\(\int \frac {x^3}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx\) [227]

Optimal result

Integrand size = 31, antiderivative size = 269 \[ \int \frac {x^3}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=-\frac {2 x \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}{5 b^2}-\frac {4 \sqrt {2+\sqrt {3}} a^2 \left (a^{2/3}-b^{2/3} x\right ) \sqrt {\frac {a^{4/3}+a^{2/3} b^{2/3} x+b^{4/3} x^2}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) a^{2/3}-b^{2/3} x}{\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{8/3} \sqrt {\frac {a^{2/3} \left (a^{2/3}-b^{2/3} x\right )}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \] Output:

-2/5*x*(a-b*x^(3/2))^(1/2)*(a+b*x^(3/2))^(1/2)/b^2-4/15*(1/2*6^(1/2)+1/2*2 
^(1/2))*a^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^(8/3)/(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 = 5.73 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.27 \[ \int \frac {x^3}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=\frac {x^4 \sqrt {1-\frac {b^2 x^3}{a^2}} \, _2F_1\left (\frac {1}{2},\frac {4}{3};\frac {7}{3};\frac {b^2 x^3}{a^2}\right )}{4 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \] Input:

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

Output:

(x^4*Sqrt[1 - (b^2*x^3)/a^2]*HypergeometricPFQ[{1/2, 4/3}, {7/3}, (b^2*x^3 
)/a^2])/(4*Sqrt[a - b*x^(3/2)]*Sqrt[a + b*x^(3/2)])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 269, 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 = {845, 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.

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

\(\Big \downarrow \) 845

\(\displaystyle \frac {2 a^2 \int \frac {1}{\sqrt {a-b x^{3/2}} \sqrt {b x^{3/2}+a}}dx}{5 b^2}-\frac {2 x \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}{5 b^2}\)

\(\Big \downarrow \) 785

\(\displaystyle \frac {2 a^2 \sqrt {a^2-b^2 x^3} \int \frac {1}{\sqrt {a^2-b^2 x^3}}dx}{5 b^2 \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}-\frac {2 x \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}{5 b^2}\)

\(\Big \downarrow \) 759

\(\displaystyle -\frac {4 \sqrt {2+\sqrt {3}} a^2 \left (a^{2/3}-b^{2/3} x\right ) \sqrt {\frac {a^{2/3} b^{2/3} x+a^{4/3}+b^{4/3} x^2}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) a^{2/3}-b^{2/3} x}{\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt [4]{3} b^{8/3} \sqrt {\frac {a^{2/3} \left (a^{2/3}-b^{2/3} x\right )}{\left (\left (1+\sqrt {3}\right ) a^{2/3}-b^{2/3} x\right )^2}} \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}-\frac {2 x \sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}}{5 b^2}\)

Input:

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

Output:

(-2*x*Sqrt[a - b*x^(3/2)]*Sqrt[a + b*x^(3/2)])/(5*b^2) - (4*Sqrt[2 + Sqrt[ 
3]]*a^2*(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 
]])/(5*3^(1/4)*b^(8/3)*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]
 

rule 845

Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^ 
(n_))^(p_), x_Symbol] :> Simp[c^(2*n - 1)*(c*x)^(m - 2*n + 1)*(a1 + b1*x^n) 
^(p + 1)*((a2 + b2*x^n)^(p + 1)/(b1*b2*(m + 2*n*p + 1))), x] - Simp[a1*a2*c 
^(2*n)*((m - 2*n + 1)/(b1*b2*(m + 2*n*p + 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] && Eq 
Q[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && GtQ[m, 2*n - 1] && NeQ[m + 2*n*p + 1 
, 0] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.20 \[ \int \frac {x^3}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=-\frac {2 \, {\left (\sqrt {b x^{\frac {3}{2}} + a} \sqrt {-b x^{\frac {3}{2}} + a} b^{2} x + 2 \, \sqrt {-b^{2}} a^{2} {\rm weierstrassPInverse}\left (0, \frac {4 \, a^{2}}{b^{2}}, x\right )\right )}}{5 \, b^{4}} \] Input:

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

Output:

-2/5*(sqrt(b*x^(3/2) + a)*sqrt(-b*x^(3/2) + a)*b^2*x + 2*sqrt(-b^2)*a^2*we 
ierstrassPInverse(0, 4*a^2/b^2, x))/b^4
 

Sympy [A] (verification not implemented)

Time = 5.96 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.51 \[ \int \frac {x^3}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=\frac {i a^{\frac {5}{3}} {G_{6, 6}^{5, 3}\left (\begin {matrix} - \frac {7}{12}, - \frac {1}{12}, 1 & - \frac {1}{3}, - \frac {1}{3}, \frac {1}{6} \\- \frac {5}{6}, - \frac {7}{12}, - \frac {1}{3}, - \frac {1}{12}, \frac {1}{6} & 0 \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{3}}} \right )}}{6 \pi ^{\frac {3}{2}} b^{\frac {8}{3}}} + \frac {a^{\frac {5}{3}} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {4}{3}, - \frac {13}{12}, - \frac {5}{6}, - \frac {7}{12}, - \frac {1}{3}, 1 & \\- \frac {13}{12}, - \frac {7}{12} & - \frac {4}{3}, - \frac {5}{6}, - \frac {5}{6}, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{3}}} \right )} e^{\frac {i \pi }{3}}}{6 \pi ^{\frac {3}{2}} b^{\frac {8}{3}}} \] Input:

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

Output:

I*a**(5/3)*meijerg(((-7/12, -1/12, 1), (-1/3, -1/3, 1/6)), ((-5/6, -7/12, 
-1/3, -1/12, 1/6), (0,)), a**2/(b**2*x**3))/(6*pi**(3/2)*b**(8/3)) + a**(5 
/3)*meijerg(((-4/3, -13/12, -5/6, -7/12, -1/3, 1), ()), ((-13/12, -7/12), 
(-4/3, -5/6, -5/6, 0)), a**2*exp_polar(-2*I*pi)/(b**2*x**3))*exp(I*pi/3)/( 
6*pi**(3/2)*b**(8/3))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x^3}{\sqrt {a-b x^{3/2}} \sqrt {a+b x^{3/2}}} \, dx=\frac {-\frac {2 \sqrt {\sqrt {x}\, b x +a}\, \sqrt {-\sqrt {x}\, b x +a}\, x}{5}+\frac {2 \left (\int \frac {\sqrt {\sqrt {x}\, b x +a}\, \sqrt {-\sqrt {x}\, b x +a}}{-b^{2} x^{3}+a^{2}}d x \right ) a^{2}}{5}}{b^{2}} \] Input:

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

Output:

(2*( - sqrt(sqrt(x)*b*x + a)*sqrt( - sqrt(x)*b*x + a)*x + int((sqrt(sqrt(x 
)*b*x + a)*sqrt( - sqrt(x)*b*x + a))/(a**2 - b**2*x**3),x)*a**2))/(5*b**2)