Integrand size = 32, antiderivative size = 40 \[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} x\right )}{\sqrt {1-x^3}}\right )}{\sqrt {3}} \] Output:
-2/3*2^(2/3)*arctan(3^(1/2)*(1-2^(1/3)*x)/(-x^3+1)^(1/2))*3^(1/2)
Time = 2.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.02 \[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {1-x^3}}{\sqrt {3} \left (-1+\sqrt [3]{2} x\right )}\right )}{\sqrt {3}} \] Input:
Integrate[(2^(2/3) + 2*x)/((2^(2/3) - x)*Sqrt[1 - x^3]),x]
Output:
(-2*2^(2/3)*ArcTan[Sqrt[1 - x^3]/(Sqrt[3]*(-1 + 2^(1/3)*x))])/Sqrt[3]
Time = 0.43 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {2562, 216}
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.
\(\displaystyle \int \frac {2 x+2^{2/3}}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx\) |
\(\Big \downarrow \) 2562 |
\(\displaystyle -2 2^{2/3} \int \frac {1}{\frac {3 \left (1-\sqrt [3]{2} x\right )^2}{1-x^3}+1}d\frac {1-\sqrt [3]{2} x}{\sqrt {1-x^3}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {2\ 2^{2/3} \arctan \left (\frac {\sqrt {3} \left (1-\sqrt [3]{2} x\right )}{\sqrt {1-x^3}}\right )}{\sqrt {3}}\) |
Input:
Int[(2^(2/3) + 2*x)/((2^(2/3) - x)*Sqrt[1 - x^3]),x]
Output:
(-2*2^(2/3)*ArcTan[(Sqrt[3]*(1 - 2^(1/3)*x))/Sqrt[1 - x^3]])/Sqrt[3]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((e_) + (f_.)*(x_))/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^3]), x_ Symbol] :> Simp[2*(e/d) Subst[Int[1/(1 + 3*a*x^2), x], x, (1 + 2*d*(x/c)) /Sqrt[a + b*x^3]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[d*e - c*f, 0] && EqQ[b*c^3 - 4*a*d^3, 0] && EqQ[2*d*e + c*f, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.38 (sec) , antiderivative size = 112, normalized size of antiderivative = 2.80
method | result | size |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+6 \,2^{\frac {1}{3}}\right ) \ln \left (\frac {12 \sqrt {-x^{3}+1}\, x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+6 \,2^{\frac {1}{3}}\right ) x^{2} 2^{\frac {2}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+6 \,2^{\frac {1}{3}}\right ) x^{3}-6 \sqrt {-x^{3}+1}\, 2^{\frac {2}{3}}+6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+6 \,2^{\frac {1}{3}}\right ) 2^{\frac {1}{3}} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+6 \,2^{\frac {1}{3}}\right )}{\left (2^{\frac {1}{3}} x -2\right )^{3}}\right )}{3}\) | \(112\) |
default | \(\frac {4 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}}+\frac {2 i 2^{\frac {2}{3}} \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-x^{3}+1}\, \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}\right )}\) | \(253\) |
elliptic | \(\frac {4 i \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{3 \sqrt {-x^{3}+1}}+\frac {2 i 2^{\frac {2}{3}} \sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \sqrt {\frac {x -1}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\, \sqrt {-i \left (x +\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}\, \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {i \left (x +\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}}{3}, \frac {i \sqrt {3}}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}}, \sqrt {\frac {i \sqrt {3}}{-\frac {3}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-x^{3}+1}\, \left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}-2^{\frac {2}{3}}\right )}\) | \(253\) |
Input:
int((2*x+2^(2/3))/(2^(2/3)-x)/(-x^3+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*RootOf(_Z^2+6*2^(1/3))*ln((12*(-x^3+1)^(1/2)*x-3*RootOf(_Z^2+6*2^(1/3) )*x^2*2^(2/3)-RootOf(_Z^2+6*2^(1/3))*x^3-6*(-x^3+1)^(1/2)*2^(2/3)+6*RootOf (_Z^2+6*2^(1/3))*2^(1/3)*x-2*RootOf(_Z^2+6*2^(1/3)))/(2^(1/3)*x-2)^3)
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (30) = 60\).
