\(\int \frac {(a+b x+c x^2)^p}{(b d+2 c d x)^4} \, dx\) [341]

Optimal result

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

2/3*(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^4/(2*c*x+b)^3
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^p}{(b d+2 c d x)^4} \, dx=-\frac {2^{-1-2 p} (a+x (b+c x))^p \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-p,-\frac {1}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{3 c d^4 (b+2 c x)^3} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - (a + b*x + c*x**2)**p*a + 24*int(((a + b*x + c*x**2)**p*x**2)/(6*a**2* 
b**4*c + 48*a**2*b**3*c**2*x + 144*a**2*b**2*c**3*x**2 + 192*a**2*b*c**4*x 
**3 + 96*a**2*c**5*x**4 - a*b**6*p - 8*a*b**5*c*p*x + 6*a*b**5*c*x - 24*a* 
b**4*c**2*p*x**2 + 54*a*b**4*c**2*x**2 - 32*a*b**3*c**3*p*x**3 + 192*a*b** 
3*c**3*x**3 - 16*a*b**2*c**4*p*x**4 + 336*a*b**2*c**4*x**4 + 288*a*b*c**5* 
x**5 + 96*a*c**6*x**6 - b**7*p*x - 9*b**6*c*p*x**2 - 32*b**5*c**2*p*x**3 - 
 56*b**4*c**3*p*x**4 - 48*b**3*c**4*p*x**5 - 16*b**2*c**5*p*x**6),x)*a**2* 
b**3*c**3*p + 144*int(((a + b*x + c*x**2)**p*x**2)/(6*a**2*b**4*c + 48*a** 
2*b**3*c**2*x + 144*a**2*b**2*c**3*x**2 + 192*a**2*b*c**4*x**3 + 96*a**2*c 
**5*x**4 - a*b**6*p - 8*a*b**5*c*p*x + 6*a*b**5*c*x - 24*a*b**4*c**2*p*x** 
2 + 54*a*b**4*c**2*x**2 - 32*a*b**3*c**3*p*x**3 + 192*a*b**3*c**3*x**3 - 1 
6*a*b**2*c**4*p*x**4 + 336*a*b**2*c**4*x**4 + 288*a*b*c**5*x**5 + 96*a*c** 
6*x**6 - b**7*p*x - 9*b**6*c*p*x**2 - 32*b**5*c**2*p*x**3 - 56*b**4*c**3*p 
*x**4 - 48*b**3*c**4*p*x**5 - 16*b**2*c**5*p*x**6),x)*a**2*b**2*c**4*p*x + 
 288*int(((a + b*x + c*x**2)**p*x**2)/(6*a**2*b**4*c + 48*a**2*b**3*c**2*x 
 + 144*a**2*b**2*c**3*x**2 + 192*a**2*b*c**4*x**3 + 96*a**2*c**5*x**4 - a* 
b**6*p - 8*a*b**5*c*p*x + 6*a*b**5*c*x - 24*a*b**4*c**2*p*x**2 + 54*a*b**4 
*c**2*x**2 - 32*a*b**3*c**3*p*x**3 + 192*a*b**3*c**3*x**3 - 16*a*b**2*c**4 
*p*x**4 + 336*a*b**2*c**4*x**4 + 288*a*b*c**5*x**5 + 96*a*c**6*x**6 - b**7 
*p*x - 9*b**6*c*p*x**2 - 32*b**5*c**2*p*x**3 - 56*b**4*c**3*p*x**4 - 48...