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

Optimal. Leaf size=80 \[ \frac{x^{m+1} \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{m+1}{2},\frac{m+3}{2},a^2 x^2\right )}{c^2 (m+1)}+\frac{a x^{m+2} \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{m+2}{2},\frac{m+4}{2},a^2 x^2\right )}{c^2 (m+2)} \]

[Out]

(x^(1 + m)*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(c^2*(1 + m)) + (a*x^(2 + m)*Hypergeometric2
F1[5/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(c^2*(2 + m))

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Rubi [A]  time = 0.105854, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {6148, 808, 364} \[ \frac{x^{m+1} \, _2F_1\left (\frac{5}{2},\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{c^2 (m+1)}+\frac{a x^{m+2} \, _2F_1\left (\frac{5}{2},\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{c^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*Hypergeometric2F1[5/2, (1 + m)/2, (3 + m)/2, a^2*x^2])/(c^2*(1 + m)) + (a*x^(2 + m)*Hypergeometric2
F1[5/2, (2 + m)/2, (4 + m)/2, a^2*x^2])/(c^2*(2 + m))

Rule 6148

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

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 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^{\tanh ^{-1}(a x)} x^m}{\left (c-a^2 c x^2\right )^2} \, dx &=\frac{\int \frac{x^m (1+a x)}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}\\ &=\frac{\int \frac{x^m}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}+\frac{a \int \frac{x^{1+m}}{\left (1-a^2 x^2\right )^{5/2}} \, dx}{c^2}\\ &=\frac{x^{1+m} \, _2F_1\left (\frac{5}{2},\frac{1+m}{2};\frac{3+m}{2};a^2 x^2\right )}{c^2 (1+m)}+\frac{a x^{2+m} \, _2F_1\left (\frac{5}{2},\frac{2+m}{2};\frac{4+m}{2};a^2 x^2\right )}{c^2 (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.0308114, size = 82, normalized size = 1.02 \[ \frac{\frac{x^{m+1} \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{m+1}{2},\frac{m+1}{2}+1,a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{m+2}{2},\frac{m+2}{2}+1,a^2 x^2\right )}{m+2}}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((x^(1 + m)*Hypergeometric2F1[5/2, (1 + m)/2, 1 + (1 + m)/2, a^2*x^2])/(1 + m) + (a*x^(2 + m)*Hypergeometric2F
1[5/2, (2 + m)/2, 1 + (2 + m)/2, a^2*x^2])/(2 + m))/c^2

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Maple [F]  time = 0.291, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ){x}^{m}}{ \left ( -{a}^{2}c{x}^{2}+c \right ) ^{2}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}}\, 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)^2,x)

[Out]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*x^m/(-a^2*c*x^2+c)^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}}{{\left (a^{2} c x^{2} - c\right )}^{2} \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)^2,x, algorithm="maxima")

[Out]

integrate((a*x + 1)*x^m/((a^2*c*x^2 - c)^2*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} x^{2} + 1} x^{m}}{a^{5} c^{2} x^{5} - a^{4} c^{2} x^{4} - 2 \, a^{3} c^{2} x^{3} + 2 \, a^{2} c^{2} x^{2} + a c^{2} x - c^{2}}, 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)^2,x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*x^m/(a^5*c^2*x^5 - a^4*c^2*x^4 - 2*a^3*c^2*x^3 + 2*a^2*c^2*x^2 + a*c^2*x - c^2),
x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{x^{m}}{a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx + \int \frac{a x x^{m}}{a^{4} x^{4} \sqrt{- a^{2} x^{2} + 1} - 2 a^{2} x^{2} \sqrt{- a^{2} x^{2} + 1} + \sqrt{- a^{2} x^{2} + 1}}\, dx}{c^{2}} \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)**2,x)

[Out]

(Integral(x**m/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1)), x)
+ Integral(a*x*x**m/(a**4*x**4*sqrt(-a**2*x**2 + 1) - 2*a**2*x**2*sqrt(-a**2*x**2 + 1) + sqrt(-a**2*x**2 + 1))
, x))/c**2

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

[Out]

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