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.90 \[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=-\frac {1}{3} \, \sqrt {6} 2^{\frac {1}{6}} \arctan \left (\frac {\sqrt {6} 2^{\frac {1}{6}} {\left (2 \, x^{5} - 2 \, x^{2} + 2^{\frac {2}{3}} {\left (7 \, x^{4} - 4 \, x\right )} - 2^{\frac {1}{3}} {\left (5 \, x^{3} - 2\right )}\right )} \sqrt {-x^{3} + 1}}{12 \, {\left (2 \, x^{6} - 3 \, x^{3} + 1\right )}}\right ) \] Input:
integrate((2^(2/3)+2*x)/(2^(2/3)-x)/(-x^3+1)^(1/2),x, algorithm="fricas")
Output:
-1/3*sqrt(6)*2^(1/6)*arctan(1/12*sqrt(6)*2^(1/6)*(2*x^5 - 2*x^2 + 2^(2/3)* (7*x^4 - 4*x) - 2^(1/3)*(5*x^3 - 2))*sqrt(-x^3 + 1)/(2*x^6 - 3*x^3 + 1))
\[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=- \int \frac {2^{\frac {2}{3}}}{x \sqrt {1 - x^{3}} - 2^{\frac {2}{3}} \sqrt {1 - x^{3}}}\, dx - \int \frac {2 x}{x \sqrt {1 - x^{3}} - 2^{\frac {2}{3}} \sqrt {1 - x^{3}}}\, dx \] Input:
integrate((2**(2/3)+2*x)/(2**(2/3)-x)/(-x**3+1)**(1/2),x)
Output:
-Integral(2**(2/3)/(x*sqrt(1 - x**3) - 2**(2/3)*sqrt(1 - x**3)), x) - Inte gral(2*x/(x*sqrt(1 - x**3) - 2**(2/3)*sqrt(1 - x**3)), x)
\[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=\int { -\frac {2 \, x + 2^{\frac {2}{3}}}{\sqrt {-x^{3} + 1} {\left (x - 2^{\frac {2}{3}}\right )}} \,d x } \] Input:
integrate((2^(2/3)+2*x)/(2^(2/3)-x)/(-x^3+1)^(1/2),x, algorithm="maxima")
Output:
-integrate((2*x + 2^(2/3))/(sqrt(-x^3 + 1)*(x - 2^(2/3))), x)
Exception generated. \[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((2^(2/3)+2*x)/(2^(2/3)-x)/(-x^3+1)^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[2]%%%} / %%%{%%{[2,0]:[1,0,0,-2]%%},[2]%%%} Error: Bad Argument
Time = 24.70 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.85 \[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=\frac {2^{2/3}\,\sqrt {3}\,\ln \left (\frac {\left (\sqrt {1-x^3}-\sqrt {3}\,1{}\mathrm {i}+2^{1/3}\,\sqrt {3}\,x\,1{}\mathrm {i}\right )\,{\left (\sqrt {3}\,1{}\mathrm {i}+\sqrt {1-x^3}-2^{1/3}\,\sqrt {3}\,x\,1{}\mathrm {i}\right )}^3}{{\left (x-2^{2/3}\right )}^6}\right )\,1{}\mathrm {i}}{3} \] Input:
int(-(2*x + 2^(2/3))/((1 - x^3)^(1/2)*(x - 2^(2/3))),x)
Output:
(2^(2/3)*3^(1/2)*log((((1 - x^3)^(1/2) - 3^(1/2)*1i + 2^(1/3)*3^(1/2)*x*1i )*(3^(1/2)*1i + (1 - x^3)^(1/2) - 2^(1/3)*3^(1/2)*x*1i)^3)/(x - 2^(2/3))^6 )*1i)/3
\[ \int \frac {2^{2/3}+2 x}{\left (2^{2/3}-x\right ) \sqrt {1-x^3}} \, dx=2^{\frac {2}{3}} \left (\int \frac {1}{\sqrt {-x^{3}+1}\, 2^{\frac {2}{3}}-\sqrt {-x^{3}+1}\, x}d x \right )+2 \left (\int \frac {x}{\sqrt {-x^{3}+1}\, 2^{\frac {2}{3}}-\sqrt {-x^{3}+1}\, x}d x \right ) \] Input:
int((2^(2/3)+2*x)/(2^(2/3)-x)/(-x^3+1)^(1/2),x)
Output:
2**(2/3)*int(1/(sqrt( - x**3 + 1)*2**(2/3) - sqrt( - x**3 + 1)*x),x) + 2*i nt(x/(sqrt( - x**3 + 1)*2**(2/3) - sqrt( - x**3 + 1)*x),x)