3.1274 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{3/2} \, dx\)

Optimal. Leaf size=45 \[ -\frac{c (1-a x)^4 \sqrt{c-a^2 c x^2}}{4 a \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0801597, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6143, 6140, 32} \[ -\frac{c (1-a x)^4 \sqrt{c-a^2 c x^2}}{4 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6143

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

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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0305275, size = 58, normalized size = 1.29 \[ \frac{c \left (-\frac{1}{4} a^3 x^4+a^2 x^3-\frac{3 a x^2}{2}+x\right ) \sqrt{c-a^2 c x^2}}{\sqrt{1-a^2 x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 64, normalized size = 1.4 \begin{align*}{\frac{x \left ({x}^{3}{a}^{3}-4\,{a}^{2}{x}^{2}+6\,ax-4 \right ) }{4\, \left ( ax-1 \right ) ^{3} \left ( ax+1 \right ) ^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(a^3*x^3-4*a^2*x^2+6*a*x-4)*(-a^2*c*x^2+c)^(3/2)*(-a^2*x^2+1)^(3/2)/(a*x-1)^3/(a*x+1)^3

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Maxima [A]  time = 1.04349, size = 95, normalized size = 2.11 \begin{align*} -\frac{{\left (a^{4} c^{\frac{3}{2}} x^{4} - 4 \, a^{3} c^{\frac{3}{2}} x^{3} + 6 \, a^{2} c^{\frac{3}{2}} x^{2} - 4 \, a c^{\frac{3}{2}} x + 4 \, c^{\frac{3}{2}}\right )}{\left (a x + 1\right )}{\left (a x - 1\right )}}{4 \,{\left (a^{3} x^{2} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^4*c^(3/2)*x^4 - 4*a^3*c^(3/2)*x^3 + 6*a^2*c^(3/2)*x^2 - 4*a*c^(3/2)*x + 4*c^(3/2))*(a*x + 1)*(a*x - 1)
/(a^3*x^2 - a)

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Fricas [A]  time = 3.08914, size = 142, normalized size = 3.16 \begin{align*} \frac{{\left (a^{3} c x^{4} - 4 \, a^{2} c x^{3} + 6 \, a c x^{2} - 4 \, c x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{4 \,{\left (a^{2} x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(a^3*c*x^4 - 4*a^2*c*x^3 + 6*a*c*x^2 - 4*c*x)*sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)/(a^2*x^2 - 1)

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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}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{\left (a x + 1\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*(-c*(a*x - 1)*(a*x + 1))**(3/2)/(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} c x^{2} + c\right )}^{\frac{3}{2}}{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a x + 1\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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