\(\int (b d+2 c d x)^4 (a+b x+c x^2)^p \, dx\) [338]

Optimal result

Integrand size = 24, antiderivative size = 90 \[ \int (b d+2 c d x)^4 \left (a+b x+c x^2\right )^p \, dx=-\frac {2 d^4 (b+2 c x)^5 \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {7}{2}+p,\frac {7}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 \left (b^2-4 a c\right )} \] Output:

-2*d^4*(2*c*x+b)^5*(a-1/4*b^2/c+1/4*(2*c*x+b)^2/c)^(p+1)*hypergeom([1, 7/2 
+p],[7/2],(2*c*x+b)^2/(-4*a*c+b^2))/(-20*a*c+5*b^2)
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int (b d+2 c d x)^4 \left (a+b x+c x^2\right )^p \, dx=\frac {2^{-1-2 p} d^4 (b+2 c x)^5 (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 c} \] Input:

Integrate[(b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^p,x]
 

Output:

(2^(-1 - 2*p)*d^4*(b + 2*c*x)^5*(a + x*(b + c*x))^p*Hypergeometric2F1[5/2, 
 -p, 7/2, (b + 2*c*x)^2/(b^2 - 4*a*c)])/(5*c*((c*(a + x*(b + c*x)))/(-b^2 
+ 4*a*c))^p)
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.42, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1118, 279, 278}

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.

\(\displaystyle \int (b d+2 c d x)^4 \left (a+b x+c x^2\right )^p \, dx\)

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

\(\displaystyle \frac {(b d+2 c d x)^5 \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b d+2 c d x)^2}{4 c d^2}\right )^p \left (1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{10 c d}\)

Input:

Int[(b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^p,x]
 

Output:

((b*d + 2*c*d*x)^5*((4*a - b^2/c)/4 + (b*d + 2*c*d*x)^2/(4*c*d^2))^p*Hyper 
geometric2F1[5/2, -p, 7/2, (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2)])/(10*c*d 
*(1 - (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2))^p)
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 1118

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

Maple [F]

\[\int \left (2 c d x +b d \right )^{4} \left (c \,x^{2}+b x +a \right )^{p}d x\]

Input:

int((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^p,x)
 

Output:

int((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^p,x)
 

Fricas [F]

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

integrate((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^p,x, algorithm="fricas")
 

Output:

integral((16*c^4*d^4*x^4 + 32*b*c^3*d^4*x^3 + 24*b^2*c^2*d^4*x^2 + 8*b^3*c 
*d^4*x + b^4*d^4)*(c*x^2 + b*x + a)^p, x)
 

Sympy [F(-1)]

Timed out. \[ \int (b d+2 c d x)^4 \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:

integrate((2*c*d*x+b*d)**4*(c*x**2+b*x+a)**p,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^p,x, algorithm="maxima")
 

Output:

integrate((2*c*d*x + b*d)^4*(c*x^2 + b*x + a)^p, x)
 

Giac [F]

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

integrate((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^p,x, algorithm="giac")
 

Output:

integrate((2*c*d*x + b*d)^4*(c*x^2 + b*x + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b d+2 c d x)^4 \left (a+b x+c x^2\right )^p \, dx=\int {\left (b\,d+2\,c\,d\,x\right )}^4\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \] Input:

int((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^p,x)
 

Output:

int((b*d + 2*c*d*x)^4*(a + b*x + c*x^2)^p, x)
 

Reduce [F]

\[ \int (b d+2 c d x)^4 \left (a+b x+c x^2\right )^p \, dx=\text {too large to display} \] Input:

int((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^p,x)
 

Output:

(d**4*(96*(a + b*x + c*x**2)**p*a**3*c**2 - 48*(a + b*x + c*x**2)**p*a**2* 
b**2*c*p - 72*(a + b*x + c*x**2)**p*a**2*b**2*c - 96*(a + b*x + c*x**2)**p 
*a**2*b*c**2*p*x + 8*(a + b*x + c*x**2)**p*a*b**4*p**2 + 28*(a + b*x + c*x 
**2)**p*a*b**4*p + 18*(a + b*x + c*x**2)**p*a*b**4 + 48*(a + b*x + c*x**2) 
**p*a*b**3*c*p**2*x + 72*(a + b*x + c*x**2)**p*a*b**3*c*p*x + 96*(a + b*x 
+ c*x**2)**p*a*b**2*c**2*p**2*x**2 + 48*(a + b*x + c*x**2)**p*a*b**2*c**2* 
p*x**2 + 64*(a + b*x + c*x**2)**p*a*b*c**3*p**2*x**3 + 32*(a + b*x + c*x** 
2)**p*a*b*c**3*p*x**3 + 8*(a + b*x + c*x**2)**p*b**5*p**2*x + 28*(a + b*x 
+ c*x**2)**p*b**5*p*x + 15*(a + b*x + c*x**2)**p*b**5*x + 56*(a + b*x + c* 
x**2)**p*b**4*c*p**2*x**2 + 148*(a + b*x + c*x**2)**p*b**4*c*p*x**2 + 60*( 
a + b*x + c*x**2)**p*b**4*c*x**2 + 144*(a + b*x + c*x**2)**p*b**3*c**2*p** 
2*x**3 + 312*(a + b*x + c*x**2)**p*b**3*c**2*p*x**3 + 120*(a + b*x + c*x** 
2)**p*b**3*c**2*x**3 + 160*(a + b*x + c*x**2)**p*b**2*c**3*p**2*x**4 + 320 
*(a + b*x + c*x**2)**p*b**2*c**3*p*x**4 + 120*(a + b*x + c*x**2)**p*b**2*c 
**3*x**4 + 64*(a + b*x + c*x**2)**p*b*c**4*p**2*x**5 + 128*(a + b*x + c*x* 
*2)**p*b*c**4*p*x**5 + 48*(a + b*x + c*x**2)**p*b*c**4*x**5 - 1536*int(((a 
 + b*x + c*x**2)**p*x)/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x 
+ 36*b*p**2*x + 46*b*p*x + 15*b*x + 8*c*p**3*x**2 + 36*c*p**2*x**2 + 46*c* 
p*x**2 + 15*c*x**2),x)*a**3*c**3*p**4 - 6912*int(((a + b*x + c*x**2)**p*x) 
/(8*a*p**3 + 36*a*p**2 + 46*a*p + 15*a + 8*b*p**3*x + 36*b*p**2*x + 46*...