Integrand size = 18, antiderivative size = 125 \[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}}{\sqrt {3}}\right )}{2\ 2^{2/3}}-\frac {\log \left (2+\sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}\right )}{2\ 2^{2/3}}+\frac {\log \left (4-2 \sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}+2^{2/3} \left (-3-2 x+x^2\right )^{2/3}\right )}{4\ 2^{2/3}} \] Output:
1/4*3^(1/2)*arctan(-1/3*3^(1/2)+1/3*2^(1/3)*(x^2-2*x-3)^(1/3)*3^(1/2))*2^( 1/3)-1/4*ln(2+2^(1/3)*(x^2-2*x-3)^(1/3))*2^(1/3)+1/8*ln(4-2*2^(1/3)*(x^2-2 *x-3)^(1/3)+2^(2/3)*(x^2-2*x-3)^(2/3))*2^(1/3)
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=\frac {-2 \sqrt {3} \arctan \left (\frac {1-\sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}}{\sqrt {3}}\right )-2 \log \left (2+\sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}\right )+\log \left (4-2 \sqrt [3]{2} \sqrt [3]{-3-2 x+x^2}+2^{2/3} \left (-3-2 x+x^2\right )^{2/3}\right )}{4\ 2^{2/3}} \] Input:
Integrate[1/((-1 + x)*(-3 - 2*x + x^2)^(1/3)),x]
Output:
(-2*Sqrt[3]*ArcTan[(1 - 2^(1/3)*(-3 - 2*x + x^2)^(1/3))/Sqrt[3]] - 2*Log[2 + 2^(1/3)*(-3 - 2*x + x^2)^(1/3)] + Log[4 - 2*2^(1/3)*(-3 - 2*x + x^2)^(1 /3) + 2^(2/3)*(-3 - 2*x + x^2)^(2/3)])/(4*2^(2/3))
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.62, 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, 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.
\(\displaystyle \int \frac {1}{(x-1) \sqrt [3]{x^2-2 x-3}} \, dx\) |
\(\Big \downarrow \) 1118 |
\(\displaystyle \int \frac {1}{\sqrt [3]{(x-1)^2-4} (x-1)}d(x-1)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt [3]{x-5} (x-1)^2}d(x-1)^2\) |
\(\Big \downarrow \) 68 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 \int \frac {1}{\sqrt [3]{x-5}+2^{2/3}}d\sqrt [3]{x-5}}{2\ 2^{2/3}}+\frac {3}{2} \int \frac {1}{(x-1)^4-2^{2/3} \sqrt [3]{x-5}+2 \sqrt [3]{2}}d\sqrt [3]{x-5}+\frac {\log \left ((x-1)^2\right )}{2\ 2^{2/3}}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{(x-1)^4-2^{2/3} \sqrt [3]{x-5}+2 \sqrt [3]{2}}d\sqrt [3]{x-5}-\frac {3 \log \left (\sqrt [3]{x-5}+2^{2/3}\right )}{2\ 2^{2/3}}+\frac {\log \left ((x-1)^2\right )}{2\ 2^{2/3}}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{2} \left (\frac {3 \int \frac {1}{-(x-1)^4-3}d\left (1-\sqrt [3]{2} \sqrt [3]{x-5}\right )}{2^{2/3}}-\frac {3 \log \left (\sqrt [3]{x-5}+2^{2/3}\right )}{2\ 2^{2/3}}+\frac {\log \left ((x-1)^2\right )}{2\ 2^{2/3}}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (-\frac {\sqrt {3} \arctan \left (\frac {1-\sqrt [3]{2} \sqrt [3]{x-5}}{\sqrt {3}}\right )}{2^{2/3}}-\frac {3 \log \left (\sqrt [3]{x-5}+2^{2/3}\right )}{2\ 2^{2/3}}+\frac {\log \left ((x-1)^2\right )}{2\ 2^{2/3}}\right )\) |
Input:
Int[1/((-1 + x)*(-3 - 2*x + x^2)^(1/3)),x]
Output:
(-((Sqrt[3]*ArcTan[(1 - 2^(1/3)*(-5 + x)^(1/3))/Sqrt[3]])/2^(2/3)) - (3*Lo g[2^(2/3) + (-5 + x)^(1/3)])/(2*2^(2/3)) + Log[(-1 + x)^2]/(2*2^(2/3)))/2
