Integrand size = 58, antiderivative size = 208 \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 x-2 x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{b^{2/3}}+\frac {\log \left (-x+x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{b^{2/3}}-\frac {\log \left (x^2-2 x^3+x^4+\left (\sqrt [3]{b} x-\sqrt [3]{b} x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 b^{2/3}} \]
3^(1/2)*arctan(3^(1/2)*b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(2/3)/(2*x-2*x^2+b^(1/ 3)*(x+(-1-k)*x^2+k*x^3)^(2/3)))/b^(2/3)+ln(-x+x^2+b^(1/3)*(x+(-1-k)*x^2+k* x^3)^(2/3))/b^(2/3)-1/2*ln(x^2-2*x^3+x^4+(b^(1/3)*x-b^(1/3)*x^2)*(x+(-1-k) *x^2+k*x^3)^(2/3)+b^(2/3)*(x+(-1-k)*x^2+k*x^3)^(4/3))/b^(2/3)
Time = 15.66 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.79 \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}}{2 x-2 x^2+\sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}}\right )+2 \log \left (-x+x^2+\sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}\right )-\log \left (x^2-2 x^3+x^4-\sqrt [3]{b} (-1+x) x ((-1+x) x (-1+k x))^{2/3}+b^{2/3} ((-1+x) x (-1+k x))^{4/3}\right )}{2 b^{2/3}} \]
Integrate[(-1 + 2*x + (-2*k + k^2)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(b - (1 + 2*b*k)*x + (1 + b*k^2)*x^2)),x]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*((-1 + x)*x*(-1 + k*x))^(2/3))/(2*x - 2 *x^2 + b^(1/3)*((-1 + x)*x*(-1 + k*x))^(2/3))] + 2*Log[-x + x^2 + b^(1/3)* ((-1 + x)*x*(-1 + k*x))^(2/3)] - Log[x^2 - 2*x^3 + x^4 - b^(1/3)*(-1 + x)* x*((-1 + x)*x*(-1 + k*x))^(2/3) + b^(2/3)*((-1 + x)*x*(-1 + k*x))^(4/3)])/ (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.
\(\displaystyle \int \frac {\left (k^2-2 k\right ) x^2+2 x-1}{((1-x) x (1-k x))^{2/3} \left (x^2 \left (b k^2+1\right )-x (2 b k+1)+b\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {(2-k) k x^2-2 x+1}{x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(2-k) k x^2-2 x+1}{x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(2-k) k x^2-2 x+1}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \left (\frac {(2-k) k}{\left (b k^2+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}+\frac {-2 b (1-k) k-\left (2 b k^3+(1-2 b) k^2-2 k+2\right ) x+1}{\left (b k^2+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3} \left (\left (b k^2+1\right ) x^2+(-2 b k-1) x+b\right )}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (-\frac {\left (2 b k^3+(1-2 b) k^2-(2-k) k \sqrt {1-4 b (1-k)}-2 k+2\right ) \int \frac {1}{\left (-2 b k+2 \left (b k^2+1\right ) x-\sqrt {4 k b-4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b k^2+1}-\frac {\left (2 b k^3+(1-2 b) k^2+(2-k) k \sqrt {1-4 b (1-k)}-2 k+2\right ) \int \frac {1}{\left (-2 b k+2 \left (b k^2+1\right ) x+\sqrt {4 k b-4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b k^2+1}+\frac {(2-k) k (1-x)^{2/3} \sqrt [3]{x} (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{\left (b k^2+1\right ) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\) |
Int[(-1 + 2*x + (-2*k + k^2)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(b - (1 + 2 *b*k)*x + (1 + b*k^2)*x^2)),x]
3.25.98.3.1 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]
\[\int \frac {-1+2 x +\left (k^{2}-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (b -\left (2 b k +1\right ) x +\left (b \,k^{2}+1\right ) x^{2}\right )}d x\]
Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\text {Timed out} \]
integrate((-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-(2*b*k+1)*x+( b*k^2+1)*x^2),x, algorithm="fricas")
Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\text {Timed out} \]
integrate((-1+2*x+(k**2-2*k)*x**2)/((1-x)*x*(-k*x+1))**(2/3)/(b-(2*b*k+1)* x+(b*k**2+1)*x**2),x)
\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int { \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b k^{2} + 1\right )} x^{2} - {\left (2 \, b k + 1\right )} x + b\right )}} \,d x } \]
integrate((-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-(2*b*k+1)*x+( b*k^2+1)*x^2),x, algorithm="maxima")
integrate(((k^2 - 2*k)*x^2 + 2*x - 1)/(((k*x - 1)*(x - 1)*x)^(2/3)*((b*k^2 + 1)*x^2 - (2*b*k + 1)*x + b)), x)
\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int { \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b k^{2} + 1\right )} x^{2} - {\left (2 \, b k + 1\right )} x + b\right )}} \,d x } \]
integrate((-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-(2*b*k+1)*x+( b*k^2+1)*x^2),x, algorithm="giac")
integrate(((k^2 - 2*k)*x^2 + 2*x - 1)/(((k*x - 1)*(x - 1)*x)^(2/3)*((b*k^2 + 1)*x^2 - (2*b*k + 1)*x + b)), x)
Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int -\frac {\left (2\,k-k^2\right )\,x^2-2\,x+1}{\left (\left (b\,k^2+1\right )\,x^2+\left (-2\,b\,k-1\right )\,x+b\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}} \,d x \]
int(-(x^2*(2*k - k^2) - 2*x + 1)/((b + x^2*(b*k^2 + 1) - x*(2*b*k + 1))*(x *(k*x - 1)*(x - 1))^(2/3)),x)