\(\int \frac {(1-\sqrt {3}) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt {a+b x^3}} \, dx\) [83]

Optimal result

Integrand size = 35, antiderivative size = 256 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt {a+b x^3}} \, dx=\frac {2 \sqrt {a+b x^3}}{\sqrt [3]{b} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{\sqrt [3]{b} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \] 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 = 10.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.35 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt {a+b x^3}} \, dx=\frac {x \sqrt {1+\frac {b x^3}{a}} \left (-2 \left (-1+\sqrt {3}\right ) \sqrt [3]{a} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a}\right )+\sqrt [3]{b} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )\right )}{2 \sqrt {a+b x^3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, 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

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

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]*EllipticE 
[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. 1002 vs. \(2 (189 ) = 378\).

Time = 0.74 (sec) , antiderivative size = 1003, normalized size of antiderivative = 3.92

method	result	size
default	\(\text {Expression too large to display}\)	\(1003\)

Input:

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

Output:

-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)/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*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)...
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.19 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt {a+b x^3}} \, dx=-\frac {2 \, {\left (a^{\frac {1}{3}} \sqrt {b} {\left (\sqrt {3} - 1\right )} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) + b^{\frac {5}{6}} {\rm weierstrassZeta}\left (0, -\frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )\right )\right )}}{b} \] Input:

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

Output:

-2*(a^(1/3)*sqrt(b)*(sqrt(3) - 1)*weierstrassPInverse(0, -4*a/b, x) + b^(5 
/6)*weierstrassZeta(0, -4*a/b, weierstrassPInverse(0, -4*a/b, x)))/b
 

Sympy [A] (verification not implemented)

Time = 2.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.48 \[ \int \frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\sqrt {a+b x^3}} \, dx=\frac {\sqrt [3]{b} x^{2} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {5}{3}\right )} - \frac {\sqrt {3} x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [6]{a} \Gamma \left (\frac {4}{3}\right )} + \frac {x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt [6]{a} \Gamma \left (\frac {4}{3}\right )} \] 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(I*pi)/ 
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(I*pi)/a)/(3*a**(1/6)*gamma(4/3)) + x*gamma(1/3)*hyper((1 
/3, 1/2), (4/3,), b*x**3*exp_polar(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="ma 
xima")
 

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="gi 
ac")
 

Output:

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

Mupad [F(-1)]

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