Integrand size = 87, antiderivative size = 383 \[ \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {\sqrt {3} \left (a-\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}+\frac {\sqrt {3} \left (a+\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}+\frac {\left (a+\sqrt {b}\right ) \log \left (-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{5/6}}+\frac {\left (-a+\sqrt {b}\right ) \log \left (\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{5/6}}+\frac {\left (a-\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{5/6}}+\frac {\left (-a-\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{5/6}} \]
-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^(5/6)+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^(5/6)+1/2*(a+b^(1/2))*ln(-b^( 1/6)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(5/6)+1/2*(-a+b^(1/2))*ln(b^(1/6)*x+( x+(-1-k)*x^2+k*x^3)^(1/3))/b^(5/6)+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^(5/6)+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^(5/6)
\[ \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \]
Integrate[((-2*x + (1 + k)*x^2)*(1 - (1 + k)*x + (a + k)*x^2))/(((1 - x)*x *(1 - k*x))^(2/3)*(1 - 2*(1 + k)*x + (1 + 4*k + k^2)*x^2 - 2*(k + k^2)*x^3 + (-b + k^2)*x^4)),x]
Integrate[((-2*x + (1 + k)*x^2)*(1 - (1 + k)*x + (a + k)*x^2))/(((1 - x)*x *(1 - k*x))^(2/3)*(1 - 2*(1 + k)*x + (1 + 4*k + k^2)*x^2 - 2*(k + k^2)*x^3 + (-b + k^2)*x^4)), 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.
| \(\displaystyle \int \frac {\left ((k+1) x^2-2 x\right ) \left (x^2 (a+k)-(k+1) x+1\right )}{((1-x) x (1-k x))^{2/3} \left (x^4 \left (k^2-b\right )-2 \left (k^2+k\right ) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )} \, dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {x ((k+1) x-2) \left (x^2 (a+k)-(k+1) x+1\right )}{((1-x) x (1-k x))^{2/3} \left (x^4 \left (k^2-b\right )-2 \left (k^2+k\right ) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int -\frac {\sqrt [3]{x} (2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {\sqrt [3]{x} (2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {x (2-(k+1) x) \left ((a+k) x^2-(k+1) x+1\right )}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \left (\frac {(k+1) (a+k)}{\left (b-k^2\right ) \left (k x^2-(k+1) x+1\right )^{2/3}}-\frac {-\left (\left (b \left (k^2+4 k+2 a+1\right )+k \left (k^3+k+2 a \left (k^2+k+1\right )\right )\right ) x^3\right )+(k+1) \left (k^3+k^2+k+3 b+a \left (k^2+4 k+1\right )\right ) x^2-2 \left (k^3+k^2+k+a (k+1)^2+b\right ) x+(k+1) (a+k)}{\left (b-k^2\right ) \left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^2\right ) x^4\right )-2 k (k+1) x^3+\left (k^2+4 k+1\right ) x^2-2 (k+1) x+1\right )}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \left (-\frac {(k+1) (a+k) \int \frac {1}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{b-k^2}+\frac {2 \left (a (k+1)^2+b+k^3+k^2+k\right ) \int \frac {x}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{b-k^2}-\frac {(k+1) \left (a \left (k^2+4 k+1\right )+3 b+k^3+k^2+k\right ) \int \frac {x^2}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{b-k^2}+\frac {\left (b \left (2 a+k^2+4 k+1\right )+k \left (2 a \left (k^2+k+1\right )+k^3+k\right )\right ) \int \frac {x^3}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-b \left (1-\frac {k^2}{b}\right ) x^4-2 k (k+1) x^3+(k (k+4)+1) x^2-2 (k+1) x+1\right )}d\sqrt [3]{x}}{b-k^2}+\frac {(k+1) (1-x)^{2/3} \sqrt [3]{x} (a+k) (1-k x)^{2/3} \sqrt [3]{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}} \left (\frac {1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}}{1-\frac {2 k x}{-\sqrt {k^2-2 k+1}+k+1}}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {\sqrt {k^2-2 k+1} x}{1-\frac {2 k x}{k-\sqrt {k^2-2 k+1}+1}}\right )}{\left (b-k^2\right ) \left (1-\frac {2 k x}{\sqrt {k^2-2 k+1}+k+1}\right )^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3}}\right )}{((1-x) x (1-k x))^{2/3}}\) |
