Integrand size = 7, antiderivative size = 74 \[ \int \frac {1}{-2+x^3} \, dx=-\frac {\arctan \left (\frac {1+2^{2/3} x}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\log \left (2^{2/3}+\sqrt [3]{2} x+x^2\right )}{6\ 2^{2/3}} \] Output:
1/6*ln(2^(1/3)-x)*2^(1/3)-1/12*ln(2^(2/3)+2^(1/3)*x+x^2)*2^(1/3)-1/6*arcta n(1/3*(1+2^(2/3)*x)*3^(1/2))*2^(1/3)*3^(1/2)
Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.88 \[ \int \frac {1}{-2+x^3} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {1+2^{2/3} x}{\sqrt {3}}\right )-2 \log \left (2-2^{2/3} x\right )+\log \left (2+2^{2/3} x+\sqrt [3]{2} x^2\right )}{6\ 2^{2/3}} \] Input:
Integrate[(-2 + x^3)^(-1),x]
Output:
-1/6*(2*Sqrt[3]*ArcTan[(1 + 2^(2/3)*x)/Sqrt[3]] - 2*Log[2 - 2^(2/3)*x] + L og[2 + 2^(2/3)*x + 2^(1/3)*x^2])/2^(2/3)
Time = 0.20 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {750, 16, 25, 1142, 27, 1082, 217, 1103}
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.
\(\displaystyle \int \frac {1}{x^3-2} \, dx\) |
\(\Big \downarrow \) 750 |
\(\displaystyle \frac {\int -\frac {x+2 \sqrt [3]{2}}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{3\ 2^{2/3}}+\frac {\int \frac {1}{x-\sqrt [3]{2}}dx}{3\ 2^{2/3}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\int -\frac {x+2 \sqrt [3]{2}}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{3\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\int \frac {x+2 \sqrt [3]{2}}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{3\ 2^{2/3}}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\frac {3 \int \frac {1}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{2^{2/3}}+\frac {1}{2} \int \frac {\sqrt [3]{2} \left (2^{2/3} x+1\right )}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{3\ 2^{2/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\frac {3 \int \frac {1}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{2^{2/3}}+\frac {\int \frac {2^{2/3} x+1}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{2^{2/3}}}{3\ 2^{2/3}}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\frac {\int \frac {2^{2/3} x+1}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{2^{2/3}}-3 \int \frac {1}{-\left (2^{2/3} x+1\right )^2-3}d\left (2^{2/3} x+1\right )}{3\ 2^{2/3}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\frac {\int \frac {2^{2/3} x+1}{x^2+\sqrt [3]{2} x+2^{2/3}}dx}{2^{2/3}}+\sqrt {3} \arctan \left (\frac {2^{2/3} x+1}{\sqrt {3}}\right )}{3\ 2^{2/3}}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {\log \left (\sqrt [3]{2}-x\right )}{3\ 2^{2/3}}-\frac {\sqrt {3} \arctan \left (\frac {2^{2/3} x+1}{\sqrt {3}}\right )+\frac {1}{2} \log \left (x^2+\sqrt [3]{2} x+2^{2/3}\right )}{3\ 2^{2/3}}\) |
Input:
Int[(-2 + x^3)^(-1),x]
Output:
Log[2^(1/3) - x]/(3*2^(2/3)) - (Sqrt[3]*ArcTan[(1 + 2^(2/3)*x)/Sqrt[3]] + Log[2^(2/3) + 2^(1/3)*x + x^2]/2)/(3*2^(2/3))
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.30
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3}-2\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{3}\) | \(22\) |
default | \(\frac {2^{\frac {1}{3}} \ln \left (x -2^{\frac {1}{3}}\right )}{6}-\frac {\ln \left (2^{\frac {2}{3}}+2^{\frac {1}{3}} x +x^{2}\right ) 2^{\frac {1}{3}}}{12}-\frac {\arctan \left (\frac {\left (1+2^{\frac {2}{3}} x \right ) \sqrt {3}}{3}\right ) 2^{\frac {1}{3}} \sqrt {3}}{6}\) | \(54\) |
meijerg | \(\frac {2^{\frac {1}{3}} x \left (\ln \left (1-\frac {2^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{2}\right )-\frac {\ln \left (1+\frac {2^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{2}+\frac {2^{\frac {1}{3}} \left (x^{3}\right )^{\frac {2}{3}}}{2}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, 2^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}{4+2^{\frac {2}{3}} \left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{6 \left (x^{3}\right )^{\frac {1}{3}}}\) | \(84\) |
