\(\int \frac {1+\sqrt {3}+\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a-b x^3}} \, dx\) [74]

Optimal result

Integrand size = 33, antiderivative size = 251 \[ \int \frac {1+\sqrt {3}+\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a-b x^3}} \, dx=-\frac {2 \left (\frac {b}{a}\right )^{2/3} \sqrt {-a-b x^3}}{b \left (1-\sqrt {3}+\sqrt [3]{\frac {b}{a}} x\right )}+\frac {\sqrt [4]{3} \sqrt {2+\sqrt {3}} \left (1+\sqrt [3]{\frac {b}{a}} x\right ) \sqrt {\frac {1-\sqrt [3]{\frac {b}{a}} x+\left (\frac {b}{a}\right )^{2/3} x^2}{\left (1-\sqrt {3}+\sqrt [3]{\frac {b}{a}} x\right )^2}} E\left (\arcsin \left (\frac {1+\sqrt {3}+\sqrt [3]{\frac {b}{a}} x}{1-\sqrt {3}+\sqrt [3]{\frac {b}{a}} x}\right )|-7+4 \sqrt {3}\right )}{\sqrt [3]{\frac {b}{a}} \sqrt {-\frac {1+\sqrt [3]{\frac {b}{a}} x}{\left (1-\sqrt {3}+\sqrt [3]{\frac {b}{a}} x\right )^2}} \sqrt {-a-b x^3}} \] 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 = 10.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.37 \[ \int \frac {1+\sqrt {3}+\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a-b x^3}} \, dx=\frac {x \sqrt {1+\frac {b x^3}{a}} \left (2 \left (1+\sqrt {3}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {b x^3}{a}\right )+\sqrt [3]{\frac {b}{a}} x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},-\frac {b x^3}{a}\right )\right )}{2 \sqrt {-a-b x^3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 251, 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.

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

\(\Big \downarrow \) 2418

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

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. 1012 vs. \(2 (208 ) = 416\).

Time = 0.49 (sec) , antiderivative size = 1013, normalized size of antiderivative = 4.04

method	result	size
default	\(\text {Expression too large to display}\)	\(1013\)

Input:

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

Output:

-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)^(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/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...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.23 \[ \int \frac {1+\sqrt {3}+\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a-b x^3}} \, dx=-\frac {2 \, {\left (\sqrt {-b} {\left (\sqrt {3} + 1\right )} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) - \sqrt {-b} \left (\frac {b}{a}\right )^{\frac {1}{3}} {\rm weierstrassZeta}\left (0, -\frac {4 \, a}{b}, {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right )\right )\right )}}{b} \] 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 = 1.56 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.52 \[ \int \frac {1+\sqrt {3}+\sqrt [3]{\frac {b}{a}} x}{\sqrt {-a-b x^3}} \, dx=- \frac {i x^{2} \sqrt [3]{\frac {b}{a}} \Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {5}{3}\right )} - \frac {\sqrt {3} i x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {4}{3}\right )} - \frac {i x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {1}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 \sqrt {a} \Gamma \left (\frac {4}{3}\right )} \] 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*exp_polar 
(I*pi)/a)/(3*sqrt(a)*gamma(5/3)) - sqrt(3)*I*x*gamma(1/3)*hyper((1/3, 1/2) 
, (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*sqrt(a)*gamma(4/3)) - I*x*gamma(1/3 
)*hyper((1/3, 1/2), (4/3,), b*x**3*exp_polar(I*pi)/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)/(- a - b*x^3)^(1/2),x)
 

Output:

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

Reduce [F]

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

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

Output:

( - i*(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)