Integrand size = 85, antiderivative size = 383 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {\sqrt {3} \left (a-\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{2/3}}-\frac {\sqrt {3} \left (a+\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{2/3}}+\frac {\left (a+\sqrt {b}\right ) \log \left (-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\left (a-\sqrt {b}\right ) \log \left (\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\left (-a+\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\left (-a-\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}} \]
-1/2*3^(1/2)*(a-b^(1/2))*arctan(3^(1/2)*b^(1/6)*x/(b^(1/6)*x-2*(x+(-1-k)*x ^2+k*x^3)^(1/3)))/b^(2/3)-1/2*3^(1/2)*(a+b^(1/2))*arctan(3^(1/2)*b^(1/6)*x /(b^(1/6)*x+2*(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(2/3)+1/2*(a+b^(1/2))*ln(-b^( 1/6)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)+1/2*(a-b^(1/2))*ln(b^(1/6)*x+(x +(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)+1/4*(-a+b^(1/2))*ln(b^(1/3)*x^2-b^(1/6)* x*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)+1/4*(-a-b ^(1/2))*ln(b^(1/3)*x^2+b^(1/6)*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 = 11.10 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.78 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\frac {2 \sqrt {3} \left (-a+\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )-2 \sqrt {3} \left (a+\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 \left (a+\sqrt {b}\right ) \log \left (-\sqrt [6]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )+2 \left (a-\sqrt {b}\right ) \log \left (\sqrt [6]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )+\left (-a+\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )-\left (a+\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{4 b^{2/3}} \]
Integrate[((-2 + (1 + k)*x)*(1 - (1 + k)*x + (a + k)*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(1 - (2 + 2*k)*x + (1 + 4*k + k^2)*x^2 - 2*(k + k^2)*x^3 + ( -b + k^2)*x^4)),x]
(2*Sqrt[3]*(-a + Sqrt[b])*ArcTan[(Sqrt[3]*b^(1/6)*x)/(b^(1/6)*x - 2*((-1 + x)*x*(-1 + k*x))^(1/3))] - 2*Sqrt[3]*(a + Sqrt[b])*ArcTan[(Sqrt[3]*b^(1/6 )*x)/(b^(1/6)*x + 2*((-1 + x)*x*(-1 + k*x))^(1/3))] + 2*(a + Sqrt[b])*Log[ -(b^(1/6)*x) + ((-1 + x)*x*(-1 + k*x))^(1/3)] + 2*(a - Sqrt[b])*Log[b^(1/6 )*x + ((-1 + x)*x*(-1 + k*x))^(1/3)] + (-a + Sqrt[b])*Log[b^(1/3)*x^2 - b^ (1/6)*x*((-1 + x)*x*(-1 + k*x))^(1/3) + ((-1 + x)*x*(-1 + k*x))^(2/3)] - ( a + Sqrt[b])*Log[b^(1/3)*x^2 + b^(1/6)*x*((-1 + x)*x*(-1 + k*x))^(1/3) + ( (-1 + x)*x*(-1 + k*x))^(2/3)])/(4*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.
| \(\displaystyle \int \frac {((k+1) x-2) \left (x^2 (a+k)-(k+1) x+1\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (x^4 \left (k^2-b\right )-2 \left (k^2+k\right ) x^3+\left (k^2+4 k+1\right ) x^2-(2 k+2) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {(2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {(2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} (2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {(-a-k) (k+1) x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}+\frac {(k+1)^2 \left (\frac {2 (a+k)}{(k+1)^2}+1\right ) x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}+\frac {3 (-k-1) x^{4/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}+\frac {2 \sqrt [3]{x}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (\left (2 a+k^2+4 k+1\right ) \int \frac {x^{7/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}-(k+1) (a+k) \int \frac {x^{10/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}+2 \int \frac {\sqrt [3]{x}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}-3 (k+1) \int \frac {x^{4/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
