3.1129 \(\int \frac{e^{2 \tanh ^{-1}(a x)} x^m}{(c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=113 \[ \frac{\left (2 m^2-4 m+1\right ) x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,a x)}{8 c^2 (m+1)}+\frac{x^{m+1} \text{Hypergeometric2F1}(1,m+1,m+2,-a x)}{8 c^2 (m+1)}+\frac{(2-m) x^{m+1}}{4 c^2 (1-a x)}+\frac{x^{m+1}}{4 c^2 (1-a x)^2} \]

[Out]

x^(1 + m)/(4*c^2*(1 - a*x)^2) + ((2 - m)*x^(1 + m))/(4*c^2*(1 - a*x)) + (x^(1 + m)*Hypergeometric2F1[1, 1 + m,
2 + m, -(a*x)])/(8*c^2*(1 + m)) + ((1 - 4*m + 2*m^2)*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, a*x])/(8*c^
2*(1 + m))

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Rubi [A]  time = 0.178045, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6150, 103, 151, 156, 64} \[ \frac{\left (2 m^2-4 m+1\right ) x^{m+1} \, _2F_1(1,m+1;m+2;a x)}{8 c^2 (m+1)}+\frac{x^{m+1} \, _2F_1(1,m+1;m+2;-a x)}{8 c^2 (m+1)}+\frac{(2-m) x^{m+1}}{4 c^2 (1-a x)}+\frac{x^{m+1}}{4 c^2 (1-a x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x^(1 + m)/(4*c^2*(1 - a*x)^2) + ((2 - m)*x^(1 + m))/(4*c^2*(1 - a*x)) + (x^(1 + m)*Hypergeometric2F1[1, 1 + m,
2 + m, -(a*x)])/(8*c^2*(1 + m)) + ((1 - 4*m + 2*m^2)*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, a*x])/(8*c^
2*(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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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}{\left (c-a^2 c x^2\right )^2} \, dx &=\frac{\int \frac{x^m}{(1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac{x^{1+m}}{4 c^2 (1-a x)^2}-\frac{\int \frac{x^m \left (-a (3-m)-a^2 (1-m) x\right )}{(1-a x)^2 (1+a x)} \, dx}{4 a c^2}\\ &=\frac{x^{1+m}}{4 c^2 (1-a x)^2}+\frac{(2-m) x^{1+m}}{4 c^2 (1-a x)}+\frac{\int \frac{x^m \left (2 a^2 (1-m)^2-2 a^3 (2-m) m x\right )}{(1-a x) (1+a x)} \, dx}{8 a^2 c^2}\\ &=\frac{x^{1+m}}{4 c^2 (1-a x)^2}+\frac{(2-m) x^{1+m}}{4 c^2 (1-a x)}+\frac{\int \frac{x^m}{1+a x} \, dx}{8 c^2}+\frac{\left (1-4 m+2 m^2\right ) \int \frac{x^m}{1-a x} \, dx}{8 c^2}\\ &=\frac{x^{1+m}}{4 c^2 (1-a x)^2}+\frac{(2-m) x^{1+m}}{4 c^2 (1-a x)}+\frac{x^{1+m} \, _2F_1(1,1+m;2+m;-a x)}{8 c^2 (1+m)}+\frac{\left (1-4 m+2 m^2\right ) x^{1+m} \, _2F_1(1,1+m;2+m;a x)}{8 c^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0680812, size = 92, normalized size = 0.81 \[ \frac{x^{m+1} \left (\left (2 m^2-4 m+1\right ) (a x-1)^2 \text{Hypergeometric2F1}(1,m+1,m+2,a x)+(a x-1)^2 \text{Hypergeometric2F1}(1,m+1,m+2,-a x)+2 (m+1) (m (a x-1)-2 a x+3)\right )}{8 c^2 (m+1) (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(2*(1 + m)*(3 - 2*a*x + m*(-1 + a*x)) + (-1 + a*x)^2*Hypergeometric2F1[1, 1 + m, 2 + m, -(a*x)] + (
1 - 4*m + 2*m^2)*(-1 + a*x)^2*Hypergeometric2F1[1, 1 + m, 2 + m, a*x]))/(8*c^2*(1 + m)*(-1 + a*x)^2)

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Maple [F]  time = 0.418, 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 ) ^{2}}}\, 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)^2,x)

[Out]

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

[Out]

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

[Out]

integral(-x^m/(a^4*c^2*x^4 - 2*a^3*c^2*x^3 + 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} - 2 a^{3} x^{3} + 2 a x - 1}\, dx}{c^{2}} \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)**2,x)

[Out]

-Integral(x**m/(a**4*x**4 - 2*a**3*x**3 + 2*a*x - 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 )}^{2} x^{m}}{{\left (a^{2} c x^{2} - c\right )}^{2}{\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)^2,x, algorithm="giac")

[Out]

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