\(\int (d+e x)^{-5-2 p} (a+b x+c x^2)^p \, dx\) [791]

Optimal result

Integrand size = 24, antiderivative size = 577 \[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {e (2 c d-b e) (3+p) (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right )^2 (2+p) (3+2 p)}-\frac {e \left (b^2 e^2 \left (6+5 p+p^2\right )+2 c^2 d^2 \left (9+8 p+2 p^2\right )-2 c e \left (a e (3+2 p)+b d \left (9+8 p+2 p^2\right )\right )\right ) (d+e x)^{-2 (1+p)} \left (a+b x+c x^2\right )^{1+p}}{4 \left (c d^2-b d e+a e^2\right )^3 (1+p) (2+p) (3+2 p)}-\frac {e (d+e x)^{-2 (2+p)} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right ) (2+p)}+\frac {(2 c d-b e) \left (b^2 e^2 (3+p)+2 c^2 d^2 (3+2 p)-2 c e (3 a e+b d (3+2 p))\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{4 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right )^3 (1+2 p) (3+2 p)} \] Output:

-1/2*e*(-b*e+2*c*d)*(3+p)*(e*x+d)^(-3-2*p)*(c*x^2+b*x+a)^(p+1)/(a*e^2-b*d* 
e+c*d^2)^2/(2+p)/(3+2*p)-1/4*e*(b^2*e^2*(p^2+5*p+6)+2*c^2*d^2*(2*p^2+8*p+9 
)-2*c*e*(a*e*(3+2*p)+b*d*(2*p^2+8*p+9)))*(c*x^2+b*x+a)^(p+1)/(a*e^2-b*d*e+ 
c*d^2)^3/(p+1)/(2+p)/(3+2*p)/((e*x+d)^(2*p+2))-1/2*e*(c*x^2+b*x+a)^(p+1)/( 
a*e^2-b*d*e+c*d^2)/(2+p)/((e*x+d)^(4+2*p))+1/4*(-b*e+2*c*d)*(b^2*e^2*(3+p) 
+2*c^2*d^2*(3+2*p)-2*c*e*(3*a*e+b*d*(3+2*p)))*(b-(-4*a*c+b^2)^(1/2)+2*c*x) 
*(e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p*hypergeom([-p, -1-2*p],[-2*p],-4*c*(-4*a 
*c+b^2)^(1/2)*(e*x+d)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e)/(b-(-4*a*c+b^2)^(1/ 
2)+2*c*x))/(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)/(a*e^2-b*d*e+c*d^2)^3/(1+2*p)/ 
(3+2*p)/(((2*c*d-(b-(-4*a*c+b^2)^(1/2))*e)*(2*c*x+(-4*a*c+b^2)^(1/2)+b)/(2 
*c*d-(b+(-4*a*c+b^2)^(1/2))*e)/(b-(-4*a*c+b^2)^(1/2)+2*c*x))^p)
 

Mathematica [A] (verified)

Time = 2.95 (sec) , antiderivative size = 731, normalized size of antiderivative = 1.27 \[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {(d+e x)^{-2 (2+p)} (a+x (b+c x))^p \left (2 e (a+x (b+c x))+\frac {2 c (d+e x)^2 \left (-\frac {e (a+x (b+c x))}{1+p}+\frac {(-2 c d+b e) \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )^{-1-p} (d+e x) \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+2 p)}\right )}{c d^2+e (-b d+a e)}-\frac {(-2 c d+b e) (3+p) (d+e x) \left (2 e (a+x (b+c x))+\frac {e (2 c d-b e) (2+p) (d+e x) (a+x (b+c x))}{\left (c d^2+e (-b d+a e)\right ) (1+p)}+\frac {\left (b^2 e^2 (2+p)+2 c^2 d^2 (3+2 p)-2 c e (a e+b d (3+2 p))\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )^{-1-p} (d+e x)^2 \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2+e (-b d+a e)\right ) (1+2 p)}\right )}{\left (c d^2+e (-b d+a e)\right ) (3+2 p)}\right )}{4 \left (c d^2+e (-b d+a e)\right ) (2+p)} \] Input:

Integrate[(d + e*x)^(-5 - 2*p)*(a + b*x + c*x^2)^p,x]
 

Output:

