\(\int (d+e x)^m (a+b x+c x^2)^p \, dx\) [136]

Optimal result

Integrand size = 20, antiderivative size = 187 \[ \int (d+e x)^m \left (a+b x+c x^2\right )^p \, dx=\frac {(d+e x)^{1+m} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (1+m,-p,-p,2+m,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m)} \] Output:

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

Time = 0.05 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.10 \[ \int (d+e x)^m \left (a+b x+c x^2\right )^p \, dx=\frac {\left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{1+m} (a+x (b+c x))^p \operatorname {AppellF1}\left (1+m,-p,-p,2+m,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {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.

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

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

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

Input:

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

Output:

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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