\(\int (d+e x)^p (c^2 d^2-b c d e+b^2 e^2+3 b c e^2 x+3 c^2 e^2 x^2)^p \, dx\) [798]

Optimal result

Integrand size = 50, antiderivative size = 140 \[ \int (d+e x)^p \left (c^2 d^2-b c d e+b^2 e^2+3 b c e^2 x+3 c^2 e^2 x^2\right )^p \, dx=\frac {(d+e x)^p (c d+b e+3 c e x) \left (c^2 d^2-b c d e+b^2 e^2+3 b c e^2 x+3 c^2 e^2 x^2\right )^p \left (1+\frac {(c d+b e+3 c e x)^3}{(2 c d-b e)^3}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},-p,\frac {4}{3},-\frac {(c d+b e+3 c e x)^3}{(2 c d-b e)^3}\right )}{3 c e} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.98 (sec) , antiderivative size = 372, normalized size of antiderivative = 2.66 \[ \int (d+e x)^p \left (c^2 d^2-b c d e+b^2 e^2+3 b c e^2 x+3 c^2 e^2 x^2\right )^p \, dx=\frac {(d+e x)^{1+p} \left (\frac {-3 b c e^2+\sqrt {3} \sqrt {-c^2 e^2 (-2 c d+b e)^2}-6 c^2 e^2 x}{6 c^2 d e-3 b c e^2+\sqrt {3} \sqrt {-c^2 e^2 (-2 c d+b e)^2}}\right )^{-p} \left (\frac {3 b c e^2+\sqrt {3} \sqrt {-c^2 e^2 (-2 c d+b e)^2}+6 c^2 e^2 x}{-6 c^2 d e+3 b c e^2+\sqrt {3} \sqrt {-c^2 e^2 (-2 c d+b e)^2}}\right )^{-p} \left (b^2 e^2+b c e (-d+3 e x)+c^2 \left (d^2+3 e^2 x^2\right )\right )^p \operatorname {AppellF1}\left (1+p,-p,-p,2+p,\frac {6 c^2 e (d+e x)}{6 c^2 d e-3 b c e^2-\sqrt {3} \sqrt {-c^2 e^2 (-2 c d+b e)^2}},\frac {6 c^2 e (d+e x)}{6 c^2 d e-3 b c e^2+\sqrt {3} \sqrt {-c^2 e^2 (-2 c d+b e)^2}}\right )}{e (1+p)} \] Input:

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

Output:

((d + e*x)^(1 + p)*(b^2*e^2 + b*c*e*(-d + 3*e*x) + c^2*(d^2 + 3*e^2*x^2))^ 
p*AppellF1[1 + p, -p, -p, 2 + p, (6*c^2*e*(d + e*x))/(6*c^2*d*e - 3*b*c*e^ 
2 - Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)]), (6*c^2*e*(d + e*x))/(6*c^2 
*d*e - 3*b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)])])/(e*(1 + p) 
*((-3*b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)] - 6*c^2*e^2*x)/( 
6*c^2*d*e - 3*b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)]))^p*((3* 
b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)] + 6*c^2*e^2*x)/(-6*c^2 
*d*e + 3*b*c*e^2 + Sqrt[3]*Sqrt[-(c^2*e^2*(-2*c*d + b*e)^2)]))^p)
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.46 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.82, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, 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.

\(\displaystyle \int (d+e x)^p \left (b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2\right )^p \, dx\)

\(\Big \downarrow \) 1179

\(\displaystyle \frac {\left (b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2\right )^p \left (1-\frac {6 c^2 (d+e x)}{\left (3 c-\sqrt {3} \sqrt {-c^2}\right ) (2 c d-b e)}\right )^{-p} \left (1-\frac {6 c^2 (d+e x)}{\left (\sqrt {3} \sqrt {-c^2}+3 c\right ) (2 c d-b e)}\right )^{-p} \int (d+e x)^p \left (1-\frac {6 c^2 (d+e x)}{\left (3 c-\sqrt {3} \sqrt {-c^2}\right ) (2 c d-b e)}\right )^p \left (1-\frac {6 c^2 (d+e x)}{\left (3 c+\sqrt {3} \sqrt {-c^2}\right ) (2 c d-b e)}\right )^pd(d+e x)}{e}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {(d+e x)^{p+1} \left (b^2 e^2-b c d e+3 b c e^2 x+c^2 d^2+3 c^2 e^2 x^2\right )^p \left (1-\frac {6 c^2 (d+e x)}{\left (3 c-\sqrt {3} \sqrt {-c^2}\right ) (2 c d-b e)}\right )^{-p} \left (1-\frac {6 c^2 (d+e x)}{\left (\sqrt {3} \sqrt {-c^2}+3 c\right ) (2 c d-b e)}\right )^{-p} \operatorname {AppellF1}\left (p+1,-p,-p,p+2,\frac {6 c^2 (d+e x)}{\left (3 c-\sqrt {3} \sqrt {-c^2}\right ) (2 c d-b e)},\frac {6 c^2 (d+e x)}{\left (3 c+\sqrt {3} \sqrt {-c^2}\right ) (2 c d-b e)}\right )}{e (p+1)}\)

