Integrand size = 18, antiderivative size = 141 \[ \int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx=\frac {3}{2} \sqrt [3]{-6-2 x+x^2}+\frac {1}{2} \sqrt {3} \sqrt [3]{7} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-6-2 x+x^2}}{\sqrt {3} \sqrt [3]{7}}\right )-\frac {1}{2} \sqrt [3]{7} \log \left (7+7^{2/3} \sqrt [3]{-6-2 x+x^2}\right )+\frac {1}{4} \sqrt [3]{7} \log \left (-7+7^{2/3} \sqrt [3]{-6-2 x+x^2}-\sqrt [3]{7} \left (-6-2 x+x^2\right )^{2/3}\right ) \]
3/2*(x^2-2*x-6)^(1/3)-1/2*3^(1/2)*7^(1/3)*arctan(-1/3*3^(1/2)+2/21*(x^2-2* x-6)^(1/3)*3^(1/2)*7^(2/3))-1/2*7^(1/3)*ln(7+7^(2/3)*(x^2-2*x-6)^(1/3))+1/ 4*7^(1/3)*ln(-7+7^(2/3)*(x^2-2*x-6)^(1/3)-7^(1/3)*(x^2-2*x-6)^(2/3))
Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx=\frac {1}{4} \left (6 \sqrt [3]{-6-2 x+x^2}+2 \sqrt {3} \sqrt [3]{7} \arctan \left (\frac {7-2\ 7^{2/3} \sqrt [3]{-6-2 x+x^2}}{7 \sqrt {3}}\right )-2 \sqrt [3]{7} \log \left (7+7^{2/3} \sqrt [3]{-6-2 x+x^2}\right )+\sqrt [3]{7} \log \left (-7+7^{2/3} \sqrt [3]{-6-2 x+x^2}-\sqrt [3]{7} \left (-6-2 x+x^2\right )^{2/3}\right )\right ) \]
(6*(-6 - 2*x + x^2)^(1/3) + 2*Sqrt[3]*7^(1/3)*ArcTan[(7 - 2*7^(2/3)*(-6 - 2*x + x^2)^(1/3))/(7*Sqrt[3])] - 2*7^(1/3)*Log[7 + 7^(2/3)*(-6 - 2*x + x^2 )^(1/3)] + 7^(1/3)*Log[-7 + 7^(2/3)*(-6 - 2*x + x^2)^(1/3) - 7^(1/3)*(-6 - 2*x + x^2)^(2/3)])/4
Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1118, 243, 60, 70, 16, 1082, 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.
\(\displaystyle \int \frac {\sqrt [3]{x^2-2 x-6}}{x-1} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \int \frac {\sqrt [3]{(x-1)^2-7}}{x-1}d(x-1)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt [3]{x-8}}{(x-1)^2}d(x-1)^2\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{2} \left (3 \sqrt [3]{x-8}-7 \int \frac {1}{(x-8)^{2/3} (x-1)^2}d(x-1)^2\right )\) |
\(\Big \downarrow \) 70 |
\(\displaystyle \frac {1}{2} \left (3 \sqrt [3]{x-8}-7 \left (\frac {3 \int \frac {1}{\sqrt [3]{x-8}+\sqrt [3]{7}}d\sqrt [3]{x-8}}{2\ 7^{2/3}}+\frac {3 \int \frac {1}{(x-1)^4-\sqrt [3]{7} \sqrt [3]{x-8}+7^{2/3}}d\sqrt [3]{x-8}}{2 \sqrt [3]{7}}-\frac {\log \left ((x-1)^2\right )}{2\ 7^{2/3}}\right )\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (3 \sqrt [3]{x-8}-7 \left (\frac {3 \int \frac {1}{(x-1)^4-\sqrt [3]{7} \sqrt [3]{x-8}+7^{2/3}}d\sqrt [3]{x-8}}{2 \sqrt [3]{7}}+\frac {3 \log \left (\sqrt [3]{x-8}+\sqrt [3]{7}\right )}{2\ 7^{2/3}}-\frac {\log \left ((x-1)^2\right )}{2\ 7^{2/3}}\right )\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (3 \sqrt [3]{x-8}-7 \left (\frac {3 \int \frac {1}{-(x-1)^4-3}d\left (1-\frac {2 \sqrt [3]{x-8}}{\sqrt [3]{7}}\right )}{7^{2/3}}+\frac {3 \log \left (\sqrt [3]{x-8}+\sqrt [3]{7}\right )}{2\ 7^{2/3}}-\frac {\log \left ((x-1)^2\right )}{2\ 7^{2/3}}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (3 \sqrt [3]{x-8}-7 \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{x-8}}{\sqrt [3]{7}}}{\sqrt {3}}\right )}{7^{2/3}}+\frac {3 \log \left (\sqrt [3]{x-8}+\sqrt [3]{7}\right )}{2\ 7^{2/3}}-\frac {\log \left ((x-1)^2\right )}{2\ 7^{2/3}}\right )\right )\) |
