\(\int \frac {1}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx\) [228]

Optimal result

Integrand size = 28, antiderivative size = 228 \[ \int \frac {1}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=-\frac {2 \sqrt {2+\sqrt {3}} \left (a^{2/3}-b^{2/3} x\right ) \sqrt {\frac {a^{4/3}+a^{2/3} b^{2/3} x+b^{4/3} x^2}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) a^{2/3}-b^{2/3} x}{\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} b^{2/3} \sqrt {\frac {a^{2/3} \left (a^{2/3}-b^{2/3} x\right )}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \] Output:

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

Mathematica [C] (verified)

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

Time = 4.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.29 \[ \int \frac {1}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=\frac {x \sqrt {1-\frac {b^2 x^3}{a^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\frac {b^2 x^3}{a^2}\right )}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {785, 759}

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.

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

\(\Big \downarrow \) 785

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

\(\Big \downarrow \) 759

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 759

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] & 
& PosQ[a]
 

rule 785

Int[((a1_.) + (b1_.)*(x_)^(n_))^(p_)*((a2_.) + (b2_.)*(x_)^(n_))^(p_), x_Sy 
mbol] :> Simp[(a1 + b1*x^n)^FracPart[p]*((a2 + b2*x^n)^FracPart[p]/(a1*a2 + 
 b1*b2*x^(2*n))^FracPart[p])   Int[(a1*a2 + b1*b2*x^(2*n))^p, x], x] /; Fre 
eQ[{a1, b1, a2, b2, n, p}, x] && EqQ[a2*b1 + a1*b2, 0] &&  !IntegerQ[p]
 

Maple [F]

\[\int \frac {1}{\sqrt {a -b \,x^{\frac {3}{2}}}\, \sqrt {a +b \,x^{\frac {3}{2}}}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.10 \[ \int \frac {1}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=-\frac {2 \, \sqrt {-b^{2}} {\rm weierstrassPInverse}\left (0, \frac {4 \, a^{2}}{b^{2}}, x\right )}{b^{2}} \] Input:

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

Output:

-2*sqrt(-b^2)*weierstrassPInverse(0, 4*a^2/b^2, x)/b^2
 

Sympy [A] (verification not implemented)

Time = 3.37 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=\frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {5}{12}, \frac {11}{12}, 1 & \frac {2}{3}, \frac {2}{3}, \frac {7}{6} \\\frac {1}{6}, \frac {5}{12}, \frac {2}{3}, \frac {11}{12}, \frac {7}{6} & 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{3}}} \right )}}{6 \pi ^{\frac {3}{2}} \sqrt [3]{a} b^{\frac {2}{3}}} + \frac {{G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{12}, \frac {1}{6}, \frac {5}{12}, \frac {2}{3}, 1 & \\- \frac {1}{12}, \frac {5}{12} & - \frac {1}{3}, \frac {1}{6}, \frac {1}{6}, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{3}}} \right )} e^{\frac {i \pi }{3}}}{6 \pi ^{\frac {3}{2}} \sqrt [3]{a} b^{\frac {2}{3}}} \] Input:

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

Output:

I*meijerg(((5/12, 11/12, 1), (2/3, 2/3, 7/6)), ((1/6, 5/12, 2/3, 11/12, 7/ 
6), (0,)), a**2/(b**2*x**3))/(6*pi**(3/2)*a**(1/3)*b**(2/3)) + meijerg(((- 
1/3, -1/12, 1/6, 5/12, 2/3, 1), ()), ((-1/12, 5/12), (-1/3, 1/6, 1/6, 0)), 
 a**2*exp_polar(-2*I*pi)/(b**2*x**3))*exp(I*pi/3)/(6*pi**(3/2)*a**(1/3)*b* 
*(2/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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