Integrand size = 84, antiderivative size = 252 \[ \int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2-4 k x+2 k^2 x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{2/3}}+\frac {\log \left (1-2 k x+k^2 x^2-\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}}-\frac {\log \left (1-4 k x+6 k^2 x^2-4 k^3 x^3+k^4 x^4+\left (\sqrt [3]{b}-2 \sqrt [3]{b} k x+\sqrt [3]{b} k^2 x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{4 b^{2/3}} \]
1/2*3^(1/2)*arctan(3^(1/2)*b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(2/3)/(2-4*k*x+2*k ^2*x^2+b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(2/3)))/b^(2/3)+1/2*ln(1-2*k*x+k^2*x^2 -b^(1/3)*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)-1/4*ln(1-4*k*x+6*k^2*x^2-4*k^ 3*x^3+k^4*x^4+(b^(1/3)-2*b^(1/3)*k*x+b^(1/3)*k^2*x^2)*(x+(-1-k)*x^2+k*x^3) ^(2/3)+b^(2/3)*(x+(-1-k)*x^2+k*x^3)^(4/3))/b^(2/3)
\[ \int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx=\int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx \]
Integrate[-(((-1 + x)*x*(-1 + 2*x + (-2*k + k^2)*x^2))/(((1 - x)*x*(1 - k* x))^(2/3)*(1 - 4*k*x + (-b + 6*k^2)*x^2 + (2*b - 4*k^3)*x^3 + (-b + k^4)*x ^4))),x]
-Integrate[((-1 + x)*x*(-1 + 2*x + (-2*k + k^2)*x^2))/(((1 - x)*x*(1 - k*x ))^(2/3)*(1 - 4*k*x + (-b + 6*k^2)*x^2 + (2*b - 4*k^3)*x^3 + (-b + k^4)*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.
| \(\displaystyle \int -\frac {(x-1) x \left (\left (k^2-2 k\right ) x^2+2 x-1\right )}{((1-x) x (1-k x))^{2/3} \left (x^4 \left (k^4-b\right )+x^3 \left (2 b-4 k^3\right )+x^2 \left (6 k^2-b\right )-4 k x+1\right )} \, dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {(1-x) x \left ((2-k) k x^2-2 x+1\right )}{((1-x) x (1-k x))^{2/3} \left (-\left (\left (b-k^4\right ) x^4\right )+2 \left (b-2 k^3\right ) x^3-\left (b-6 k^2\right ) x^2-4 k x+1\right )}dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle -\frac {x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(1-x) \sqrt [3]{x} \left ((2-k) k x^2-2 x+1\right )}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^4\right ) x^4\right )+2 \left (b-2 k^3\right ) x^3-\left (b-6 k^2\right ) x^2-4 k x+1\right )}dx}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 x^{2/3} \left (k x^2-(k+1) x+1\right )^{2/3} \int \frac {(1-x) x \left ((2-k) k x^2-2 x+1\right )}{\left (k x^2-(k+1) x+1\right )^{2/3} \left (-\left (\left (b-k^4\right ) x^4\right )+2 \left (b-2 k^3\right ) x^3-\left (b-6 k^2\right ) x^2-4 k x+1\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \int \frac {\sqrt [3]{1-x} x \left ((2-k) k x^2-2 x+1\right )}{(1-k x)^{2/3} \left (-\left (\left (b-k^4\right ) x^4\right )+2 \left (b-2 k^3\right ) x^3-\left (b-6 k^2\right ) x^2-4 k x+1\right )}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 1387 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \int \frac {\sqrt [3]{1-x} x ((k-2) x+1) \sqrt [3]{1-k x}}{-\left (\left (b-k^4\right ) x^4\right )+2 \left (b-2 k^3\right ) x^3-\left (b-6 k^2\right ) x^2-4 k x+1}d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \int \left (\frac {(k-2) \sqrt [3]{1-x} \sqrt [3]{1-k x} x^2}{-b \left (1-\frac {k^4}{b}\right ) x^4+2 b \left (1-\frac {2 k^3}{b}\right ) x^3-b \left (1-\frac {6 k^2}{b}\right ) x^2-4 k x+1}+\frac {\sqrt [3]{1-x} \sqrt [3]{1-k x} x}{-b \left (1-\frac {k^4}{b}\right ) x^4+2 b \left (1-\frac {2 k^3}{b}\right ) x^3-b \left (1-\frac {6 k^2}{b}\right ) x^2-4 k x+1}\right )d\sqrt [3]{x}}{((1-x) x (1-k x))^{2/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \left (\int \frac {\sqrt [3]{1-x} x \sqrt [3]{1-k x}}{-b \left (1-\frac {k^4}{b}\right ) x^4+2 b \left (1-\frac {2 k^3}{b}\right ) x^3-b \left (1-\frac {6 k^2}{b}\right ) x^2-4 k x+1}d\sqrt [3]{x}-(2-k) \int \frac {\sqrt [3]{1-x} x^2 \sqrt [3]{1-k x}}{-b \left (1-\frac {k^4}{b}\right ) x^4+2 b \left (1-\frac {2 k^3}{b}\right ) x^3-b \left (1-\frac {6 k^2}{b}\right ) x^2-4 k x+1}d\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\) |
