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

Optimal. Leaf size=95 \[ \frac{(1-a x)^4 \left (c-a^2 c x^2\right )^{3/2}}{4 a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{2 (1-a x)^3 \left (c-a^2 c x^2\right )^{3/2}}{3 a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.180267, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {6192, 6193, 43} \[ \frac{(1-a x)^4 \left (c-a^2 c x^2\right )^{3/2}}{4 a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}}-\frac{2 (1-a x)^3 \left (c-a^2 c x^2\right )^{3/2}}{3 a^4 x^3 \left (1-\frac{1}{a^2 x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6192

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

Rule 6193

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

Rubi steps

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

Mathematica [A]  time = 0.0324301, size = 53, normalized size = 0.56 \[ -\frac{c (a x-1)^3 (3 a x+5) \sqrt{c-a^2 c x^2}}{12 a^2 x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*(-1 + a*x)^3*(5 + 3*a*x)*Sqrt[c - a^2*c*x^2])/(12*a^2*Sqrt[1 - 1/(a^2*x^2)]*x)

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Maple [A]  time = 0.041, size = 68, normalized size = 0.7 \begin{align*}{\frac{x \left ( 3\,{x}^{3}{a}^{3}-4\,{a}^{2}{x}^{2}-6\,ax+12 \right ) }{ \left ( 12\,ax+12 \right ) \left ( ax-1 \right ) ^{2}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{3}{2}}}\sqrt{{\frac{ax-1}{ax+1}}}} \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)/(a*x+1))^(1/2),x)

[Out]

1/12*x*(3*a^3*x^3-4*a^2*x^2-6*a*x+12)*(-a^2*c*x^2+c)^(3/2)*((a*x-1)/(a*x+1))^(1/2)/(a*x+1)/(a*x-1)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \sqrt{\frac{a x - 1}{a x + 1}}\,{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)/(a*x+1))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.88801, size = 96, normalized size = 1.01 \begin{align*} -\frac{{\left (3 \, a^{3} c x^{4} - 4 \, a^{2} c x^{3} - 6 \, a c x^{2} + 12 \, c x\right )} \sqrt{-a^{2} c}}{12 \, a} \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)/(a*x+1))^(1/2),x, algorithm="fricas")

[Out]

-1/12*(3*a^3*c*x^4 - 4*a^2*c*x^3 - 6*a*c*x^2 + 12*c*x)*sqrt(-a^2*c)/a

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)/(a*x+1))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-a^{2} c x^{2} + c\right )}^{\frac{3}{2}} \sqrt{\frac{a x - 1}{a x + 1}}\,{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)/(a*x+1))^(1/2),x, algorithm="giac")

[Out]

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