Integrand size = 44, antiderivative size = 161 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \]
-3^(1/2)*arctan(3^(1/2)*b^(1/3)*x/(b^(1/3)*x+2*(x+(-1-k)*x^2+k*x^3)^(1/3)) )/b^(1/3)+ln(-b^(1/3)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(1/3)-1/2*ln(b^(2/3) *x^2+b^(1/3)*x*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(1 /3)
Time = 15.25 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.80 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )-2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{2 \sqrt [3]{b}} \]
-1/2*(2*Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3)*x + 2*((-1 + x)*x*(-1 + k*x))^(1/3))] - 2*Log[-(b^(1/3)*x) + ((-1 + x)*x*(-1 + k*x))^(1/3)] + Lo g[b^(2/3)*x^2 + b^(1/3)*x*((-1 + x)*x*(-1 + k*x))^(1/3) + ((-1 + x)*x*(-1 + k*x))^(2/3)])/b^(1/3)
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.
| \(\displaystyle \int \frac {(k+1) x-2}{\sqrt [3]{(1-x) x (1-k x)} \left (x^2 (k-b)-(k+1) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {2-(k+1) x}{\sqrt [3]{x} \left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \sqrt [3]{k x^2-(k+1) x+1}}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {2-(k+1) x}{\sqrt [3]{x} \left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \sqrt [3]{k x^2-(k+1) x+1}}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} (2-(k+1) x)}{\left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \sqrt [3]{k x^2-(k+1) x+1}}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {(-k-1) x^{4/3}}{\left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \sqrt [3]{k x^2-(k+1) x+1}}+\frac {2 \sqrt [3]{x}}{\left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \sqrt [3]{k x^2-(k+1) x+1}}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\frac {4 (b-k) \int \frac {\sqrt [3]{x}}{\left (-k-2 (b-k) x-\sqrt {k^2-2 k+4 b+1}-1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {4 b+(k-1)^2}}-\frac {4 (b-k) \int \frac {\sqrt [3]{x}}{\left (k+2 (b-k) x-\sqrt {k^2-2 k+4 b+1}+1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {4 b+(k-1)^2}}-\frac {(k+1) \left (-\sqrt {4 b+(k-1)^2}+k+1\right ) \int \frac {\sqrt [3]{x}}{\left (k+2 (b-k) x-\sqrt {k^2-2 k+4 b+1}+1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {4 b+(k-1)^2}}+\frac {(k+1) \left (\sqrt {4 b+(k-1)^2}+k+1\right ) \int \frac {\sqrt [3]{x}}{\left (k+2 (b-k) x+\sqrt {k^2-2 k+4 b+1}+1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {4 b+(k-1)^2}}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
3.22.74.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.60 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 b^{\frac {1}{3}}}\) | \(114\) |
1/2*(2*3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)*x+2*((-1+x)*x*(k*x-1))^(1/3))/b ^(1/3)/x)+2*ln((-b^(1/3)*x+((-1+x)*x*(k*x-1))^(1/3))/x)-ln((b^(2/3)*x^2+b^ (1/3)*((-1+x)*x*(k*x-1))^(1/3)*x+((-1+x)*x*(k*x-1))^(2/3))/x^2))/b^(1/3)
Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int { -\frac {{\left (k + 1\right )} x - 2}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.75 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {\log \left (b^{\frac {2}{3}} + b^{\frac {1}{3}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}}\right )}{2 \, b^{\frac {1}{3}}} + \frac {\log \left ({\left | -b^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \right |}\right )}{b^{\frac {1}{3}}} \]
sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(k - k/x - 1/x + 1/x^2)^(1/3))/b^( 1/3))/b^(1/3) - 1/2*log(b^(2/3) + b^(1/3)*(k - k/x - 1/x + 1/x^2)^(1/3) + (k - k/x - 1/x + 1/x^2)^(2/3))/b^(1/3) + log(abs(-b^(1/3) + (k - k/x - 1/x + 1/x^2)^(1/3)))/b^(1/3)
Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int -\frac {x\,\left (k+1\right )-2}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b-k\right )\,x^2+\left (k+1\right )\,x-1\right )} \,d x \]