3.1273 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^{5/2} \, dx\)

Optimal. Leaf size=95 \[ \frac{c^2 (1-a x)^6 \sqrt{c-a^2 c x^2}}{6 a \sqrt{1-a^2 x^2}}-\frac{2 c^2 (1-a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}} \]

[Out]

(-2*c^2*(1 - a*x)^5*Sqrt[c - a^2*c*x^2])/(5*a*Sqrt[1 - a^2*x^2]) + (c^2*(1 - a*x)^6*Sqrt[c - a^2*c*x^2])/(6*a*
Sqrt[1 - a^2*x^2])

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Rubi [A]  time = 0.0888439, 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 = {6143, 6140, 43} \[ \frac{c^2 (1-a x)^6 \sqrt{c-a^2 c x^2}}{6 a \sqrt{1-a^2 x^2}}-\frac{2 c^2 (1-a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c - a^2*c*x^2)^(5/2)/E^(3*ArcTanh[a*x]),x]

[Out]

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

Mathematica [A]  time = 0.0392596, size = 52, normalized size = 0.55 \[ \frac{c^2 (a x-1)^5 (5 a x+7) \sqrt{c-a^2 c x^2}}{30 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a^2*c*x^2)^(5/2)/E^(3*ArcTanh[a*x]),x]

[Out]

(c^2*(-1 + a*x)^5*(7 + 5*a*x)*Sqrt[c - a^2*c*x^2])/(30*a*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.028, size = 81, normalized size = 0.9 \begin{align*}{\frac{x \left ( 5\,{x}^{5}{a}^{5}-18\,{x}^{4}{a}^{4}+15\,{x}^{3}{a}^{3}+20\,{a}^{2}{x}^{2}-45\,ax+30 \right ) }{30\, \left ( ax+1 \right ) ^{4} \left ( ax-1 \right ) ^{4}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{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)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

1/30*x*(5*a^5*x^5-18*a^4*x^4+15*a^3*x^3+20*a^2*x^2-45*a*x+30)*(-a^2*c*x^2+c)^(5/2)*(-a^2*x^2+1)^(3/2)/(a*x+1)^
4/(a*x-1)^4

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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}}{\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)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="maxima")

[Out]

integrate((-a^2*c*x^2 + c)^(5/2)*(-a^2*x^2 + 1)^(3/2)/(a*x + 1)^3, x)

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Fricas [A]  time = 3.05641, size = 208, normalized size = 2.19 \begin{align*} -\frac{{\left (5 \, a^{5} c^{2} x^{6} - 18 \, a^{4} c^{2} x^{5} + 15 \, a^{3} c^{2} x^{4} + 20 \, a^{2} c^{2} x^{3} - 45 \, a c^{2} x^{2} + 30 \, c^{2} x\right )} \sqrt{-a^{2} c x^{2} + c} \sqrt{-a^{2} x^{2} + 1}}{30 \,{\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)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

-1/30*(5*a^5*c^2*x^6 - 18*a^4*c^2*x^5 + 15*a^3*c^2*x^4 + 20*a^2*c^2*x^3 - 45*a*c^2*x^2 + 30*c^2*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(-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)**(5/2)/(a*x+1)**3*(-a**2*x**2+1)**(3/2),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}}{\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)^(5/2)/(a*x+1)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate((-a^2*c*x^2 + c)^(5/2)*(-a^2*x^2 + 1)^(3/2)/(a*x + 1)^3, x)