\(\int \frac {1}{(b e-c e x)^{2/3} (b^2+b c x+c^2 x^2)^{2/3}} \, dx\) [721]

Optimal result

Integrand size = 33, antiderivative size = 71 \[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=\frac {x \left (1-\frac {c^3 x^3}{b^3}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {c^3 x^3}{b^3}\right )}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(258\) vs. \(2(71)=142\).

Time = 10.19 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.63 \[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=-\frac {3 \sqrt [3]{e (b-c x)} \left (b-\sqrt {3} \sqrt {-b^2}+2 c x\right ) \left (\frac {3 b^2+\sqrt {3} b \sqrt {-b^2}+3 b c x-\sqrt {3} \sqrt {-b^2} c x}{3 b^2-\sqrt {3} b \sqrt {-b^2}+3 b c x+\sqrt {3} \sqrt {-b^2} c x}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {4 \sqrt {3} \sqrt {-b^2} (b-c x)}{\left (3 b+\sqrt {3} \sqrt {-b^2}\right ) \left (-b+\sqrt {3} \sqrt {-b^2}-2 c x\right )}\right )}{\left (3 b-\sqrt {3} \sqrt {-b^2}\right ) c e \left (b^2+b c x+c^2 x^2\right )^{2/3}} \] Input:

Integrate[1/((b*e - c*e*x)^(2/3)*(b^2 + b*c*x + c^2*x^2)^(2/3)),x]
 

Output:

(-3*(e*(b - c*x))^(1/3)*(b - Sqrt[3]*Sqrt[-b^2] + 2*c*x)*((3*b^2 + Sqrt[3] 
*b*Sqrt[-b^2] + 3*b*c*x - Sqrt[3]*Sqrt[-b^2]*c*x)/(3*b^2 - Sqrt[3]*b*Sqrt[ 
-b^2] + 3*b*c*x + Sqrt[3]*Sqrt[-b^2]*c*x))^(2/3)*Hypergeometric2F1[1/3, 2/ 
3, 4/3, (4*Sqrt[3]*Sqrt[-b^2]*(b - c*x))/((3*b + Sqrt[3]*Sqrt[-b^2])*(-b + 
 Sqrt[3]*Sqrt[-b^2] - 2*c*x))])/((3*b - Sqrt[3]*Sqrt[-b^2])*c*e*(b^2 + b*c 
*x + c^2*x^2)^(2/3))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1151, 779, 778}

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.

\(\displaystyle \int \frac {1}{\left (b^2+b c x+c^2 x^2\right )^{2/3} (b e-c e x)^{2/3}} \, dx\)

\(\Big \downarrow \) 1151

\(\displaystyle \frac {\left (b^3 e-c^3 e x^3\right )^{2/3} \int \frac {1}{\left (b^3 e-c^3 e x^3\right )^{2/3}}dx}{\left (b^2+b c x+c^2 x^2\right )^{2/3} (b e-c e x)^{2/3}}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {\left (1-\frac {c^3 x^3}{b^3}\right )^{2/3} \int \frac {1}{\left (1-\frac {c^3 x^3}{b^3}\right )^{2/3}}dx}{\left (b^2+b c x+c^2 x^2\right )^{2/3} (b e-c e x)^{2/3}}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {x \left (1-\frac {c^3 x^3}{b^3}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {c^3 x^3}{b^3}\right )}{\left (b^2+b c x+c^2 x^2\right )^{2/3} (b e-c e x)^{2/3}}\)

Input:

Int[1/((b*e - c*e*x)^(2/3)*(b^2 + b*c*x + c^2*x^2)^(2/3)),x]
 

Output:

(x*(1 - (c^3*x^3)/b^3)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, (c^3*x^3)/b^ 
3])/((b*e - c*e*x)^(2/3)*(b^2 + b*c*x + c^2*x^2)^(2/3))
 

Defintions of rubi rules used

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x 
^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(1 + b*(x^n/a))^p, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1151

