\(\int \frac {(a+b x+c x^2)^p}{d+f x^2} \, dx\) [173]

Optimal result

Integrand size = 22, antiderivative size = 541 \[ \int \frac {\left (a+b x+c x^2\right )^p}{d+f x^2} \, dx=-\frac {4^{-1+p} \left (\frac {\sqrt {f} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c \left (\sqrt {-d}+\sqrt {f} x\right )}\right )^{-p} \left (\frac {\sqrt {f} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c \left (\sqrt {-d}+\sqrt {f} x\right )}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c \sqrt {-d}-\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {f}}{2 c \left (\sqrt {-d}+\sqrt {f} x\right )},\frac {2 \sqrt {-d}-\frac {\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {f}}{c}}{2 \left (\sqrt {-d}+\sqrt {f} x\right )}\right )}{\sqrt {-d} \sqrt {f} p}+\frac {4^{-1+p} \left (\frac {\sqrt {-d} \sqrt {f} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c \left (d+\sqrt {-d} \sqrt {f} x\right )}\right )^{-p} \left (\frac {\sqrt {-d} \sqrt {f} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c \left (d+\sqrt {-d} \sqrt {f} x\right )}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) \sqrt {-d} \sqrt {f}}{2 c \left (d+\sqrt {-d} \sqrt {f} x\right )},\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) \sqrt {-d} \sqrt {f}}{2 c \left (d+\sqrt {-d} \sqrt {f} x\right )}\right )}{\sqrt {-d} \sqrt {f} p} \] Output:

-4^(-1+p)*(c*x^2+b*x+a)^p*AppellF1(-2*p,-p,-p,1-2*p,(2*(-d)^(1/2)-(b+(-4*a 
*c+b^2)^(1/2))*f^(1/2)/c)/(2*(-d)^(1/2)+2*f^(1/2)*x),1/2*(2*c*(-d)^(1/2)-( 
b-(-4*a*c+b^2)^(1/2))*f^(1/2))/c/((-d)^(1/2)+f^(1/2)*x))/(-d)^(1/2)/f^(1/2 
)/p/((f^(1/2)*(b-(-4*a*c+b^2)^(1/2)+2*c*x)/c/((-d)^(1/2)+f^(1/2)*x))^p)/(( 
f^(1/2)*(b+(-4*a*c+b^2)^(1/2)+2*c*x)/c/((-d)^(1/2)+f^(1/2)*x))^p)+4^(-1+p) 
*(c*x^2+b*x+a)^p*AppellF1(-2*p,-p,-p,1-2*p,1/2*(2*c*d-(b-(-4*a*c+b^2)^(1/2 
))*(-d)^(1/2)*f^(1/2))/c/(d+(-d)^(1/2)*f^(1/2)*x),1/2*(2*c*d-(b+(-4*a*c+b^ 
2)^(1/2))*(-d)^(1/2)*f^(1/2))/c/(d+(-d)^(1/2)*f^(1/2)*x))/(-d)^(1/2)/f^(1/ 
2)/p/(((-d)^(1/2)*f^(1/2)*(b-(-4*a*c+b^2)^(1/2)+2*c*x)/c/(d+(-d)^(1/2)*f^( 
1/2)*x))^p)/(((-d)^(1/2)*f^(1/2)*(b+(-4*a*c+b^2)^(1/2)+2*c*x)/c/(d+(-d)^(1 
/2)*f^(1/2)*x))^p)
 

Mathematica [F]

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

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

Output:

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

Rubi [F]

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.

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

\(\Big \downarrow \) 1326

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 1326

Int[((a_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_ 
Symbol] :> Unintegrable[(a + c*x^2)^p*(d + e*x + f*x^2)^q, x] /; FreeQ[{a, 
c, d, e, f, p, q}, x] &&  !IGtQ[p, 0] &&  !IGtQ[q, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((c*x^2 + b*x + a)^p/(f*x^2 + d), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((a + b*x + c*x**2)**p/(d + f*x**2),x)