\(\int \sqrt {d+e x} (a+b x+c x^2)^p \, dx\) [781]

Optimal result

Integrand size = 22, antiderivative size = 185 \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\frac {2 (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e} \] Output:

2/3*(e*x+d)^(3/2)*(c*x^2+b*x+a)^p*AppellF1(3/2,-p,-p,5/2,2*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 = 0.67 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.14 \[ \int \sqrt {d+e x} \left (a+b x+c x^2\right )^p \, dx=\frac {2^{1-2 p} \left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{8 c d+4 \left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{3/2} (a+x (b+c x))^p \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e} \] Input:

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

Output:

(2^(1 - 2*p)*(d + e*x)^(3/2)*(a + x*(b + c*x))^p*AppellF1[3/2, -p, -p, 5/2 
, (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)])/(3*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 = 185, 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.

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

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {2 (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (\frac {3}{2},-p,-p,\frac {5}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{3 e}\)

Input:

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

Output:

(2*(d + e*x)^(3/2)*(a + b*x + c*x^2)^p*AppellF1[3/2, -p, -p, 5/2, (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)])/(3*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*(4*sqrt(d + e*x)*(a + b*x + c*x**2)**p*a*e*p + 2*sqrt(d + e*x)*(a + b*x 
 + c*x**2)**p*b*d*p + sqrt(d + e*x)*(a + b*x + c*x**2)**p*b*d + 2*sqrt(d + 
 e*x)*(a + b*x + c*x**2)**p*b*e*p*x + sqrt(d + e*x)*(a + b*x + c*x**2)**p* 
b*e*x + 4*sqrt(d + e*x)*(a + b*x + c*x**2)**p*c*d*p*x - 64*int((sqrt(d + e 
*x)*(a + b*x + c*x**2)**p*x**2)/(8*a*b*d*e*p**2 + 10*a*b*d*e*p + 3*a*b*d*e 
 + 8*a*b*e**2*p**2*x + 10*a*b*e**2*p*x + 3*a*b*e**2*x + 16*a*c*d**2*p**2 + 
 12*a*c*d**2*p + 16*a*c*d*e*p**2*x + 12*a*c*d*e*p*x + 8*b**2*d*e*p**2*x + 
10*b**2*d*e*p*x + 3*b**2*d*e*x + 8*b**2*e**2*p**2*x**2 + 10*b**2*e**2*p*x* 
*2 + 3*b**2*e**2*x**2 + 16*b*c*d**2*p**2*x + 12*b*c*d**2*p*x + 24*b*c*d*e* 
p**2*x**2 + 22*b*c*d*e*p*x**2 + 3*b*c*d*e*x**2 + 8*b*c*e**2*p**2*x**3 + 10 
*b*c*e**2*p*x**3 + 3*b*c*e**2*x**3 + 16*c**2*d**2*p**2*x**2 + 12*c**2*d**2 
*p*x**2 + 16*c**2*d*e*p**2*x**3 + 12*c**2*d*e*p*x**3),x)*a*b*c*e**3*p**4 - 
 96*int((sqrt(d + e*x)*(a + b*x + c*x**2)**p*x**2)/(8*a*b*d*e*p**2 + 10*a* 
b*d*e*p + 3*a*b*d*e + 8*a*b*e**2*p**2*x + 10*a*b*e**2*p*x + 3*a*b*e**2*x + 
 16*a*c*d**2*p**2 + 12*a*c*d**2*p + 16*a*c*d*e*p**2*x + 12*a*c*d*e*p*x + 8 
*b**2*d*e*p**2*x + 10*b**2*d*e*p*x + 3*b**2*d*e*x + 8*b**2*e**2*p**2*x**2 
+ 10*b**2*e**2*p*x**2 + 3*b**2*e**2*x**2 + 16*b*c*d**2*p**2*x + 12*b*c*d** 
2*p*x + 24*b*c*d*e*p**2*x**2 + 22*b*c*d*e*p*x**2 + 3*b*c*d*e*x**2 + 8*b*c* 
e**2*p**2*x**3 + 10*b*c*e**2*p*x**3 + 3*b*c*e**2*x**3 + 16*c**2*d**2*p**2* 
x**2 + 12*c**2*d**2*p*x**2 + 16*c**2*d*e*p**2*x**3 + 12*c**2*d*e*p*x**3...