3.1128 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^m}{c-a^2 c x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac{x^{m+1} \text{Hypergeometric2F1}(2,m+1,m+2,a x)}{c (m+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0812267, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {6150, 64} \[ \frac{x^{m+1} \, _2F_1(2,m+1;m+2;a x)}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0128715, size = 25, normalized size = 1. \[ \frac{x^{m+1} \text{Hypergeometric2F1}(2,m+1,m+2,a x)}{c (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.372, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ax+1 \right ) ^{2}{x}^{m}}{ \left ( -{a}^{2}{x}^{2}+1 \right ) \left ( -{a}^{2}c{x}^{2}+c \right ) }}\, 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),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{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),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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