Input:
int(1/(x^3-2),x,method=_RETURNVERBOSE)
Output:
1/3*sum(1/_R^2*ln(x-_R),_R=RootOf(_Z^3-2))
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.86 \[ \int \frac {1}{-2+x^3} \, dx=-\frac {1}{2} \cdot 4^{\frac {1}{6}} \sqrt {\frac {1}{3}} \arctan \left (\frac {1}{2} \cdot 4^{\frac {1}{6}} \sqrt {\frac {1}{3}} {\left (4^{\frac {2}{3}} x + 4^{\frac {1}{3}}\right )}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \log \left (2 \, x^{2} + 4^{\frac {2}{3}} x + 2 \cdot 4^{\frac {1}{3}}\right ) + \frac {1}{12} \cdot 4^{\frac {2}{3}} \log \left (2 \, x - 4^{\frac {2}{3}}\right ) \] Input:
integrate(1/(x^3-2),x, algorithm="fricas")
Output:
-1/2*4^(1/6)*sqrt(1/3)*arctan(1/2*4^(1/6)*sqrt(1/3)*(4^(2/3)*x + 4^(1/3))) - 1/24*4^(2/3)*log(2*x^2 + 4^(2/3)*x + 2*4^(1/3)) + 1/12*4^(2/3)*log(2*x - 4^(2/3))
Time = 0.14 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.96 \[ \int \frac {1}{-2+x^3} \, dx=\frac {\sqrt [3]{2} \log {\left (x - \sqrt [3]{2} \right )}}{6} - \frac {\sqrt [3]{2} \log {\left (x^{2} + \sqrt [3]{2} x + 2^{\frac {2}{3}} \right )}}{12} - \frac {\sqrt [3]{2} \sqrt {3} \operatorname {atan}{\left (\frac {2^{\frac {2}{3}} \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \] Input:
integrate(1/(x**3-2),x)
Output:
2**(1/3)*log(x - 2**(1/3))/6 - 2**(1/3)*log(x**2 + 2**(1/3)*x + 2**(2/3))/ 12 - 2**(1/3)*sqrt(3)*atan(2**(2/3)*sqrt(3)*x/3 + sqrt(3)/3)/6
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.76 \[ \int \frac {1}{-2+x^3} \, dx=-\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2 \, x + 2^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {1}{3}} \log \left (x^{2} + 2^{\frac {1}{3}} x + 2^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{3}} \log \left (x - 2^{\frac {1}{3}}\right ) \] Input:
integrate(1/(x^3-2),x, algorithm="maxima")
Output:
-1/6*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2*x + 2^(1/3))) - 1/12*2^ (1/3)*log(x^2 + 2^(1/3)*x + 2^(2/3)) + 1/6*2^(1/3)*log(x - 2^(1/3))
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.77 \[ \int \frac {1}{-2+x^3} \, dx=-\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2 \, x + 2^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {1}{3}} \log \left (x^{2} + 2^{\frac {1}{3}} x + 2^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{3}} \log \left ({\left | x - 2^{\frac {1}{3}} \right |}\right ) \] Input:
integrate(1/(x^3-2),x, algorithm="giac")
Output:
-1/6*sqrt(3)*2^(1/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2*x + 2^(1/3))) - 1/12*2^ (1/3)*log(x^2 + 2^(1/3)*x + 2^(2/3)) + 1/6*2^(1/3)*log(abs(x - 2^(1/3)))
Time = 0.16 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97 \[ \int \frac {1}{-2+x^3} \, dx=\frac {2^{1/3}\,\ln \left (x-2^{1/3}\right )}{6}+\frac {2^{1/3}\,\ln \left (x-\frac {2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{12}-\frac {2^{1/3}\,\ln \left (x+\frac {2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \] Input:
int(1/(x^3 - 2),x)
Output:
(2^(1/3)*log(x - 2^(1/3)))/6 + (2^(1/3)*log(x - (2^(1/3)*(3^(1/2)*1i - 1)) /2)*(3^(1/2)*1i - 1))/12 - (2^(1/3)*log(x + (2^(1/3)*(3^(1/2)*1i + 1))/2)* (3^(1/2)*1i + 1))/12
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.70 \[ \int \frac {1}{-2+x^3} \, dx=\frac {2^{\frac {1}{3}} \left (-2 \sqrt {3}\, \mathit {atan} \left (\frac {\left (2^{\frac {1}{3}}+2 x \right ) 2^{\frac {2}{3}}}{2 \sqrt {3}}\right )-\mathrm {log}\left (2^{\frac {2}{3}}+2^{\frac {1}{3}} x +x^{2}\right )+2 \,\mathrm {log}\left (-2^{\frac {1}{3}}+x \right )\right )}{12} \] Input:
int(1/(x^3-2),x)
Output:
(2**(1/3)*( - 2*sqrt(3)*atan((2**(1/3) + 2*x)/(2**(1/3)*sqrt(3))) - log(2* *(2/3) + 2**(1/3)*x + x**2) + 2*log( - 2**(1/3) + x)))/12