\(\int \frac {(a+b x+c x^2)^p}{e x+f x^2} \, dx\) [174]

Optimal result

Integrand size = 24, antiderivative size = 337 \[ \int \frac {\left (a+b x+c x^2\right )^p}{e x+f x^2} \, dx=\frac {2^{-1+2 p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b-\sqrt {b^2-4 a c}}{2 c x},-\frac {b+\sqrt {b^2-4 a c}}{2 c x}\right )}{e p}-\frac {2^{-1+2 p} \left (\frac {f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (e+f x)}\right )^{-p} \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (e+f x)}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}{2 c (e+f x)},\frac {2 e-\frac {\left (b+\sqrt {b^2-4 a c}\right ) f}{c}}{2 (e+f x)}\right )}{e p} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 1.53 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x+c x^2\right )^p}{e x+f x^2} \, dx=\frac {2^{-1+2 p} (a+x (b+c x))^p \left (\left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{c x}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,-\frac {b+\sqrt {b^2-4 a c}}{2 c x},\frac {-b+\sqrt {b^2-4 a c}}{2 c x}\right )-\left (\frac {f \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (e+f x)}\right )^{-p} \left (\frac {f \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (e+f x)}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}{2 c (e+f x)},\frac {2 c e-b f+\sqrt {b^2-4 a c} f}{2 c e+2 c f x}\right )\right )}{e p} \] Input:

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

Output:

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

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.

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

\(\Big \downarrow \) 1325

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

Input:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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