∫ (a+b tanh−1(cx))3 ___________________________

3.18 \(\int \frac{(a+b \tanh ^{-1}(c x))^3}{d+e x} \, dx\)

Optimal. Leaf size=272 \[ -\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e}+\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e}+\frac{3 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}-\frac{3 b^3 \text{PolyLog}\left (4,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{4 e}+\frac{3 b^3 \text{PolyLog}\left (4,1-\frac{2}{c x+1}\right )}{4 e}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{e} \]

[Out]

-(((a + b*ArcTanh[c*x])^3*Log[2/(1 + c*x)])/e) + ((a + b*ArcTanh[c*x])^3*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c

*x))])/e + (3*b*(a + b*ArcTanh[c*x])^2*PolyLog[2, 1 - 2/(1 + c*x)])/(2*e) - (3*b*(a + b*ArcTanh[c*x])^2*PolyLo

g[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e) + (3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - 2/(1 + c*x

)])/(2*e) - (3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e) + (3*b^3*

PolyLog[4, 1 - 2/(1 + c*x)])/(4*e) - (3*b^3*PolyLog[4, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(4*e)

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Rubi [A] time = 0.0566941, antiderivative size = 272, 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 = {5924} \[ -\frac{3 b^2 \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e}+\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 e}-\frac{3 b \left (a+b \tanh ^{-1}(c x)\right )^2 \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e}+\frac{3 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{2 e}-\frac{3 b^3 \text{PolyLog}\left (4,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{4 e}+\frac{3 b^3 \text{PolyLog}\left (4,1-\frac{2}{c x+1}\right )}{4 e}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^3}{e} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(((a + b*ArcTanh[c*x])^3*Log[2/(1 + c*x)])/e) + ((a + b*ArcTanh[c*x])^3*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c

*x))])/e + (3*b*(a + b*ArcTanh[c*x])^2*PolyLog[2, 1 - 2/(1 + c*x)])/(2*e) - (3*b*(a + b*ArcTanh[c*x])^2*PolyLo

g[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e) + (3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - 2/(1 + c*x

)])/(2*e) - (3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e) + (3*b^3*

PolyLog[4, 1 - 2/(1 + c*x)])/(4*e) - (3*b^3*PolyLog[4, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(4*e)

Rule 5924

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^3*Log[

2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^3*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(

3*b*(a + b*ArcTanh[c*x])^2*PolyLog[2, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(3*b*(a + b*ArcTanh[c*x])^2*PolyLog[2

, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x] + Simp[(3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - 2/(1

+ c*x)])/(2*e), x] - Simp[(3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/

(2*e), x] + Simp[(3*b^3*PolyLog[4, 1 - 2/(1 + c*x)])/(4*e), x] - Simp[(3*b^3*PolyLog[4, 1 - (2*c*(d + e*x))/((

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

Mathematica [F] time = 109.64, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{d+e x} \, dx \]

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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Maple [C] time = 0.43, size = 2367, normalized size = 8.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^3/(e*x+d),x)

[Out]

3*c*a*b^2/e*d/(c*d+e)*arctanh(c*x)^2*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+3*c*a*b^2/e*d/(c*d+e)*arcta

nh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-3/2*I*a*b^2/e*arctanh(c*x)^2*Pi*csgn(I/((c*x+1)^2/(

-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1

))^2-3/2*I*a*b^2/e*arctanh(c*x)^2*Pi*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*

x+1)^2/(-c^2*x^2+1)+1))^2*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1)))+1/2*I*b^3/e*ar

ctanh(c*x)^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x

^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1)))-3/

4*b^3/e*polylog(4,-(c*x+1)^2/(-c^2*x^2+1))+3/4*b^3/(c*d+e)*polylog(4,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+

a^3*ln(c*e*x+c*d)/e+3/2*I*a*b^2/e*arctanh(c*x)^2*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^

2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e

+c*d*((c*x+1)^2/(-c^2*x^2+1)+1)))+3/2*a*b^2/e*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-3/2*a*b^2/(c*d+e)*polylog(3,(

c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+3/2*a^2*b/e*dilog((c*e*x-e)/(-c*d-e))-3/2*a^2*b/e*dilog((c*e*x+e)/(-c*

d+e))-b^3/e*arctanh(c*x)^3*ln(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))-3/2*b^3/e*arctanh(c

*x)^2*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+3/2*b^3/e*arctanh(c*x)*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+b^3/(c*d+e)

*arctanh(c*x)^3*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+3/2*b^3/(c*d+e)*arctanh(c*x)^2*polylog(2,(c*d+e)

*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-3/2*b^3/(c*d+e)*arctanh(c*x)*polylog(3,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+

e))+b^3*ln(c*e*x+c*d)/e*arctanh(c*x)^3-3/2*c*a*b^2/e*d/(c*d+e)*polylog(3,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+

e))+c*b^3/e*d/(c*d+e)*arctanh(c*x)^3*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+3/2*c*b^3/e*d/(c*d+e)*arcta

nh(c*x)^2*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-3/2*c*b^3/e*d/(c*d+e)*arctanh(c*x)*polylog(3,(c*d

+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))-1/2*I*b^3/e*arctanh(c*x)^3*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(

((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^2-1/2*I*b^3/e*arctanh

(c*x)^3*Pi*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^2*

csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1)))+3/2*I*a*b^2/e*arctanh(c*x)^2*Pi*csgn(I*(

((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^3-3*a*b^2/e*arctanh(c

*x)^2*ln(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))-3*a*b^2/e*arctanh(c*x)*polylog(2,-(c*x+1

)^2/(-c^2*x^2+1))+3*a*b^2/(c*d+e)*arctanh(c*x)^2*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+3*a*b^2/(c*d+e)

*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+3*a^2*b*ln(c*e*x+c*d)/e*arctanh(c*x)+3/2*a^2*

b/e*ln(c*e*x+c*d)*ln((c*e*x-e)/(-c*d-e))-3/2*a^2*b/e*ln(c*e*x+c*d)*ln((c*e*x+e)/(-c*d+e))+3*a*b^2*ln(c*e*x+c*d

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

h(c*x)^3*Pi*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^3

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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{3} \log \left (e x + d\right )}{e} + \int \frac{b^{3}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}^{3}}{8 \,{\left (e x + d\right )}} + \frac{3 \, a b^{2}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}^{2}}{4 \,{\left (e x + d\right )}} + \frac{3 \, a^{2} b{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{2 \,{\left (e x + d\right )}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*log(e*x + d)/e + integrate(1/8*b^3*(log(c*x + 1) - log(-c*x + 1))^3/(e*x + d) + 3/4*a*b^2*(log(c*x + 1) -

log(-c*x + 1))^2/(e*x + d) + 3/2*a^2*b*(log(c*x + 1) - log(-c*x + 1))/(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} \operatorname{artanh}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\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*arctanh(c*x))^3/(e*x+d),x, algorithm="fricas")

[Out]

integral((b^3*arctanh(c*x)^3 + 3*a*b^2*arctanh(c*x)^2 + 3*a^2*b*arctanh(c*x) + a^3)/(e*x + d), x)

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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\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*atanh(c*x))**3/(e*x+d),x)

[Out]

Integral((a + b*atanh(c*x))**3/(d + e*x), x)

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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\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*arctanh(c*x))^3/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3/(e*x + d), x)

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