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

Optimal. Leaf size=68 \[ \frac{a x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}} \]

[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]

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Rubi [A]  time = 0.143545, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6160, 6150, 43} \[ \frac{a x^2 \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}}+\frac{x \log (x) \sqrt{c-\frac{c}{a^2 x^2}}}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{\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^{\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}{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}\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}}\\ \end{align*}

Mathematica [A]  time = 0.0206847, size = 37, normalized size = 0.54 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} (a x+\log (x))}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.15, size = 52, normalized size = 0.8 \begin{align*} -{\frac{x \left ( ax+\ln \left ( x \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((a*x+1)/(-a^2*x^2+1)^(1/2)*(c-c/a^2/x^2)^(1/2),x)

[Out]

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

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Maxima [C]  time = 1.18066, size = 23, normalized size = 0.34 \begin{align*} -i \, \sqrt{c} x - \frac{i \, \sqrt{c} \log \left (x\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*sqrt(c)*x - I*sqrt(c)*log(x)/a

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Fricas [B]  time = 2.42968, size = 648, normalized size = 9.53 \begin{align*} \left [\frac{{\left (a^{2} x^{2} - 1\right )} \sqrt{-c} \log \left (\frac{a^{2} c x^{6} + a^{2} c x^{2} - c x^{4} +{\left (a x^{5} - a x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{-c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}} - c}{a^{2} x^{4} - x^{2}}\right ) - 2 \,{\left (a^{2} x^{2} - a^{2} x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \,{\left (a^{3} x^{2} - a\right )}}, -\frac{{\left (a^{2} x^{2} - 1\right )} \sqrt{c} \arctan \left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left (a x^{3} + a x\right )} \sqrt{c} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{4} -{\left (a^{2} + 1\right )} c x^{2} + c}\right ) +{\left (a^{2} x^{2} - a^{2} x\right )} \sqrt{-a^{2} x^{2} + 1} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{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)*(c-c/a^2/x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*((a^2*x^2 - 1)*sqrt(-c)*log((a^2*c*x^6 + a^2*c*x^2 - c*x^4 + (a*x^5 - a*x)*sqrt(-a^2*x^2 + 1)*sqrt(-c)*sq
rt((a^2*c*x^2 - c)/(a^2*x^2)) - c)/(a^2*x^4 - x^2)) - 2*(a^2*x^2 - a^2*x)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c*x^2 -
c)/(a^2*x^2)))/(a^3*x^2 - a), -((a^2*x^2 - 1)*sqrt(c)*arctan(sqrt(-a^2*x^2 + 1)*(a*x^3 + a*x)*sqrt(c)*sqrt((a
^2*c*x^2 - c)/(a^2*x^2))/(a^2*c*x^4 - (a^2 + 1)*c*x^2 + c)) + (a^2*x^2 - a^2*x)*sqrt(-a^2*x^2 + 1)*sqrt((a^2*c
*x^2 - c)/(a^2*x^2)))/(a^3*x^2 - a)]

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

[Out]

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

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

[Out]

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