-\frac{3 i b^3 \text{PolyLog}\left (4,1-\frac{2}{1-i c x}\right )}{4 e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{e} \]
[Out]
-(((a + b*ArcTan[c*x])^3*Log[2/(1 - I*c*x)])/e) + ((a + b*ArcTan[c*x])^3*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 -
I*c*x))])/e + (((3*I)/2)*b*(a + b*ArcTan[c*x])^2*PolyLog[2, 1 - 2/(1 - I*c*x)])/e - (((3*I)/2)*b*(a + b*ArcTa
n[c*x])^2*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e - (3*b^2*(a + b*ArcTan[c*x])*PolyLog[3,
1 - 2/(1 - I*c*x)])/(2*e) + (3*b^2*(a + b*ArcTan[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x
))])/(2*e) - (((3*I)/4)*b^3*PolyLog[4, 1 - 2/(1 - I*c*x)])/e + (((3*I)/4)*b^3*PolyLog[4, 1 - (2*c*(d + e*x))/(
(c*d + I*e)*(1 - I*c*x))])/e
________________________________________________________________________________________
Rubi [A] time = 0.0568978, antiderivative size = 320, normalized size of antiderivative = 1.,
number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used =
{4860} \[ \frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e}-\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 e}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e}+\frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 e}+\frac{3 i b^3 \text{PolyLog}\left (4,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{4 e}-\frac{3 i b^3 \text{PolyLog}\left (4,1-\frac{2}{1-i c x}\right )}{4 e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e}-\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{e} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTan[c*x])^3/(d + e*x),x]
[Out]
-(((a + b*ArcTan[c*x])^3*Log[2/(1 - I*c*x)])/e) + ((a + b*ArcTan[c*x])^3*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 -
I*c*x))])/e + (((3*I)/2)*b*(a + b*ArcTan[c*x])^2*PolyLog[2, 1 - 2/(1 - I*c*x)])/e - (((3*I)/2)*b*(a + b*ArcTa
n[c*x])^2*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e - (3*b^2*(a + b*ArcTan[c*x])*PolyLog[3,
1 - 2/(1 - I*c*x)])/(2*e) + (3*b^2*(a + b*ArcTan[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x
))])/(2*e) - (((3*I)/4)*b^3*PolyLog[4, 1 - 2/(1 - I*c*x)])/e + (((3*I)/4)*b^3*PolyLog[4, 1 - (2*c*(d + e*x))/(
(c*d + I*e)*(1 - I*c*x))])/e
Rule 4860
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^3*Log[2/
(1 - I*c*x)])/e, x] + (Simp[((a + b*ArcTan[c*x])^3*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e, x] + Sim
p[(3*I*b*(a + b*ArcTan[c*x])^2*PolyLog[2, 1 - 2/(1 - I*c*x)])/(2*e), x] - Simp[(3*I*b*(a + b*ArcTan[c*x])^2*Po
lyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(2*e), x] - Simp[(3*b^2*(a + b*ArcTan[c*x])*PolyLog[3
, 1 - 2/(1 - I*c*x)])/(2*e), x] + Simp[(3*b^2*(a + b*ArcTan[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*
(1 - I*c*x))])/(2*e), x] - Simp[(3*I*b^3*PolyLog[4, 1 - 2/(1 - I*c*x)])/(4*e), x] + Simp[(3*I*b^3*PolyLog[4, 1
- (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/(4*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0
]
Rubi steps
\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d+e x} \, dx &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2}{1-i c x}\right )}{e}+\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e}+\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 e}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\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 e}+\frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e}-\frac{3 i b^3 \text{Li}_4\left (1-\frac{2}{1-i c x}\right )}{4 e}+\frac{3 i b^3 \text{Li}_4\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{4 e}\\ \end{align*}
Mathematica [F] time = 180.004, size = 0, normalized size = 0. \[ \text{\$Aborted} \]
Verification is Not applicable to the result.
