3.1203 \(\int e^{-\tanh ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6153, 6150, 43} \[ \frac {x^2 \sqrt {c-a^2 c x^2}}{2 \sqrt {1-a^2 x^2}}-\frac {a x^3 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Sqrt[c - a^2*c*x^2])/(2*Sqrt[1 - a^2*x^2]) - (a*x^3*Sqrt[c - a^2*c*x^2])/(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 6153

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.57 \[ -\frac {x^2 (2 a x-3) \sqrt {c-a^2 c x^2}}{6 \sqrt {1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 50, normalized size = 0.68 \[ \frac {\sqrt {-a^{2} c x^{2} + c} \sqrt {-a^{2} x^{2} + 1} {\left (2 \, a x^{3} - 3 \, x^{2}\right )}}{6 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 51, normalized size = 0.69 \[ \frac {x^{2} \left (2 a x -3\right ) \sqrt {-a^{2} c \,x^{2}+c}\, \sqrt {-a^{2} x^{2}+1}}{6 \left (a x -1\right ) \left (a x +1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 41, normalized size = 0.55 \[ -\frac {{\left (2 \, a \sqrt {c} x^{3} - 3 \, \sqrt {c} x^{2}\right )} {\left (a x + 1\right )} {\left (a x - 1\right )}}{6 \, {\left (a^{2} x^{2} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\sqrt {c-a^2\,c\,x^2}\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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