\(\int \frac {(1-\sqrt {3}) \sqrt [3]{a}-\sqrt [3]{b} x}{\sqrt {a-b x^3}} \, dx\) [84]

Optimal result

Integrand size = 37, antiderivative size = 263 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x}{\sqrt {a-b x^3}} \, dx=-\frac {2 \sqrt {a-b x^3}}{\sqrt [3]{b} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x\right )}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}+\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{\sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}-\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x\right )^2}} \sqrt {a-b x^3}} \] Output:

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

Mathematica [C] (verified)

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

Time = 10.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.34 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x}{\sqrt {a-b x^3}} \, dx=-\frac {x \sqrt {1-\frac {b x^3}{a}} \left (2 \left (-1+\sqrt {3}\right ) \sqrt [3]{a} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\frac {b x^3}{a}\right )+\sqrt [3]{b} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\frac {b x^3}{a}\right )\right )}{2 \sqrt {a-b x^3}} \] Input:

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

Output:

-1/2*(x*Sqrt[1 - (b*x^3)/a]*(2*(-1 + Sqrt[3])*a^(1/3)*Hypergeometric2F1[1/ 
3, 1/2, 4/3, (b*x^3)/a] + b^(1/3)*x*Hypergeometric2F1[1/2, 2/3, 5/3, (b*x^ 
3)/a]))/Sqrt[a - b*x^3]
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {2416}

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.

\(\displaystyle \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x}{\sqrt {a-b x^3}} \, dx\)

\(\Big \downarrow \) 2416

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2416

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = N 
umer[Simplify[(1 - Sqrt[3])*(d/c)]], s = Denom[Simplify[(1 - Sqrt[3])*(d/c) 
]]}, Simp[2*d*s^3*(Sqrt[a + b*x^3]/(a*r^2*((1 + Sqrt[3])*s + r*x))), x] - S 
imp[3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/( 
(1 + Sqrt[3])*s + r*x)^2]/(r^2*Sqrt[a + b*x^3]*Sqrt[s*((s + r*x)/((1 + Sqrt 
[3])*s + r*x)^2)]))*EllipticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3]) 
*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && Eq 
Q[b*c^3 - 2*(5 - 3*Sqrt[3])*a*d^3, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 948 vs. \(2 (196 ) = 392\).

Time = 0.78 (sec) , antiderivative size = 949, normalized size of antiderivative = 3.61

method	result	size
default	\(\text {Expression too large to display}\)	\(949\)

Input:

int(((1-3^(1/2))*a^(1/3)-b^(1/3)*x)/(-b*x^3+a)^(1/2),x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.21 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x}{\sqrt {a-b x^3}} \, dx=\frac {2 \, {\left (a^{\frac {1}{3}} \sqrt {-b} {\left (\sqrt {3} - 1\right )} {\rm weierstrassPInverse}\left (0, \frac {4 \, a}{b}, x\right ) - \sqrt {-b} b^{\frac {1}{3}} {\rm weierstrassZeta}\left (0, \frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, \frac {4 \, a}{b}, x\right )\right )\right )}}{b} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 3.02 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.49 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}-\sqrt [3]{b} x}{\sqrt {a-b x^3}} \, dx=- \frac {\sqrt [3]{b} x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{2 i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {5}{3}\right )} - \frac {\sqrt {3} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{2 i \pi }}{a}} \right )}}{3 \sqrt [6]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{2 i \pi }}{a}} \right )}}{3 \sqrt [6]{a} \Gamma \left (\frac {4}{3}\right )} \] Input:

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

Output:

-b**(1/3)*x**2*gamma(2/3)*hyper((1/2, 2/3), (5/3,), b*x**3*exp_polar(2*I*p 
i)/a)/(3*sqrt(a)*gamma(5/3)) - sqrt(3)*x*gamma(1/3)*hyper((1/3, 1/2), (4/3 
,), b*x**3*exp_polar(2*I*pi)/a)/(3*a**(1/6)*gamma(4/3)) + x*gamma(1/3)*hyp 
er((1/3, 1/2), (4/3,), b*x**3*exp_polar(2*I*pi)/a)/(3*a**(1/6)*gamma(4/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

 - a**(1/3)*sqrt(3)*int(sqrt(a - b*x**3)/(a - b*x**3),x) + a**(1/3)*int(sq 
rt(a - b*x**3)/(a - b*x**3),x) - b**(1/3)*int((sqrt(a - b*x**3)*x)/(a - b* 
x**3),x)