\(\int \frac {(a+b x+c x^2)^p}{\sqrt {d+e x}} \, dx\) [782]

Optimal result

Integrand size = 22, antiderivative size = 183 \[ \int \frac {\left (a+b x+c x^2\right )^p}{\sqrt {d+e x}} \, dx=\frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e} \] Output:

2*(e*x+d)^(1/2)*(c*x^2+b*x+a)^p*AppellF1(1/2,-p,-p,3/2,2*c*(e*x+d)/(2*c*d- 
(b-(-4*a*c+b^2)^(1/2))*e),2*c*(e*x+d)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e))/e/ 
((1-2*c*(e*x+d)/(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e))^p)/((1-2*c*(e*x+d)/(2*c* 
d-(b+(-4*a*c+b^2)^(1/2))*e))^p)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x+c x^2\right )^p}{\sqrt {d+e x}} \, dx=\frac {2^{1-2 p} \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{8 c d+4 \left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \sqrt {d+e x} (a+x (b+c x))^p \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{e} \] Input:

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

Output:

(2^(1 - 2*p)*Sqrt[d + e*x]*(a + x*(b + c*x))^p*AppellF1[1/2, -p, -p, 3/2, 
(2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c* 
d + (-b + Sqrt[b^2 - 4*a*c])*e)])/(e*((e*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x)) 
/(8*c*d + 4*(-b + Sqrt[b^2 - 4*a*c])*e))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2* 
c*x))/(-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e))^p)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1179, 150}

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.

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

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {2 \sqrt {d+e x} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (\frac {1}{2},-p,-p,\frac {3}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e}\)

Input:

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

Output:

(2*Sqrt[d + e*x]*(a + b*x + c*x^2)^p*AppellF1[1/2, -p, -p, 3/2, (2*c*(d + 
e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + S 
qrt[b^2 - 4*a*c])*e)])/(e*(1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4* 
a*c])*e))^p*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^p)
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 1179

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

Maple [F]

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

Input:

int((c*x^2+b*x+a)^p/(e*x+d)^(1/2),x)
 

Output:

int((c*x^2+b*x+a)^p/(e*x+d)^(1/2),x)
 

Fricas [F]

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

integrate((c*x^2+b*x+a)^p/(e*x+d)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((c*x**2+b*x+a)**p/(e*x+d)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((c*x^2+b*x+a)^p/(e*x+d)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((c*x^2+b*x+a)^p/(e*x+d)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((a + b*x + c*x^2)^p/(d + e*x)^(1/2),x)
 

Output:

int((a + b*x + c*x^2)^p/(d + e*x)^(1/2), x)
 

Reduce [F]

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

int((c*x^2+b*x+a)^p/(e*x+d)^(1/2),x)
 

Output:

(2*(sqrt(d + e*x)*(a + b*x + c*x**2)**p*b - 2*int((sqrt(d + e*x)*(a + b*x 
+ c*x**2)**p*x**2)/(2*a*b*d*e*p + a*b*d*e + 2*a*b*e**2*p*x + a*b*e**2*x + 
4*a*c*d**2*p + 4*a*c*d*e*p*x + 2*b**2*d*e*p*x + b**2*d*e*x + 2*b**2*e**2*p 
*x**2 + b**2*e**2*x**2 + 4*b*c*d**2*p*x + 6*b*c*d*e*p*x**2 + b*c*d*e*x**2 
+ 2*b*c*e**2*p*x**3 + b*c*e**2*x**3 + 4*c**2*d**2*p*x**2 + 4*c**2*d*e*p*x* 
*3),x)*b**2*c*e**2*p**2 - int((sqrt(d + e*x)*(a + b*x + c*x**2)**p*x**2)/( 
2*a*b*d*e*p + a*b*d*e + 2*a*b*e**2*p*x + a*b*e**2*x + 4*a*c*d**2*p + 4*a*c 
*d*e*p*x + 2*b**2*d*e*p*x + b**2*d*e*x + 2*b**2*e**2*p*x**2 + b**2*e**2*x* 
*2 + 4*b*c*d**2*p*x + 6*b*c*d*e*p*x**2 + b*c*d*e*x**2 + 2*b*c*e**2*p*x**3 
+ b*c*e**2*x**3 + 4*c**2*d**2*p*x**2 + 4*c**2*d*e*p*x**3),x)*b**2*c*e**2*p 
 + 2*int((sqrt(d + e*x)*(a + b*x + c*x**2)**p*x**2)/(2*a*b*d*e*p + a*b*d*e 
 + 2*a*b*e**2*p*x + a*b*e**2*x + 4*a*c*d**2*p + 4*a*c*d*e*p*x + 2*b**2*d*e 
*p*x + b**2*d*e*x + 2*b**2*e**2*p*x**2 + b**2*e**2*x**2 + 4*b*c*d**2*p*x + 
 6*b*c*d*e*p*x**2 + b*c*d*e*x**2 + 2*b*c*e**2*p*x**3 + b*c*e**2*x**3 + 4*c 
**2*d**2*p*x**2 + 4*c**2*d*e*p*x**3),x)*b*c**2*d*e*p + 8*int((sqrt(d + e*x 
)*(a + b*x + c*x**2)**p*x**2)/(2*a*b*d*e*p + a*b*d*e + 2*a*b*e**2*p*x + a* 
b*e**2*x + 4*a*c*d**2*p + 4*a*c*d*e*p*x + 2*b**2*d*e*p*x + b**2*d*e*x + 2* 
b**2*e**2*p*x**2 + b**2*e**2*x**2 + 4*b*c*d**2*p*x + 6*b*c*d*e*p*x**2 + b* 
c*d*e*x**2 + 2*b*c*e**2*p*x**3 + b*c*e**2*x**3 + 4*c**2*d**2*p*x**2 + 4*c* 
*2*d*e*p*x**3),x)*c**3*d**2*p**2 + 2*int((sqrt(d + e*x)*(a + b*x + c*x*...