(3*(-8 + x)^(1/3) - 7*(-((Sqrt[3]*ArcTan[(1 - (2*(-8 + x)^(1/3))/7^(1/3))/ Sqrt[3]])/7^(2/3)) + (3*Log[7^(1/3) + (-8 + x)^(1/3)])/(2*7^(2/3)) - Log[( -1 + x)^2]/(2*7^(2/3))))/2
3.20.93.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ {q = Rt[-(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2) , x] + (Simp[3/(2*b*q) Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(1 /3)], x] + Simp[3/(2*b*q^2) 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] :> 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_))^(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 = 9.53 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.72
method | result | size |
pseudoelliptic | \(\frac {3 \left (x^{2}-2 x -6\right )^{\frac {1}{3}}}{2}-\frac {7^{\frac {1}{3}} \ln \left (\left (x^{2}-2 x -6\right )^{\frac {1}{3}}+7^{\frac {1}{3}}\right )}{2}+\frac {7^{\frac {1}{3}} \ln \left (\left (x^{2}-2 x -6\right )^{\frac {2}{3}}-7^{\frac {1}{3}} \left (x^{2}-2 x -6\right )^{\frac {1}{3}}+7^{\frac {2}{3}}\right )}{4}-\frac {\sqrt {3}\, 7^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3}}{3}+\frac {2 \left (x^{2}-2 x -6\right )^{\frac {1}{3}} \sqrt {3}\, 7^{\frac {2}{3}}}{21}\right )}{2}\) | \(102\) |
risch | \(\text {Expression too large to display}\) | \(2720\) |
3/2*(x^2-2*x-6)^(1/3)-1/2*7^(1/3)*ln((x^2-2*x-6)^(1/3)+7^(1/3))+1/4*7^(1/3 )*ln((x^2-2*x-6)^(2/3)-7^(1/3)*(x^2-2*x-6)^(1/3)+7^(2/3))-1/2*3^(1/2)*7^(1 /3)*arctan(-1/3*3^(1/2)+2/21*(x^2-2*x-6)^(1/3)*3^(1/2)*7^(2/3))
Time = 0.43 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx=\frac {1}{2} \, \sqrt {3} \left (-7\right )^{\frac {1}{3}} \arctan \left (\frac {2}{21} \, \sqrt {3} \left (-7\right )^{\frac {2}{3}} {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \, \left (-7\right )^{\frac {1}{3}} \log \left (\left (-7\right )^{\frac {2}{3}} + \left (-7\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 6\right )}^{\frac {2}{3}}\right ) + \frac {1}{2} \, \left (-7\right )^{\frac {1}{3}} \log \left (-\left (-7\right )^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}}\right ) + \frac {3}{2} \, {\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}} \]
1/2*sqrt(3)*(-7)^(1/3)*arctan(2/21*sqrt(3)*(-7)^(2/3)*(x^2 - 2*x - 6)^(1/3 ) - 1/3*sqrt(3)) - 1/4*(-7)^(1/3)*log((-7)^(2/3) + (-7)^(1/3)*(x^2 - 2*x - 6)^(1/3) + (x^2 - 2*x - 6)^(2/3)) + 1/2*(-7)^(1/3)*log(-(-7)^(1/3) + (x^2 - 2*x - 6)^(1/3)) + 3/2*(x^2 - 2*x - 6)^(1/3)
\[ \int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx=\int \frac {\sqrt [3]{x^{2} - 2 x - 6}}{x - 1}\, dx \]
\[ \int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx=\int { \frac {{\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}}}{x - 1} \,d x } \]
\[ \int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx=\int { \frac {{\left (x^{2} - 2 \, x - 6\right )}^{\frac {1}{3}}}{x - 1} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{-6-2 x+x^2}}{-1+x} \, dx=\int \frac {{\left (x^2-2\,x-6\right )}^{1/3}}{x-1} \,d x \]