3.616 \(\int e^{\coth ^{-1}(a x)} (c-a^2 c x^2)^{5/2} \, dx\)

Optimal. Leaf size=136 \[ \frac{(a x+1)^6 \left (c-a^2 c x^2\right )^{5/2}}{6 a^6 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{4 (a x+1)^5 \left (c-a^2 c x^2\right )^{5/2}}{5 a^6 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}+\frac{(a x+1)^4 \left (c-a^2 c x^2\right )^{5/2}}{a^6 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}} \]

[Out]

((1 + a*x)^4*(c - a^2*c*x^2)^(5/2))/(a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5) - (4*(1 + a*x)^5*(c - a^2*c*x^2)^(5/2))/
(5*a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5) + ((1 + a*x)^6*(c - a^2*c*x^2)^(5/2))/(6*a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5)

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Rubi [A]  time = 0.175529, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {6192, 6193, 43} \[ \frac{(a x+1)^6 \left (c-a^2 c x^2\right )^{5/2}}{6 a^6 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}-\frac{4 (a x+1)^5 \left (c-a^2 c x^2\right )^{5/2}}{5 a^6 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}}+\frac{(a x+1)^4 \left (c-a^2 c x^2\right )^{5/2}}{a^6 x^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*(c - a^2*c*x^2)^(5/2),x]

[Out]

((1 + a*x)^4*(c - a^2*c*x^2)^(5/2))/(a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5) - (4*(1 + a*x)^5*(c - a^2*c*x^2)^(5/2))/
(5*a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5) + ((1 + a*x)^6*(c - a^2*c*x^2)^(5/2))/(6*a^6*(1 - 1/(a^2*x^2))^(5/2)*x^5)

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 )^{5/2} \, dx &=\frac{\left (c-a^2 c x^2\right )^{5/2} \int e^{\coth ^{-1}(a x)} \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5 \, dx}{\left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac{\left (c-a^2 c x^2\right )^{5/2} \int (-1+a x)^2 (1+a x)^3 \, dx}{a^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac{\left (c-a^2 c x^2\right )^{5/2} \int \left (4 (1+a x)^3-4 (1+a x)^4+(1+a x)^5\right ) \, dx}{a^5 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}\\ &=\frac{(1+a x)^4 \left (c-a^2 c x^2\right )^{5/2}}{a^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}-\frac{4 (1+a x)^5 \left (c-a^2 c x^2\right )^{5/2}}{5 a^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}+\frac{(1+a x)^6 \left (c-a^2 c x^2\right )^{5/2}}{6 a^6 \left (1-\frac{1}{a^2 x^2}\right )^{5/2} x^5}\\ \end{align*}

Mathematica [A]  time = 0.0405933, size = 63, normalized size = 0.46 \[ \frac{c^2 (a x+1)^4 \left (5 a^2 x^2-14 a x+11\right ) \sqrt{c-a^2 c x^2}}{30 a^2 x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[a*x]*(c - a^2*c*x^2)^(5/2),x]

[Out]

(c^2*(1 + a*x)^4*(11 - 14*a*x + 5*a^2*x^2)*Sqrt[c - a^2*c*x^2])/(30*a^2*Sqrt[1 - 1/(a^2*x^2)]*x)

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Maple [A]  time = 0.043, size = 84, normalized size = 0.6 \begin{align*}{\frac{x \left ( 5\,{x}^{5}{a}^{5}+6\,{x}^{4}{a}^{4}-15\,{x}^{3}{a}^{3}-20\,{a}^{2}{x}^{2}+15\,ax+30 \right ) }{30\, \left ( ax-1 \right ) ^{2} \left ( ax+1 \right ) ^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*(-a^2*c*x^2+c)^(5/2),x)

[Out]

1/30*x*(5*a^5*x^5+6*a^4*x^4-15*a^3*x^3-20*a^2*x^2+15*a*x+30)*(-a^2*c*x^2+c)^(5/2)/(a*x-1)^2/(a*x+1)^3/((a*x-1)
/(a*x+1))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} c x^{2} + c\right )}^{\frac{5}{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(1/((a*x-1)/(a*x+1))^(1/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((-a^2*c*x^2 + c)^(5/2)/sqrt((a*x - 1)/(a*x + 1)), x)

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Fricas [A]  time = 1.60498, size = 153, normalized size = 1.12 \begin{align*} \frac{{\left (5 \, a^{5} c^{2} x^{6} + 6 \, a^{4} c^{2} x^{5} - 15 \, a^{3} c^{2} x^{4} - 20 \, a^{2} c^{2} x^{3} + 15 \, a c^{2} x^{2} + 30 \, c^{2} x\right )} \sqrt{-a^{2} c}}{30 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/30*(5*a^5*c^2*x^6 + 6*a^4*c^2*x^5 - 15*a^3*c^2*x^4 - 20*a^2*c^2*x^3 + 15*a*c^2*x^2 + 30*c^2*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(1/((a*x-1)/(a*x+1))**(1/2)*(-a**2*c*x**2+c)**(5/2),x)

[Out]

Timed out

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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{5}{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(1/((a*x-1)/(a*x+1))^(1/2)*(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(5/2)/sqrt((a*x - 1)/(a*x + 1)), x)