Integrand size = 13, antiderivative size = 67 \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\sqrt {3} \arctan \left (\frac {1+\frac {2 (-1+x)}{\sqrt [3]{(-1+x)^2 (1+x)}}}{\sqrt {3}}\right )-\frac {1}{2} \log (1+x)-\frac {3}{2} \log \left (1-\frac {-1+x}{\sqrt [3]{(-1+x)^2 (1+x)}}\right ) \]
-1/2*ln(1+x)-3/2*ln(1+(1-x)/((-1+x)^2*(1+x))^(1/3))+arctan(1/3*(1+2*(-1+x) /((-1+x)^2*(1+x))^(1/3))*3^(1/2))*3^(1/2)
Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.84 \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\frac {(-1+x)^{2/3} \sqrt [3]{1+x} \left (-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{1+x}}\right )-2 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{1+x}\right )+\log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{2 \sqrt [3]{(-1+x)^2 (1+x)}} \]
((-1 + x)^(2/3)*(1 + x)^(1/3)*(-2*Sqrt[3]*ArcTan[(Sqrt[3]*(1 + x)^(1/3))/( 2*(-1 + x)^(1/3) + (1 + x)^(1/3))] - 2*Log[(-1 + x)^(1/3) - (1 + x)^(1/3)] + Log[(-1 + x)^(2/3) + (-1 + x)^(1/3)*(1 + x)^(1/3) + (1 + x)^(2/3)]))/(2 *((-1 + x)^2*(1 + x))^(1/3))
Time = 0.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.70, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2477, 474, 473, 71}
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.
\(\displaystyle \int \frac {1}{\sqrt [3]{(x-1)^2 (x+1)}} \, dx\) |
\(\Big \downarrow \) 2477 |
\(\displaystyle \frac {\sqrt [3]{x-1} \sqrt [3]{x^2-1} \int \frac {1}{\sqrt [3]{x-1} \sqrt [3]{x^2-1}}dx}{\sqrt [3]{(1-x)^2 (x+1)}}\) |
\(\Big \downarrow \) 474 |
\(\displaystyle \frac {\sqrt [3]{1-x} \sqrt [3]{x^2-1} \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{x^2-1}}dx}{\sqrt [3]{(1-x)^2 (x+1)}}\) |
\(\Big \downarrow \) 473 |
\(\displaystyle \frac {\left (x^2-1\right ) \int \frac {1}{\sqrt [3]{-x-1} (1-x)^{2/3}}dx}{(-x-1)^{2/3} \sqrt [3]{1-x} \sqrt [3]{(1-x)^2 (x+1)}}\) |
\(\Big \downarrow \) 71 |
\(\displaystyle \frac {\left (x^2-1\right ) \left (\sqrt {3} \arctan \left (\frac {2 \sqrt [3]{-x-1}}{\sqrt {3} \sqrt [3]{1-x}}+\frac {1}{\sqrt {3}}\right )+\frac {3}{2} \log \left (\frac {\sqrt [3]{-x-1}}{\sqrt [3]{1-x}}-1\right )+\frac {1}{2} \log (1-x)\right )}{(-x-1)^{2/3} \sqrt [3]{1-x} \sqrt [3]{(1-x)^2 (x+1)}}\) |
((-1 + x^2)*(Sqrt[3]*ArcTan[1/Sqrt[3] + (2*(-1 - x)^(1/3))/(Sqrt[3]*(1 - x )^(1/3))] + (3*Log[-1 + (-1 - x)^(1/3)/(1 - x)^(1/3)])/2 + Log[1 - x]/2))/ ((-1 - x)^(2/3)*(1 - x)^(1/3)*((1 - x)^2*(1 + x))^(1/3))
3.3.26.3.1 Defintions of rubi rules used
Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, Simp[(-Sqrt[3])*(q/d)*ArcTan[2*q*((a + b*x)^(1/3)/( Sqrt[3]*(c + d*x)^(1/3))) + 1/Sqrt[3]], x] + (-Simp[3*(q/(2*d))*Log[q*((a + b*x)^(1/3)/(c + d*x)^(1/3)) - 1], x] - Simp[(q/(2*d))*Log[c + d*x], x])] / ; FreeQ[{a, b, c, d}, x] && PosQ[d/b]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ c^(n - 1)*((a + b*x^2)^(p + 1)/((1 + d*(x/c))^(p + 1)*(a/c + (b*x)/d)^(p + 1))) Int[(1 + d*(x/c))^(n + p)*(a/c + (b/d)*x)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[n] || GtQ[c, 0]) && !Gt Q[a, 0] && !(IntegerQ[n] && (IntegerQ[3*p] || IntegerQ[4*p]))