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] :> 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 = 6.65 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {1}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {1}{3}} \left (x^{2}-2 x -3\right )^{\frac {1}{3}}-1\right )}{3}\right )-2 \ln \left (\left (x^{2}-2 x -3\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )+\ln \left (\left (x^{2}-2 x -3\right )^{\frac {2}{3}}-2^{\frac {2}{3}} \left (x^{2}-2 x -3\right )^{\frac {1}{3}}+2 \,2^{\frac {1}{3}}\right )\right )}{8}\) | \(83\) |
trager | \(\text {Expression too large to display}\) | \(1214\) |
Input:
int(1/(-1+x)/(x^2-2*x-3)^(1/3),x,method=_RETURNVERBOSE)
Output:
1/8*2^(1/3)*(2*3^(1/2)*arctan(1/3*3^(1/2)*(2^(1/3)*(x^2-2*x-3)^(1/3)-1))-2 *ln((x^2-2*x-3)^(1/3)+2^(2/3))+ln((x^2-2*x-3)^(2/3)-2^(2/3)*(x^2-2*x-3)^(1 /3)+2*2^(1/3)))
Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=\frac {1}{4} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (-\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} - 2 \, {\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}}\right )}\right ) + \frac {1}{16} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} - 4^{\frac {1}{3}} {\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 3\right )}^{\frac {2}{3}}\right ) - \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}}\right ) \] Input:
integrate(1/(-1+x)/(x^2-2*x-3)^(1/3),x, algorithm="fricas")
Output:
1/4*4^(1/6)*sqrt(3)*arctan(-1/6*4^(1/6)*sqrt(3)*(4^(1/3) - 2*(x^2 - 2*x - 3)^(1/3))) + 1/16*4^(2/3)*log(4^(2/3) - 4^(1/3)*(x^2 - 2*x - 3)^(1/3) + (x ^2 - 2*x - 3)^(2/3)) - 1/8*4^(2/3)*log(4^(1/3) + (x^2 - 2*x - 3)^(1/3))
\[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=\int \frac {1}{\sqrt [3]{\left (x - 3\right ) \left (x + 1\right )} \left (x - 1\right )}\, dx \] Input:
integrate(1/(-1+x)/(x**2-2*x-3)**(1/3),x)
Output:
Integral(1/(((x - 3)*(x + 1))**(1/3)*(x - 1)), x)
\[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}} {\left (x - 1\right )}} \,d x } \] Input:
integrate(1/(-1+x)/(x^2-2*x-3)^(1/3),x, algorithm="maxima")
Output:
integrate(1/((x^2 - 2*x - 3)^(1/3)*(x - 1)), x)
\[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} - 2 \, x - 3\right )}^{\frac {1}{3}} {\left (x - 1\right )}} \,d x } \] Input:
integrate(1/(-1+x)/(x^2-2*x-3)^(1/3),x, algorithm="giac")
Output:
integrate(1/((x^2 - 2*x - 3)^(1/3)*(x - 1)), x)
Timed out. \[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=\int \frac {1}{\left (x-1\right )\,{\left (x^2-2\,x-3\right )}^{1/3}} \,d x \] Input:
int(1/((x - 1)*(x^2 - 2*x - 3)^(1/3)),x)
Output:
int(1/((x - 1)*(x^2 - 2*x - 3)^(1/3)), x)
\[ \int \frac {1}{(-1+x) \sqrt [3]{-3-2 x+x^2}} \, dx=\int \frac {1}{\left (x^{2}-2 x -3\right )^{\frac {1}{3}} x -\left (x^{2}-2 x -3\right )^{\frac {1}{3}}}d x \] Input:
int(1/(-1+x)/(x^2-2*x-3)^(1/3),x)
Output:
int(1/((x**2 - 2*x - 3)**(1/3)*x - (x**2 - 2*x - 3)**(1/3)),x)