Integrand size = 84, antiderivative size = 270 \[ \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{2/3}}+\frac {\log \left (x-\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\log \left (x+\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}-\frac {\log \left (x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}-\frac {\log \left (x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}} \]
-1/2*3^(1/2)*arctan(3^(1/2)*x^2/(x^2+2*b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(2/3)) )/b^(2/3)+1/2*ln(x-b^(1/6)*(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)+1/2*ln(x+b^ (1/6)*(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)-1/4*ln(x^2-b^(1/6)*x*(x+(-1-k)*x ^2+k*x^3)^(1/3)+b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)-1/4*ln(x^2+b^( 1/6)*x*(x+(-1-k)*x^2+k*x^3)^(1/3)+b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2 /3)
Time = 10.03 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.73 \[ \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}}\right )-2 \log \left (x-\sqrt [6]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )-2 \log \left (x+\sqrt [6]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (x^2-\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+\sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}\right )+\log \left (x^2+\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+\sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}\right )}{4 b^{2/3}} \]
Integrate[((-1 + x)*(-1 + k*x)*(-2*x + (1 + k)*x^2))/(((1 - x)*x*(1 - k*x) )^(2/3)*(b - 2*b*(1 + k)*x + (b + 4*b*k + b*k^2)*x^2 - 2*b*k*(1 + k)*x^3 + (-1 + b*k^2)*x^4)),x]
-1/4*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^2)/(x^2 + 2*b^(1/3)*((-1 + x)*x*(-1 + k* x))^(2/3))] - 2*Log[x - b^(1/6)*((-1 + x)*x*(-1 + k*x))^(1/3)] - 2*Log[x + b^(1/6)*((-1 + x)*x*(-1 + k*x))^(1/3)] + Log[x^2 - b^(1/6)*x*((-1 + x)*x* (-1 + k*x))^(1/3) + b^(1/3)*((-1 + x)*x*(-1 + k*x))^(2/3)] + Log[x^2 + b^( 1/6)*x*((-1 + x)*x*(-1 + k*x))^(1/3) + b^(1/3)*((-1 + x)*x*(-1 + k*x))^(2/ 3)])/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.
| \(\displaystyle \int \frac {(x-1) (k x-1) \left ((k+1) x^2-2 x\right )}{((1-x) x (1-k x))^{2/3} \left (x^4 \left (b k^2-1\right )+x^2 \left (b k^2+4 b k+b\right )-2 b k (k+1) x^3-2 b (k+1) x+b\right )} \, dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {(x-1) x (k x-1) ((k+1) x-2)}{((1-x) x (1-k x))^{2/3} \left (x^4 \left (b k^2-1\right )+x^2 \left (b k^2+4 b k+b\right )-2 b k (k+1) x^3-2 b (k+1) x+b\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {(1-x) \sqrt [3]{x} (1-k x) (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (1-b k^2\right ) x^4\right )-2 b k (k+1) x^3+b \left (k^2+4 k+1\right ) x^2-2 b (k+1) x+b\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(1-x) \sqrt [3]{x} (1-k x) (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (1-b k^2\right ) x^4\right )-2 b k (k+1) x^3+b \left (k^2+4 k+1\right ) x^2-2 b (k+1) x+b\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(1-x) x (1-k x) (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (1-b k^2\right ) x^4\right )-2 b k (k+1) x^3+b \left (k^2+4 k+1\right ) x^2-2 b (k+1) x+b\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \int \frac {\sqrt [3]{1-x} x \sqrt [3]{1-k x} (2-(k+1) x)}{-\left (\left (1-b k^2\right ) x^4\right )-2 b k (k+1) x^3+b \left (k^2+4 k+1\right ) x^2-2 b (k+1) x+b}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \int \frac {\sqrt [3]{1-x} x \sqrt [3]{1-k x} ((k+1) x-2)}{x^4-b (x-1)^2 (k x-1)^2}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \int \left (\frac {(-k-1) \sqrt [3]{1-x} \sqrt [3]{1-k x} x^2}{-\left (\left (1-b k^2\right ) x^4\right )-2 b k (k+1) x^3+b (k (k+4)+1) x^2-2 b (k+1) x+b}+\frac {2 \sqrt [3]{1-x} \sqrt [3]{1-k x} x}{-\left (\left (1-b k^2\right ) x^4\right )-2 b k (k+1) x^3+b (k (k+4)+1) x^2-2 b (k+1) x+b}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \left (2 \int \frac {\sqrt [3]{1-x} x \sqrt [3]{1-k x}}{b (x-1)^2 (k x-1)^2-x^4}d\sqrt [3]{x}-(k+1) \int \frac {\sqrt [3]{1-x} x^2 \sqrt [3]{1-k x}}{b (x-1)^2 (k x-1)^2-x^4}d\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\) |
Int[((-1 + x)*(-1 + k*x)*(-2*x + (1 + k)*x^2))/(((1 - x)*x*(1 - k*x))^(2/3 )*(b - 2*b*(1 + k)*x + (b + 4*b*k + b*k^2)*x^2 - 2*b*k*(1 + k)*x^3 + (-1 + b*k^2)*x^4)),x]
