3.4.52 \(\int \frac {x^m}{(1+\frac {a x^2}{b})^2} \, dx\) [352]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 36, normalized size of antiderivative = 1.00, 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 = {371} \begin {gather*} \frac {x^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {a x^2}{b}\right )}{m+1} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], 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*}

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Mathematica [A]
time = 0.03, size = 38, normalized size = 1.06 \begin {gather*} \frac {x^{1+m} \, _2F_1\left (2,\frac {1+m}{2};1+\frac {1+m}{2};-\frac {a x^2}{b}\right )}{1+m} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 5.
time = 0.06, size = 92, normalized size = 2.56

method	result	size
meijerg	\(\frac {\left (\frac {a}{b}\right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {2 x^{1+m} \left (\frac {a}{b}\right )^{\frac {1}{2}+\frac {m}{2}}}{2+\frac {2 a \,x^{2}}{b}}+\frac {2 x^{1+m} \left (\frac {a}{b}\right )^{\frac {1}{2}+\frac {m}{2}} \left (-\frac {m^{2}}{4}+\frac {1}{4}\right ) \Phi \left (-\frac {a \,x^{2}}{b}, 1, \frac {1}{2}+\frac {m}{2}\right )}{1+m}\right )}{2}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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] Result contains complex when optimal does not.
time = 3.01, size = 343, normalized size = 9.53 \begin {gather*} - \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^m}{{\left (\frac {a\,x^2}{b}+1\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/((a*x^2)/b + 1)^2,x)

[Out]

int(x^m/((a*x^2)/b + 1)^2, x)

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