Int[((-2 + (1 + k)*x)*(1 - (1 + k)*x + (a + k)*x^2))/(((1 - x)*x*(1 - k*x) )^(1/3)*(1 - (2 + 2*k)*x + (1 + 4*k + k^2)*x^2 - 2*(k + k^2)*x^3 + (-b + k ^2)*x^4)),x]
3.30.79.3.1 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]
Time = 0.76 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.67
| method | result | size |
| pseudoelliptic | \(\frac {\frac {\left (-a +\sqrt {b}\right ) \ln \left (\frac {b^{\frac {1}{3}} x^{2}-b^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\frac {\left (-a -\sqrt {b}\right ) \ln \left (\frac {b^{\frac {1}{3}} x^{2}+b^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\sqrt {3}\, \left (a -\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{6}} x -2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{6}} x}\right )+\sqrt {3}\, \left (a +\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{6}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{6}} x}\right )+\left (a +\sqrt {b}\right ) \ln \left (\frac {b^{\frac {1}{6}} x -\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )+\ln \left (\frac {b^{\frac {1}{6}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right ) \left (a -\sqrt {b}\right )}{2 b^{\frac {2}{3}}}\) | \(258\) |
int((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(1-(2+2*k) *x+(k^2+4*k+1)*x^2-2*(k^2+k)*x^3+(k^2-b)*x^4),x,method=_RETURNVERBOSE)
1/2*(1/2*(-a+b^(1/2))*ln((b^(1/3)*x^2-b^(1/6)*((-1+x)*x*(k*x-1))^(1/3)*x+( (-1+x)*x*(k*x-1))^(2/3))/x^2)+1/2*(-a-b^(1/2))*ln((b^(1/3)*x^2+b^(1/6)*((- 1+x)*x*(k*x-1))^(1/3)*x+((-1+x)*x*(k*x-1))^(2/3))/x^2)+3^(1/2)*(a-b^(1/2)) *arctan(1/3*3^(1/2)*(b^(1/6)*x-2*((-1+x)*x*(k*x-1))^(1/3))/b^(1/6)/x)+3^(1 /2)*(a+b^(1/2))*arctan(1/3*3^(1/2)*(b^(1/6)*x+2*((-1+x)*x*(k*x-1))^(1/3))/ b^(1/6)/x)+(a+b^(1/2))*ln((b^(1/6)*x-((-1+x)*x*(k*x-1))^(1/3))/x)+ln((b^(1 /6)*x+((-1+x)*x*(k*x-1))^(1/3))/x)*(a-b^(1/2)))/b^(2/3)
Exception generated. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Exception raised: TypeError} \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(1-( 2+2*k)*x+(k^2+4*k+1)*x^2-2*(k^2+k)*x^3+(k^2-b)*x^4),x, algorithm="fricas")
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (trace 0)
Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x**2)/((1-x)*x*(-k*x+1))**(1/3)/(1 -(2+2*k)*x+(k**2+4*k+1)*x**2-2*(k**2+k)*x**3+(k**2-b)*x**4),x)
\[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} - 2 \, {\left (k^{2} + k\right )} x^{3} + {\left (k^{2} + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(1-( 2+2*k)*x+(k^2+4*k+1)*x^2-2*(k^2+k)*x^3+(k^2-b)*x^4),x, algorithm="maxima")
integrate(((a + k)*x^2 - (k + 1)*x + 1)*((k + 1)*x - 2)/(((k^2 - b)*x^4 - 2*(k^2 + k)*x^3 + (k^2 + 4*k + 1)*x^2 - 2*(k + 1)*x + 1)*((k*x - 1)*(x - 1 )*x)^(1/3)), x)
\[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} - 2 \, {\left (k^{2} + k\right )} x^{3} + {\left (k^{2} + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(1/3)/(1-( 2+2*k)*x+(k^2+4*k+1)*x^2-2*(k^2+k)*x^3+(k^2-b)*x^4),x, algorithm="giac")
integrate(((a + k)*x^2 - (k + 1)*x + 1)*((k + 1)*x - 2)/(((k^2 - b)*x^4 - 2*(k^2 + k)*x^3 + (k^2 + 4*k + 1)*x^2 - 2*(k + 1)*x + 1)*((k*x - 1)*(x - 1 )*x)^(1/3)), x)
Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int -\frac {\left (x\,\left (k+1\right )-2\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (x\,\left (2\,k+2\right )+x^4\,\left (b-k^2\right )-x^2\,\left (k^2+4\,k+1\right )+2\,x^3\,\left (k^2+k\right )-1\right )} \,d x \]
int(-((x*(k + 1) - 2)*(x^2*(a + k) - x*(k + 1) + 1))/((x*(k*x - 1)*(x - 1) )^(1/3)*(x*(2*k + 2) + x^4*(b - k^2) - x^2*(4*k + k^2 + 1) + 2*x^3*(k + k^ 2) - 1)),x)