3.1279 \(\int e^{-3 \tanh ^{-1}(a x)} x^m \sqrt{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=136 \[ \frac{4 x^{m+1} \sqrt{c-a^2 c x^2} \text{Hypergeometric2F1}(1,m+1,m+2,-a x)}{(m+1) \sqrt{1-a^2 x^2}}-\frac{3 x^{m+1} \sqrt{c-a^2 c x^2}}{(m+1) \sqrt{1-a^2 x^2}}+\frac{a x^{m+2} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.201518, antiderivative size = 136, 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 = {6153, 6150, 88, 64} \[ \frac{4 x^{m+1} \sqrt{c-a^2 c x^2} \, _2F_1(1,m+1;m+2;-a x)}{(m+1) \sqrt{1-a^2 x^2}}-\frac{3 x^{m+1} \sqrt{c-a^2 c x^2}}{(m+1) \sqrt{1-a^2 x^2}}+\frac{a x^{m+2} \sqrt{c-a^2 c x^2}}{(m+2) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x^(1 + m)*Sqrt[c - a^2*c*x^2])/((1 + m)*Sqrt[1 - a^2*x^2]) + (a*x^(2 + m)*Sqrt[c - a^2*c*x^2])/((2 + m)*Sq
rt[1 - a^2*x^2]) + (4*x^(1 + m)*Sqrt[c - a^2*c*x^2]*Hypergeometric2F1[1, 1 + m, 2 + m, -(a*x)])/((1 + m)*Sqrt[
1 - a^2*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 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 \tanh ^{-1}(a x)} x^m \sqrt{c-a^2 c x^2} \, dx &=\frac{\sqrt{c-a^2 c x^2} \int e^{-3 \tanh ^{-1}(a x)} x^m \sqrt{1-a^2 x^2} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\frac{\sqrt{c-a^2 c x^2} \int \frac{x^m (1-a x)^2}{1+a x} \, dx}{\sqrt{1-a^2 x^2}}\\ &=\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}{\sqrt{1-a^2 x^2}}\\ &=-\frac{3 x^{1+m} \sqrt{c-a^2 c x^2}}{(1+m) \sqrt{1-a^2 x^2}}+\frac{a x^{2+m} \sqrt{c-a^2 c x^2}}{(2+m) \sqrt{1-a^2 x^2}}+\frac{\left (4 \sqrt{c-a^2 c x^2}\right ) \int \frac{x^m}{1+a x} \, dx}{\sqrt{1-a^2 x^2}}\\ &=-\frac{3 x^{1+m} \sqrt{c-a^2 c x^2}}{(1+m) \sqrt{1-a^2 x^2}}+\frac{a x^{2+m} \sqrt{c-a^2 c x^2}}{(2+m) \sqrt{1-a^2 x^2}}+\frac{4 x^{1+m} \sqrt{c-a^2 c x^2} \, _2F_1(1,1+m;2+m;-a x)}{(1+m) \sqrt{1-a^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0500204, size = 74, normalized size = 0.54 \[ \frac{x^{m+1} \sqrt{c-a^2 c x^2} (4 (m+2) \text{Hypergeometric2F1}(1,m+1,m+2,-a x)+m (a x-3)+a x-6)}{(m+1) (m+2) \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.435, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( ax+1 \right ) ^{3}}\sqrt{-{a}^{2}c{x}^{2}+c} \left ( -{a}^{2}{x}^{2}+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)^3*(-a^2*x^2+1)^(3/2),x)

[Out]

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

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

[Out]

integrate(sqrt(-a^2*c*x^2 + c)*(-a^2*x^2 + 1)^(3/2)*x^m/(a*x + 1)^3, 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}{\left (a x - 1\right )} x^{m}}{a^{2} x^{2} + 2 \, 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)^3*(-a^2*x^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*c*x^2 + c)*sqrt(-a^2*x^2 + 1)*(a*x - 1)*x^m/(a^2*x^2 + 2*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)**3*(-a**2*x**2+1)**(3/2),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}{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}} x^{m}}{{\left (a x + 1\right )}^{3}}\,{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)^3*(-a^2*x^2+1)^(3/2),x, algorithm="giac")

[Out]

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