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

Optimal. Leaf size=41 \[ \frac {\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)}{c}-\frac {\text {Li}_2\left (\frac {2}{a x+1}-1\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1593, 5932, 2447} \[ \frac {\log \left (2-\frac {2}{a x+1}\right ) \tanh ^{-1}(a x)}{c}-\frac {\text {PolyLog}\left (2,\frac {2}{a x+1}-1\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[a*x]/(c*x + a*c*x^2),x]

[Out]

(ArcTanh[a*x]*Log[2 - 2/(1 + a*x)])/c - PolyLog[2, -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 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

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]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(a x)}{c x+a c x^2} \, dx &=\int \frac {\tanh ^{-1}(a x)}{x (c+a c x)} \, dx\\ &=\frac {\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {a \int \frac {\log \left (2-\frac {2}{1+a x}\right )}{1-a^2 x^2} \, dx}{c}\\ &=\frac {\tanh ^{-1}(a x) \log \left (2-\frac {2}{1+a x}\right )}{c}-\frac {\text {Li}_2\left (-1+\frac {2}{1+a x}\right )}{2 c}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 39, normalized size = 0.95 \[ \frac {\tanh ^{-1}(a x) \log \left (1-e^{-2 \tanh ^{-1}(a x)}\right )}{c}-\frac {\text {Li}_2\left (e^{-2 \tanh ^{-1}(a x)}\right )}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (a x\right )}{a c x^{2} + c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(a*c*x^2+c*x),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(a*c*x^2+c*x),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 126, normalized size = 3.07 \[ \frac {\arctanh \left (a x \right ) \ln \left (a x \right )}{c}-\frac {\arctanh \left (a x \right ) \ln \left (a x +1\right )}{c}+\frac {\ln \left (a x +1\right )^{2}}{4 c}-\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (a x +1\right )}{2 c}+\frac {\ln \left (-\frac {a x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {a x}{2}\right )}{2 c}+\frac {\dilog \left (\frac {1}{2}+\frac {a x}{2}\right )}{2 c}-\frac {\dilog \left (a x \right )}{2 c}-\frac {\dilog \left (a x +1\right )}{2 c}-\frac {\ln \left (a x \right ) \ln \left (a x +1\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*arctanh(a*x)*ln(a*x)-1/c*arctanh(a*x)*ln(a*x+1)+1/4/c*ln(a*x+1)^2-1/2/c*ln(-1/2*a*x+1/2)*ln(a*x+1)+1/2/c*l
n(-1/2*a*x+1/2)*ln(1/2+1/2*a*x)+1/2/c*dilog(1/2+1/2*a*x)-1/2/c*dilog(a*x)-1/2/c*dilog(a*x+1)-1/2/c*ln(a*x)*ln(
a*x+1)

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maxima [B]  time = 0.32, size = 120, normalized size = 2.93 \[ \frac {1}{4} \, a {\left (\frac {\log \left (a x + 1\right )^{2}}{a c} - \frac {2 \, {\left (\log \left (a x + 1\right ) \log \left (-\frac {1}{2} \, a x + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, a x + \frac {1}{2}\right )\right )}}{a c} - \frac {2 \, {\left (\log \left (a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-a x\right )\right )}}{a c} + \frac {2 \, {\left (\log \left (-a x + 1\right ) \log \relax (x) + {\rm Li}_2\left (a x\right )\right )}}{a c}\right )} - {\left (\frac {\log \left (a x + 1\right )}{c} - \frac {\log \relax (x)}{c}\right )} \operatorname {artanh}\left (a x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(a*x)/(a*c*x^2+c*x),x, algorithm="maxima")

[Out]

1/4*a*(log(a*x + 1)^2/(a*c) - 2*(log(a*x + 1)*log(-1/2*a*x + 1/2) + dilog(1/2*a*x + 1/2))/(a*c) - 2*(log(a*x +
1)*log(x) + dilog(-a*x))/(a*c) + 2*(log(-a*x + 1)*log(x) + dilog(a*x))/(a*c)) - (log(a*x + 1)/c - log(x)/c)*a
rctanh(a*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {atanh}\left (a\,x\right )}{a\,c\,x^2+c\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(a*x)/(c*x + a*c*x^2),x)

[Out]

int(atanh(a*x)/(c*x + a*c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(a*x)/(a*c*x**2+c*x),x)

[Out]

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

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