3.500 \(\int (\frac{x^m}{(1-a^2 x^2)^2}+\frac{a x^{1+m}}{(1-a^2 x^2)^2}) \, dx\)

Optimal. Leaf size=70 \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.0231833, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 1, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.028, Rules used = {364} \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}+\frac{a x^{m+2} \, _2F_1\left (2,\frac{m+2}{2};\frac{m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0059114, size = 67, normalized size = 0.96 \[ x^{m+1} \left (\frac{a x \, _2F_1\left (2,\frac{m}{2}+1;\frac{m}{2}+2;a^2 x^2\right )}{m+2}+\frac{\, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};a^2 x^2\right )}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.079, size = 177, normalized size = 2.5 \begin{align*} -{\frac{1}{2\,a} \left ( -{a}^{2} \right ) ^{-{\frac{m}{2}}} \left ({\frac{{x}^{m} \left ( -2-m \right ) }{ \left ( 2+m \right ) \left ( -{a}^{2}{x}^{2}+1 \right ) } \left ( -{a}^{2} \right ) ^{{\frac{m}{2}}}}+{\frac{{x}^{m}m}{2} \left ( -{a}^{2} \right ) ^{{\frac{m}{2}}}{\it LerchPhi} \left ({a}^{2}{x}^{2},1,{\frac{m}{2}} \right ) } \right ) }+{\frac{1}{2} \left ( -{a}^{2} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( -2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{1/2+m/2} \left ( -1-m \right ) }{ \left ( 1+m \right ) \left ( -2\,{a}^{2}{x}^{2}+2 \right ) }}+2\,{\frac{{x}^{1+m} \left ( -{a}^{2} \right ) ^{1/2+m/2} \left ( -1/4\,{m}^{2}+1/4 \right ){\it LerchPhi} \left ({a}^{2}{x}^{2},1,1/2+m/2 \right ) }{1+m}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a*(-a^2)^(-1/2*m)*(1/(2+m)*x^m*(-a^2)^(1/2*m)*(-2-m)/(-a^2*x^2+1)+1/2*x^m*(-a^2)^(1/2*m)*m*LerchPhi(a^2*x
^2,1,1/2*m))+1/2*(-a^2)^(-1/2-1/2*m)*(-2/(1+m)*x^(1+m)*(-a^2)^(1/2+1/2*m)*(-1-m)/(-2*a^2*x^2+2)+2/(1+m)*x^(1+m
)*(-a^2)^(1/2+1/2*m)*(-1/4*m^2+1/4)*LerchPhi(a^2*x^2,1,1/2+1/2*m))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a x^{m + 1} + x^{m}}{a^{4} x^{4} - 2 \, a^{2} x^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*x^(m + 1) + x^m)/(a^4*x^4 - 2*a^2*x^2 + 1), x)

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Sympy [C]  time = 4.71473, size = 673, normalized size = 9.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*m**2*x**3*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m
/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a**2*x**3*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2
+ 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) + a*(-a**2*m**2*x**4*x**m*lerchphi(a**2*x**2*exp_p
olar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*a**2*m*x**4*x**m*
lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2
)) + m**2*x**2*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 +
2) - 8*gamma(m/2 + 2)) + 2*m*x**2*x**m*lerchphi(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1)*gamma(m/2 + 1)/(8*a**
2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2)) - 2*m*x**2*x**m*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma
(m/2 + 2)) - 4*x**2*x**m*gamma(m/2 + 1)/(8*a**2*x**2*gamma(m/2 + 2) - 8*gamma(m/2 + 2))) + m**2*x*x**m*lerchph
i(a**2*x**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/
2)) - 2*m*x*x**m*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - x*x**m*lerchphi(a**2*x
**2*exp_polar(2*I*pi), 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2)) - 2*
x*x**m*gamma(m/2 + 1/2)/(8*a**2*x**2*gamma(m/2 + 3/2) - 8*gamma(m/2 + 3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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