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Sy 
mbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + c*x^2)^FracPart[p]/(a*d + c 
*e*x^3)^FracPart[p])   Int[(d + e*x)^(m - p)*(a*d + c*e*x^3)^p, x], x] /; F 
reeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*d + a*e, 0] && EqQ[c*d + b*e, 0] && 
 IGtQ[m - p + 1, 0] &&  !IntegerQ[p]
 

Maple [F]

\[\int \frac {1}{\left (-c e x +b e \right )^{\frac {2}{3}} \left (c^{2} x^{2}+c b x +b^{2}\right )^{\frac {2}{3}}}d x\]

Input:

int(1/(-c*e*x+b*e)^(2/3)/(c^2*x^2+b*c*x+b^2)^(2/3),x)
 

Output:

int(1/(-c*e*x+b*e)^(2/3)/(c^2*x^2+b*c*x+b^2)^(2/3),x)
 

Fricas [F]

\[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (c^{2} x^{2} + b c x + b^{2}\right )}^{\frac {2}{3}} {\left (-c e x + b e\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate(1/(-c*e*x+b*e)^(2/3)/(c^2*x^2+b*c*x+b^2)^(2/3),x, algorithm="fri 
cas")
 

Output:

integral(-(c^2*x^2 + b*c*x + b^2)^(1/3)*(-c*e*x + b*e)^(1/3)/(c^3*e*x^3 - 
b^3*e), x)
 

Sympy [F]

\[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=\int \frac {1}{\left (- e \left (- b + c x\right )\right )^{\frac {2}{3}} \left (b^{2} + b c x + c^{2} x^{2}\right )^{\frac {2}{3}}}\, dx \] Input:

integrate(1/(-c*e*x+b*e)**(2/3)/(c**2*x**2+b*c*x+b**2)**(2/3),x)
 

Output:

Integral(1/((-e*(-b + c*x))**(2/3)*(b**2 + b*c*x + c**2*x**2)**(2/3)), x)
 

Maxima [F]

\[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (c^{2} x^{2} + b c x + b^{2}\right )}^{\frac {2}{3}} {\left (-c e x + b e\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate(1/(-c*e*x+b*e)^(2/3)/(c^2*x^2+b*c*x+b^2)^(2/3),x, algorithm="max 
ima")
 

Output:

integrate(1/((c^2*x^2 + b*c*x + b^2)^(2/3)*(-c*e*x + b*e)^(2/3)), x)
 

Giac [F]

\[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=\int { \frac {1}{{\left (c^{2} x^{2} + b c x + b^{2}\right )}^{\frac {2}{3}} {\left (-c e x + b e\right )}^{\frac {2}{3}}} \,d x } \] Input:

integrate(1/(-c*e*x+b*e)^(2/3)/(c^2*x^2+b*c*x+b^2)^(2/3),x, algorithm="gia 
c")
 

Output:

integrate(1/((c^2*x^2 + b*c*x + b^2)^(2/3)*(-c*e*x + b*e)^(2/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=\int \frac {1}{{\left (b\,e-c\,e\,x\right )}^{2/3}\,{\left (b^2+b\,c\,x+c^2\,x^2\right )}^{2/3}} \,d x \] Input:

int(1/((b*e - c*e*x)^(2/3)*(b^2 + c^2*x^2 + b*c*x)^(2/3)),x)
 

Output:

int(1/((b*e - c*e*x)^(2/3)*(b^2 + c^2*x^2 + b*c*x)^(2/3)), x)
 

Reduce [F]

\[ \int \frac {1}{(b e-c e x)^{2/3} \left (b^2+b c x+c^2 x^2\right )^{2/3}} \, dx=\frac {\int \frac {1}{\left (-c x +b \right )^{\frac {2}{3}} \left (c^{2} x^{2}+b c x +b^{2}\right )^{\frac {2}{3}}}d x}{e^{\frac {2}{3}}} \] Input:

int(1/(-c*e*x+b*e)^(2/3)/(c^2*x^2+b*c*x+b^2)^(2/3),x)
 

Output:

int(1/((b - c*x)**(2/3)*(b**2 + b*c*x + c**2*x**2)**(2/3)),x)/e**(2/3)