3.735 \(\int e^{-3 \coth ^{-1}(a x)} x^m \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=137 \[ \frac{4 x^m \sqrt{c-a^2 c x^2} \text{Hypergeometric2F1}(1,m+1,m+2,-a x)}{a (m+1) \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{3 x^m \sqrt{c-a^2 c x^2}}{a (m+1) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{m+1} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

(-3*x^m*Sqrt[c - a^2*c*x^2])/(a*(1 + m)*Sqrt[1 - 1/(a^2*x^2)]) + (x^(1 + m)*Sqrt[c - a^2*c*x^2])/((2 + m)*Sqrt
[1 - 1/(a^2*x^2)]) + (4*x^m*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[1, 1 + m, 2 + m, -(a*x)])/(a*(1 + m)*Sqrt[1
- 1/(a^2*x^2)])

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Rubi [A]  time = 0.243495, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {6192, 6193, 88, 64} \[ \frac{4 x^m \sqrt{c-a^2 c x^2} \, _2F_1(1,m+1;m+2;-a x)}{a (m+1) \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{3 x^m \sqrt{c-a^2 c x^2}}{a (m+1) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{m+1} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x^m*Sqrt[c - a^2*c*x^2])/(a*(1 + m)*Sqrt[1 - 1/(a^2*x^2)]) + (x^(1 + m)*Sqrt[c - a^2*c*x^2])/((2 + m)*Sqrt
[1 - 1/(a^2*x^2)]) + (4*x^m*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[1, 1 + m, 2 + m, -(a*x)])/(a*(1 + m)*Sqrt[1
- 1/(a^2*x^2)])

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int e^{-3 \coth ^{-1}(a x)} x^m \sqrt{c-a^2 c x^2} \, dx &=\frac{\sqrt{c-a^2 c x^2} \int e^{-3 \coth ^{-1}(a x)} \sqrt{1-\frac{1}{a^2 x^2}} x^{1+m} \, dx}{\sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=\frac{\sqrt{c-a^2 c x^2} \int \frac{x^m (-1+a x)^2}{1+a x} \, dx}{a \sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=\frac{\sqrt{c-a^2 c x^2} \int \left (-3 x^m+a x^{1+m}+\frac{4 x^m}{1+a x}\right ) \, dx}{a \sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=-\frac{3 x^m \sqrt{c-a^2 c x^2}}{a (1+m) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{1+m} \sqrt{c-a^2 c x^2}}{(2+m) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\left (4 \sqrt{c-a^2 c x^2}\right ) \int \frac{x^m}{1+a x} \, dx}{a \sqrt{1-\frac{1}{a^2 x^2}} x}\\ &=-\frac{3 x^m \sqrt{c-a^2 c x^2}}{a (1+m) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{1+m} \sqrt{c-a^2 c x^2}}{(2+m) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{4 x^m \sqrt{c-a^2 c x^2} \, _2F_1(1,1+m;2+m;-a x)}{a (1+m) \sqrt{1-\frac{1}{a^2 x^2}}}\\ \end{align*}

Mathematica [A]  time = 0.051405, size = 75, normalized size = 0.55 \[ \frac{x^m \sqrt{c-a^2 c x^2} (4 (m+2) \text{Hypergeometric2F1}(1,m+1,m+2,-a x)+m (a x-3)+a x-6)}{a (m+1) (m+2) \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^m*Sqrt[c - a^2*c*x^2]*(-6 + a*x + m*(-3 + a*x) + 4*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -(a*x)]))/(a*
(1 + m)*(2 + m)*Sqrt[1 - 1/(a^2*x^2)])

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Maple [F]  time = 0.357, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}\sqrt{-{a}^{2}c{x}^{2}+c} \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} c x^{2} + c}{\left (a x - 1\right )} x^{m} \sqrt{\frac{a x - 1}{a x + 1}}}{a x + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*c*x^2 + c)*(a*x - 1)*x^m*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1), x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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