3.131 \(\int \frac{\tanh ^{-1}(a x)^3}{c x+a c x^2} \, dx\)
Optimal. Leaf size=93 \[ -\frac{3 \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )}{4 c}-\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{2 c}-\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{2 c}+\frac{\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]
[Out]
(ArcTanh[a*x]^3*Log[2 - 2/(1 + a*x)])/c - (3*ArcTanh[a*x]^2*PolyLog[2, -1 + 2/(1 + a*x)])/(2*c) - (3*ArcTanh[a
*x]*PolyLog[3, -1 + 2/(1 + a*x)])/(2*c) - (3*PolyLog[4, -1 + 2/(1 + a*x)])/(4*c)
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Rubi [A] time = 0.175708, antiderivative size = 93, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used =
{1593, 5932, 5948, 6056, 6060, 6610} \[ -\frac{3 \text{PolyLog}\left (4,\frac{2}{a x+1}-1\right )}{4 c}-\frac{3 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,\frac{2}{a x+1}-1\right )}{2 c}-\frac{3 \tanh ^{-1}(a x) \text{PolyLog}\left (3,\frac{2}{a x+1}-1\right )}{2 c}+\frac{\log \left (2-\frac{2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \]
Antiderivative was successfully verified.
[In]
Int[ArcTanh[a*x]^3/(c*x + a*c*x^2),x]
[Out]
(ArcTanh[a*x]^3*Log[2 - 2/(1 + a*x)])/c - (3*ArcTanh[a*x]^2*PolyLog[2, -1 + 2/(1 + a*x)])/(2*c) - (3*ArcTanh[a
*x]*PolyLog[3, -1 + 2/(1 + a*x)])/(2*c) - (3*PolyLog[4, -1 + 2/(1 + a*x)])/(4*c)
Rule 1593
Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]
Rule 5932
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTanh[c*
x])^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTanh[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 5948
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Rule 6056
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(b*p)/2, Int[((a + b*ArcTanh[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[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2
/(1 + c*x))^2, 0]
Rule 6060
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a
+ b*ArcTanh[c*x])^p*PolyLog[k + 1, u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[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[c^2*d + e, 0] && EqQ[u^2 -
(1 - 2/(1 + 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{\tanh ^{-1}(a x)^3}{c x+a c x^2} \, dx &=\int \frac{\tanh ^{-1}(a x)^3}{x (c+a c x)} \, dx\\ &=\frac{\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{(3 a) \int \frac{\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}+\frac{(3 a) \int \frac{\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{2 c}+\frac{(3 a) \int \frac{\text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{1-a^2 x^2} \, dx}{2 c}\\ &=\frac{\tanh ^{-1}(a x)^3 \log \left (2-\frac{2}{1+a x}\right )}{c}-\frac{3 \tanh ^{-1}(a x)^2 \text{Li}_2\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 \tanh ^{-1}(a x) \text{Li}_3\left (-1+\frac{2}{1+a x}\right )}{2 c}-\frac{3 \text{Li}_4\left (-1+\frac{2}{1+a x}\right )}{4 c}\\ \end{align*}
Mathematica [A] time = 0.0726874, size = 86, normalized size = 0.92 \[ \frac{96 \tanh ^{-1}(a x)^2 \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )-96 \tanh ^{-1}(a x) \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )+48 \text{PolyLog}\left (4,e^{2 \tanh ^{-1}(a x)}\right )-32 \tanh ^{-1}(a x)^4+64 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+\pi ^4}{64 c} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[ArcTanh[a*x]^3/(c*x + a*c*x^2),x]
[Out]
(Pi^4 - 32*ArcTanh[a*x]^4 + 64*ArcTanh[a*x]^3*Log[1 - E^(2*ArcTanh[a*x])] + 96*ArcTanh[a*x]^2*PolyLog[2, E^(2*
ArcTanh[a*x])] - 96*ArcTanh[a*x]*PolyLog[3, E^(2*ArcTanh[a*x])] + 48*PolyLog[4, E^(2*ArcTanh[a*x])])/(64*c)