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ c^IntPart[n]*((c + d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]) Int[(1 + d *(x/c))^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c^2 + a*d^2, 0] && !(IntegerQ[n] || GtQ[c, 0])
Int[(Px_)^(p_), x_Symbol] :> With[{a = Coeff[Px, x, 0], b = Coeff[Px, x, 1] , c = Coeff[Px, x, 2], d = Coeff[Px, x, 3]}, Simp[Px^p/((c + d*x)^p*(b + d* x^2)^p) Int[(c + d*x)^p*(b + d*x^2)^p, x], x] /; EqQ[b*c - a*d, 0]] /; Fr eeQ[p, x] && PolyQ[Px, x, 3] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.53 (sec) , antiderivative size = 370, normalized size of antiderivative = 5.52
method | result | size |
trager | \(-\ln \left (\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +x^{2}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-1}{-1+x}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x +6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +2 x^{2}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-4 x +2}{-1+x}\right )\) | \(370\) |
-ln((4*RootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(2/3)-3*R ootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(1/3)*x-4*RootOf(_Z^2-_Z+1)^2*x-4*RootOf(_Z ^2-_Z+1)*x^2+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(1/3)+3*(x^3-x^2-x+1)^(1/3) *x+2*RootOf(_Z^2-_Z+1)*x+x^2-3*(x^3-x^2-x+1)^(1/3)+2*RootOf(_Z^2-_Z+1)-1)/ (-1+x))+RootOf(_Z^2-_Z+1)*ln(-(2*RootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+ 1)*(x^3-x^2-x+1)^(2/3)-2*RootOf(_Z^2-_Z+1)^2*x-5*RootOf(_Z^2-_Z+1)*x^2-3*( x^3-x^2-x+1)^(2/3)+3*(x^3-x^2-x+1)^(1/3)*x+6*RootOf(_Z^2-_Z+1)*x+2*x^2-3*( x^3-x^2-x+1)^(1/3)-RootOf(_Z^2-_Z+1)-4*x+2)/(-1+x))
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (56) = 112\).
Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.91 \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x - 1\right )} + 2 \, \sqrt {3} {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}}}{3 \, {\left (x - 1\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 2 \, x + {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {2}{3}} + 1}{x^{2} - 2 \, x + 1}\right ) - \log \left (-\frac {x - {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}} - 1}{x - 1}\right ) \]
-sqrt(3)*arctan(1/3*(sqrt(3)*(x - 1) + 2*sqrt(3)*(x^3 - x^2 - x + 1)^(1/3) )/(x - 1)) + 1/2*log((x^2 + (x^3 - x^2 - x + 1)^(1/3)*(x - 1) - 2*x + (x^3 - x^2 - x + 1)^(2/3) + 1)/(x^2 - 2*x + 1)) - log(-(x - (x^3 - x^2 - x + 1 )^(1/3) - 1)/(x - 1))
\[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\int \frac {1}{\sqrt [3]{\left (x - 1\right )^{2} \left (x + 1\right )}}\, dx \]
\[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\int { \frac {1}{\left ({\left (x + 1\right )} {\left (x - 1\right )}^{2}\right )^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\int { \frac {1}{\left ({\left (x + 1\right )} {\left (x - 1\right )}^{2}\right )^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\int \frac {1}{{\left ({\left (x-1\right )}^2\,\left (x+1\right )\right )}^{1/3}} \,d x \]
\[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\int \frac {1}{\left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}}d x \]