-1/4*((a + x*(b + c*x))^p*(2*e*(a + x*(b + c*x)) + (2*c*(d + e*x)^2*(-((e* 
(a + x*(b + c*x)))/(1 + p)) + ((-2*c*d + b*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c 
*x)*(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x) 
)/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))) 
^(-1 - p)*(d + e*x)*Hypergeometric2F1[-1 - 2*p, -p, -2*p, (-4*c*Sqrt[b^2 - 
 4*a*c]*(d + e*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 
4*a*c] - 2*c*x))])/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(1 + 2*p))))/(c*d 
^2 + e*(-(b*d) + a*e)) - ((-2*c*d + b*e)*(3 + p)*(d + e*x)*(2*e*(a + x*(b 
+ c*x)) + (e*(2*c*d - b*e)*(2 + p)*(d + e*x)*(a + x*(b + c*x)))/((c*d^2 + 
e*(-(b*d) + a*e))*(1 + p)) + ((b^2*e^2*(2 + p) + 2*c^2*d^2*(3 + 2*p) - 2*c 
*e*(a*e + b*d*(3 + 2*p)))*(b + Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d + (-b + 
 Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((-2*c*d + (b + Sq 
rt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)))^(-1 - p)*(d + e*x)^ 
2*Hypergeometric2F1[-1 - 2*p, -p, -2*p, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x)) 
/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))]) 
/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(c*d^2 + e*(-(b*d) + a*e))*(1 + 2*p 
))))/((c*d^2 + e*(-(b*d) + a*e))*(3 + 2*p))))/((c*d^2 + e*(-(b*d) + a*e))* 
(2 + p)*(d + e*x)^(2*(2 + p)))
 

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 609, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1167, 25, 1238, 25, 1228, 1155}

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.

\(\displaystyle \int (d+e x)^{-2 p-5} \left (a+b x+c x^2\right )^p \, dx\)

\(\Big \downarrow \) 1167

\(\displaystyle -\frac {\int -(d+e x)^{-2 (p+2)} (2 c d (p+2)-b e (p+3)-2 c e x) \left (c x^2+b x+a\right )^pdx}{2 (p+2) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+2)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int (d+e x)^{-2 (p+2)} (2 c d (p+2)-b e (p+3)-2 c e x) \left (c x^2+b x+a\right )^pdx}{2 (p+2) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+2)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 1238

\(\displaystyle \frac {-\frac {\int -(d+e x)^{-2 p-3} \left (2 c^2 \left (2 p^2+7 p+6\right ) d^2+b^2 e^2 \left (p^2+5 p+6\right )-c e \left (2 a e (2 p+3)+b d \left (4 p^2+15 p+15\right )\right )-c e (2 c d-b e) (p+3) x\right ) \left (c x^2+b x+a\right )^pdx}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac {e (p+3) (2 c d-b e) (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+2)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int (d+e x)^{-2 p-3} \left (2 c^2 \left (2 p^2+7 p+6\right ) d^2+b^2 e^2 \left (p^2+5 p+6\right )-c e \left (2 a e (2 p+3)+b d \left (4 p^2+15 p+15\right )\right )-c e (2 c d-b e) (p+3) x\right ) \left (c x^2+b x+a\right )^pdx}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac {e (p+3) (2 c d-b e) (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+2)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 1228

\(\displaystyle \frac {\frac {\frac {(p+2) (2 c d-b e) \left (-2 c e (3 a e+b d (2 p+3))+b^2 e^2 (p+3)+2 c^2 d^2 (2 p+3)\right ) \int (d+e x)^{-2 (p+1)} \left (c x^2+b x+a\right )^pdx}{2 \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1} \left (-2 c e \left (a e (2 p+3)+b d \left (2 p^2+8 p+9\right )\right )+b^2 e^2 \left (p^2+5 p+6\right )+2 c^2 d^2 \left (2 p^2+8 p+9\right )\right )}{2 (p+1) \left (a e^2-b d e+c d^2\right )}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac {e (p+3) (2 c d-b e) (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+2)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}\)

\(\Big \downarrow \) 1155

\(\displaystyle \frac {\frac {\frac {(p+2) \left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (2 c d-b e) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (-2 c e (3 a e+b d (2 p+3))+b^2 e^2 (p+3)+2 c^2 d^2 (2 p+3)\right ) \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-2 p-1,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{2 (2 p+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1} \left (-2 c e \left (a e (2 p+3)+b d \left (2 p^2+8 p+9\right )\right )+b^2 e^2 \left (p^2+5 p+6\right )+2 c^2 d^2 \left (2 p^2+8 p+9\right )\right )}{2 (p+1) \left (a e^2-b d e+c d^2\right )}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}-\frac {e (p+3) (2 c d-b e) (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^{p+1}}{(2 p+3) \left (a e^2-b d e+c d^2\right )}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+2)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+2) \left (a e^2-b d e+c d^2\right )}\)

Input:

Int[(d + e*x)^(-5 - 2*p)*(a + b*x + c*x^2)^p,x]
 

Output:

-1/2*(e*(a + b*x + c*x^2)^(1 + p))/((c*d^2 - b*d*e + a*e^2)*(2 + p)*(d + e 
*x)^(2*(2 + p))) + (-((e*(2*c*d - b*e)*(3 + p)*(d + e*x)^(-3 - 2*p)*(a + b 
*x + c*x^2)^(1 + p))/((c*d^2 - b*d*e + a*e^2)*(3 + 2*p))) + (-1/2*(e*(b^2* 
e^2*(6 + 5*p + p^2) + 2*c^2*d^2*(9 + 8*p + 2*p^2) - 2*c*e*(a*e*(3 + 2*p) + 
 b*d*(9 + 8*p + 2*p^2)))*(a + b*x + c*x^2)^(1 + p))/((c*d^2 - b*d*e + a*e^ 
2)*(1 + p)*(d + e*x)^(2*(1 + p))) + ((2*c*d - b*e)*(2 + p)*(b^2*e^2*(3 + p 
) + 2*c^2*d^2*(3 + 2*p) - 2*c*e*(3*a*e + b*d*(3 + 2*p)))*(b - Sqrt[b^2 - 4 
*a*c] + 2*c*x)*(d + e*x)^(-1 - 2*p)*(a + b*x + c*x^2)^p*Hypergeometric2F1[ 
-1 - 2*p, -p, -2*p, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((2*c*d - (b + Sqrt 
[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))])/(2*(2*c*d - (b - Sqrt 
[b^2 - 4*a*c])*e)*(c*d^2 - b*d*e + a*e^2)*(1 + 2*p)*(((2*c*d - (b - Sqrt[b 
^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((2*c*d - (b + Sqrt[b^2 - 
 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)))^p))/((c*d^2 - b*d*e + a*e^2) 
*(3 + 2*p)))/(2*(c*d^2 - b*d*e + a*e^2)*(2 + p))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 1155

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[(-(b - q + 2*c*x))*(d + e*x)^ 
(m + 1)*((a + b*x + c*x^2)^p/((m + 1)*(2*c*d - b*e + e*q)*((2*c*d - b*e + e 
*q)*((b + q + 2*c*x)/((2*c*d - b*e - e*q)*(b - q + 2*c*x))))^p))*Hypergeome 
tric2F1[m + 1, -p, m + 2, -4*c*q*((d + e*x)/((2*c*d - b*e - e*q)*(b - q + 2 
*c*x)))], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[m + 2*p + 2, 0]
 

rule 1167

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2))   Int[ 
(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, 
 x]*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[m 
, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp 
lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 0])
 

rule 1228

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + 
 b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e 
*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^ 
(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
] && EqQ[Simplify[m + 2*p + 3], 0]
 

rule 1238

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* 
x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) 
*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ 
(c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m 
+ 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && ILtQ[S 
implify[m + 2*p + 3], 0] && NeQ[m, -1]
 

Maple [F]

\[\int \left (e x +d \right )^{-5-2 p} \left (c \,x^{2}+b x +a \right )^{p}d x\]

Input:

int((e*x+d)^(-5-2*p)*(c*x^2+b*x+a)^p,x)
 

Output:

int((e*x+d)^(-5-2*p)*(c*x^2+b*x+a)^p,x)
 

Fricas [F]

\[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 5} \,d x } \] Input:

integrate((e*x+d)^(-5-2*p)*(c*x^2+b*x+a)^p,x, algorithm="fricas")
 

Output:

integral((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 5), x)
 

Sympy [F(-1)]

Timed out. \[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(-5-2*p)*(c*x**2+b*x+a)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 5} \,d x } \] Input:

integrate((e*x+d)^(-5-2*p)*(c*x^2+b*x+a)^p,x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 5), x)
 

Giac [F]

\[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 5} \,d x } \] Input:

integrate((e*x+d)^(-5-2*p)*(c*x^2+b*x+a)^p,x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 5), x)
 

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+5}} \,d x \] Input:

int((a + b*x + c*x^2)^p/(d + e*x)^(2*p + 5),x)
 

Output:

int((a + b*x + c*x^2)^p/(d + e*x)^(2*p + 5), x)
 

Reduce [F]

\[ \int (d+e x)^{-5-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (e x +d \right )^{2 p} d^{5}+5 \left (e x +d \right )^{2 p} d^{4} e x +10 \left (e x +d \right )^{2 p} d^{3} e^{2} x^{2}+10 \left (e x +d \right )^{2 p} d^{2} e^{3} x^{3}+5 \left (e x +d \right )^{2 p} d \,e^{4} x^{4}+\left (e x +d \right )^{2 p} e^{5} x^{5}}d x \] Input:

int((e*x+d)^(-5-2*p)*(c*x^2+b*x+a)^p,x)
 

Output:

int((a + b*x + c*x**2)**p/((d + e*x)**(2*p)*d**5 + 5*(d + e*x)**(2*p)*d**4 
*e*x + 10*(d + e*x)**(2*p)*d**3*e**2*x**2 + 10*(d + e*x)**(2*p)*d**2*e**3* 
x**3 + 5*(d + e*x)**(2*p)*d*e**4*x**4 + (d + e*x)**(2*p)*e**5*x**5),x)