Input:

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

Output:

((d + e*x)^(1 + p)*(c^2*d^2 - b*c*d*e + b^2*e^2 + 3*b*c*e^2*x + 3*c^2*e^2* 
x^2)^p*AppellF1[1 + p, -p, -p, 2 + p, (6*c^2*(d + e*x))/((3*c - Sqrt[3]*Sq 
rt[-c^2])*(2*c*d - b*e)), (6*c^2*(d + e*x))/((3*c + Sqrt[3]*Sqrt[-c^2])*(2 
*c*d - b*e))])/(e*(1 + p)*(1 - (6*c^2*(d + e*x))/((3*c - Sqrt[3]*Sqrt[-c^2 
])*(2*c*d - b*e)))^p*(1 - (6*c^2*(d + e*x))/((3*c + Sqrt[3]*Sqrt[-c^2])*(2 
*c*d - b*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(3*(d + e*x)**p*(b**2*e**2 - b*c*d*e + 3*b*c*e**2*x + c**2*d**2 + 3*c**2*e 
**2*x**2)**p*b**2*d*e**2 + (d + e*x)**p*(b**2*e**2 - b*c*d*e + 3*b*c*e**2* 
x + c**2*d**2 + 3*c**2*e**2*x**2)**p*b**2*e**3*x - 3*(d + e*x)**p*(b**2*e* 
*2 - b*c*d*e + 3*b*c*e**2*x + c**2*d**2 + 3*c**2*e**2*x**2)**p*b*c*d**2*e 
+ 2*(d + e*x)**p*(b**2*e**2 - b*c*d*e + 3*b*c*e**2*x + c**2*d**2 + 3*c**2* 
e**2*x**2)**p*b*c*d*e**2*x + 3*(d + e*x)**p*(b**2*e**2 - b*c*d*e + 3*b*c*e 
**2*x + c**2*d**2 + 3*c**2*e**2*x**2)**p*c**2*d**3 + (d + e*x)**p*(b**2*e* 
*2 - b*c*d*e + 3*b*c*e**2*x + c**2*d**2 + 3*c**2*e**2*x**2)**p*c**2*d**2*e 
*x + 9*int(((d + e*x)**p*(b**2*e**2 - b*c*d*e + 3*b*c*e**2*x + c**2*d**2 + 
 3*c**2*e**2*x**2)**p*x**2)/(3*b**4*d*e**4*p + b**4*d*e**4 + 3*b**4*e**5*p 
*x + b**4*e**5*x + 3*b**3*c*d**2*e**3*p + b**3*c*d**2*e**3 + 12*b**3*c*d*e 
**4*p*x + 4*b**3*c*d*e**4*x + 9*b**3*c*e**5*p*x**2 + 3*b**3*c*e**5*x**2 + 
18*b**2*c**2*d**2*e**3*p*x + 6*b**2*c**2*d**2*e**3*x + 27*b**2*c**2*d*e**4 
*p*x**2 + 9*b**2*c**2*d*e**4*x**2 + 9*b**2*c**2*e**5*p*x**3 + 3*b**2*c**2* 
e**5*x**3 + 3*b*c**3*d**4*e*p + b*c**3*d**4*e + 12*b*c**3*d**3*e**2*p*x + 
4*b*c**3*d**3*e**2*x + 27*b*c**3*d**2*e**3*p*x**2 + 9*b*c**3*d**2*e**3*x** 
2 + 18*b*c**3*d*e**4*p*x**3 + 6*b*c**3*d*e**4*x**3 + 3*c**4*d**5*p + c**4* 
d**5 + 3*c**4*d**4*e*p*x + c**4*d**4*e*x + 9*c**4*d**3*e**2*p*x**2 + 3*c** 
4*d**3*e**2*x**2 + 9*c**4*d**2*e**3*p*x**3 + 3*c**4*d**2*e**3*x**3),x)*b** 
5*c*e**8*p**2 + 3*int(((d + e*x)**p*(b**2*e**2 - b*c*d*e + 3*b*c*e**2*x...