Integrand size = 49, antiderivative size = 154 \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \]
3^(1/2)*arctan(3^(1/2)*x/(x+2*b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(1/3) +ln(x-b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(1/3)-1/2*ln(x^2+b^(1/3)*x*(x+ (-1-k)*x^2+k*x^3)^(1/3)+b^(2/3)*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(1/3)
Time = 15.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.81 \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 \log \left (x-\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )-\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+b^{2/3} ((-1+x) x (-1+k x))^{2/3}\right )}{2 \sqrt [3]{b}} \]
Integrate[(-2*x + (1 + k)*x^2)/(((1 - x)*x*(1 - k*x))^(2/3)*(b - b*(1 + k) *x + (-1 + b*k)*x^2)),x]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*b^(1/3)*((-1 + x)*x*(-1 + k*x))^(1/3) )] + 2*Log[x - b^(1/3)*((-1 + x)*x*(-1 + k*x))^(1/3)] - Log[x^2 + b^(1/3)* x*((-1 + x)*x*(-1 + k*x))^(1/3) + b^(2/3)*((-1 + x)*x*(-1 + k*x))^(2/3)])/ (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.
| \(\displaystyle \int \frac {(k+1) x^2-2 x}{((1-x) x (1-k x))^{2/3} \left (x^2 (b k-1)-b (k+1) x+b\right )} \, dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {x ((k+1) x-2)}{((1-x) x (1-k x))^{2/3} \left (x^2 (b k-1)-b (k+1) x+b\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {\sqrt [3]{x} (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {\sqrt [3]{x} (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {x (2-(k+1) x)}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \left (\frac {k+1}{(1-b k) \left (k x^2-(k+1) x+1\right )^{2/3}}+\frac {b (k+1)-\left (b k^2+b+2\right ) x}{(b k-1) \left (k x^2-(k+1) x+1\right )^{2/3} \left ((b k-1) x^2-b (k+1) x+b\right )}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (\frac {\left (b \left (k^2+1\right )+\sqrt {b} (k+1) \sqrt {b (1-k)^2+4}+2\right ) \int \frac {1}{\left (-b (k+1)+2 (b k-1) x-\sqrt {b} \sqrt {b k^2-2 b k+b+4}\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{1-b k}+\frac {\left (b \left (k^2+1\right )-\sqrt {b} (k+1) \sqrt {b (1-k)^2+4}+2\right ) \int \frac {1}{\left (-b (k+1)+2 (b k-1) x+\sqrt {b} \sqrt {b k^2-2 b k+b+4}\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{1-b k}+\frac {(k+1) (1-x)^{2/3} \sqrt [3]{x} (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{(1-b k) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\) |
3.22.18.3.1 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 = 1.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.84
| method | result | size |
| pseudoelliptic | \(\frac {-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {1}{b}\right )^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{b}\right )^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-\left (\frac {1}{b}\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {1}{b}\right )^{\frac {2}{3}} x^{2}+\left (\frac {1}{b}\right )^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 \left (\frac {1}{b}\right )^{\frac {2}{3}} b}\) | \(129\) |
int((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-b*(1+k)*x+(b*k-1)*x^2),x, method=_RETURNVERBOSE)
1/2*(-2*3^(1/2)*arctan(1/3*3^(1/2)*((1/b)^(1/3)*x+2*((-1+x)*x*(k*x-1))^(1/ 3))/(1/b)^(1/3)/x)+2*ln((-(1/b)^(1/3)*x+((-1+x)*x*(k*x-1))^(1/3))/x)-ln((( 1/b)^(2/3)*x^2+(1/b)^(1/3)*((-1+x)*x*(k*x-1))^(1/3)*x+((-1+x)*x*(k*x-1))^( 2/3))/x^2))/(1/b)^(2/3)/b
Timed out. \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\text {Timed out} \]
integrate((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-b*(1+k)*x+(b*k-1)*x ^2),x, algorithm="fricas")
Timed out. \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\int { -\frac {{\left (k + 1\right )} x^{2} - 2 \, x}{{\left (b {\left (k + 1\right )} x - {\left (b k - 1\right )} x^{2} - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}} \,d x } \]
integrate((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-b*(1+k)*x+(b*k-1)*x ^2),x, algorithm="maxima")
-integrate(((k + 1)*x^2 - 2*x)/((b*(k + 1)*x - (b*k - 1)*x^2 - b)*((k*x - 1)*(x - 1)*x)^(2/3)), x)
Time = 0.49 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=-\frac {\sqrt {3} {\left | b \right |}^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + \frac {1}{b^{\frac {1}{3}}}\right )}\right )}{b} - \frac {{\left | b \right |}^{\frac {2}{3}} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{b^{\frac {1}{3}}} + \frac {1}{b^{\frac {2}{3}}}\right )}{2 \, b} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} - \frac {1}{b^{\frac {1}{3}}} \right |}\right )}{b^{\frac {1}{3}}} \]
integrate((-2*x+(1+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(b-b*(1+k)*x+(b*k-1)*x ^2),x, algorithm="giac")
-sqrt(3)*abs(b)^(2/3)*arctan(1/3*sqrt(3)*b^(1/3)*(2*(k - k/x - 1/x + 1/x^2 )^(1/3) + 1/b^(1/3)))/b - 1/2*abs(b)^(2/3)*log((k - k/x - 1/x + 1/x^2)^(2/ 3) + (k - k/x - 1/x + 1/x^2)^(1/3)/b^(1/3) + 1/b^(2/3))/b + log(abs((k - k /x - 1/x + 1/x^2)^(1/3) - 1/b^(1/3)))/b^(1/3)
Timed out. \[ \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\int -\frac {2\,x-x^2\,\left (k+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b\,k-1\right )\,x^2-b\,\left (k+1\right )\,x+b\right )} \,d x \]