3.619 \(\int \frac{e^{\coth ^{-1}(a x)}}{\sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=38 \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{\sqrt{c-a^2 c x^2}} \]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x*Log[1 - a*x])/Sqrt[c - a^2*c*x^2]

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Rubi [A]  time = 0.146069, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6192, 6193, 31} \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{\sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]/Sqrt[c - a^2*c*x^2],x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x*Log[1 - a*x])/Sqrt[c - a^2*c*x^2]

Rule 6192

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c + d*x^2)^p/(x^(2*p)*(
1 - 1/(a^2*x^2))^p), Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !
IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{e^{\coth ^{-1}(a x)}}{\sqrt{c-a^2 c x^2}} \, dx &=\frac{\left (\sqrt{1-\frac{1}{a^2 x^2}} x\right ) \int \frac{e^{\coth ^{-1}(a x)}}{\sqrt{1-\frac{1}{a^2 x^2}} x} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=\frac{\left (a \sqrt{1-\frac{1}{a^2 x^2}} x\right ) \int \frac{1}{-1+a x} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-\frac{1}{a^2 x^2}} x \log (1-a x)}{\sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0186762, size = 38, normalized size = 1. \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{\sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[a*x]/Sqrt[c - a^2*c*x^2],x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x*Log[1 - a*x])/Sqrt[c - a^2*c*x^2]

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Maple [A]  time = 0.134, size = 51, normalized size = 1.3 \begin{align*} -{\frac{\ln \left ( ax-1 \right ) }{ca \left ( ax+1 \right ) }\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(1/2),x)

[Out]

-ln(a*x-1)*(-c*(a^2*x^2-1))^(1/2)/a/c/(a*x+1)/((a*x-1)/(a*x+1))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-a^{2} c x^{2} + c} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-a^2*c*x^2 + c)*sqrt((a*x - 1)/(a*x + 1))), x)

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Fricas [A]  time = 1.46157, size = 49, normalized size = 1.29 \begin{align*} -\frac{\sqrt{-a^{2} c} \log \left (a x - 1\right )}{a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(-a^2*c)*log(a*x - 1)/(a^2*c)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/2)/(-a**2*c*x**2+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-a^{2} c x^{2} + c} \sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)/(-a^2*c*x^2+c)^(1/2),x, algorithm="giac")

[Out]

integrate(1/(sqrt(-a^2*c*x^2 + c)*sqrt((a*x - 1)/(a*x + 1))), x)