3.1001 \(\int \frac{e^{\tanh ^{-1}(a x)} x^m}{\sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=51 \[ \frac{\sqrt{1-a^2 x^2} x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,a x)}{(m+1) \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.183227, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {6153, 6150, 64} \[ \frac{\sqrt{1-a^2 x^2} x^{m+1} \, _2F_1(1,m+1;m+2;a x)}{(m+1) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6153

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c +
d*x^2)^FracPart[p])/(1 - a^2*x^2)^FracPart[p], Int[x^m*(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), x], x] /; FreeQ[{a,
c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] &&  !(IntegerQ[p] || GtQ[c, 0]) &&  !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 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 \frac{e^{\tanh ^{-1}(a x)} x^m}{\sqrt{c-a^2 c x^2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{\tanh ^{-1}(a x)} x^m}{\sqrt{1-a^2 x^2}} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int \frac{x^m}{1-a x} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=\frac{x^{1+m} \sqrt{1-a^2 x^2} \, _2F_1(1,1+m;2+m;a x)}{(1+m) \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0249566, size = 51, normalized size = 1. \[ \frac{\sqrt{1-a^2 x^2} x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,a x)}{(m+1) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.333, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ax+1 \right ){x}^{m}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + 1)*x^m/(sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)), 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} \sqrt{-a^{2} x^{2} + 1} x^{m}}{a^{3} c x^{3} - a^{2} c x^{2} - a c x + c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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