Int[-(((-1 + x)*x*(-1 + 2*x + (-2*k + k^2)*x^2))/(((1 - x)*x*(1 - k*x))^(2 /3)*(1 - 4*k*x + (-b + 6*k^2)*x^2 + (2*b - 4*k^3)*x^3 + (-b + k^4)*x^4))), x]
3.28.32.3.1 Defintions of rubi rules used
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)* (x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^ p, x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p]) Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && !IntegerQ[p] && !(EqQ[q, 1] && EqQ[n, 2])
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]
\[\int -\frac {\left (-1+x \right ) x \left (-1+2 x +\left (k^{2}-2 k \right ) x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-4 k x +\left (6 k^{2}-b \right ) x^{2}+\left (-4 k^{3}+2 b \right ) x^{3}+\left (k^{4}-b \right ) x^{4}\right )}d x\]
int(-(-1+x)*x*(-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-4*k*x+(6* k^2-b)*x^2+(-4*k^3+2*b)*x^3+(k^4-b)*x^4),x)
int(-(-1+x)*x*(-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-4*k*x+(6* k^2-b)*x^2+(-4*k^3+2*b)*x^3+(k^4-b)*x^4),x)
Timed out. \[ \int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate(-(-1+x)*x*(-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-4*k *x+(6*k^2-b)*x^2+(-4*k^3+2*b)*x^3+(k^4-b)*x^4),x, algorithm="fricas")
Timed out. \[ \int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate(-(-1+x)*x*(-1+2*x+(k**2-2*k)*x**2)/((1-x)*x*(-k*x+1))**(2/3)/(1- 4*k*x+(6*k**2-b)*x**2+(-4*k**3+2*b)*x**3+(k**4-b)*x**4),x)
\[ \int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx=\int { -\frac {{\left ({\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1\right )} {\left (x - 1\right )} x}{{\left ({\left (k^{4} - b\right )} x^{4} - 2 \, {\left (2 \, k^{3} - b\right )} x^{3} + {\left (6 \, k^{2} - b\right )} x^{2} - 4 \, k x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}} \,d x } \]
integrate(-(-1+x)*x*(-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-4*k *x+(6*k^2-b)*x^2+(-4*k^3+2*b)*x^3+(k^4-b)*x^4),x, algorithm="maxima")
-integrate(((k^2 - 2*k)*x^2 + 2*x - 1)*(x - 1)*x/(((k^4 - b)*x^4 - 2*(2*k^ 3 - b)*x^3 + (6*k^2 - b)*x^2 - 4*k*x + 1)*((k*x - 1)*(x - 1)*x)^(2/3)), x)
\[ \int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx=\int { -\frac {{\left ({\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1\right )} {\left (x - 1\right )} x}{{\left ({\left (k^{4} - b\right )} x^{4} - 2 \, {\left (2 \, k^{3} - b\right )} x^{3} + {\left (6 \, k^{2} - b\right )} x^{2} - 4 \, k x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}} \,d x } \]
integrate(-(-1+x)*x*(-1+2*x+(k^2-2*k)*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(1-4*k *x+(6*k^2-b)*x^2+(-4*k^3+2*b)*x^3+(k^4-b)*x^4),x, algorithm="giac")
integrate(-((k^2 - 2*k)*x^2 + 2*x - 1)*(x - 1)*x/(((k^4 - b)*x^4 - 2*(2*k^ 3 - b)*x^3 + (6*k^2 - b)*x^2 - 4*k*x + 1)*((k*x - 1)*(x - 1)*x)^(2/3)), x)
Timed out. \[ \int -\frac {(-1+x) x \left (-1+2 x+\left (-2 k+k^2\right ) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-4 k x+\left (-b+6 k^2\right ) x^2+\left (2 b-4 k^3\right ) x^3+\left (-b+k^4\right ) x^4\right )} \, dx=-\int \frac {x\,\left (x-1\right )\,\left (\left (2\,k-k^2\right )\,x^2-2\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b-k^4\right )\,x^4+\left (4\,k^3-2\,b\right )\,x^3+\left (b-6\,k^2\right )\,x^2+4\,k\,x-1\right )} \,d x \]
int(-(x*(x - 1)*(x^2*(2*k - k^2) - 2*x + 1))/((x*(k*x - 1)*(x - 1))^(2/3)* (x^4*(b - k^4) + x^2*(b - 6*k^2) + 4*k*x - x^3*(2*b - 4*k^3) - 1)),x)