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

Optimal. Leaf size=36 \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{a x^2}{b}\right )}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0078092, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.062, Rules used = {364} \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{a x^2}{b}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(1 + (a*x^2)/b)^2,x]

[Out]

(x^(1 + m)*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((a*x^2)/b)])/(1 + 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 \frac{x^m}{\left (1+\frac{a x^2}{b}\right )^2} \, dx &=\frac{x^{1+m} \, _2F_1\left (2,\frac{1+m}{2};\frac{3+m}{2};-\frac{a x^2}{b}\right )}{1+m}\\ \end{align*}

Mathematica [A]  time = 0.0085049, size = 38, normalized size = 1.06 \[ \frac{x^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+1}{2}+1;-\frac{a x^2}{b}\right )}{m+1} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/(1 + (a*x^2)/b)^2,x]

[Out]

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

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Maple [C]  time = 0.05, size = 92, normalized size = 2.6 \begin{align*}{\frac{1}{2} \left ({\frac{a}{b}} \right ) ^{-{\frac{1}{2}}-{\frac{m}{2}}} \left ( 2\,{{x}^{1+m} \left ({\frac{a}{b}} \right ) ^{1/2+m/2} \left ( 2+2\,{\frac{a{x}^{2}}{b}} \right ) ^{-1}}+2\,{\frac{{x}^{1+m} \left ( -1/4\,{m}^{2}+1/4 \right ) }{1+m} \left ({\frac{a}{b}} \right ) ^{1/2+m/2}{\it LerchPhi} \left ( -{\frac{a{x}^{2}}{b}},1,1/2+m/2 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1/b*a)^(-1/2-1/2*m)*(2*x^(1+m)*(1/b*a)^(1/2+1/2*m)/(2+2/b*a*x^2)+2/(1+m)*x^(1+m)*(1/b*a)^(1/2+1/2*m)*(-1/
4*m^2+1/4)*LerchPhi(-1/b*a*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{x^{m}}{{\left (\frac{a x^{2}}{b} + 1\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(a*x^2/b + 1)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 7.30806, size = 343, normalized size = 9.53 \begin{align*} - \frac{a m^{2} x^{3} x^{m} \Phi \left (\frac{a x^{2} e^{i \pi }}{b}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{8 a x^{2} \Gamma \left (\frac{m}{2} + \frac{3}{2}\right ) + 8 b \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{a x^{3} x^{m} \Phi \left (\frac{a x^{2} e^{i \pi }}{b}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{8 a x^{2} \Gamma \left (\frac{m}{2} + \frac{3}{2}\right ) + 8 b \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} - \frac{b m^{2} x x^{m} \Phi \left (\frac{a x^{2} e^{i \pi }}{b}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{8 a x^{2} \Gamma \left (\frac{m}{2} + \frac{3}{2}\right ) + 8 b \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{2 b m x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{8 a x^{2} \Gamma \left (\frac{m}{2} + \frac{3}{2}\right ) + 8 b \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{b x x^{m} \Phi \left (\frac{a x^{2} e^{i \pi }}{b}, 1, \frac{m}{2} + \frac{1}{2}\right ) \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{8 a x^{2} \Gamma \left (\frac{m}{2} + \frac{3}{2}\right ) + 8 b \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} + \frac{2 b x x^{m} \Gamma \left (\frac{m}{2} + \frac{1}{2}\right )}{8 a x^{2} \Gamma \left (\frac{m}{2} + \frac{3}{2}\right ) + 8 b \Gamma \left (\frac{m}{2} + \frac{3}{2}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*m**2*x**3*x**m*lerchphi(a*x**2*exp_polar(I*pi)/b, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a*x**2*gamma(m/2 + 3/2)
+ 8*b*gamma(m/2 + 3/2)) + a*x**3*x**m*lerchphi(a*x**2*exp_polar(I*pi)/b, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(8*a*
x**2*gamma(m/2 + 3/2) + 8*b*gamma(m/2 + 3/2)) - b*m**2*x*x**m*lerchphi(a*x**2*exp_polar(I*pi)/b, 1, m/2 + 1/2)
*gamma(m/2 + 1/2)/(8*a*x**2*gamma(m/2 + 3/2) + 8*b*gamma(m/2 + 3/2)) + 2*b*m*x*x**m*gamma(m/2 + 1/2)/(8*a*x**2
*gamma(m/2 + 3/2) + 8*b*gamma(m/2 + 3/2)) + b*x*x**m*lerchphi(a*x**2*exp_polar(I*pi)/b, 1, m/2 + 1/2)*gamma(m/
2 + 1/2)/(8*a*x**2*gamma(m/2 + 3/2) + 8*b*gamma(m/2 + 3/2)) + 2*b*x*x**m*gamma(m/2 + 1/2)/(8*a*x**2*gamma(m/2
+ 3/2) + 8*b*gamma(m/2 + 3/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(a*x^2/b + 1)^2, x)