\(\int \frac {x^m}{(a+b x^2)^2 (c+d x^2)} \, dx\) [1604]

Optimal result

Integrand size = 22, antiderivative size = 156 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right )}+\frac {b (b c (1-m)-a d (3-m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^2 (1+m)}+\frac {d^2 x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{c (b c-a d)^2 (1+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\frac {x^{1+m} \left (-a b c d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^2 d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )+b c (b c-a d) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )\right )}{a^2 c (b c-a d)^2 (1+m)} \] Input:

Integrate[x^m/((a + b*x^2)^2*(c + d*x^2)),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {374, 446, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx\)

\(\Big \downarrow \) 374

\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)}-\frac {\int \frac {x^m \left (-b d (1-m) x^2+2 a d-b c (1-m)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 a (b c-a d)}\)

\(\Big \downarrow \) 446

\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)}-\frac {\int \left (\frac {b (a d (3-m)-b c (1-m)) x^m}{(b c-a d) \left (b x^2+a\right )}+\frac {2 a d^2 x^m}{(a d-b c) \left (d x^2+c\right )}\right )dx}{2 a (b c-a d)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)}-\frac {\frac {b x^{m+1} (a d (3-m)-b (c-c m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a (m+1) (b c-a d)}-\frac {2 a d^2 x^{m+1} \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)}}{2 a (b c-a d)}\)

Input:

Int[x^m/((a + b*x^2)^2*(c + d*x^2)),x]
 

Output:

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

Defintions of rubi rules used

rule 374

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
 + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - 
a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, 
c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, 
 c, d, e, m, 2, p, q, x]
 

rule 446

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((e_) + (f_.)*(x_)^2))/( 
(c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^2)^ 
p*((e + f*x^2)/(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^m/(b^2*d*x^6 + (b^2*c + 2*a*b*d)*x^4 + a^2*c + (2*a*b*c + a^2*d 
)*x^2), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 39.42 (sec) , antiderivative size = 3245, normalized size of antiderivative = 20.80 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

a**2*d*m**2*x**(m - 1)*lerchphi(a*exp_polar(I*pi)/(b*x**2), 1, 1/2 - m/2)* 
gamma(1/2 - m/2)/(8*a**4*d**2*gamma(3/2 - m/2) - 16*a**3*b*c*d*gamma(3/2 - 
 m/2) + 8*a**3*b*d**2*x**2*gamma(3/2 - m/2) + 8*a**2*b**2*c**2*gamma(3/2 - 
 m/2) - 16*a**2*b**2*c*d*x**2*gamma(3/2 - m/2) + 8*a*b**3*c**2*x**2*gamma( 
3/2 - m/2)) - 4*a**2*d*m*x**(m - 1)*lerchphi(a*exp_polar(I*pi)/(b*x**2), 1 
, 1/2 - m/2)*gamma(1/2 - m/2)/(8*a**4*d**2*gamma(3/2 - m/2) - 16*a**3*b*c* 
d*gamma(3/2 - m/2) + 8*a**3*b*d**2*x**2*gamma(3/2 - m/2) + 8*a**2*b**2*c** 
2*gamma(3/2 - m/2) - 16*a**2*b**2*c*d*x**2*gamma(3/2 - m/2) + 8*a*b**3*c** 
2*x**2*gamma(3/2 - m/2)) + 2*a**2*d*m*x**(m - 1)*lerchphi(c*exp_polar(I*pi 
)/(d*x**2), 1, 1/2 - m/2)*gamma(1/2 - m/2)/(8*a**4*d**2*gamma(3/2 - m/2) - 
 16*a**3*b*c*d*gamma(3/2 - m/2) + 8*a**3*b*d**2*x**2*gamma(3/2 - m/2) + 8* 
a**2*b**2*c**2*gamma(3/2 - m/2) - 16*a**2*b**2*c*d*x**2*gamma(3/2 - m/2) + 
 8*a*b**3*c**2*x**2*gamma(3/2 - m/2)) + 3*a**2*d*x**(m - 1)*lerchphi(a*exp 
_polar(I*pi)/(b*x**2), 1, 1/2 - m/2)*gamma(1/2 - m/2)/(8*a**4*d**2*gamma(3 
/2 - m/2) - 16*a**3*b*c*d*gamma(3/2 - m/2) + 8*a**3*b*d**2*x**2*gamma(3/2 
- m/2) + 8*a**2*b**2*c**2*gamma(3/2 - m/2) - 16*a**2*b**2*c*d*x**2*gamma(3 
/2 - m/2) + 8*a*b**3*c**2*x**2*gamma(3/2 - m/2)) - 2*a**2*d*x**(m - 1)*ler 
chphi(c*exp_polar(I*pi)/(d*x**2), 1, 1/2 - m/2)*gamma(1/2 - m/2)/(8*a**4*d 
**2*gamma(3/2 - m/2) - 16*a**3*b*c*d*gamma(3/2 - m/2) + 8*a**3*b*d**2*x**2 
*gamma(3/2 - m/2) + 8*a**2*b**2*c**2*gamma(3/2 - m/2) - 16*a**2*b**2*c*...
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx=\int \frac {x^{m}}{b^{2} d \,x^{6}+2 a b d \,x^{4}+b^{2} c \,x^{4}+a^{2} d \,x^{2}+2 a b c \,x^{2}+a^{2} c}d x \] Input:

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

Output:

int(x**m/(a**2*c + a**2*d*x**2 + 2*a*b*c*x**2 + 2*a*b*d*x**4 + b**2*c*x**4 
 + b**2*d*x**6),x)