\(\int \frac {x^m}{(a+b x^2)^2 (c+d x^2)^2} \, dx\) [1605]

Optimal result

Integrand size = 22, antiderivative size = 230 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {d (b c+a d) x^{1+m}}{2 a c (b c-a d)^2 \left (c+d x^2\right )}+\frac {b x^{1+m}}{2 a (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {b^2 (a d (5-m)-b (c-c m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{2 a^2 (b c-a d)^3 (1+m)}-\frac {d^2 (a d (1-m)-b c (5-m)) x^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )}{2 c^2 (b c-a d)^3 (1+m)} \] Output:

1/2*d*(a*d+b*c)*x^(1+m)/a/c/(-a*d+b*c)^2/(d*x^2+c)+1/2*b*x^(1+m)/a/(-a*d+b 
*c)/(b*x^2+a)/(d*x^2+c)-1/2*b^2*(a*d*(5-m)-b*(-c*m+c))*x^(1+m)*hypergeom([ 
1, 1/2+1/2*m],[3/2+1/2*m],-b*x^2/a)/a^2/(-a*d+b*c)^3/(1+m)-1/2*d^2*(a*d*(1 
-m)-b*c*(5-m))*x^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c^2/ 
(-a*d+b*c)^3/(1+m)
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.75 \[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {x^{1+m} \left (2 a b^2 c^2 d \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )-2 a^2 b c d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )-(b c-a d) \left (b^2 c^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+a^2 d^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {d x^2}{c}\right )\right )\right )}{a^2 c^2 (-b c+a d)^3 (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {374, 441, 27, 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.

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

\(\Big \downarrow \) 374

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

\(\Big \downarrow \) 441

\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}-\frac {\frac {\int \frac {2 x^m \left (-b^2 (1-m) c^2+4 a b d c-b d (b c+a d) (1-m) x^2-a^2 d^2 (1-m)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {d x^{m+1} (a d+b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}-\frac {\frac {\int \frac {x^m \left (-b^2 (1-m) c^2+4 a b d c-b d (b c+a d) (1-m) x^2-a^2 d^2 (1-m)\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )}dx}{c (b c-a d)}-\frac {d x^{m+1} (a d+b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 446

\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}-\frac {\frac {\int \left (\frac {b^2 c (a d (5-m)-b c (1-m)) x^m}{(b c-a d) \left (b x^2+a\right )}+\frac {a d^2 (a d (1-m)-b c (5-m)) x^m}{(b c-a d) \left (d x^2+c\right )}\right )dx}{c (b c-a d)}-\frac {d x^{m+1} (a d+b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b x^{m+1}}{2 a \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}-\frac {\frac {\frac {b^2 c x^{m+1} (a d (5-m)-b (c-c m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a (m+1) (b c-a d)}+\frac {a d^2 x^{m+1} (a d (1-m)-b c (5-m)) \operatorname {Hypergeometric2F1}\left (1,\frac {m+1}{2},\frac {m+3}{2},-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)}}{c (b c-a d)}-\frac {d x^{m+1} (a d+b c)}{c \left (c+d x^2\right ) (b c-a d)}}{2 a (b c-a d)}\)

Input:

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

Output:

(b*x^(1 + m))/(2*a*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)) - (-((d*(b*c + a*d 
)*x^(1 + m))/(c*(b*c - a*d)*(c + d*x^2))) + ((b^2*c*(a*d*(5 - 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)) + (a*d^2*(a*d*(1 - m) - b*c*(5 - m))*x^(1 + m)*Hyp 
ergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(b*c - a*d)*(1 + 
 m)))/(c*(b*c - a*d)))/(2*a*(b*c - a*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 441

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^m/(b^2*d^2*x^8 + 2*(b^2*c*d + a*b*d^2)*x^6 + (b^2*c^2 + 4*a*b*c 
*d + a^2*d^2)*x^4 + a^2*c^2 + 2*(a*b*c^2 + a^2*c*d)*x^2), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^m}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\int \frac {x^{m}}{b^{2} d^{2} x^{8}+2 a b \,d^{2} x^{6}+2 b^{2} c d \,x^{6}+a^{2} d^{2} x^{4}+4 a b c d \,x^{4}+b^{2} c^{2} x^{4}+2 a^{2} c d \,x^{2}+2 a b \,c^{2} x^{2}+a^{2} c^{2}}d x \] Input:

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

Output:

int(x**m/(a**2*c**2 + 2*a**2*c*d*x**2 + a**2*d**2*x**4 + 2*a*b*c**2*x**2 + 
 4*a*b*c*d*x**4 + 2*a*b*d**2*x**6 + b**2*c**2*x**4 + 2*b**2*c*d*x**6 + b** 
2*d**2*x**8),x)