3.1285 \(\int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{(1-a^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=37 \[ -\frac{2 (1-2 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{3 a \sqrt{1-a^2 x^2}} \]

[Out]

(-2*E^(ArcTanh[a*x]/2)*(1 - 2*a*x))/(3*a*Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.0399511, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {6135} \[ -\frac{2 (1-2 a x) e^{\frac{1}{2} \tanh ^{-1}(a x)}}{3 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^(ArcTanh[a*x]/2)/(1 - a^2*x^2)^(3/2),x]

[Out]

(-2*E^(ArcTanh[a*x]/2)*(1 - 2*a*x))/(3*a*Sqrt[1 - a^2*x^2])

Rule 6135

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((n - a*x)*E^(n*ArcTanh[a*x]))
/(a*c*(n^2 - 1)*Sqrt[c + d*x^2]), x] /; FreeQ[{a, c, d, n}, x] && EqQ[a^2*c + d, 0] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{e^{\frac{1}{2} \tanh ^{-1}(a x)}}{\left (1-a^2 x^2\right )^{3/2}} \, dx &=-\frac{2 e^{\frac{1}{2} \tanh ^{-1}(a x)} (1-2 a x)}{3 a \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0128866, size = 32, normalized size = 0.86 \[ \frac{2 (2 a x-1)}{3 a (1-a x)^{3/4} \sqrt [4]{a x+1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(ArcTanh[a*x]/2)/(1 - a^2*x^2)^(3/2),x]

[Out]

(2*(-1 + 2*a*x))/(3*a*(1 - a*x)^(3/4)*(1 + a*x)^(1/4))

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Maple [A]  time = 0.031, size = 54, normalized size = 1.5 \begin{align*} -{\frac{ \left ( 2\,ax-2 \right ) \left ( ax+1 \right ) \left ( 2\,ax-1 \right ) }{3\,a}\sqrt{{(ax+1){\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*x^2+1)^(3/2),x)

[Out]

-2/3*(a*x-1)*(a*x+1)*(2*a*x-1)*((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/a/(-a^2*x^2+1)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*x^2 + 1)^(3/2), x)

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Fricas [A]  time = 2.45104, size = 117, normalized size = 3.16 \begin{align*} -\frac{2 \, \sqrt{-a^{2} x^{2} + 1}{\left (2 \, a x - 1\right )} \sqrt{-\frac{\sqrt{-a^{2} x^{2} + 1}}{a x - 1}}}{3 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

-2/3*sqrt(-a^2*x^2 + 1)*(2*a*x - 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))/(a^3*x^2 - a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x + 1}{\sqrt{- a^{2} x^{2} + 1}}}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a**2*x**2+1)**(1/2))**(1/2)/(-a**2*x**2+1)**(3/2),x)

[Out]

Integral(sqrt((a*x + 1)/sqrt(-a**2*x**2 + 1))/(-(a*x - 1)*(a*x + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x + 1}{\sqrt{-a^{2} x^{2} + 1}}}}{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x+1)/(-a^2*x^2+1)^(1/2))^(1/2)/(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate(sqrt((a*x + 1)/sqrt(-a^2*x^2 + 1))/(-a^2*x^2 + 1)^(3/2), x)