\(\int \frac {1}{(a-b x)^{2/3} (a+b x)^{2/3}} \, dx\) [207]

Optimal result

Integrand size = 20, antiderivative size = 281 \[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{2/3}} \, dx=\frac {3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (1-\frac {b^2 x^2}{a^2}\right )^{2/3} \left (1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right ) \sqrt {\frac {1+\sqrt [3]{1-\frac {b^2 x^2}{a^2}}+\left (1-\frac {b^2 x^2}{a^2}\right )^{2/3}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}\right ),-7+4 \sqrt {3}\right )}{b^2 x (a-b x)^{2/3} (a+b x)^{2/3} \sqrt {-\frac {1-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}}{\left (1-\sqrt {3}-\sqrt [3]{1-\frac {b^2 x^2}{a^2}}\right )^2}}} \] Output:

3^(3/4)*(1/2*6^(1/2)-1/2*2^(1/2))*a^2*(1-b^2*x^2/a^2)^(2/3)*(1-(1-b^2*x^2/ 
a^2)^(1/3))*((1+(1-b^2*x^2/a^2)^(1/3)+(1-b^2*x^2/a^2)^(2/3))/(1-3^(1/2)-(1 
-b^2*x^2/a^2)^(1/3))^2)^(1/2)*EllipticF((1+3^(1/2)-(1-b^2*x^2/a^2)^(1/3))/ 
(1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3)),2*I-I*3^(1/2))/b^2/x/(-b*x+a)^(2/3)/(b*x 
+a)^(2/3)/(-(1-(1-b^2*x^2/a^2)^(1/3))/(1-3^(1/2)-(1-b^2*x^2/a^2)^(1/3))^2) 
^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.23 \[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{a-b x} \left (\frac {a+b x}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},\frac {a-b x}{2 a}\right )}{2^{2/3} b (a+b x)^{2/3}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.26 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {46, 234, 760}

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.

\(\displaystyle \int \frac {1}{(a-b x)^{2/3} (a+b x)^{2/3}} \, dx\)

\(\Big \downarrow \) 46

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

\(\Big \downarrow \) 234

\(\displaystyle -\frac {3 \sqrt {-b^2 x^2} \left (a^2-b^2 x^2\right )^{2/3} \int \frac {1}{\sqrt {-b^2 x^2}}d\sqrt [3]{a^2-b^2 x^2}}{2 b^2 x (a-b x)^{2/3} (a+b x)^{2/3}}\)

\(\Big \downarrow \) 760

\(\displaystyle \frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (a^2-b^2 x^2\right )^{2/3} \left (a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right ) \sqrt {\frac {a^{4/3}+\left (a^2-b^2 x^2\right )^{2/3}+a^{2/3} \sqrt [3]{a^2-b^2 x^2}}{\left (\left (1-\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}}{\left (1-\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}}\right ),-7+4 \sqrt {3}\right )}{b^2 x (a-b x)^{2/3} (a+b x)^{2/3} \sqrt {-\frac {a^{2/3} \left (a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right )}{\left (\left (1-\sqrt {3}\right ) a^{2/3}-\sqrt [3]{a^2-b^2 x^2}\right )^2}}}\)

Input:

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

Output:

(3^(3/4)*Sqrt[2 - Sqrt[3]]*(a^2 - b^2*x^2)^(2/3)*(a^(2/3) - (a^2 - b^2*x^2 
)^(1/3))*Sqrt[(a^(4/3) + a^(2/3)*(a^2 - b^2*x^2)^(1/3) + (a^2 - b^2*x^2)^( 
2/3))/((1 - Sqrt[3])*a^(2/3) - (a^2 - b^2*x^2)^(1/3))^2]*EllipticF[ArcSin[ 
((1 + Sqrt[3])*a^(2/3) - (a^2 - b^2*x^2)^(1/3))/((1 - Sqrt[3])*a^(2/3) - ( 
a^2 - b^2*x^2)^(1/3))], -7 + 4*Sqrt[3]])/(b^2*x*(a - b*x)^(2/3)*(a + b*x)^ 
(2/3)*Sqrt[-((a^(2/3)*(a^(2/3) - (a^2 - b^2*x^2)^(1/3)))/((1 - Sqrt[3])*a^ 
(2/3) - (a^2 - b^2*x^2)^(1/3))^2)])
 

Defintions of rubi rules used

rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(a 
 + b*x)^FracPart[m]*((c + d*x)^FracPart[m]/(a*c + b*d*x^2)^FracPart[m])   I 
nt[(a*c + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 
 0] &&  !IntegerQ[2*m]
 

rule 234

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Simp[3*(Sqrt[b*x^2]/(2*b*x)) 
   Subst[Int[1/Sqrt[-a + x^3], x], x, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b 
}, x]
 

rule 760

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], 
s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s 
*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- 
s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) 
*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x 
] && NegQ[a]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(-b*x+a)^(2/3)/(b*x+a)^(2/3),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 1.71 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.38 \[ \int \frac {1}{(a-b x)^{2/3} (a+b x)^{2/3}} \, dx=\frac {{G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {1}{3}, \frac {5}{6}, 1 & \frac {1}{2}, \frac {2}{3}, \frac {7}{6} \\\frac {1}{6}, \frac {1}{3}, \frac {2}{3}, \frac {5}{6}, \frac {7}{6} & 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )} e^{\frac {2 i \pi }{3}}}{4 \pi \sqrt [3]{a} b \Gamma \left (\frac {2}{3}\right )} - \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{6}, 0, \frac {1}{3}, \frac {1}{2}, 1 & \\- \frac {1}{6}, \frac {1}{3} & - \frac {1}{2}, 0, \frac {1}{6}, 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{4 \pi \sqrt [3]{a} b \Gamma \left (\frac {2}{3}\right )} \] Input:

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

Output:

meijerg(((1/3, 5/6, 1), (1/2, 2/3, 7/6)), ((1/6, 1/3, 2/3, 5/6, 7/6), (0,) 
), a**2*exp_polar(-2*I*pi)/(b**2*x**2))*exp(2*I*pi/3)/(4*pi*a**(1/3)*b*gam 
ma(2/3)) - meijerg(((-1/2, -1/6, 0, 1/3, 1/2, 1), ()), ((-1/6, 1/3), (-1/2 
, 0, 1/6, 0)), a**2/(b**2*x**2))/(4*pi*a**(1/3)*b*gamma(2/3))
 

Maxima [F]

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

integrate(1/(-b*x+a)^(2/3)/(b*x+a)^(2/3),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate(1/(-b*x+a)^(2/3)/(b*x+a)^(2/3),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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