Integrand size = 52, antiderivative size = 230 \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 \sqrt [3]{b} x-2 \sqrt [3]{b} k x^2+\left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt {b} x+\sqrt {b} k x^2+\sqrt [6]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b x^2-2 b k x^3+b k^2 x^4+\left (b^{2/3} x-b^{2/3} k x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 \sqrt [3]{b}} \]
3^(1/2)*arctan(3^(1/2)*(x+(-1-k)*x^2+k*x^3)^(2/3)/(2*b^(1/3)*x-2*b^(1/3)*k *x^2+(x+(-1-k)*x^2+k*x^3)^(2/3)))/b^(1/3)+ln(-x*b^(1/2)+b^(1/2)*k*x^2+b^(1 /6)*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(1/3)-1/2*ln(b*x^2-2*b*k*x^3+b*k^2*x^4+( b^(2/3)*x-b^(2/3)*k*x^2)*(x+(-1-k)*x^2+k*x^3)^(2/3)+b^(1/3)*(x+(-1-k)*x^2+ k*x^3)^(4/3))/b^(1/3)
Time = 14.20 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.76 \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} ((-1+x) x (-1+k x))^{2/3}}{-2 \sqrt [3]{b} x (-1+k x)+((-1+x) x (-1+k x))^{2/3}}\right )+2 \log \left (\sqrt [6]{b} \left (\sqrt [3]{b} x (-1+k x)+((-1+x) x (-1+k x))^{2/3}\right )\right )-\log \left (b x^2-2 b k x^3+b k^2 x^4+b^{2/3} x (1-k x) ((-1+x) x (-1+k x))^{2/3}+\sqrt [3]{b} ((-1+x) x (-1+k x))^{4/3}\right )}{2 \sqrt [3]{b}} \]
Integrate[(-1 + 2*k*x + (1 - 2*k)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(1 - ( 2 + b)*x + (1 + b*k)*x^2)),x]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*((-1 + x)*x*(-1 + k*x))^(2/3))/(-2*b^(1/3)*x*(- 1 + k*x) + ((-1 + x)*x*(-1 + k*x))^(2/3))] + 2*Log[b^(1/6)*(b^(1/3)*x*(-1 + k*x) + ((-1 + x)*x*(-1 + k*x))^(2/3))] - Log[b*x^2 - 2*b*k*x^3 + b*k^2*x ^4 + b^(2/3)*x*(1 - k*x)*((-1 + x)*x*(-1 + k*x))^(2/3) + b^(1/3)*((-1 + x) *x*(-1 + k*x))^(4/3)])/(2*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.
| \(\displaystyle \int \frac {(1-2 k) x^2+2 k x-1}{((1-x) x (1-k x))^{2/3} \left (x^2 (b k+1)-(b+2) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {-\left ((1-2 k) x^2\right )-2 k x+1}{x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left ((b k+1) x^2-(b+2) x+1\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {-\left ((1-2 k) x^2\right )-2 k x+1}{x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left ((b k+1) x^2-(b+2) x+1\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {-\left ((1-2 k) x^2\right )-2 k x+1}{\left (k x^2-(k+1) x+1\right )^{2/3} \left ((b k+1) x^2-(b+2) x+1\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \left (\frac {-((2-b) k)-\left (2 b k^2-2 b k-2 k+b+2\right ) x+2}{(b k+1) \left (k x^2-(k+1) x+1\right )^{2/3} \left ((b k+1) x^2+(-b-2) x+1\right )}-\frac {1-2 k}{(b k+1) \left (k x^2-(k+1) x+1\right )^{2/3}}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (-\frac {\left (b \left (2 k^2-2 k+1\right )+\sqrt {b} (1-2 k) \sqrt {b-4 k+4}-2 k+2\right ) \int \frac {1}{\left (-b-\sqrt {b-4 k+4} \sqrt {b}+2 (b k+1) x-2\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b k+1}-\frac {\left (b \left (2 k^2-2 k+1\right )-\sqrt {b} (1-2 k) \sqrt {b-4 k+4}-2 k+2\right ) \int \frac {1}{\left (-b+\sqrt {b-4 k+4} \sqrt {b}+2 (b k+1) x-2\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b k+1}-\frac {(1-2 k) (1-x)^{2/3} \sqrt [3]{x} (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{(b k+1) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\) |
Int[(-1 + 2*k*x + (1 - 2*k)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(1 - (2 + b) *x + (1 + b*k)*x^2)),x]
3.27.21.3.1 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]
\[\int \frac {-1+2 k x +\left (1-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (2+b \right ) x +\left (b k +1\right ) x^{2}\right )}d x\]
Timed out. \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\text {Timed out} \]
integrate((-1+2*k*x+(1-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(2+b)*x+(b*k+ 1)*x^2),x, algorithm="fricas")
Timed out. \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\int { -\frac {{\left (2 \, k - 1\right )} x^{2} - 2 \, k x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b k + 1\right )} x^{2} - {\left (b + 2\right )} x + 1\right )}} \,d x } \]
integrate((-1+2*k*x+(1-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(2+b)*x+(b*k+ 1)*x^2),x, algorithm="maxima")
-integrate(((2*k - 1)*x^2 - 2*k*x + 1)/(((k*x - 1)*(x - 1)*x)^(2/3)*((b*k + 1)*x^2 - (b + 2)*x + 1)), x)
\[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\int { -\frac {{\left (2 \, k - 1\right )} x^{2} - 2 \, k x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b k + 1\right )} x^{2} - {\left (b + 2\right )} x + 1\right )}} \,d x } \]
integrate((-1+2*k*x+(1-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-(2+b)*x+(b*k+ 1)*x^2),x, algorithm="giac")
integrate(-((2*k - 1)*x^2 - 2*k*x + 1)/(((k*x - 1)*(x - 1)*x)^(2/3)*((b*k + 1)*x^2 - (b + 2)*x + 1)), x)
Timed out. \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(2+b) x+(1+b k) x^2\right )} \, dx=\int -\frac {\left (2\,k-1\right )\,x^2-2\,k\,x+1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b\,k+1\right )\,x^2+\left (-b-2\right )\,x+1\right )} \,d x \]
int(-(x^2*(2*k - 1) - 2*k*x + 1)/((x*(k*x - 1)*(x - 1))^(2/3)*(x^2*(b*k + 1) - x*(b + 2) + 1)),x)