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

Optimal. Leaf size=80 \[ \frac{x^m \sqrt{c-\frac{c}{a^2 x^2}}}{a m \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{m+1} \sqrt{c-\frac{c}{a^2 x^2}}}{(m+1) \sqrt{1-\frac{1}{a^2 x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.244871, antiderivative size = 80, 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 = {6197, 6193, 43} \[ \frac{x^m \sqrt{c-\frac{c}{a^2 x^2}}}{a m \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{x^{m+1} \sqrt{c-\frac{c}{a^2 x^2}}}{(m+1) \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6197

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d/x^2
)^FracPart[p])/(1 - 1/(a^2*x^2))^FracPart[p], Int[u*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a
, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !IntegerQ[n/2] &&  !(IntegerQ[p] || GtQ[c, 0])

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

Mathematica [A]  time = 0.03545, size = 53, normalized size = 0.66 \[ \frac{\sqrt{c-\frac{c}{a^2 x^2}} \left (\frac{a x^{m+1}}{m+1}+\frac{x^m}{m}\right )}{a \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.13, size = 63, normalized size = 0.8 \begin{align*}{\frac{{x}^{1+m} \left ( axm+m+1 \right ) }{ \left ( ax+1 \right ) \left ( 1+m \right ) m}\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{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)*x^m*(c-c/a^2/x^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.24026, size = 59, normalized size = 0.74 \begin{align*} \frac{{\left (a \sqrt{c} m x + \sqrt{c}{\left (m + 1\right )}\right )}{\left (a x + 1\right )} x^{m}}{{\left (m^{2} + m\right )} a^{2} x +{\left (m^{2} + m\right )} a} \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*(c-c/a^2/x^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75595, size = 154, normalized size = 1.92 \begin{align*} -\frac{{\left (a m x^{2} +{\left (m + 1\right )} x\right )} x^{m} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{m^{2} -{\left (a m^{2} + a m\right )} x + m} \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*(c-c/a^2/x^2)^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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

[Out]

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