\(\int (d+e x)^{3/2} (a+b x+c x^2)^p \, dx\) [780]

Optimal result

Integrand size = 22, antiderivative size = 185 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\frac {2 (d+e x)^{5/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{5 e} \] Output:

2/5*(e*x+d)^(5/2)*(c*x^2+b*x+a)^p*AppellF1(5/2,-p,-p,7/2,2*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 = 1.65 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.29 \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\frac {2 \left (\frac {e \left (-b+\sqrt {\frac {b^2-4 a c}{e^2}} e-2 c x\right )}{2 c d+e \left (-b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )}\right )^{-p} \left (\frac {e \left (b+\sqrt {\frac {b^2-4 a c}{e^2}} e+2 c x\right )}{-2 c d+e \left (b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )}\right )^{-p} (d+e x)^{5/2} (a+x (b+c x))^p \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {2 c (d+e x)}{2 c d-e \left (b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )},\frac {2 c (d+e x)}{2 c d+e \left (-b+\sqrt {\frac {b^2-4 a c}{e^2}} e\right )}\right )}{5 e} \] Input:

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

Output:

(2*(d + e*x)^(5/2)*(a + x*(b + c*x))^p*AppellF1[5/2, -p, -p, 7/2, (2*c*(d 
+ e*x))/(2*c*d - e*(b + Sqrt[(b^2 - 4*a*c)/e^2]*e)), (2*c*(d + e*x))/(2*c* 
d + e*(-b + Sqrt[(b^2 - 4*a*c)/e^2]*e))])/(5*e*((e*(-b + Sqrt[(b^2 - 4*a*c 
)/e^2]*e - 2*c*x))/(2*c*d + e*(-b + Sqrt[(b^2 - 4*a*c)/e^2]*e)))^p*((e*(b 
+ Sqrt[(b^2 - 4*a*c)/e^2]*e + 2*c*x))/(-2*c*d + e*(b + Sqrt[(b^2 - 4*a*c)/ 
e^2]*e)))^p)
 

Rubi [A] (verified)

Time = 0.35 (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.

\(\displaystyle \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx\)

\(\Big \downarrow \) 1179

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

\(\Big \downarrow \) 150

\(\displaystyle \frac {2 (d+e x)^{5/2} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (\frac {5}{2},-p,-p,\frac {7}{2},\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{5 e}\)

Input:

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

Output:

(2*(d + e*x)^(5/2)*(a + b*x + c*x^2)^p*AppellF1[5/2, -p, -p, 7/2, (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)])/(5*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]

\[\int \left (e x +d \right )^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{p}d x\]

Input:

int((e*x+d)^(3/2)*(c*x^2+b*x+a)^p,x)
 

Output:

int((e*x+d)^(3/2)*(c*x^2+b*x+a)^p,x)
 

Fricas [F]

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

integrate((e*x+d)^(3/2)*(c*x^2+b*x+a)^p,x, algorithm="fricas")
 

Output:

integral((e*x + d)^(3/2)*(c*x^2 + b*x + a)^p, x)
 

Sympy [F(-1)]

Timed out. \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(3/2)*(c*x**2+b*x+a)**p,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((e*x+d)^(3/2)*(c*x^2+b*x+a)^p,x, algorithm="maxima")
 

Output:

integrate((e*x + d)^(3/2)*(c*x^2 + b*x + a)^p, x)
 

Giac [F]

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

integrate((e*x+d)^(3/2)*(c*x^2+b*x+a)^p,x, algorithm="giac")
 

Output:

integrate((e*x + d)^(3/2)*(c*x^2 + b*x + a)^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\int {\left (d+e\,x\right )}^{3/2}\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \] Input:

int((d + e*x)^(3/2)*(a + b*x + c*x^2)^p,x)
 

Output:

int((d + e*x)^(3/2)*(a + b*x + c*x^2)^p, x)
 

Reduce [F]

\[ \int (d+e x)^{3/2} \left (a+b x+c x^2\right )^p \, dx=\text {too large to display} \] Input:

int((e*x+d)^(3/2)*(c*x^2+b*x+a)^p,x)
 

Output:

(2*( - 6*sqrt(d + e*x)*(a + b*x + c*x**2)**p*a*b*e**2*p + 32*sqrt(d + e*x) 
*(a + b*x + c*x**2)**p*a*c*d*e*p**2 + 36*sqrt(d + e*x)*(a + b*x + c*x**2)* 
*p*a*c*d*e*p - 4*sqrt(d + e*x)*(a + b*x + c*x**2)**p*b**2*d*e*p**2 - 4*sqr 
t(d + e*x)*(a + b*x + c*x**2)**p*b**2*d*e*p + 4*sqrt(d + e*x)*(a + b*x + c 
*x**2)**p*b**2*e**2*p**2*x + 2*sqrt(d + e*x)*(a + b*x + c*x**2)**p*b**2*e* 
*2*p*x + 8*sqrt(d + e*x)*(a + b*x + c*x**2)**p*b*c*d**2*p**2 + 12*sqrt(d + 
 e*x)*(a + b*x + c*x**2)**p*b*c*d**2*p + 3*sqrt(d + e*x)*(a + b*x + c*x**2 
)**p*b*c*d**2 + 16*sqrt(d + e*x)*(a + b*x + c*x**2)**p*b*c*d*e*p**2*x + 16 
*sqrt(d + e*x)*(a + b*x + c*x**2)**p*b*c*d*e*p*x + 6*sqrt(d + e*x)*(a + b* 
x + c*x**2)**p*b*c*d*e*x + 8*sqrt(d + e*x)*(a + b*x + c*x**2)**p*b*c*e**2* 
p**2*x**2 + 10*sqrt(d + e*x)*(a + b*x + c*x**2)**p*b*c*e**2*p*x**2 + 3*sqr 
t(d + e*x)*(a + b*x + c*x**2)**p*b*c*e**2*x**2 + 16*sqrt(d + e*x)*(a + b*x 
 + c*x**2)**p*c**2*d**2*p**2*x + 24*sqrt(d + e*x)*(a + b*x + c*x**2)**p*c* 
*2*d**2*p*x + 16*sqrt(d + e*x)*(a + b*x + c*x**2)**p*c**2*d*e*p**2*x**2 + 
12*sqrt(d + e*x)*(a + b*x + c*x**2)**p*c**2*d*e*p*x**2 + 512*int((sqrt(d + 
 e*x)*(a + b*x + c*x**2)**p*x**2)/(32*a*b*d*e*p**3 + 80*a*b*d*e*p**2 + 62* 
a*b*d*e*p + 15*a*b*d*e + 32*a*b*e**2*p**3*x + 80*a*b*e**2*p**2*x + 62*a*b* 
e**2*p*x + 15*a*b*e**2*x + 64*a*c*d**2*p**3 + 128*a*c*d**2*p**2 + 60*a*c*d 
**2*p + 64*a*c*d*e*p**3*x + 128*a*c*d*e*p**2*x + 60*a*c*d*e*p*x + 32*b**2* 
d*e*p**3*x + 80*b**2*d*e*p**2*x + 62*b**2*d*e*p*x + 15*b**2*d*e*x + 32*...