\(\int \frac {(a+b x+c x^2)^p}{d+e x+f x^2} \, dx\) [172]

Optimal result

Integrand size = 25, antiderivative size = 525 \[ \int \frac {\left (a+b x+c x^2\right )^p}{d+e x+f x^2} \, dx=\frac {\left (\frac {f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c \left (e-\sqrt {e^2-4 d f}+2 f x\right )}\right )^{-p} \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c \left (e-\sqrt {e^2-4 d f}+2 f x\right )}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {e-\frac {\left (b-\sqrt {b^2-4 a c}\right ) f}{c}-\sqrt {e^2-4 d f}}{e-\sqrt {e^2-4 d f}+2 f x},\frac {e-\frac {\left (b+\sqrt {b^2-4 a c}\right ) f}{c}-\sqrt {e^2-4 d f}}{e-\sqrt {e^2-4 d f}+2 f x}\right )}{2 \sqrt {e^2-4 d f} p}-\frac {\left (\frac {f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}\right )^{-p} \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {e-\frac {\left (b+\sqrt {b^2-4 a c}\right ) f}{c}+\sqrt {e^2-4 d f}}{e+\sqrt {e^2-4 d f}+2 f x},-\frac {\left (b-\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}{c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}\right )}{2 \sqrt {e^2-4 d f} p} \] Output:

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

Mathematica [F]

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

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

Output:

Integrate[(a + b*x + c*x^2)^p/(d + e*x + 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.

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

\(\Big \downarrow \) 1325

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 1325

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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