\(\int \frac {(a+b x+c x^2)^p}{(b d+2 c d x)^2} \, dx\) [340]

Optimal result

Integrand size = 24, antiderivative size = 88 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=\frac {2 \left (\frac {1}{4} \left (4 a-\frac {b^2}{c}\right )+\frac {(b+2 c x)^2}{4 c}\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+p,\frac {1}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d^2 (b+2 c x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^2} \, dx=-\frac {2^{-1-2 p} (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{c d^2 (b+2 c x)} \] Input:

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

Output:

-((2^(-1 - 2*p)*(a + x*(b + c*x))^p*Hypergeometric2F1[-1/2, -p, 1/2, (b + 
2*c*x)^2/(b^2 - 4*a*c)])/(c*d^2*(b + 2*c*x)*((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 = 1.45, 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.

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - (a + b*x + c*x**2)**p*a + 8*int(((a + b*x + c*x**2)**p*x**2)/(2*a**2*b 
**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4*p - 4*a*b**3*c*p*x + 2 
*a*b**3*c*x - 4*a*b**2*c**2*p*x**2 + 10*a*b**2*c**2*x**2 + 16*a*b*c**3*x** 
3 + 8*a*c**4*x**4 - b**5*p*x - 5*b**4*c*p*x**2 - 8*b**3*c**2*p*x**3 - 4*b* 
*2*c**3*p*x**4),x)*a**2*b*c**3*p + 16*int(((a + b*x + c*x**2)**p*x**2)/(2* 
a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4*p - 4*a*b**3*c*p 
*x + 2*a*b**3*c*x - 4*a*b**2*c**2*p*x**2 + 10*a*b**2*c**2*x**2 + 16*a*b*c* 
*3*x**3 + 8*a*c**4*x**4 - b**5*p*x - 5*b**4*c*p*x**2 - 8*b**3*c**2*p*x**3 
- 4*b**2*c**3*p*x**4),x)*a**2*c**4*p*x - 4*int(((a + b*x + c*x**2)**p*x**2 
)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4*p - 4*a*b** 
3*c*p*x + 2*a*b**3*c*x - 4*a*b**2*c**2*p*x**2 + 10*a*b**2*c**2*x**2 + 16*a 
*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*p*x - 5*b**4*c*p*x**2 - 8*b**3*c**2*p* 
x**3 - 4*b**2*c**3*p*x**4),x)*a*b**3*c**2*p**2 - 2*int(((a + b*x + c*x**2) 
**p*x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b**4*p - 
 4*a*b**3*c*p*x + 2*a*b**3*c*x - 4*a*b**2*c**2*p*x**2 + 10*a*b**2*c**2*x** 
2 + 16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*p*x - 5*b**4*c*p*x**2 - 8*b**3 
*c**2*p*x**3 - 4*b**2*c**3*p*x**4),x)*a*b**3*c**2*p - 8*int(((a + b*x + c* 
x**2)**p*x**2)/(2*a**2*b**2*c + 8*a**2*b*c**2*x + 8*a**2*c**3*x**2 - a*b** 
4*p - 4*a*b**3*c*p*x + 2*a*b**3*c*x - 4*a*b**2*c**2*p*x**2 + 10*a*b**2*c** 
2*x**2 + 16*a*b*c**3*x**3 + 8*a*c**4*x**4 - b**5*p*x - 5*b**4*c*p*x**2 ...