[In]
Integrate[(a + b*ArcTan[c*x])^3/(d + e*x),x]
[Out]
$Aborted
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Maple [C] time = 0.573, size = 2616, normalized size = 8.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctan(c*x))^3/(e*x+d),x)
[Out]
b^3*ln(c*e*x+c*d)/e*arctan(c*x)^3-b^3/e*arctan(c*x)^3*ln(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2
+1)+I*e+d*c)-3/2*b^3/e*arctan(c*x)*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+b^3*arctan(c*x)^3*ln(1-(I*e-d*c)/(d*c+I
*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*d*c)+3/2*b^3*arctan(c*x)*polylog(3,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+
1))/(e+I*d*c)+3/2*a*b^2*polylog(3,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*d*c)-3/2*a*b^2/e*polylog(3
,-(1+I*c*x)^2/(c^2*x^2+1))-3/4*I*b^3/e*polylog(4,-(1+I*c*x)^2/(c^2*x^2+1))+3/4*I*b^3*polylog(4,(I*e-d*c)/(d*c+
I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*d*c)+a^3*ln(c*e*x+c*d)/e-3*I*c*a*b^2/e*d/(d*c-I*e)*arctan(c*x)*polylog(2,(I
*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+3/2*I*a*b^2/e*arctan(c*x)^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*
csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*
(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c))+3*a*b^2*ln(c*e*x+c*d)/e*arctan(c*x)^2-3*a*
b^2/e*arctan(c*x)^2*ln(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)+3*a*b^2*arctan(c*x)^2
*ln(1-(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*d*c)+3*a^2*b*ln(c*e*x+c*d)/e*arctan(c*x)+3/2*I*b^3/e*a
rctan(c*x)^2*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-3/2*I*b^3*arctan(c*x)^2*polylog(2,(I*e-d*c)/(d*c+I*e)*(1+I*c*
x)^2/(c^2*x^2+1))/(e+I*d*c)+3/2*I*a^2*b/e*dilog((I*e-e*c*x)/(d*c+I*e))-3/2*I*a^2*b/e*dilog((I*e+e*c*x)/(I*e-d*
c))+3/2*c*a*b^2/e*d/(d*c-I*e)*polylog(3,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+c*b^3/e*d/(d*c-I*e)*arcta
n(c*x)^3*ln(1-(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+3/2*c*b^3/e*d/(d*c-I*e)*arctan(c*x)*polylog(3,(I*e-
d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+3/2*I*a*b^2/e*arctan(c*x)^2*Pi*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*
d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3-1/2*I*b^3/e*arctan(c*x)^3*Pi*csgn(I*(-I*(1+I
*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*(-I*(1+I*c*x)
^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c))-3/2*I*a^2*b*ln(c*e*x+c*d)/e*ln((I*e+e*c*x)/(I*e-d*c))+3
*I*a*b^2/e*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-3*I*a*b^2*arctan(c*x)*polylog(2,(I*e-d*c)/(d*c+I*e)
*(1+I*c*x)^2/(c^2*x^2+1))/(e+I*d*c)+3*c*a*b^2/e*d/(d*c-I*e)*arctan(c*x)^2*ln(1-(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2
/(c^2*x^2+1))-3/2*I*a*b^2/e*arctan(c*x)^2*Pi*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+
I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d
*c))+1/2*I*b^3/e*arctan(c*x)^3*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d
*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c
*x)^2/(c^2*x^2+1)+I*e+d*c))-3/2*I*a*b^2/e*arctan(c*x)^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(-I*(1+I
*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2-3/2*I*c*b^3/e*d/(d*c
-I*e)*arctan(c*x)^2*polylog(2,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+1/2*I*b^3/e*arctan(c*x)^3*Pi*csgn(I
*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3+3/2*I*a^2*b
*ln(c*e*x+c*d)/e*ln((I*e-e*c*x)/(d*c+I*e))-1/2*I*b^3/e*arctan(c*x)^3*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*cs
gn(I*(-I*(1+I*c*x)^2/(c^2*x^2+1)*e+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+d*c)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2+3/4*I*c
*b^3/e*d/(d*c-I*e)*polylog(4,(I*e-d*c)/(d*c+I*e)*(1+I*c*x)^2/(c^2*x^2+1))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{3} \log \left (e x + d\right )}{e} + \int \frac{28 \, b^{3} \arctan \left (c x\right )^{3} + 3 \, b^{3} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2} + 96 \, a b^{2} \arctan \left (c x\right )^{2} + 96 \, a^{2} b \arctan \left (c x\right )}{32 \,{\left (e x + d\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x))^3/(e*x+d),x, algorithm="maxima")
[Out]
a^3*log(e*x + d)/e + integrate(1/32*(28*b^3*arctan(c*x)^3 + 3*b^3*arctan(c*x)*log(c^2*x^2 + 1)^2 + 96*a*b^2*ar
ctan(c*x)^2 + 96*a^2*b*arctan(c*x))/(e*x + d), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \arctan \left (c x\right )^{3} + 3 \, a b^{2} \arctan \left (c x\right )^{2} + 3 \, a^{2} b \arctan \left (c x\right ) + a^{3}}{e x + d}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x))^3/(e*x+d),x, algorithm="fricas")
[Out]
integral((b^3*arctan(c*x)^3 + 3*a*b^2*arctan(c*x)^2 + 3*a^2*b*arctan(c*x) + a^3)/(e*x + d), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right )^{3}}{d + e x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atan(c*x))**3/(e*x+d),x)
[Out]
Integral((a + b*atan(c*x))**3/(d + e*x), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{3}}{e x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctan(c*x))^3/(e*x+d),x, algorithm="giac")
[Out]
integrate((b*arctan(c*x) + a)^3/(e*x + d), x)