\(\int \frac {x^m (A+B x^2)}{b x^2+c x^4} \, dx\) [276]

Optimal result

Integrand size = 24, antiderivative size = 71 \[ \int \frac {x^m \left (A+B x^2\right )}{b x^2+c x^4} \, dx=-\frac {B x^{-1+m}}{c (1-m)}+\frac {(b B-A c) x^{-1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\frac {c x^2}{b}\right )}{b c (1-m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.77 \[ \int \frac {x^m \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {x^{-1+m} \left (b B+(-b B+A c) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1+m),\frac {1+m}{2},-\frac {c x^2}{b}\right )\right )}{b c (-1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 71, 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, 363, 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.

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

\(\Big \downarrow \) 9

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

\(\Big \downarrow \) 363

\(\displaystyle -\frac {(b B-A c) \int \frac {x^{m-2}}{c x^2+b}dx}{c}-\frac {B x^{m-1}}{c (1-m)}\)

\(\Big \downarrow \) 278

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

Input:

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

Output:

-((B*x^(-1 + m))/(c*(1 - m))) + ((b*B - A*c)*x^(-1 + m)*Hypergeometric2F1[ 
1, (-1 + m)/2, (1 + m)/2, -((c*x^2)/b)])/(b*c*(1 - 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 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^m \left (A+B x^2\right )}{b x^2+c x^4} \, dx=\frac {x^{m} b +\left (\int \frac {x^{m}}{c \,x^{4}+b \,x^{2}}d x \right ) a c m x -\left (\int \frac {x^{m}}{c \,x^{4}+b \,x^{2}}d x \right ) a c x -\left (\int \frac {x^{m}}{c \,x^{4}+b \,x^{2}}d x \right ) b^{2} m x +\left (\int \frac {x^{m}}{c \,x^{4}+b \,x^{2}}d x \right ) b^{2} x}{c x \left (m -1\right )} \] Input:

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

Output:

(x**m*b + int(x**m/(b*x**2 + c*x**4),x)*a*c*m*x - int(x**m/(b*x**2 + c*x** 
4),x)*a*c*x - int(x**m/(b*x**2 + c*x**4),x)*b**2*m*x + int(x**m/(b*x**2 + 
c*x**4),x)*b**2*x)/(c*x*(m - 1))