\(\int \frac {(d+e x+f x^2)^q}{a+b x+c x^2} \, dx\) [170]

Optimal result

Integrand size = 25, antiderivative size = 527 \[ \int \frac {\left (d+e x+f x^2\right )^q}{a+b x+c x^2} \, dx=\frac {\left (\frac {c \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-q} \left (\frac {c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-q} \left (d+e x+f x^2\right )^q \operatorname {AppellF1}\left (-2 q,-q,-q,1-2 q,\frac {b-\sqrt {b^2-4 a c}-\frac {c \left (e-\sqrt {e^2-4 d f}\right )}{f}}{b-\sqrt {b^2-4 a c}+2 c x},\frac {b-\sqrt {b^2-4 a c}-\frac {c \left (e+\sqrt {e^2-4 d f}\right )}{f}}{b-\sqrt {b^2-4 a c}+2 c x}\right )}{2 \sqrt {b^2-4 a c} q}-\frac {\left (\frac {c \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-q} \left (\frac {c \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-q} \left (d+e x+f x^2\right )^q \operatorname {AppellF1}\left (-2 q,-q,-q,1-2 q,\frac {\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e-\sqrt {e^2-4 d f}\right )}{f \left (b+\sqrt {b^2-4 a c}+2 c x\right )},\frac {\left (b+\sqrt {b^2-4 a c}\right ) f-c \left (e+\sqrt {e^2-4 d f}\right )}{f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}\right )}{2 \sqrt {b^2-4 a c} q} \] Output:

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

Mathematica [F]

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

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

Output:

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

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

\(\Big \downarrow \) 1325

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

Input:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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