Integrand size = 45, antiderivative size = 155 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]
-3^(1/2)*arctan(3^(1/2)*x/(x+2*b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(2/3 )+ln(x-b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)-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^(2/3)
Time = 15.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.79 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )-2 \log \left (x-\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+b^{2/3} ((-1+x) x (-1+k x))^{2/3}\right )}{2 b^{2/3}} \]
Integrate[(-2 + (1 + k)*x)/(((1 - x)*x*(1 - k*x))^(1/3)*(b - b*(1 + k)*x + (-1 + b*k)*x^2)),x]
-1/2*(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)])/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.
| \(\displaystyle \int \frac {(k+1) x-2}{\sqrt [3]{(1-x) x (1-k x)} \left (x^2 (b k-1)-b (k+1) x+b\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {2-(k+1) x}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {2-(k+1) x}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} (2-(k+1) x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {(-k-1) x^{4/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}+\frac {2 \sqrt [3]{x}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left ((1-b k) x^2\right )-b (k+1) x+b\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\frac {4 (1-b k) \int \frac {\sqrt [3]{x}}{\left (-k b-b-\sqrt {b k^2-2 b k+b+4} \sqrt {b}-2 (1-b k) x\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {b} \sqrt {b (1-k)^2+4}}-\frac {4 (1-b k) \int \frac {\sqrt [3]{x}}{\left (b (k+1)+2 (1-b k) x-\sqrt {b} \sqrt {b k^2-2 b k+b+4}\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {b} \sqrt {b (1-k)^2+4}}+(k+1) \left (1-\frac {\sqrt {b} (k+1)}{\sqrt {b (1-k)^2+4}}\right ) \int \frac {\sqrt [3]{x}}{\left (b (k+1)+2 (1-b k) x-\sqrt {b} \sqrt {b k^2-2 b k+b+4}\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}+(k+1) \left (\frac {\sqrt {b} (k+1)}{\sqrt {b (1-k)^2+4}}+1\right ) \int \frac {\sqrt [3]{x}}{\left (b (k+1)+2 (1-b k) x+\sqrt {b} \sqrt {b k^2-2 b k+b+4}\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
3.22.33.3.1 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]
Time = 0.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83
| method | result | size |
| pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {1}{b}\right )^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{b}\right )^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-\left (\frac {1}{b}\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {1}{b}\right )^{\frac {2}{3}} x^{2}+\left (\frac {1}{b}\right )^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 \left (\frac {1}{b}\right )^{\frac {1}{3}} b}\) | \(129\) |
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)^(1/3)/b
Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\text {Timed out} \]
integrate((-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-b*(1+k)*x+(b*k-1)*x^2), x, algorithm="fricas")
Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\int { -\frac {{\left (k + 1\right )} x - 2}{{\left (b {\left (k + 1\right )} x - {\left (b k - 1\right )} x^{2} - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-b*(1+k)*x+(b*k-1)*x^2), x, algorithm="maxima")
-integrate(((k + 1)*x - 2)/((b*(k + 1)*x - (b*k - 1)*x^2 - b)*((k*x - 1)*( x - 1)*x)^(1/3)), x)
Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\frac {\sqrt {3} {\left | b \right |}^{\frac {4}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + \frac {1}{b^{\frac {1}{3}}}\right )}\right )}{b^{2}} - \frac {{\left | b \right |}^{\frac {4}{3}} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{b^{\frac {1}{3}}} + \frac {1}{b^{\frac {2}{3}}}\right )}{2 \, b^{2}} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} - \frac {1}{b^{\frac {1}{3}}} \right |}\right )}{b^{\frac {2}{3}}} \]
integrate((-2+(1+k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-b*(1+k)*x+(b*k-1)*x^2), x, algorithm="giac")
sqrt(3)*abs(b)^(4/3)*arctan(1/3*sqrt(3)*b^(1/3)*(2*(k - k/x - 1/x + 1/x^2) ^(1/3) + 1/b^(1/3)))/b^2 - 1/2*abs(b)^(4/3)*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^2 + log(abs((k - k/x - 1/x + 1/x^2)^(1/3) - 1/b^(1/3)))/b^(2/3)
Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\int \frac {x\,\left (k+1\right )-2}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b\,k-1\right )\,x^2-b\,\left (k+1\right )\,x+b\right )} \,d x \]