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

Optimal. Leaf size=82 \[ \frac{x^m \sqrt{c-a^2 c x^2}}{a (m+1) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{m+1} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

(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)])

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Rubi [A]  time = 0.212945, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6192, 6193, 43} \[ \frac{x^m \sqrt{c-a^2 c x^2}}{a (m+1) \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{m+1} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*x^m*Sqrt[c - a^2*c*x^2],x]

[Out]

(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)])

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

Mathematica [A]  time = 0.0330308, size = 56, normalized size = 0.68 \[ \frac{x^m \sqrt{c-a^2 c x^2} (a m x+a x+m+2)}{a (m+1) (m+2) \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[a*x]*x^m*Sqrt[c - a^2*c*x^2],x]

[Out]

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

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Maple [A]  time = 0.039, size = 62, normalized size = 0.8 \begin{align*}{\frac{{x}^{1+m} \left ( amx+ax+m+2 \right ) }{ \left ( ax+1 \right ) \left ( 2+m \right ) \left ( 1+m \right ) }\sqrt{-{a}^{2}c{x}^{2}+c}{\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)*x^m*(-a^2*c*x^2+c)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.08742, size = 73, normalized size = 0.89 \begin{align*} \frac{{\left (a \sqrt{-c}{\left (m + 1\right )} x^{2} + \sqrt{-c}{\left (m + 2\right )} x\right )}{\left (a x + 1\right )} x^{m}}{{\left (m^{2} + 3 \, m + 2\right )} a x + m^{2} + 3 \, m + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.96989, size = 166, normalized size = 2.02 \begin{align*} -\frac{\sqrt{-a^{2} c x^{2} + c}{\left ({\left (a m + a\right )} x^{2} +{\left (m + 2\right )} x\right )} x^{m} \sqrt{\frac{a x - 1}{a x + 1}}}{m^{2} -{\left (a m^{2} + 3 \, a m + 2 \, a\right )} x + 3 \, m + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

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