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

Optimal. Leaf size=74 \[ \frac {x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-a^2 x^2}}-\frac {a x^4 \sqrt {c-\frac {c}{a^2 x^2}}}{3 \sqrt {1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6160, 6150, 43} \[ \frac {x^3 \sqrt {c-\frac {c}{a^2 x^2}}}{2 \sqrt {1-a^2 x^2}}-\frac {a x^4 \sqrt {c-\frac {c}{a^2 x^2}}}{3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c - c/(a^2*x^2)]*x^3)/(2*Sqrt[1 - a^2*x^2]) - (a*Sqrt[c - c/(a^2*x^2)]*x^4)/(3*Sqrt[1 - a^2*x^2])

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])

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

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.57 \[ -\frac {x^3 (2 a x-3) \sqrt {c-\frac {c}{a^2 x^2}}}{6 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(Sqrt[c - c/(a^2*x^2)]*x^3*(-3 + 2*a*x))/Sqrt[1 - a^2*x^2]

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fricas [A]  time = 2.44, size = 58, normalized size = 0.78 \[ \frac {{\left (2 \, a x^{4} - 3 \, x^{3}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{6 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*x^4 - 3*x^3)*sqrt(-a^2*x^2 + 1)*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 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} \sqrt {c - \frac {c}{a^{2} x^{2}}} x^{2}}{a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(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^2/(a*x + 1), x)

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maple [A]  time = 0.03, size = 57, normalized size = 0.77 \[ \frac {x^{3} \left (2 a x -3\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {-a^{2} x^{2}+1}}{6 \left (a x -1\right ) \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3*(2*a*x-3)*(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 = 0.38, size = 43, normalized size = 0.58 \[ \frac {{\left (2 i \, a \sqrt {c} x^{3} - 3 i \, \sqrt {c} x^{2}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )}}{6 \, {\left (a^{3} x^{2} - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*I*a*sqrt(c)*x^3 - 3*I*sqrt(c)*x^2)*(a*x + 1)*(a*x - 1)/(a^3*x^2 - a)

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mupad [B]  time = 0.97, size = 67, normalized size = 0.91 \[ -\frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,\left (\frac {x^4\,\sqrt {1-a^2\,x^2}}{3\,a}-\frac {x^3\,\sqrt {1-a^2\,x^2}}{2\,a^2}\right )}{\frac {1}{a^2}-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )}}{a x + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*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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