\(\int (d+e x)^{-1-2 p} (f+g x) (a+b x+c x^2)^p \, dx\) [1130]

Optimal result

Integrand size = 29, antiderivative size = 395 \[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=\frac {g (d+e x)^{1-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e^2 (1-2 p)}-\frac {(e f-d g) (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 e^2 p} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.75 (sec) , antiderivative size = 381, normalized size of antiderivative = 0.96 \[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=-\frac {2^{-1-p} \left (\frac {b-\sqrt {b^2-4 a c}}{2 c}+x\right )^{-p} \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{c}\right )^p \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{-2 p} (a+x (b+c x))^p \left (2 g p (d+e x) \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )+(e f-d g) (-1+2 p) \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )\right )}{e^2 p (-1+2 p)} \] Input:

Integrate[(d + e*x)^(-1 - 2*p)*(f + g*x)*(a + b*x + c*x^2)^p,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1269, 1179, 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.

\(\displaystyle \int (f+g x) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \, dx\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {(e f-d g) \int (d+e x)^{-2 p-1} \left (c x^2+b x+a\right )^pdx}{e}+\frac {g \int (d+e x)^{-2 p} \left (c x^2+b x+a\right )^pdx}{e}\)

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

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

Input:

Int[(d + e*x)^(-1 - 2*p)*(f + g*x)*(a + b*x + c*x^2)^p,x]
 

Output:

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

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 1179

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[(a + b*x + c*x^2)^p/(e*(1 - ( 
d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) 
^p)   Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d 
- e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m 
, p}, x]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [F]

\[\int \left (e x +d \right )^{-1-2 p} \left (g x +f \right ) \left (c \,x^{2}+b x +a \right )^{p}d x\]

Input:

int((e*x+d)^(-1-2*p)*(g*x+f)*(c*x^2+b*x+a)^p,x)
 

Output:

int((e*x+d)^(-1-2*p)*(g*x+f)*(c*x^2+b*x+a)^p,x)
 

Fricas [F]

\[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (g x + f\right )} {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 1} \,d x } \] Input:

integrate((e*x+d)^(-1-2*p)*(g*x+f)*(c*x^2+b*x+a)^p,x, algorithm="fricas")
 

Output:

integral((g*x + f)*(c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 1), x)
 

Sympy [F(-1)]

Timed out. \[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(-1-2*p)*(g*x+f)*(c*x**2+b*x+a)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (g x + f\right )} {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 1} \,d x } \] Input:

integrate((e*x+d)^(-1-2*p)*(g*x+f)*(c*x^2+b*x+a)^p,x, algorithm="maxima")
 

Output:

integrate((g*x + f)*(c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 1), x)
 

Giac [F]

\[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (g x + f\right )} {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 1} \,d x } \] Input:

integrate((e*x+d)^(-1-2*p)*(g*x+f)*(c*x^2+b*x+a)^p,x, algorithm="giac")
 

Output:

integrate((g*x + f)*(c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 1), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=\int \frac {\left (f+g\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+1}} \,d x \] Input:

int(((f + g*x)*(a + b*x + c*x^2)^p)/(d + e*x)^(2*p + 1),x)
 

Output:

int(((f + g*x)*(a + b*x + c*x^2)^p)/(d + e*x)^(2*p + 1), x)
 

Reduce [F]

\[ \int (d+e x)^{-1-2 p} (f+g x) \left (a+b x+c x^2\right )^p \, dx=\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (e x +d \right )^{2 p} d +\left (e x +d \right )^{2 p} e x}d x \right ) f +\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p} x}{\left (e x +d \right )^{2 p} d +\left (e x +d \right )^{2 p} e x}d x \right ) g \] Input:

int((e*x+d)^(-1-2*p)*(g*x+f)*(c*x^2+b*x+a)^p,x)
 

Output:

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