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

Optimal. Leaf size=81 \[ \frac{x^{m+1} \sqrt{c-\frac{c}{a^2 x^2}}}{m \sqrt{1-a^2 x^2}}-\frac{a x^{m+2} \sqrt{c-\frac{c}{a^2 x^2}}}{(m+1) \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.234089, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {6160, 6150, 43} \[ \frac{x^{m+1} \sqrt{c-\frac{c}{a^2 x^2}}}{m \sqrt{1-a^2 x^2}}-\frac{a x^{m+2} \sqrt{c-\frac{c}{a^2 x^2}}}{(m+1) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6160

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

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, 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)} \sqrt{c-\frac{c}{a^2 x^2}} x^m \, dx &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int e^{-\tanh ^{-1}(a x)} x^{-1+m} \sqrt{1-a^2 x^2} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int x^{-1+m} (1-a x) \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\left (\sqrt{c-\frac{c}{a^2 x^2}} x\right ) \int \left (x^{-1+m}-a x^m\right ) \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\sqrt{c-\frac{c}{a^2 x^2}} x^{1+m}}{m \sqrt{1-a^2 x^2}}-\frac{a \sqrt{c-\frac{c}{a^2 x^2}} x^{2+m}}{(1+m) \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0387454, size = 52, normalized size = 0.64 \[ \frac{x \sqrt{c-\frac{c}{a^2 x^2}} \left (\frac{x^m}{m}-\frac{a x^{m+1}}{m+1}\right )}{\sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.08, size = 69, normalized size = 0.9 \begin{align*}{\frac{{x}^{1+m} \left ( axm-m-1 \right ) }{ \left ( 1+m \right ) m \left ( ax-1 \right ) \left ( ax+1 \right ) }\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}}\sqrt{-{a}^{2}{x}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.14975, size = 74, normalized size = 0.91 \begin{align*} \frac{{\left (i \, a \sqrt{c} m x + \sqrt{c}{\left (-i \, m - i\right )}\right )}{\left (a x + 1\right )}{\left (a x - 1\right )} x^{m}}{{\left (m^{2} + m\right )} a^{3} x^{2} -{\left (m^{2} + m\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.20564, size = 151, normalized size = 1.86 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1}{\left (a m x^{2} -{\left (m + 1\right )} x\right )} x^{m} \sqrt{\frac{a^{2} c x^{2} - c}{a^{2} x^{2}}}}{{\left (a^{2} m^{2} + a^{2} m\right )} x^{2} - m^{2} - m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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