Integrand size = 18, antiderivative size = 86 \[ \int \frac {1}{(-1+x) \sqrt [3]{2-2 x+x^2}} \, dx=\frac {1}{2} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{2-2 x+x^2}}{\sqrt {3}}\right )+\frac {1}{2} \log \left (-1+\sqrt [3]{2-2 x+x^2}\right )-\frac {1}{4} \log \left (1+\sqrt [3]{2-2 x+x^2}+\left (2-2 x+x^2\right )^{2/3}\right ) \]
1/2*3^(1/2)*arctan(1/3*3^(1/2)+2/3*(x^2-2*x+2)^(1/3)*3^(1/2))+1/2*ln(-1+(x ^2-2*x+2)^(1/3))-1/4*ln(1+(x^2-2*x+2)^(1/3)+(x^2-2*x+2)^(2/3))
Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94 \[ \int \frac {1}{(-1+x) \sqrt [3]{2-2 x+x^2}} \, dx=\frac {1}{4} \left (2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{2-2 x+x^2}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{2-2 x+x^2}\right )-\log \left (1+\sqrt [3]{2-2 x+x^2}+\left (2-2 x+x^2\right )^{2/3}\right )\right ) \]
(2*Sqrt[3]*ArcTan[(1 + 2*(2 - 2*x + x^2)^(1/3))/Sqrt[3]] + 2*Log[-1 + (2 - 2*x + x^2)^(1/3)] - Log[1 + (2 - 2*x + x^2)^(1/3) + (2 - 2*x + x^2)^(2/3) ])/4
Time = 0.21 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.55, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1118, 243, 67, 16, 1083, 217}
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.
\(\displaystyle \int \frac {1}{(x-1) \sqrt [3]{x^2-2 x+2}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \int \frac {1}{\sqrt [3]{(x-1)^2+1} (x-1)}d(x-1)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{(x-1)^2 \sqrt [3]{x}}d(x-1)^2\) |
\(\Big \downarrow \) 67 |
\(\displaystyle \frac {1}{2} \left (-\frac {3}{2} \int \frac {1}{2-x}d\sqrt [3]{x}+\frac {3}{2} \int \frac {1}{(x-1)^4+x}d\sqrt [3]{x}-\frac {1}{2} \log \left ((x-1)^2\right )\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{(x-1)^4+x}d\sqrt [3]{x}+\frac {3}{2} \log (2-x)-\frac {1}{2} \log \left ((x-1)^2\right )\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (-3 \int \frac {1}{-(x-1)^4-3}d\left (2 \sqrt [3]{x}+1\right )+\frac {3}{2} \log (2-x)-\frac {1}{2} \log \left ((x-1)^2\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x}+1}{\sqrt {3}}\right )+\frac {3}{2} \log (2-x)-\frac {1}{2} \log \left ((x-1)^2\right )\right )\) |
3.12.55.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ {q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x ] + (Simp[3/(2*b) Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], x] - Simp[3/(2*b*q) Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[1/e Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
Time = 4.39 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(\frac {\ln \left (-1+\left (x^{2}-2 x +2\right )^{\frac {1}{3}}\right )}{2}-\frac {\ln \left (1+\left (x^{2}-2 x +2\right )^{\frac {1}{3}}+\left (x^{2}-2 x +2\right )^{\frac {2}{3}}\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{2}-2 x +2\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{2}\) | \(67\) |
trager | \(\frac {\ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x -15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {2}{3}}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+24 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {1}{3}}-6 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +9 \left (x^{2}-2 x +2\right )^{\frac {2}{3}}-10 x^{2}-7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+15 \left (x^{2}-2 x +2\right )^{\frac {1}{3}}+20 x -35}{\left (x -1\right )^{2}}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}-10 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x -15 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {2}{3}}+16 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-9 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {1}{3}}-32 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -24 \left (x^{2}-2 x +2\right )^{\frac {2}{3}}+3 x^{2}+35 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+15 \left (x^{2}-2 x +2\right )^{\frac {1}{3}}-6 x +7}{\left (x -1\right )^{2}}\right )}{2}\) | \(285\) |
1/2*ln(-1+(x^2-2*x+2)^(1/3))-1/4*ln(1+(x^2-2*x+2)^(1/3)+(x^2-2*x+2)^(2/3)) +1/2*3^(1/2)*arctan(1/3*(2*(x^2-2*x+2)^(1/3)+1)*3^(1/2))
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(-1+x) \sqrt [3]{2-2 x+x^2}} \, dx=\frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {2}{3}} + {\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} - 1\right ) \]
1/2*sqrt(3)*arctan(2/3*sqrt(3)*(x^2 - 2*x + 2)^(1/3) + 1/3*sqrt(3)) - 1/4* log((x^2 - 2*x + 2)^(2/3) + (x^2 - 2*x + 2)^(1/3) + 1) + 1/2*log((x^2 - 2* x + 2)^(1/3) - 1)
\[ \int \frac {1}{(-1+x) \sqrt [3]{2-2 x+x^2}} \, dx=\int \frac {1}{\left (x - 1\right ) \sqrt [3]{x^{2} - 2 x + 2}}\, dx \]
\[ \int \frac {1}{(-1+x) \sqrt [3]{2-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} {\left (x - 1\right )}} \,d x } \]
\[ \int \frac {1}{(-1+x) \sqrt [3]{2-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} {\left (x - 1\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(-1+x) \sqrt [3]{2-2 x+x^2}} \, dx=\int \frac {1}{\left (x-1\right )\,{\left (x^2-2\,x+2\right )}^{1/3}} \,d x \]