3.736 \(\int e^{-3 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}} \, dx\)

Optimal. Leaf size=106 \[ \frac{a x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

[Out]

(a*Sqrt[c - c/(a^2*x^2)]*x^2)/Sqrt[1 - a^2*x^2] + (Sqrt[c - c/(a^2*x^2)]*x*Log[x])/Sqrt[1 - a^2*x^2] - (4*Sqrt
[c - c/(a^2*x^2)]*x*Log[1 + a*x])/Sqrt[1 - a^2*x^2]

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Rubi [A]  time = 0.147256, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6160, 6150, 72} \[ \frac{a x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}-\frac{4 x \sqrt{c-\frac{c}{a^2 x^2}} \log (a x+1)}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[c - c/(a^2*x^2)]*x^2)/Sqrt[1 - a^2*x^2] + (Sqrt[c - c/(a^2*x^2)]*x*Log[x])/Sqrt[1 - a^2*x^2] - (4*Sqrt
[c - c/(a^2*x^2)]*x*Log[1 + a*x])/Sqrt[1 - a^2*x^2]

Rule 6160

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

Rule 6150

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int e^{-3 \tanh ^{-1}(a x)} \sqrt{c-\frac{c}{a^2 x^2}} \, dx &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int \frac{e^{-3 \tanh ^{-1}(a x)} \sqrt{1-a^2 x^2}}{x} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int \frac{(1-a x)^2}{x (1+a x)} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int \left (a+\frac{1}{x}-\frac{4 a}{1+a x}\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{a \sqrt{c-\frac{c}{a^2 x^2}} x^2}{\sqrt{1-a^2 x^2}}+\frac{\sqrt{c-\frac{c}{a^2 x^2}} x \log (x)}{\sqrt{1-a^2 x^2}}-\frac{4 \sqrt{c-\frac{c}{a^2 x^2}} x \log (1+a x)}{\sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0260811, size = 45, normalized size = 0.42 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} (a x-4 \log (a x+1)+\log (x))}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x*(a*x + Log[x] - 4*Log[1 + a*x]))/Sqrt[1 - a^2*x^2]

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Maple [A]  time = 0.154, size = 60, normalized size = 0.6 \begin{align*} -{\frac{x \left ( ax+\ln \left ( x \right ) -4\,\ln \left ( ax+1 \right ) \right ) }{{a}^{2}{x}^{2}-1}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} x^{2} + 1}{\left (a x - 1\right )} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} x^{2} + 2 \, a x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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