\(\int \frac {b^2 x^m}{(b+a x^2)^2} \, dx\) [385]

Optimal result

Integrand size = 16, antiderivative size = 36 \[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {a x^2}{b}\right )}{1+m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 371} \[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=\frac {x^{m+1} \operatorname {Hypergeometric2F1}\left (2,\frac {m+1}{2},\frac {m+3}{2},-\frac {a x^2}{b}\right )}{m+1} \]

[In]

Int[(b^2*x^m)/(b + a*x^2)^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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*} \text {integral}& = b^2 \int \frac {x^m}{\left (b+a x^2\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] (verified)

Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.06 \[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=\frac {x^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},1+\frac {1+m}{2},-\frac {a x^2}{b}\right )}{1+m} \]

[In]

Integrate[(b^2*x^m)/(b + a*x^2)^2,x]

[Out]

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

Maple [F]

\[\int \frac {b^{2} x^{m}}{\left (a \,x^{2}+b \right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=\int { \frac {b^{2} x^{m}}{{\left (a x^{2} + b\right )}^{2}} \,d x } \]

[In]

integrate(b^2*x^m/(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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.89 (sec) , antiderivative size = 381, normalized size of antiderivative = 10.58 \[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=b^{2} \left (- \frac {a m^{2} x^{2} x^{m + 1} \Phi \left (\frac {a x^{2} e^{i \pi }}{b}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a b^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) + 8 b^{3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a x^{2} x^{m + 1} \Phi \left (\frac {a x^{2} e^{i \pi }}{b}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a b^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) + 8 b^{3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} - \frac {b m^{2} x^{m + 1} \Phi \left (\frac {a x^{2} e^{i \pi }}{b}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a b^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) + 8 b^{3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {2 b m x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a b^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) + 8 b^{3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {b x^{m + 1} \Phi \left (\frac {a x^{2} e^{i \pi }}{b}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a b^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) + 8 b^{3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {2 b x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{8 a b^{2} x^{2} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) + 8 b^{3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}\right ) \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=\int { \frac {b^{2} x^{m}}{{\left (a x^{2} + b\right )}^{2}} \,d x } \]

[In]

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

[Out]

b^2*integrate(x^m/(a*x^2 + b)^2, x)

Giac [F]

\[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=\int { \frac {b^{2} x^{m}}{{\left (a x^{2} + b\right )}^{2}} \,d x } \]

[In]

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

[Out]

integrate(b^2*x^m/(a*x^2 + b)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {b^2 x^m}{\left (b+a x^2\right )^2} \, dx=\int \frac {b^2\,x^m}{{\left (a\,x^2+b\right )}^2} \,d x \]

[In]

int((b^2*x^m)/(b + a*x^2)^2,x)

[Out]

int((b^2*x^m)/(b + a*x^2)^2, x)