\(\int \frac {(a+b x+c x^2)^p}{b d+2 c d x} \, dx\) [335]

Optimal result

Integrand size = 24, antiderivative size = 63 \[ \int \frac {\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx=\frac {\left (a+b x+c x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1-\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{\left (b^2-4 a c\right ) d (1+p)} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx=\frac {(a+x (b+c x))^{1+p} \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,\frac {4 c (a+x (b+c x))}{-b^2+4 a c}\right )}{\left (b^2-4 a c\right ) d (1+p)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.54, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1118, 243, 75}

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.

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

\(\Big \downarrow \) 1118

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b x+c x^2\right )^p}{b d+2 c d x} \, dx=\frac {\left (c \,x^{2}+b x +a \right )^{p} a -4 \left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p} x^{2}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) a \,c^{2} p +\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p} x^{2}}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) b^{2} c p -4 \left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p} x}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) a b c p +\left (\int \frac {\left (c \,x^{2}+b x +a \right )^{p} x}{2 c^{2} x^{3}+3 b c \,x^{2}+2 a c x +b^{2} x +a b}d x \right ) b^{3} p}{b^{2} d p} \] Input:

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

Output:

((a + b*x + c*x**2)**p*a - 4*int(((a + b*x + c*x**2)**p*x**2)/(a*b + 2*a*c 
*x + b**2*x + 3*b*c*x**2 + 2*c**2*x**3),x)*a*c**2*p + int(((a + b*x + c*x* 
*2)**p*x**2)/(a*b + 2*a*c*x + b**2*x + 3*b*c*x**2 + 2*c**2*x**3),x)*b**2*c 
*p - 4*int(((a + b*x + c*x**2)**p*x)/(a*b + 2*a*c*x + b**2*x + 3*b*c*x**2 
+ 2*c**2*x**3),x)*a*b*c*p + int(((a + b*x + c*x**2)**p*x)/(a*b + 2*a*c*x + 
 b**2*x + 3*b*c*x**2 + 2*c**2*x**3),x)*b**3*p)/(b**2*d*p)