\(\int \frac {(a+b x+c x^2)^p}{(b d+2 c d x)^6} \, dx\) [342]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [F]
Sympy [F(-1)]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 90 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\frac {2 \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,-\frac {3}{2}+p,-\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 \left (b^2-4 a c\right ) d^6 (b+2 c x)^5} \] Output:

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

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=-\frac {2^{-1-2 p} (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-p,-\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{5 c d^6 (b+2 c x)^5} \] Input:

Integrate[(a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6,x]
 

Output:

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

\(\displaystyle \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx\)

\(\Big \downarrow \) 1118

\(\displaystyle \frac {\int \frac {\left (\frac {(b d+2 c x d)^2}{4 c d^2}+\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )\right )^p}{(b d+2 c x d)^6}d(b d+2 c x d)}{2 c d}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {\left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b d+2 c d x)^2}{4 c d^2}\right )^p \left (1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}\right )^{-p} \int \frac {\left (1-\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )^p}{(b d+2 c x d)^6}d(b d+2 c x d)}{2 c d}\)

\(\Big \downarrow \) 278

\(\displaystyle -\frac {\left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b d+2 c d x)^2}{4 c d^2}\right )^p \left (1-\frac {(b d+2 c d x)^2}{d^2 \left (b^2-4 a c\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-p,-\frac {3}{2},\frac {(b d+2 c x d)^2}{\left (b^2-4 a c\right ) d^2}\right )}{10 c d (b d+2 c d x)^5}\)

Input:

Int[(a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6,x]
 

Output:

-1/10*(((4*a - b^2/c)/4 + (b*d + 2*c*d*x)^2/(4*c*d^2))^p*Hypergeometric2F1 
[-5/2, -p, -3/2, (b*d + 2*c*d*x)^2/((b^2 - 4*a*c)*d^2)])/(c*d*(b*d + 2*c*d 
*x)^5*(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]

\[\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (2 c d x +b d \right )^{6}}d x\]

Input:

int((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x)
 

Output:

int((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x)
 

Fricas [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{6}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x, algorithm="fricas")
 

Output:

integral((c*x^2 + b*x + a)^p/(64*c^6*d^6*x^6 + 192*b*c^5*d^6*x^5 + 240*b^2 
*c^4*d^6*x^4 + 160*b^3*c^3*d^6*x^3 + 60*b^4*c^2*d^6*x^2 + 12*b^5*c*d^6*x + 
 b^6*d^6), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)**p/(2*c*d*x+b*d)**6,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{6}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x, algorithm="maxima")
 

Output:

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d)^6, x)
 

Giac [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (2 \, c d x + b d\right )}^{6}} \,d x } \] Input:

integrate((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x, algorithm="giac")
 

Output:

integrate((c*x^2 + b*x + a)^p/(2*c*d*x + b*d)^6, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (b\,d+2\,c\,d\,x\right )}^6} \,d x \] Input:

int((a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6,x)
 

Output:

int((a + b*x + c*x^2)^p/(b*d + 2*c*d*x)^6, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^6} \, dx=\text {too large to display} \] Input:

int((c*x^2+b*x+a)^p/(2*c*d*x+b*d)^6,x)
 

Output:

( - (a + b*x + c*x**2)**p*a + 40*int(((a + b*x + c*x**2)**p*x**2)/(10*a**2 
*b**6*c + 120*a**2*b**5*c**2*x + 600*a**2*b**4*c**3*x**2 + 1600*a**2*b**3* 
c**4*x**3 + 2400*a**2*b**2*c**5*x**4 + 1920*a**2*b*c**6*x**5 + 640*a**2*c* 
*7*x**6 - a*b**8*p - 12*a*b**7*c*p*x + 10*a*b**7*c*x - 60*a*b**6*c**2*p*x* 
*2 + 130*a*b**6*c**2*x**2 - 160*a*b**5*c**3*p*x**3 + 720*a*b**5*c**3*x**3 
- 240*a*b**4*c**4*p*x**4 + 2200*a*b**4*c**4*x**4 - 192*a*b**3*c**5*p*x**5 
+ 4000*a*b**3*c**5*x**5 - 64*a*b**2*c**6*p*x**6 + 4320*a*b**2*c**6*x**6 + 
2560*a*b*c**7*x**7 + 640*a*c**8*x**8 - b**9*p*x - 13*b**8*c*p*x**2 - 72*b* 
*7*c**2*p*x**3 - 220*b**6*c**3*p*x**4 - 400*b**5*c**4*p*x**5 - 432*b**4*c* 
*5*p*x**6 - 256*b**3*c**6*p*x**7 - 64*b**2*c**7*p*x**8),x)*a**2*b**5*c**3* 
p + 400*int(((a + b*x + c*x**2)**p*x**2)/(10*a**2*b**6*c + 120*a**2*b**5*c 
**2*x + 600*a**2*b**4*c**3*x**2 + 1600*a**2*b**3*c**4*x**3 + 2400*a**2*b** 
2*c**5*x**4 + 1920*a**2*b*c**6*x**5 + 640*a**2*c**7*x**6 - a*b**8*p - 12*a 
*b**7*c*p*x + 10*a*b**7*c*x - 60*a*b**6*c**2*p*x**2 + 130*a*b**6*c**2*x**2 
 - 160*a*b**5*c**3*p*x**3 + 720*a*b**5*c**3*x**3 - 240*a*b**4*c**4*p*x**4 
+ 2200*a*b**4*c**4*x**4 - 192*a*b**3*c**5*p*x**5 + 4000*a*b**3*c**5*x**5 - 
 64*a*b**2*c**6*p*x**6 + 4320*a*b**2*c**6*x**6 + 2560*a*b*c**7*x**7 + 640* 
a*c**8*x**8 - b**9*p*x - 13*b**8*c*p*x**2 - 72*b**7*c**2*p*x**3 - 220*b**6 
*c**3*p*x**4 - 400*b**5*c**4*p*x**5 - 432*b**4*c**5*p*x**6 - 256*b**3*c**6 
*p*x**7 - 64*b**2*c**7*p*x**8),x)*a**2*b**4*c**4*p*x + 1600*int(((a + b...