3.119 \(\int \frac{\tanh ^{-1}(a x)^2}{c x-a c x^2} \, dx\)

Optimal. Leaf size=67 \[ -\frac{\text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )}{2 c}+\frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )}{c}+\frac{\log \left (2-\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{c} \]

[Out]

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

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Rubi [A]  time = 0.138105, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1593, 5932, 5948, 6058, 6610} \[ -\frac{\text{PolyLog}\left (3,\frac{2}{1-a x}-1\right )}{2 c}+\frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,\frac{2}{1-a x}-1\right )}{c}+\frac{\log \left (2-\frac{2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]^2/(c*x - a*c*x^2),x]

[Out]

(ArcTanh[a*x]^2*Log[2 - 2/(1 - a*x)])/c + (ArcTanh[a*x]*PolyLog[2, -1 + 2/(1 - a*x)])/c - PolyLog[3, -1 + 2/(1
- a*x)]/(2*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 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[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 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)^2}{c x-a c x^2} \, dx &=\int \frac{\tanh ^{-1}(a x)^2}{x (c-a c x)} \, dx\\ &=\frac{\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1-a x}\right )}{c}-\frac{(2 a) \int \frac{\tanh ^{-1}(a x) \log \left (2-\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{a \int \frac{\text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac{\tanh ^{-1}(a x)^2 \log \left (2-\frac{2}{1-a x}\right )}{c}+\frac{\tanh ^{-1}(a x) \text{Li}_2\left (-1+\frac{2}{1-a x}\right )}{c}-\frac{\text{Li}_3\left (-1+\frac{2}{1-a x}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.150091, size = 59, normalized size = 0.88 \[ \frac{\tanh ^{-1}(a x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(a x)}\right )}{c}-\frac{\text{PolyLog}\left (3,e^{2 \tanh ^{-1}(a x)}\right )}{2 c}+\frac{\tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )}{c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcTanh[a*x]^2/(c*x - a*c*x^2),x]

[Out]

(ArcTanh[a*x]^2*Log[1 - E^(2*ArcTanh[a*x])])/c + (ArcTanh[a*x]*PolyLog[2, E^(2*ArcTanh[a*x])])/c - PolyLog[3,
E^(2*ArcTanh[a*x])]/(2*c)

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Maple [C]  time = 0.32, size = 717, normalized size = 10.7 \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)^2/(-a*c*x^2+c*x),x)

[Out]

-1/c*arctanh(a*x)^2*ln(a*x-1)+1/c*arctanh(a*x)^2*ln(a*x)-1/c*arctanh(a*x)^2*ln((a*x+1)^2/(-a^2*x^2+1)-1)+1/c*a
rctanh(a*x)^2*ln(1-(a*x+1)/(-a^2*x^2+1)^(1/2))+2/c*arctanh(a*x)*polylog(2,(a*x+1)/(-a^2*x^2+1)^(1/2))-2/c*poly
log(3,(a*x+1)/(-a^2*x^2+1)^(1/2))+1/c*arctanh(a*x)^2*ln(1+(a*x+1)/(-a^2*x^2+1)^(1/2))+2/c*arctanh(a*x)*polylog
(2,-(a*x+1)/(-a^2*x^2+1)^(1/2))-2/c*polylog(3,-(a*x+1)/(-a^2*x^2+1)^(1/2))+1/2*I/c*arctanh(a*x)^2*Pi*csgn(I*((
a*x+1)^2/(-a^2*x^2+1)-1)/((a*x+1)^2/(-a^2*x^2+1)+1))^3-I/c*arctanh(a*x)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1)
)^2+I/c*arctanh(a*x)^2*Pi*csgn(I/((a*x+1)^2/(-a^2*x^2+1)+1))^3+I/c*arctanh(a*x)^2*Pi-1/2*I/c*arctanh(a*x)^2*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))^2-1/2*I/c*ar
ctanh(a*x)^2*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+1/2*I/c*arctanh(a*x)^2*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))+1/c*arctanh(a*x)^2*ln(2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\log \left (-a x + 1\right )^{3}}{12 \, c} + \frac{1}{4} \, \int -\frac{\log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{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)^2/(-a*c*x^2+c*x),x, algorithm="maxima")

[Out]

-1/12*log(-a*x + 1)^3/c + 1/4*integrate(-(log(a*x + 1)^2 - 2*log(a*x + 1)*log(-a*x + 1))/(a*c*x^2 - 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 )^{2}}{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)^2/(-a*c*x^2+c*x),x, algorithm="fricas")

[Out]

integral(-arctanh(a*x)^2/(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}^{2}{\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)**2/(-a*c*x**2+c*x),x)

[Out]

-Integral(atanh(a*x)**2/(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 )^{2}}{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)^2/(-a*c*x^2+c*x),x, algorithm="giac")

[Out]

integrate(-arctanh(a*x)^2/(a*c*x^2 - c*x), x)