Int[((-2*x + (1 + k)*x^2)*(1 - (1 + k)*x + (a + k)*x^2))/(((1 - x)*x*(1 - k*x))^(2/3)*(1 - 2*(1 + k)*x + (1 + 4*k + k^2)*x^2 - 2*(k + k^2)*x^3 + (-b + k^2)*x^4)),x]
3.30.78.3.1 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]
Time = 1.01 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.68
| method | result | size |
| pseudoelliptic | \(\frac {\frac {\left (a -\sqrt {b}\right ) \ln \left (\frac {b^{\frac {1}{3}} x^{2}-b^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\frac {\left (-a -\sqrt {b}\right ) \ln \left (\frac {b^{\frac {1}{3}} x^{2}+b^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}+\sqrt {3}\, \left (a -\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{6}} x -2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{6}} x}\right )-\sqrt {3}\, \left (a +\sqrt {b}\right ) \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{6}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{6}} x}\right )+\left (a +\sqrt {b}\right ) \ln \left (\frac {b^{\frac {1}{6}} x -\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {b^{\frac {1}{6}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right ) \left (a -\sqrt {b}\right )}{2 b^{\frac {5}{6}}}\) | \(260\) |
int((-2*x+(1+k)*x^2)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-2*( 1+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/b^(5/6)*(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)-l n((b^(1/6)*x+((-1+x)*x*(k*x-1))^(1/3))/x)*(a-b^(1/2)))
Exception generated. \[ \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Exception raised: TypeError} \]
integrate((-2*x+(1+k)*x^2)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/ (1-2*(1+k)*x+(k^2+4*k+1)*x^2-2*(k^2+k)*x^3+(k^2-b)*x^4),x, algorithm="fric as")
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (trace 0)
Timed out. \[ \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate((-2*x+(1+k)*x**2)*(1-(1+k)*x+(a+k)*x**2)/((1-x)*x*(-k*x+1))**(2/ 3)/(1-2*(1+k)*x+(k**2+4*k+1)*x**2-2*(k**2+k)*x**3+(k**2-b)*x**4),x)
\[ \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x^{2} - 2 \, x\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} - 2 \, {\left (k^{2} + k\right )} x^{3} + {\left (k^{2} + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}} \,d x } \]
integrate((-2*x+(1+k)*x^2)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/ (1-2*(1+k)*x+(k^2+4*k+1)*x^2-2*(k^2+k)*x^3+(k^2-b)*x^4),x, algorithm="maxi ma")
integrate(((a + k)*x^2 - (k + 1)*x + 1)*((k + 1)*x^2 - 2*x)/(((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)^(2/3)), x)
\[ \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x^{2} - 2 \, x\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} - 2 \, {\left (k^{2} + k\right )} x^{3} + {\left (k^{2} + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}} \,d x } \]
integrate((-2*x+(1+k)*x^2)*(1-(1+k)*x+(a+k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/ (1-2*(1+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 - 2*x)/(((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)^(2/3)), x)
Timed out. \[ \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int \frac {\left (2\,x-x^2\,\left (k+1\right )\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (x^4\,\left (b-k^2\right )-x^2\,\left (k^2+4\,k+1\right )+2\,x\,\left (k+1\right )+2\,x^3\,\left (k^2+k\right )-1\right )} \,d x \]
int(((2*x - x^2*(k + 1))*(x^2*(a + k) - x*(k + 1) + 1))/((x*(k*x - 1)*(x - 1))^(2/3)*(x^4*(b - k^2) - x^2*(4*k + k^2 + 1) + 2*x*(k + 1) + 2*x^3*(k + k^2) - 1)),x)