3.1208 \(\int e^{-\tanh ^{-1}(a x)} (c-a^2 c x^2)^{5/2} \, dx\)

Optimal. Leaf size=140 \[ -\frac{c^2 (1-a x)^6 \sqrt{c-a^2 c x^2}}{6 a \sqrt{1-a^2 x^2}}+\frac{4 c^2 (1-a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}}-\frac{c^2 (1-a x)^4 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}} \]

[Out]

-((c^2*(1 - a*x)^4*Sqrt[c - a^2*c*x^2])/(a*Sqrt[1 - a^2*x^2])) + (4*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 = 0.0974888, antiderivative size = 140, 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} \[ -\frac{c^2 (1-a x)^6 \sqrt{c-a^2 c x^2}}{6 a \sqrt{1-a^2 x^2}}+\frac{4 c^2 (1-a x)^5 \sqrt{c-a^2 c x^2}}{5 a \sqrt{1-a^2 x^2}}-\frac{c^2 (1-a x)^4 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c^2*(1 - a*x)^4*Sqrt[c - a^2*c*x^2])/(a*Sqrt[1 - a^2*x^2])) + (4*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^{-\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^{-\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)^3 (1+a x)^2 \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (c^2 \sqrt{c-a^2 c x^2}\right ) \int \left (4 (1-a x)^3-4 (1-a x)^4+(1-a x)^5\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{c^2 (1-a x)^4 \sqrt{c-a^2 c x^2}}{a \sqrt{1-a^2 x^2}}+\frac{4 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 = 0.0421286, size = 60, normalized size = 0.43 \[ -\frac{c^2 (a x-1)^4 \left (5 a^2 x^2+14 a x+11\right ) \sqrt{c-a^2 c x^2}}{30 a \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - a^2*c*x^2)^(5/2)/E^ArcTanh[a*x],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*Sqrt[1 - a^2*x^2])

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Maple [A]  time = 0.029, size = 81, 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 ) ^{3} \left ( ax-1 \right ) ^{3}} \left ( -{a}^{2}c{x}^{2}+c \right ) ^{{\frac{5}{2}}}\sqrt{-{a}^{2}{x}^{2}+1}} \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)*(-a^2*x^2+1)^(1/2),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^2*x^2+1)^(1/2)/(a*x+1)^3
/(a*x-1)^3

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

[Out]

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

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Fricas [A]  time = 2.34548, size = 205, normalized size = 1.46 \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 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)*(-a^2*x^2+1)^(1/2),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*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{\sqrt{- \left (a x - 1\right ) \left (a x + 1\right )} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{5}{2}}}{a x + 1}\, dx \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)*(-a**2*x**2+1)**(1/2),x)

[Out]

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

[Out]

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