\(\int \frac {1}{x^3 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx\) [229]

Optimal result

Integrand size = 31, antiderivative size = 271 \[ \int \frac {1}{x^3 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=-\frac {\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}{2 a^2 x^2}-\frac {\sqrt {2+\sqrt {3}} b^{4/3} \left (a^{2/3}-b^{2/3} x\right ) \sqrt {\frac {a^{4/3}+a^{2/3} b^{2/3} x+b^{4/3} x^2}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) a^{2/3}-b^{2/3} x}{\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x}\right ),-7-4 \sqrt {3}\right )}{2 \sqrt [4]{3} a^2 \sqrt {\frac {a^{2/3} \left (a^{2/3}-b^{2/3} x\right )}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \] Output:

-1/2*(a-b*x^(3/2))^(1/2)*(a+b*x^(3/2))^(1/2)/a^2/x^2-1/6*(1/2*6^(1/2)+1/2* 
2^(1/2))*b^(4/3)*(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)/a^2/(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 = 5.65 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.27 \[ \int \frac {1}{x^3 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=-\frac {\sqrt {1-\frac {b^2 x^3}{a^2}} \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\frac {b^2 x^3}{a^2}\right )}{2 x^2 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {849, 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.

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

\(\Big \downarrow \) 849

\(\displaystyle \frac {b^2 \int \frac {1}{\sqrt {a-b x^{3/2}} \sqrt {b x^{3/2}+a}}dx}{4 a^2}-\frac {\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}{2 a^2 x^2}\)

\(\Big \downarrow \) 785

\(\displaystyle \frac {b^2 \sqrt {a^2-b^2 x^3} \int \frac {1}{\sqrt {a^2-b^2 x^3}}dx}{4 a^2 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}-\frac {\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}{2 a^2 x^2}\)

\(\Big \downarrow \) 759

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

Input:

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

Output:

-1/2*(Sqrt[a - b*x^(3/2)]*Sqrt[a + b*x^(3/2)])/(a^2*x^2) - (Sqrt[2 + Sqrt[ 
3]]*b^(4/3)*(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 - Sqr 
t[3])*a^(2/3) - b^(2/3)*x)/((1 + Sqrt[3])*a^(2/3) - b^(2/3)*x)], -7 - 4*Sq 
rt[3]])/(2*3^(1/4)*a^2*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]
 

rule 849

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^ 
(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*(a1 + b1*x^n)^(p + 1)*((a2 + b2 
*x^n)^(p + 1)/(a1*a2*c*(m + 1))), x] - Simp[b1*b2*((m + 2*n*(p + 1) + 1)/(a 
1*a2*c^(2*n)*(m + 1)))   Int[(c*x)^(m + 2*n)*(a1 + b1*x^n)^p*(a2 + b2*x^n)^ 
p, x], x] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2, 0] && I 
GtQ[2*n, 0] && LtQ[m, -1] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.19 \[ \int \frac {1}{x^3 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=-\frac {\sqrt {-b^{2}} x^{2} {\rm weierstrassPInverse}\left (0, \frac {4 \, a^{2}}{b^{2}}, x\right ) + \sqrt {b x^{\frac {3}{2}} + a} \sqrt {-b x^{\frac {3}{2}} + a}}{2 \, a^{2} x^{2}} \] Input:

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

Output:

-1/2*(sqrt(-b^2)*x^2*weierstrassPInverse(0, 4*a^2/b^2, x) + sqrt(b*x^(3/2) 
 + a)*sqrt(-b*x^(3/2) + a))/(a^2*x^2)
 

Sympy [A] (verification not implemented)

Time = 15.34 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.39 \[ \int \frac {1}{x^3 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=\frac {i b^{\frac {4}{3}} {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {17}{12}, \frac {23}{12}, 1 & \frac {5}{3}, \frac {5}{3}, \frac {13}{6} \\\frac {7}{6}, \frac {17}{12}, \frac {5}{3}, \frac {23}{12}, \frac {13}{6} & 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{3}}} \right )}}{6 \pi ^{\frac {3}{2}} a^{\frac {7}{3}}} + \frac {b^{\frac {4}{3}} {G_{6, 6}^{2, 6}\left (\begin {matrix} \frac {2}{3}, \frac {11}{12}, \frac {7}{6}, \frac {17}{12}, \frac {5}{3}, 1 & \\\frac {11}{12}, \frac {17}{12} & \frac {2}{3}, \frac {7}{6}, \frac {7}{6}, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{3}}} \right )} e^{\frac {i \pi }{3}}}{6 \pi ^{\frac {3}{2}} a^{\frac {7}{3}}} \] Input:

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

Output:

I*b**(4/3)*meijerg(((17/12, 23/12, 1), (5/3, 5/3, 13/6)), ((7/6, 17/12, 5/ 
3, 23/12, 13/6), (0,)), a**2/(b**2*x**3))/(6*pi**(3/2)*a**(7/3)) + b**(4/3 
)*meijerg(((2/3, 11/12, 7/6, 17/12, 5/3, 1), ()), ((11/12, 17/12), (2/3, 7 
/6, 7/6, 0)), a**2*exp_polar(-2*I*pi)/(b**2*x**3))*exp(I*pi/3)/(6*pi**(3/2 
)*a**(7/3))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

integrate(1/x^3/(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^3), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{x^3 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=\int \frac {\sqrt {\sqrt {x}\, b x +a}\, \sqrt {-\sqrt {x}\, b x +a}}{-b^{2} x^{6}+a^{2} x^{3}}d x \] Input:

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