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

Optimal. Leaf size=71 \[ \frac {x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {a x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6160, 6140} \[ \frac {x^2 \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}}-\frac {a x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6140

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.58 \[ \frac {x \left (x-\frac {a x^2}{2}\right ) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.16, size = 57, normalized size = 0.80 \[ \frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x^{3} - 2 \, x^{2}\right )} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{2 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {c - \frac {c}{a^{2} x^{2}}} x}{a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 56, normalized size = 0.79 \[ \frac {x^{2} \left (a x -2\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {-a^{2} x^{2}+1}}{2 \left (a x -1\right ) \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.39, size = 51, normalized size = 0.72 \[ \frac {{\left (i \, a^{2} \sqrt {c} x^{2} - 2 i \, a \sqrt {c} x + 2 i \, \sqrt {c}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )}}{2 \, {\left (a^{4} x^{2} - a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.93, size = 67, normalized size = 0.94 \[ -\frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,\left (\frac {x^3\,\sqrt {1-a^2\,x^2}}{2\,a}-\frac {x^2\,\sqrt {1-a^2\,x^2}}{a^2}\right )}{\frac {1}{a^2}-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )}}{a x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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