\(\int \frac {(a+b x+c x^2)^p}{(d+e x)^2} \, dx\) [778]

Optimal result

Integrand size = 20, antiderivative size = 196 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(d+e x)^2} \, dx=-\frac {4^p \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2 (1-p),\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-2 p) (d+e x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.97 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(d+e x)^2} \, dx=\frac {4^p \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} (a+x (b+c x))^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 c d-b e+\sqrt {b^2-4 a c} e}{2 c d+2 c e x}\right )}{e (-1+2 p) (d+e x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1178, 150}

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.

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

\(\Big \downarrow \) 1178

\(\displaystyle -\frac {4^p \left (\frac {1}{d+e x}\right )^{2 p} \left (a+b x+c x^2\right )^p \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \int \left (\frac {1}{d+e x}\right )^{-2 p} \left (1-\frac {2 d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )^p \left (1-\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )^pd\frac {1}{d+e x}}{e}\)

\(\Big \downarrow \) 150

\(\displaystyle -\frac {4^p \left (a+b x+c x^2\right )^p \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-2 p) (d+e x)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 1178

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(1/(d + e*x))^(2*p))*((a + 
b*x + c*x^2)^p/(e*(e*((b - q + 2*c*x)/(2*c*(d + e*x))))^p*(e*((b + q + 2*c* 
x)/(2*c*(d + e*x))))^p))   Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - e*((b 
 - q)/(2*c)))*x, x]^p*Simp[1 - (d - e*((b + q)/(2*c)))*x, x]^p, x], x, 1/(d 
 + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && ILtQ[m, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

((a + b*x + c*x**2)**p*b + int((a + b*x + c*x**2)**p/(a*b*d**2*e*p - a*b*d 
**2*e + 2*a*b*d*e**2*p*x - 2*a*b*d*e**2*x + a*b*e**3*p*x**2 - a*b*e**3*x** 
2 + 2*a*c*d**3*p + 4*a*c*d**2*e*p*x + 2*a*c*d*e**2*p*x**2 + b**2*d**2*e*p* 
x - b**2*d**2*e*x + 2*b**2*d*e**2*p*x**2 - 2*b**2*d*e**2*x**2 + b**2*e**3* 
p*x**3 - b**2*e**3*x**3 + 2*b*c*d**3*p*x + 5*b*c*d**2*e*p*x**2 - b*c*d**2* 
e*x**2 + 4*b*c*d*e**2*p*x**3 - 2*b*c*d*e**2*x**3 + b*c*e**3*p*x**4 - b*c*e 
**3*x**4 + 2*c**2*d**3*p*x**2 + 4*c**2*d**2*e*p*x**3 + 2*c**2*d*e**2*p*x** 
4),x)*a*b**2*d*e**2*p**2 - int((a + b*x + c*x**2)**p/(a*b*d**2*e*p - a*b*d 
**2*e + 2*a*b*d*e**2*p*x - 2*a*b*d*e**2*x + a*b*e**3*p*x**2 - a*b*e**3*x** 
2 + 2*a*c*d**3*p + 4*a*c*d**2*e*p*x + 2*a*c*d*e**2*p*x**2 + b**2*d**2*e*p* 
x - b**2*d**2*e*x + 2*b**2*d*e**2*p*x**2 - 2*b**2*d*e**2*x**2 + b**2*e**3* 
p*x**3 - b**2*e**3*x**3 + 2*b*c*d**3*p*x + 5*b*c*d**2*e*p*x**2 - b*c*d**2* 
e*x**2 + 4*b*c*d*e**2*p*x**3 - 2*b*c*d*e**2*x**3 + b*c*e**3*p*x**4 - b*c*e 
**3*x**4 + 2*c**2*d**3*p*x**2 + 4*c**2*d**2*e*p*x**3 + 2*c**2*d*e**2*p*x** 
4),x)*a*b**2*d*e**2*p + int((a + b*x + c*x**2)**p/(a*b*d**2*e*p - a*b*d**2 
*e + 2*a*b*d*e**2*p*x - 2*a*b*d*e**2*x + a*b*e**3*p*x**2 - a*b*e**3*x**2 + 
 2*a*c*d**3*p + 4*a*c*d**2*e*p*x + 2*a*c*d*e**2*p*x**2 + b**2*d**2*e*p*x - 
 b**2*d**2*e*x + 2*b**2*d*e**2*p*x**2 - 2*b**2*d*e**2*x**2 + b**2*e**3*p*x 
**3 - b**2*e**3*x**3 + 2*b*c*d**3*p*x + 5*b*c*d**2*e*p*x**2 - b*c*d**2*e*x 
**2 + 4*b*c*d*e**2*p*x**3 - 2*b*c*d*e**2*x**3 + b*c*e**3*p*x**4 - b*c*e...