3.2.30 \(\int \frac {\tanh ^{-1}(a x)^3}{x (c+a c x)} \, dx\) [130]

Optimal. Leaf size=93 \[ \frac {\tanh ^{-1}(a x)^3 \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {3 \tanh ^{-1}(a x)^2 \text {PolyLog}\left (2,-1+\frac {2}{1+a x}\right )}{2 c}-\frac {3 \tanh ^{-1}(a x) \text {PolyLog}\left (3,-1+\frac {2}{1+a x}\right )}{2 c}-\frac {3 \text {PolyLog}\left (4,-1+\frac {2}{1+a x}\right )}{4 c} \]

[Out]

arctanh(a*x)^3*ln(2-2/(a*x+1))/c-3/2*arctanh(a*x)^2*polylog(2,-1+2/(a*x+1))/c-3/2*arctanh(a*x)*polylog(3,-1+2/
(a*x+1))/c-3/4*polylog(4,-1+2/(a*x+1))/c

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Rubi [A]
time = 0.13, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6079, 6095, 6203, 6207, 6745} \begin {gather*} -\frac {3 \text {Li}_4\left (\frac {2}{a x+1}-1\right )}{4 c}-\frac {3 \text {Li}_2\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)^2}{2 c}-\frac {3 \text {Li}_3\left (\frac {2}{a x+1}-1\right ) \tanh ^{-1}(a x)}{2 c}+\frac {\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)^3}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^3/(x*(c + a*c*x)),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 6079

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 6095

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 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[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 6207

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 6745

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}{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*}

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Mathematica [A]
time = 0.07, size = 86, normalized size = 0.92 \begin {gather*} \frac {\pi ^4-32 \tanh ^{-1}(a x)^4+64 \tanh ^{-1}(a x)^3 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )+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 )}{64 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[a*x]^3/(x*(c + a*c*x)),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] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 23.77, size = 1157, normalized size = 12.44

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1157\)
default	\(\text {Expression too large to display}\)	\(1157\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(a*x)^3/x/(a*c*x+c),x,method=_RETURNVERBOSE)

[Out]

1/c*arctanh(a*x)^3*ln(a*x)-1/c*arctanh(a*x)^3*ln(a*x+1)-3/c*(1/6*I*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)^3-1/6*I*arctanh
(a*x)^3*Pi*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3-2/3*arctanh(a*x)^3*ln((a*x+1)/(-a^2
*x^2+1)^(1/2))-1/3*ln(2)*arctanh(a*x)^3-1/6*I*Pi*csgn(I*(a*x+1)^2/(a^2*x^2-1))^3*arctanh(a*x)^3+1/6*I*arctanh(
a*x)^3*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))^2-1
/6*I*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)^3-1/6*I*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+1/6*I*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)^3-1/6*I*arctanh(a*x
)^3*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)^2*polylog(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))+2*arctanh(a*x)*polylog(3,-(
a*x+1)/(-a^2*x^2+1)^(1/2))+1/3*arctanh(a*x)^3*ln((a*x+1)^2/(-a^2*x^2+1)-1)-1/3*arctanh(a*x)^3*ln(1-(a*x+1)/(-a
^2*x^2+1)^(1/2))-arctanh(a*x)^2*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))+2*arctanh(a*x)*polylog(3,(a*x+1)/(-a^2*x
^2+1)^(1/2))-1/3*arctanh(a*x)^3*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+1/6*arctanh(a*x)^4-2*polylog(4,(a*x+1)/(-a^2*
x^2+1)^(1/2))-2*polylog(4,-(a*x+1)/(-a^2*x^2+1)^(1/2))-1/6*I*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)^3-1/3*I*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+1/6*I*arctanh(a*x)^3*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)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x/(a*c*x+c),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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x/(a*c*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\operatorname {atanh}^{3}{\left (a x \right )}}{a x^{2} + x}\, dx}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)**3/x/(a*c*x+c),x)

[Out]

Integral(atanh(a*x)**3/(a*x**2 + x), x)/c

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)^3/x/(a*c*x+c),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {atanh}\left (a\,x\right )}^3}{x\,\left (c+a\,c\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^3/(x*(c + a*c*x)),x)

[Out]

int(atanh(a*x)^3/(x*(c + a*c*x)), x)

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