Integrand size = 48, antiderivative size = 160 \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]
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^(2/3)+ln(-b^(1/3)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)-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^(2/ 3)
Time = 15.30 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.81 \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )-\log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{2 b^{2/3}} \]
Integrate[(-2*x + (1 + k)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(1 - (1 + k)*x + (-b + k)*x^2)),x]
(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)] - Log[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)])/(2*b^(2/3))
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.
| \(\displaystyle \int \frac {(k+1) x^2-2 x}{((1-x) x (1-k x))^{2/3} \left (x^2 (k-b)-(k+1) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {x ((k+1) x-2)}{((1-x) x (1-k x))^{2/3} \left (x^2 (k-b)-(k+1) x+1\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {\sqrt [3]{x} (2-(k+1) x)}{\left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {\sqrt [3]{x} (2-(k+1) x)}{\left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {x (2-(k+1) x)}{\left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \left (\frac {k+1}{(b-k) \left (k x^2-(k+1) x+1\right )^{2/3}}-\frac {k-\left (k^2+2 b+1\right ) x+1}{(b-k) \left (k x^2-(k+1) x+1\right )^{2/3} \left ((k-b) x^2+(-k-1) x+1\right )}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (\frac {\left ((k+1) \sqrt {4 b+(k-1)^2}+2 b+k^2+1\right ) \int \frac {1}{\left (-k+2 (k-b) x-\sqrt {k^2-2 k+4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b-k}+\frac {\left (-(k+1) \sqrt {4 b+(k-1)^2}+2 b+k^2+1\right ) \int \frac {1}{\left (-k+2 (k-b) x+\sqrt {k^2-2 k+4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b-k}+\frac {(k+1) (1-x)^{2/3} \sqrt [3]{x} (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{(b-k) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\) |
3.22.72.3.1 Defintions of rubi rules used
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
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 = 1.11 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(\frac {-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 b^{\frac {2}{3}}}\) | \(114\) |
int((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(1+k)*x+(-b+k)*x^2),x,met hod=_RETURNVERBOSE)
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^(2/3)
Timed out. \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
integrate((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(1+k)*x+(-b+k)*x^2) ,x, algorithm="fricas")
Timed out. \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int { -\frac {{\left (k + 1\right )} x^{2} - 2 \, x}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )}} \,d x } \]
integrate((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(1+k)*x+(-b+k)*x^2) ,x, algorithm="maxima")
-integrate(((k + 1)*x^2 - 2*x)/(((k*x - 1)*(x - 1)*x)^(2/3)*((b - k)*x^2 + (k + 1)*x - 1)), x)
Time = 0.47 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76 \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {2}{3}}} - \frac {\log \left (b^{\frac {2}{3}} + b^{\frac {1}{3}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}}\right )}{2 \, b^{\frac {2}{3}}} + \frac {\log \left ({\left | -b^{\frac {1}{3}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \right |}\right )}{b^{\frac {2}{3}}} \]
integrate((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(1+k)*x+(-b+k)*x^2) ,x, algorithm="giac")
-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^(2/3) - 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^(2/3) + log(abs(-b^(1/3) + (k - k/x - 1/ x + 1/x^2)^(1/3)))/b^(2/3)
Timed out. \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int \frac {2\,x-x^2\,\left (k+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b-k\right )\,x^2+\left (k+1\right )\,x-1\right )} \,d x \]