3.1134 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^m}{\sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac{(2 m+1) \sqrt{1-a^2 x^2} x^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{(m+1) \sqrt{c-a^2 c x^2}}-\frac{2 a (m+1) \sqrt{1-a^2 x^2} x^{m+2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{(m+2) \sqrt{c-a^2 c x^2}}+\frac{2 (a x+1) x^{m+1}}{\sqrt{c-a^2 c x^2}} \]

[Out]

(2*x^(1 + m)*(1 + a*x))/Sqrt[c - a^2*c*x^2] - ((1 + 2*m)*x^(1 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, (1
+ m)/2, (3 + m)/2, a^2*x^2])/((1 + m)*Sqrt[c - a^2*c*x^2]) - (2*a*(1 + m)*x^(2 + m)*Sqrt[1 - a^2*x^2]*Hyperge
ometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 + m)*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.293024, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6151, 1806, 808, 365, 364} \[ -\frac{(2 m+1) \sqrt{1-a^2 x^2} x^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{(m+1) \sqrt{c-a^2 c x^2}}-\frac{2 a (m+1) \sqrt{1-a^2 x^2} x^{m+2} \, _2F_1\left (\frac{1}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{(m+2) \sqrt{c-a^2 c x^2}}+\frac{2 (a x+1) x^{m+1}}{\sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(1 + m)*(1 + a*x))/Sqrt[c - a^2*c*x^2] - ((1 + 2*m)*x^(1 + m)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[1/2, (1
+ m)/2, (3 + m)/2, a^2*x^2])/((1 + m)*Sqrt[c - a^2*c*x^2]) - (2*a*(1 + m)*x^(2 + m)*Sqrt[1 - a^2*x^2]*Hyperge
ometric2F1[1/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/((2 + m)*Sqrt[c - a^2*c*x^2])

Rule 6151

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

Rule 1806

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, -Simp[((c*x)^(m + 1)*(f + g*x)*(a + b*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int
[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /; F
reeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 365

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])
/(1 + (b*x^n)/a)^FracPart[p], Int[(c*x)^m*(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

Mathematica [C]  time = 0.114016, size = 66, normalized size = 0.39 \[ \frac{\sqrt{1-a^2 x^2} x^{m+1} F_1\left (m+1;\frac{3}{2},-\frac{1}{2};m+2;a x,-a x\right )}{(m+1) \sqrt{a x-1} \sqrt{-c (a x+1)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^(1 + m)*Sqrt[1 - a^2*x^2]*AppellF1[1 + m, 3/2, -1/2, 2 + m, a*x, -(a*x)])/((1 + m)*Sqrt[-1 + a*x]*Sqrt[-(c*
(1 + a*x))])

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Maple [F]  time = 0.397, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{2}{x}^{m}}{-{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)^2/(-a^2*x^2+1)*x^m/(-a^2*c*x^2+c)^(1/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*c*x^2 + c)*x^m/(a^2*c*x^2 - 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}}{a x \sqrt{- a^{2} c x^{2} + c} - \sqrt{- a^{2} c x^{2} + c}}\, dx - \int \frac{a x x^{m}}{a x \sqrt{- a^{2} c x^{2} + c} - \sqrt{- a^{2} c x^{2} + c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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