\(\int \frac {1+\sqrt {3}-\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a+b x^3}} \, dx\) [73]

Optimal result

Integrand size = 33, antiderivative size = 256 \[ \int \frac {1+\sqrt {3}-\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a+b x^3}} \, dx=\frac {2 \left (\frac {b}{a}\right )^{2/3} \sqrt {-a+b x^3}}{b \left (1-\sqrt {3}-\sqrt [3]{\frac {b}{a}} x\right )}-\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (1-\sqrt [3]{\frac {b}{a}} x\right ) \sqrt {\frac {1+\sqrt [3]{\frac {b}{a}} x+\left (\frac {b}{a}\right )^{2/3} x^2}{\left (1-\sqrt {3}-\sqrt [3]{\frac {b}{a}} x\right )^2}} E\left (\arcsin \left (\frac {1+\sqrt {3}-\sqrt [3]{\frac {b}{a}} x}{1-\sqrt {3}-\sqrt [3]{\frac {b}{a}} x}\right )|-7+4 \sqrt {3}\right )}{\sqrt [3]{\frac {b}{a}} \sqrt {-\frac {1-\sqrt [3]{\frac {b}{a}} x}{\left (1-\sqrt {3}-\sqrt [3]{\frac {b}{a}} x\right )^2}} \sqrt {-a+b x^3}} \] Output:

2*(b/a)^(2/3)*(b*x^3-a)^(1/2)/b/(1-3^(1/2)-(b/a)^(1/3)*x)-3^(1/4)*(1/2*6^( 
1/2)+1/2*2^(1/2))*(1-(b/a)^(1/3)*x)*((1+(b/a)^(1/3)*x+(b/a)^(2/3)*x^2)/(1- 
3^(1/2)-(b/a)^(1/3)*x)^2)^(1/2)*EllipticE((1+3^(1/2)-(b/a)^(1/3)*x)/(1-3^( 
1/2)-(b/a)^(1/3)*x),2*I-I*3^(1/2))/(b/a)^(1/3)/(-(1-(b/a)^(1/3)*x)/(1-3^(1 
/2)-(b/a)^(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 = 10.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.35 \[ \int \frac {1+\sqrt {3}-\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a+b x^3}} \, dx=-\frac {x \sqrt {1-\frac {b x^3}{a}} \left (-2 \left (1+\sqrt {3}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},\frac {b x^3}{a}\right )+\sqrt [3]{\frac {b}{a}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\frac {b x^3}{a}\right )\right )}{2 \sqrt {-a+b x^3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2418}

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.

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

\(\Big \downarrow \) 2418

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2418

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 - S 
qrt[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] && NegQ[a] && 
 EqQ[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. 952 vs. \(2 (213 ) = 426\).

Time = 0.48 (sec) , antiderivative size = 953, normalized size of antiderivative = 3.72

method	result	size
default	\(\text {Expression too large to display}\)	\(953\)

Input:

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

Output:

-2/3*I*(b/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)/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)*E 
llipticF(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*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)/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*I/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)^(...
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.20 \[ \int \frac {1+\sqrt {3}-\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a+b x^3}} \, dx=\frac {2 \, {\left (\sqrt {b} {\left (\sqrt {3} + 1\right )} {\rm weierstrassPInverse}\left (0, \frac {4 \, a}{b}, x\right ) + \sqrt {b} \left (\frac {b}{a}\right )^{\frac {1}{3}} {\rm weierstrassZeta}\left (0, \frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, \frac {4 \, a}{b}, x\right )\right )\right )}}{b} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 1.94 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.45 \[ \int \frac {1+\sqrt {3}-\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a+b x^3}} \, dx=\frac {i x^{2} \sqrt [3]{\frac {b}{a}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3}}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {5}{3}\right )} - \frac {\sqrt {3} i x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3}}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {4}{3}\right )} - \frac {i x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3}}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {4}{3}\right )} \] Input:

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

Output:

I*x**2*(b/a)**(1/3)*gamma(2/3)*hyper((1/2, 2/3), (5/3,), b*x**3/a)/(3*sqrt 
(a)*gamma(5/3)) - sqrt(3)*I*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), b*x**3/ 
a)/(3*sqrt(a)*gamma(4/3)) - I*x*gamma(1/3)*hyper((1/3, 1/2), (4/3,), b*x** 
3/a)/(3*sqrt(a)*gamma(4/3))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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