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Maple [C] time = 0.207, size = 1217, normalized size = 13.1 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(arctanh(a*x)^3/(a*c*x^2+c*x),x)
[Out]
1/2*I/c*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*((a*x+1)^2/(-a^2*x^2+1
)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^3-1/2*I/c*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))*csgn(I*(a*x+1)^2
/(a^2*x^2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))*arctanh(a*x)^3+I/c*Pi*csgn(I*(a*x+1)/(-
a^2*x^2+1)^(1/2))*csgn(I*(a*x+1)^2/(a^2*x^2-1))^2*arctanh(a*x)^3+1/2*I/c*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))
*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^3-1/2*I/c*Pi*csgn(I*(a*x+1)^2/(a^2*x^
2-1))*csgn(I*(a*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^3+1/2*I/c*Pi*csgn(I*(a*x+1)/(-a^
2*x^2+1)^(1/2))^2*csgn(I*(a*x+1)^2/(a^2*x^2-1))*arctanh(a*x)^3-1/2*I/c*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*c
sgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^3-1/2*I/c*Pi*csgn(I/((a*x+1)^2/(-a
^2*x^2+1)+1))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^2*arctanh(a*x)^3-6/c*arctanh(a*x)*
polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+1/c*arctanh(a*x)^3*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+3/c*arctanh(a*x)^2*
polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-6/c*arctanh(a*x)*polylog(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+1/c*arctanh(a*x)^
3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+3/c*arctanh(a*x)^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-1/c*arctanh(a*x)^
3*ln((a*x+1)^2/(-a^2*x^2+1)-1)+2/c*arctanh(a*x)^3*ln((a*x+1)/(-a^2*x^2+1)^(1/2))+1/c*ln(2)*arctanh(a*x)^3+1/c*
arctanh(a*x)^3*ln(a*x)-1/c*arctanh(a*x)^3*ln(a*x+1)+1/2*I/c*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)^3+
1/2*I/c*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^3+1/2*I/c*Pi*csgn(I*(a
*x+1)^2/(a^2*x^2-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3*arctanh(a*x)^3-1/2/c*arctanh(a*x)^4+6/c*polylog(4,(a*x+1)/(-
a^2*x^2+1)^(1/2))+6/c*polylog(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\log \left (a x + 1\right ) \log \left (-a x + 1\right )^{3}}{8 \, c} - \frac{1}{8} \, \int -\frac{{\left (a x - 1\right )} \log \left (a x + 1\right )^{3} - 3 \,{\left (a x - 1\right )} \log \left (a x + 1\right )^{2} \log \left (-a x + 1\right ) - 3 \,{\left (a^{2} x^{2} + 1\right )} \log \left (a x + 1\right ) \log \left (-a x + 1\right )^{2}}{a^{2} c x^{3} - c x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^3/(a*c*x^2+c*x),x, algorithm="maxima")
[Out]
1/8*log(a*x + 1)*log(-a*x + 1)^3/c - 1/8*integrate(-((a*x - 1)*log(a*x + 1)^3 - 3*(a*x - 1)*log(a*x + 1)^2*log
(-a*x + 1) - 3*(a^2*x^2 + 1)*log(a*x + 1)*log(-a*x + 1)^2)/(a^2*c*x^3 - c*x), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{artanh}\left (a x\right )^{3}}{a c x^{2} + c x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^3/(a*c*x^2+c*x),x, algorithm="fricas")
[Out]
integral(arctanh(a*x)^3/(a*c*x^2 + c*x), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{atanh}^{3}{\left (a x \right )}}{a x^{2} + x}\, dx}{c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(atanh(a*x)**3/(a*c*x**2+c*x),x)
[Out]
Integral(atanh(a*x)**3/(a*x**2 + x), x)/c
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{artanh}\left (a x\right )^{3}}{a c x^{2} + c x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(arctanh(a*x)^3/(a*c*x^2+c*x),x, algorithm="giac")
[Out]
integrate(arctanh(a*x)^3/(a*c*x^2 + c*x), x)