Integrand size = 18, antiderivative size = 88 \[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=-\frac {1}{4} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{-4-4 x+x^2}}{\sqrt {3}}\right )-\frac {1}{4} \log \left (2+\sqrt [3]{-4-4 x+x^2}\right )+\frac {1}{8} \log \left (4-2 \sqrt [3]{-4-4 x+x^2}+\left (-4-4 x+x^2\right )^{2/3}\right ) \]
1/4*3^(1/2)*arctan(-1/3*3^(1/2)+1/3*(x^2-4*x-4)^(1/3)*3^(1/2))-1/4*ln(2+(x ^2-4*x-4)^(1/3))+1/8*ln(4-2*(x^2-4*x-4)^(1/3)+(x^2-4*x-4)^(2/3))
Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\frac {1}{8} \left (2 \sqrt {3} \arctan \left (\frac {-1+\sqrt [3]{-4-4 x+x^2}}{\sqrt {3}}\right )-2 \log \left (2+\sqrt [3]{-4-4 x+x^2}\right )+\log \left (4-2 \sqrt [3]{-4-4 x+x^2}+\left (-4-4 x+x^2\right )^{2/3}\right )\right ) \]
(2*Sqrt[3]*ArcTan[(-1 + (-4 - 4*x + x^2)^(1/3))/Sqrt[3]] - 2*Log[2 + (-4 - 4*x + x^2)^(1/3)] + Log[4 - 2*(-4 - 4*x + x^2)^(1/3) + (-4 - 4*x + x^2)^( 2/3)])/8
Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.58, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1118, 243, 68, 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.
\(\displaystyle \int \frac {1}{(x-2) \sqrt [3]{x^2-4 x-4}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \int \frac {1}{\sqrt [3]{(x-2)^2-8} (x-2)}d(x-2)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt [3]{x-10} (x-2)^2}d(x-2)^2\) |
\(\Big \downarrow \) 68 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{(x-2)^4-2 \sqrt [3]{x-10}+4}d\sqrt [3]{x-10}-\frac {3}{4} \int \frac {1}{x}d\sqrt [3]{x-10}+\frac {1}{4} \log \left ((x-2)^2\right )\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{(x-2)^4-2 \sqrt [3]{x-10}+4}d\sqrt [3]{x-10}+\frac {1}{4} \log \left ((x-2)^2\right )-\frac {3 \log (x)}{4}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (-3 \int \frac {1}{-(x-2)^4-12}d\left (2 \sqrt [3]{x-10}-2\right )+\frac {1}{4} \log \left ((x-2)^2\right )-\frac {3 \log (x)}{4}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x-10}-2}{2 \sqrt {3}}\right )+\frac {1}{4} \log \left ((x-2)^2\right )-\frac {3 \log (x)}{4}\right )\) |
((Sqrt[3]*ArcTan[(-2 + 2*(-10 + x)^(1/3))/(2*Sqrt[3])])/2 + Log[(-2 + x)^2 ]/4 - (3*Log[x])/4)/2
3.12.99.3.1 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] && NegQ[(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 = 4.72 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76
method | result | size |
pseudoelliptic | \(-\frac {\ln \left (2+\left (x^{2}-4 x -4\right )^{\frac {1}{3}}\right )}{4}+\frac {\ln \left (4-2 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+\left (x^{2}-4 x -4\right )^{\frac {2}{3}}\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (\left (x^{2}-4 x -4\right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right )}{4}\) | \(67\) |
trager | \(-\frac {\ln \left (-\frac {4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x^{2}-16 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x +48 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {2}{3}}+9 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x^{2}+60 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {1}{3}}-36 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x -9 \left (x^{2}-4 x -4\right )^{\frac {2}{3}}+2 x^{2}-28 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-48 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}-8 x -56}{\left (x -2\right )^{2}}\right )}{4}+\frac {\operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \ln \left (-\frac {32 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x^{2}-128 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x -96 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-20 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x^{2}-72 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+80 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x +30 \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-3 x^{2}+224 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )+96 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+12 x +28}{\left (x -2\right )^{2}}\right )}{2}\) | \(347\) |
-1/4*ln(2+(x^2-4*x-4)^(1/3))+1/8*ln(4-2*(x^2-4*x-4)^(1/3)+(x^2-4*x-4)^(2/3 ))+1/4*3^(1/2)*arctan(1/3*((x^2-4*x-4)^(1/3)-1)*3^(1/2))
Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {1}{8} \, \log \left ({\left (x^{2} - 4 \, x - 4\right )}^{\frac {2}{3}} - 2 \, {\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} + 4\right ) - \frac {1}{4} \, \log \left ({\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} + 2\right ) \]
1/4*sqrt(3)*arctan(1/3*sqrt(3)*(x^2 - 4*x - 4)^(1/3) - 1/3*sqrt(3)) + 1/8* log((x^2 - 4*x - 4)^(2/3) - 2*(x^2 - 4*x - 4)^(1/3) + 4) - 1/4*log((x^2 - 4*x - 4)^(1/3) + 2)
\[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\int \frac {1}{\left (x - 2\right ) \sqrt [3]{x^{2} - 4 x - 4}}\, dx \]
\[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} {\left (x - 2\right )}} \,d x } \]
\[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} {\left (x - 2\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\int \frac {1}{\left (x-2\right )\,{\left (x^2-4\,x-4\right )}^{1/3}} \,d x \]