Integrand size = 62, antiderivative size = 205 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\frac {3 \sqrt [3]{x+(-1-k) x^2+k x^3}}{x}+\frac {\left (\sqrt {3} a+\sqrt {3} b\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {(a+b) \log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}+\frac {(-a-b) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]
3*(x+(-1-k)*x^2+k*x^3)^(1/3)/x+(3^(1/2)*a+3^(1/2)*b)*arctan(3^(1/2)*b^(1/3 )*x/(b^(1/3)*x+2*(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(2/3)+(a+b)*ln(-b^(1/3)*x+ (x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)+1/2*(-a-b)*ln(b^(2/3)*x^2+b^(1/3)*x*(x +(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)
Time = 15.39 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.81 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\frac {6 b^{2/3} \sqrt [3]{(-1+x) x (-1+k x)}+2 \sqrt {3} (a+b) x \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 (a+b) x \log \left (-\sqrt [3]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )-(a+b) x \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{2 b^{2/3} x} \]
Integrate[((-2 + (1 + k)*x)*(1 - (1 + k)*x + (a + k)*x^2))/(x*((1 - x)*x*( 1 - k*x))^(2/3)*(1 - (1 + k)*x + (-b + k)*x^2)),x]
(6*b^(2/3)*((-1 + x)*x*(-1 + k*x))^(1/3) + 2*Sqrt[3]*(a + b)*x*ArcTan[(Sqr t[3]*b^(1/3)*x)/(b^(1/3)*x + 2*((-1 + x)*x*(-1 + k*x))^(1/3))] + 2*(a + b) *x*Log[-(b^(1/3)*x) + ((-1 + x)*x*(-1 + k*x))^(1/3)] - (a + b)*x*Log[b^(2/ 3)*x^2 + b^(1/3)*x*((-1 + x)*x*(-1 + k*x))^(1/3) + ((-1 + x)*x*(-1 + k*x)) ^(2/3)])/(2*b^(2/3)*x)
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.
| \(\displaystyle \int \frac {((k+1) x-2) \left (x^2 (a+k)-(k+1) x+1\right )}{x ((1-x) x (1-k x))^{2/3} \left (x^2 (k-b)-(k+1) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {(2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{x^{5/3} \left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{x^{5/3} \left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{x \left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \left (-\frac {(k+1) (a+k)}{(k-b) \left (k x^2-(k+1) x+1\right )^{2/3}}+\frac {2}{x \left (k x^2-(k+1) x+1\right )^{2/3}}+\frac {(a+b) \left (-k+\left (k^2+2 b+1\right ) x-1\right )}{(b-k) \left (-\left ((b-k) x^2\right )-(k+1) x+1\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (\frac {(a+b) \left ((k+1) \sqrt {4 b+(k-1)^2}+2 b+k^2+1\right ) \int \frac {1}{\left (-k+2 (k-b) x-\sqrt {k^2-2 k+4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b-k}+\frac {(a+b) \left (-(k+1) \sqrt {4 b+(k-1)^2}+2 b+k^2+1\right ) \int \frac {1}{\left (-k+2 (k-b) x+\sqrt {k^2-2 k+4 b+1}-1\right ) \left (k x^2+(-k-1) x+1\right )^{2/3}}d\sqrt [3]{x}}{b-k}+\frac {(k+1) (1-x)^{2/3} \sqrt [3]{x} (a+k) (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{(b-k) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}-\frac {3 \sqrt [3]{2} (1-x)^{2/3} \operatorname {Gamma}\left (\frac {2}{3}\right ) \operatorname {Gamma}\left (\frac {7}{6}\right ) (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{k-\sqrt {(k-1)^2}+1}} \left (\left (1-\frac {6 k x}{k-\sqrt {(k-1)^2}+1}\right ) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {4}{3},\frac {2 \sqrt {(k-1)^2} x}{2-\left (k-\sqrt {(k-1)^2}+1\right ) x}\right )+\frac {12 \sqrt {(k-1)^2} k (1-x) x (1-k x) \operatorname {Hypergeometric2F1}\left (\frac {5}{3},2,\frac {7}{3},\frac {2 \sqrt {(k-1)^2} x}{2-\left (k-\sqrt {(k-1)^2}+1\right ) x}\right )}{\left (k-\sqrt {(k-1)^2}+1\right ) \left (-2 k x+k+\sqrt {(k-1)^2}+1\right ) \left (2-\left (k-\sqrt {(k-1)^2}+1\right ) x\right )}\right )}{\sqrt {\pi } x^{2/3} \operatorname {Gamma}\left (\frac {1}{3}\right ) \left (1-\frac {2 k x}{k+\sqrt {(k-1)^2}+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\) |
Int[((-2 + (1 + k)*x)*(1 - (1 + k)*x + (a + k)*x^2))/(x*((1 - x)*x*(1 - k* x))^(2/3)*(1 - (1 + k)*x + (-b + k)*x^2)),x]
3.25.89.3.1 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 = 1.07 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.67
| method | result | size |
| pseudoelliptic | \(-\frac {-6 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} b^{\frac {2}{3}}+\left (2 \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )\right ) x \left (a +b \right )}{2 b^{\frac {2}{3}} x}\) | \(138\) |
int((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/x/((1-x)*x*(-k*x+1))^(2/3)/(1-(1+k) *x+(-b+k)*x^2),x,method=_RETURNVERBOSE)
-1/2*(-6*((-1+x)*x*(k*x-1))^(1/3)*b^(2/3)+(2*arctan(1/3*3^(1/2)*(b^(1/3)*x +2*((-1+x)*x*(k*x-1))^(1/3))/b^(1/3)/x)*3^(1/2)+ln((b^(2/3)*x^2+b^(1/3)*(( -1+x)*x*(k*x-1))^(1/3)*x+((-1+x)*x*(k*x-1))^(2/3))/x^2)-2*ln((-b^(1/3)*x+( (-1+x)*x*(k*x-1))^(1/3))/x))*x*(a+b))/b^(2/3)/x
Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/x/((1-x)*x*(-k*x+1))^(2/3)/(1 -(1+k)*x+(-b+k)*x^2),x, algorithm="fricas")
Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x**2)/x/((1-x)*x*(-k*x+1))**(2/3)/ (1-(1+k)*x+(-b+k)*x**2),x)
\[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int { -\frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )} x} \,d x } \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/x/((1-x)*x*(-k*x+1))^(2/3)/(1 -(1+k)*x+(-b+k)*x^2),x, algorithm="maxima")
-integrate(((a + k)*x^2 - (k + 1)*x + 1)*((k + 1)*x - 2)/(((k*x - 1)*(x - 1)*x)^(2/3)*((b - k)*x^2 + (k + 1)*x - 1)*x), x)
\[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int { -\frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )} x} \,d x } \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/x/((1-x)*x*(-k*x+1))^(2/3)/(1 -(1+k)*x+(-b+k)*x^2),x, algorithm="giac")
integrate(-((a + k)*x^2 - (k + 1)*x + 1)*((k + 1)*x - 2)/(((k*x - 1)*(x - 1)*x)^(2/3)*((b - k)*x^2 + (k + 1)*x - 1)*x), x)
Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x ((1-x) x (1-k x))^{2/3} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int -\frac {\left (x\,\left (k+1\right )-2\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{x\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b-k\right )\,x^2+\left (k+1\right )\,x-1\right )} \,d x \]
int(-((x*(k + 1) - 2)*(x^2*(a + k) - x*(k + 1) + 1))/(x*(x*(k*x - 1)*(x - 1))^(2/3)*(x*(k + 1) + x^2*(b - k) - 1)),x)