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

Optimal. Leaf size=67 \[ -\frac {\text {Li}_3\left (\frac {2}{1-a x}-1\right )}{2 c}+\frac {\text {Li}_2\left (\frac {2}{1-a x}-1\right ) \tanh ^{-1}(a x)}{c}+\frac {\log \left (2-\frac {2}{1-a x}\right ) \tanh ^{-1}(a x)^2}{c} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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*}

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Mathematica [A]  time = 0.15, size = 59, normalized size = 0.88 \[ \frac {\tanh ^{-1}(a x) \text {Li}_2\left (e^{2 \tanh ^{-1}(a x)}\right )}{c}-\frac {\text {Li}_3\left (e^{2 \tanh ^{-1}(a x)}\right )}{2 c}+\frac {\tanh ^{-1}(a x)^2 \log \left (1-e^{2 \tanh ^{-1}(a x)}\right )}{c} \]

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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fricas [F]  time = 1.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\operatorname {artanh}\left (a x\right )^{2}}{a c x^{2} - c x}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {\operatorname {artanh}\left (a x\right )^{2}}{a c x^{2} - c x}\,{d x} \]

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)

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maple [C]  time = 0.52, size = 717, normalized size = 10.70 \[ \frac {\arctanh \left (a x \right )^{2} \ln \left (a x \right )}{c}-\frac {\arctanh \left (a x \right )^{2} \ln \left (a x -1\right )}{c}-\frac {\arctanh \left (a x \right )^{2} \ln \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{c}+\frac {\arctanh \left (a x \right )^{2} \ln \left (1-\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {2 \arctanh \left (a x \right ) \polylog \left (2, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {2 \polylog \left (3, \frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {\arctanh \left (a x \right )^{2} \ln \left (1+\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}+\frac {2 \arctanh \left (a x \right ) \polylog \left (2, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {2 \polylog \left (3, -\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )}{c}-\frac {i \arctanh \left (a x \right )^{2} \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{2} \pi }{2 c}+\frac {i \arctanh \left (a x \right )^{2} \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \mathrm {csgn}\left (i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right ) \pi }{2 c}+\frac {i \arctanh \left (a x \right )^{2} \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{3} \pi }{2 c}+\frac {i \arctanh \left (a x \right )^{2} \pi }{c}+\frac {i \arctanh \left (a x \right )^{2} \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{3} \pi }{c}-\frac {i \arctanh \left (a x \right )^{2} \mathrm {csgn}\left (i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}-1\right )}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{2} \pi }{2 c}-\frac {i \arctanh \left (a x \right )^{2} \mathrm {csgn}\left (\frac {i}{1+\frac {\left (a x +1\right )^{2}}{-a^{2} x^{2}+1}}\right )^{2} \pi }{c}+\frac {\arctanh \left (a x \right )^{2} \ln \relax (2)}{c} \]

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/c*arctanh(a*x)^2*ln(a*x-1)-1/c*arctanh(a*x)^2*ln((a*x+1)^2/(-a^2*x^2+1)-1)+1/c*ar
ctanh(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*polyl
og(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))+I/c*arctanh(a*x)^2*csgn(I/(1+(a*x+1)
^2/(-a^2*x^2+1)))^3*Pi+I/c*arctanh(a*x)^2*Pi+1/2*I/c*arctanh(a*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+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)/(1+(a*x+1)^2/(-a^2*x^2+1)))*Pi-1/2*I/c*arctanh(a
*x)^2*csgn(I/(1+(a*x+1)^2/(-a^2*x^2+1)))*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi+1/
2*I/c*arctanh(a*x)^2*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^3*Pi-I/c*arctanh(a*x)^2*csg
n(I/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi-1/2*I/c*arctanh(a*x)^2*csgn(I*((a*x+1)^2/(-a^2*x^2+1)-1))*csgn(I*((a*x+1)
^2/(-a^2*x^2+1)-1)/(1+(a*x+1)^2/(-a^2*x^2+1)))^2*Pi+1/c*arctanh(a*x)^2*ln(2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (-a x + 1\right )^{3}}{12 \, c} + \frac {1}{4} \, \int -\frac {\log \left (a x + 1\right )^{2} - 2 \, \log \left (a x + 1\right ) \log \left (-a x + 1\right )}{a c x^{2} - c x}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {atanh}\left (a\,x\right )}^2}{c\,x-a\,c\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)^2/(c*x - a*c*x^2),x)

[Out]

int(atanh(a*x)^2/(c*x - a*c*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {\operatorname {atanh}^{2}{\left (a x \right )}}{a x^{2} - x}\, dx}{c} \]

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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