\(\int \frac {x^m (A+B x^2)}{(b x^2+c x^4)^2} \, dx\) [277]

Optimal result

Integrand size = 24, antiderivative size = 98 \[ \int \frac {x^m \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=-\frac {(b B-A c) x^{-3+m}}{2 b c \left (b+c x^2\right )}+\frac {(b B (3-m)-A c (5-m)) x^{-3+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),-\frac {c x^2}{b}\right )}{2 b^2 c (3-m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.82 \[ \int \frac {x^m \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {x^{-3+m} \left (b B \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),-\frac {c x^2}{b}\right )+(-b B+A c) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),-\frac {c x^2}{b}\right )\right )}{b^2 c (-3+m)} \] Input:

Integrate[(x^m*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {9, 362, 278}

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.

\(\displaystyle \int \frac {x^m \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx\)

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 362

\(\displaystyle -\frac {(b B (3-m)-A c (5-m)) \int \frac {x^{m-4}}{c x^2+b}dx}{2 b c}-\frac {x^{m-3} (b B-A c)}{2 b c \left (b+c x^2\right )}\)

\(\Big \downarrow \) 278

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

Input:

Int[(x^m*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]
 

Output:

-1/2*((b*B - A*c)*x^(-3 + m))/(b*c*(b + c*x^2)) + ((b*B*(3 - m) - A*c*(5 - 
 m))*x^(-3 + m)*Hypergeometric2F1[1, (-3 + m)/2, (-1 + m)/2, -((c*x^2)/b)] 
)/(2*b^2*c*(3 - m))
 

Defintions of rubi rules used

rule 9

Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, 
x, Min]}, Simp[1/e^(p*r)   Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, 
x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && 
  !MonomialQ[Px, x]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 362

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*b*e 
*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(2*a*b*(p + 1))   I 
nt[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && N 
eQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) || 
  !RationalQ[m] || (ILtQ[p + 1/2, 0] && LeQ[-1, m, -2*(p + 1)]))
 

Maple [F]

\[\int \frac {x^{m} \left (B \,x^{2}+A \right )}{\left (c \,x^{4}+b \,x^{2}\right )^{2}}d x\]

Input:

int(x^m*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
 

Output:

int(x^m*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
 

Fricas [F]

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

integrate(x^m*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(x**m*(B*x**2+A)/(c*x**4+b*x**2)**2,x)
 

Output:

Integral(x**m*(A + B*x**2)/(x**4*(b + c*x**2)**2), x)
 

Maxima [F]

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

integrate(x^m*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")
 

Output:

integrate((B*x^2 + A)*x^m/(c*x^4 + b*x^2)^2, x)
 

Giac [F]

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

integrate(x^m*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")
 

Output:

integrate((B*x^2 + A)*x^m/(c*x^4 + b*x^2)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^m \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\int \frac {x^m\,\left (B\,x^2+A\right )}{{\left (c\,x^4+b\,x^2\right )}^2} \,d x \] Input:

int((x^m*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)
 

Output:

int((x^m*(A + B*x^2))/(b*x^2 + c*x^4)^2, x)
 

Reduce [F]

\[ \int \frac {x^m \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx=\frac {x^{m} b +\left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) a b c \,m^{2} x^{3}-10 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) a b c m \,x^{3}+25 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) a b c \,x^{3}+\left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) a \,c^{2} m^{2} x^{5}-10 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) a \,c^{2} m \,x^{5}+25 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) a \,c^{2} x^{5}-\left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) b^{3} m^{2} x^{3}+8 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) b^{3} m \,x^{3}-15 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) b^{3} x^{3}-\left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) b^{2} c \,m^{2} x^{5}+8 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) b^{2} c m \,x^{5}-15 \left (\int \frac {x^{m}}{c^{2} m \,x^{8}-5 c^{2} x^{8}+2 b c m \,x^{6}-10 b c \,x^{6}+b^{2} m \,x^{4}-5 b^{2} x^{4}}d x \right ) b^{2} c \,x^{5}}{c \,x^{3} \left (c m \,x^{2}-5 c \,x^{2}+b m -5 b \right )} \] Input:

int(x^m*(B*x^2+A)/(c*x^4+b*x^2)^2,x)
 

Output:

(x**m*b + int(x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 10*b*c*x**6 
 + c**2*m*x**8 - 5*c**2*x**8),x)*a*b*c*m**2*x**3 - 10*int(x**m/(b**2*m*x** 
4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 10*b*c*x**6 + c**2*m*x**8 - 5*c**2*x**8), 
x)*a*b*c*m*x**3 + 25*int(x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 
10*b*c*x**6 + c**2*m*x**8 - 5*c**2*x**8),x)*a*b*c*x**3 + int(x**m/(b**2*m* 
x**4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 10*b*c*x**6 + c**2*m*x**8 - 5*c**2*x** 
8),x)*a*c**2*m**2*x**5 - 10*int(x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b*c*m* 
x**6 - 10*b*c*x**6 + c**2*m*x**8 - 5*c**2*x**8),x)*a*c**2*m*x**5 + 25*int( 
x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 10*b*c*x**6 + c**2*m*x**8 
 - 5*c**2*x**8),x)*a*c**2*x**5 - int(x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b 
*c*m*x**6 - 10*b*c*x**6 + c**2*m*x**8 - 5*c**2*x**8),x)*b**3*m**2*x**3 + 8 
*int(x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 10*b*c*x**6 + c**2*m 
*x**8 - 5*c**2*x**8),x)*b**3*m*x**3 - 15*int(x**m/(b**2*m*x**4 - 5*b**2*x* 
*4 + 2*b*c*m*x**6 - 10*b*c*x**6 + c**2*m*x**8 - 5*c**2*x**8),x)*b**3*x**3 
- int(x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 10*b*c*x**6 + c**2* 
m*x**8 - 5*c**2*x**8),x)*b**2*c*m**2*x**5 + 8*int(x**m/(b**2*m*x**4 - 5*b* 
*2*x**4 + 2*b*c*m*x**6 - 10*b*c*x**6 + c**2*m*x**8 - 5*c**2*x**8),x)*b**2* 
c*m*x**5 - 15*int(x**m/(b**2*m*x**4 - 5*b**2*x**4 + 2*b*c*m*x**6 - 10*b*c* 
x**6 + c**2*m*x**8 - 5*c**2*x**8),x)*b**2*c*x**5)/(c*x**3*(b*m - 5*b + c*m 
*x**2 - 5*c*x**2))