3.28.93.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.72 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.53
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {1}{b}\right )^{\frac {1}{3}} x^{2}+2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}\right )}{3 \left (\frac {1}{b}\right )^{\frac {1}{3}} x^{2}}\right )+2 \ln \left (\frac {-\left (\frac {1}{b}\right )^{\frac {1}{3}} x^{2}+\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )-\ln \left (\frac {\left (\frac {1}{b}\right )^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}} x +\left (-1+x \right ) \left (k x -1\right ) \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}+\left (\frac {1}{b}\right )^{\frac {2}{3}} x^{3}}{x^{3}}\right )}{4 \left (\frac {1}{b}\right )^{\frac {1}{3}} b}\) | \(142\) |
int((-1+x)*(k*x-1)*(-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-2*b*(1+k)* x+(b*k^2+4*b*k+b)*x^2-2*b*k*(1+k)*x^3+(b*k^2-1)*x^4),x,method=_RETURNVERBO SE)
1/4*(2*3^(1/2)*arctan(1/3*3^(1/2)*((1/b)^(1/3)*x^2+2*((-1+x)*x*(k*x-1))^(2 /3))/(1/b)^(1/3)/x^2)+2*ln((-(1/b)^(1/3)*x^2+((-1+x)*x*(k*x-1))^(2/3))/x^2 )-ln(((1/b)^(1/3)*((-1+x)*x*(k*x-1))^(2/3)*x+(-1+x)*(k*x-1)*((-1+x)*x*(k*x -1))^(1/3)+(1/b)^(2/3)*x^3)/x^3))/(1/b)^(1/3)/b
Timed out. \[ \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate((-1+x)*(k*x-1)*(-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-2*b* (1+k)*x+(b*k^2+4*b*k+b)*x^2-2*b*k*(1+k)*x^3+(b*k^2-1)*x^4),x, algorithm="f ricas")
Timed out. \[ \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate((-1+x)*(k*x-1)*(-2*x+(1+k)*x**2)/((1-x)*x*(-k*x+1))**(2/3)/(b-2* b*(1+k)*x+(b*k**2+4*b*k+b)*x**2-2*b*k*(1+k)*x**3+(b*k**2-1)*x**4),x)
\[ \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\int { -\frac {{\left ({\left (k + 1\right )} x^{2} - 2 \, x\right )} {\left (k x - 1\right )} {\left (x - 1\right )}}{{\left (2 \, b {\left (k + 1\right )} k x^{3} - {\left (b k^{2} - 1\right )} x^{4} + 2 \, b {\left (k + 1\right )} x - {\left (b k^{2} + 4 \, b k + b\right )} x^{2} - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}} \,d x } \]
integrate((-1+x)*(k*x-1)*(-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-2*b* (1+k)*x+(b*k^2+4*b*k+b)*x^2-2*b*k*(1+k)*x^3+(b*k^2-1)*x^4),x, algorithm="m axima")
-integrate(((k + 1)*x^2 - 2*x)*(k*x - 1)*(x - 1)/((2*b*(k + 1)*k*x^3 - (b* k^2 - 1)*x^4 + 2*b*(k + 1)*x - (b*k^2 + 4*b*k + b)*x^2 - b)*((k*x - 1)*(x - 1)*x)^(2/3)), x)
Time = 0.60 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.07 \[ \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=-\frac {{\left | b \right |} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{2 \, \left (-b^{5}\right )^{\frac {1}{3}}} + \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{4}} - \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {2}{3}} \arctan \left (-\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} - 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{4}} - \frac {\left (-b^{5}\right )^{\frac {2}{3}} \log \left (\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{4}} - \frac {\left (-b^{5}\right )^{\frac {2}{3}} \log \left (-\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{4}} \]
integrate((-1+x)*(k*x-1)*(-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-2*b* (1+k)*x+(b*k^2+4*b*k+b)*x^2-2*b*k*(1+k)*x^3+(b*k^2-1)*x^4),x, algorithm="g iac")
-1/2*abs(b)*log((k - k/x - 1/x + 1/x^2)^(2/3) + (-1/b)^(1/3))/(-b^5)^(1/3) + 1/2*sqrt(3)*(-b^5)^(2/3)*arctan((sqrt(3)*(-1/b)^(1/6) + 2*(k - k/x - 1/ x + 1/x^2)^(1/3))/(-1/b)^(1/6))/b^4 - 1/2*sqrt(3)*(-b^5)^(2/3)*arctan(-(sq rt(3)*(-1/b)^(1/6) - 2*(k - k/x - 1/x + 1/x^2)^(1/3))/(-1/b)^(1/6))/b^4 - 1/4*(-b^5)^(2/3)*log(sqrt(3)*(k - k/x - 1/x + 1/x^2)^(1/3)*(-1/b)^(1/6) + (k - k/x - 1/x + 1/x^2)^(2/3) + (-1/b)^(1/3))/b^4 - 1/4*(-b^5)^(2/3)*log(- sqrt(3)*(k - k/x - 1/x + 1/x^2)^(1/3)*(-1/b)^(1/6) + (k - k/x - 1/x + 1/x^ 2)^(2/3) + (-1/b)^(1/3))/b^4
Timed out. \[ \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\int -\frac {\left (2\,x-x^2\,\left (k+1\right )\right )\,\left (k\,x-1\right )\,\left (x-1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b\,k^2-1\right )\,x^4-2\,b\,k\,\left (k+1\right )\,x^3+\left (b\,k^2+4\,b\,k+b\right )\,x^2-2\,b\,\left (k+1\right )\,x+b\right )} \,d x \]
int(-((2*x - x^2*(k + 1))*(k*x - 1)*(x - 1))/((x*(k*x - 1)*(x - 1))^(2/3)* (b + x^4*(b*k^2 - 1) + x^2*(b + 4*b*k + b*k^2) - 2*b*x*(k + 1) - 2*b*k*x^3 *(k + 1))),x)