Integrand size = 82, antiderivative size = 404 \[ \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\frac {1}{\sqrt {3}}-\frac {2 k x}{\sqrt {3}}+\frac {k^2 x^2}{\sqrt {3}}+\frac {2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{\sqrt {3} \sqrt [3]{b}}}{(-1+k x)^2}\right )}{2 b^{2/3}}-\frac {\log \left (-\sqrt [6]{b} k+\sqrt [6]{b} k^2 x-k \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}-\frac {\log \left (-\sqrt [6]{b} k+\sqrt [6]{b} k^2 x+k \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [3]{b} k^2-2 \sqrt [3]{b} k^3 x+\sqrt [3]{b} k^4 x^2+\left (\sqrt [6]{b} k^2-\sqrt [6]{b} k^3 x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+k^2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\log \left (\sqrt [3]{b} k^2-2 \sqrt [3]{b} k^3 x+\sqrt [3]{b} k^4 x^2+\left (-\sqrt [6]{b} k^2+\sqrt [6]{b} k^3 x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+k^2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}} \]
1/2*3^(1/2)*arctan((1/3*3^(1/2)-2/3*3^(1/2)*k*x+1/3*k^2*x^2*3^(1/2)+2/3*(x +(-1-k)*x^2+k*x^3)^(2/3)*3^(1/2)/b^(1/3))/(k*x-1)^2)/b^(2/3)-1/2*ln(-b^(1/ 6)*k+b^(1/6)*k^2*x-k*(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)-1/2*ln(-b^(1/6)*k +b^(1/6)*k^2*x+k*(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(2/3)+1/4*ln(b^(1/3)*k^2-2* b^(1/3)*k^3*x+b^(1/3)*k^4*x^2+(b^(1/6)*k^2-b^(1/6)*k^3*x)*(x+(-1-k)*x^2+k* x^3)^(1/3)+k^2*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)+1/4*ln(b^(1/3)*k^2-2*b^ (1/3)*k^3*x+b^(1/3)*k^4*x^2+(-b^(1/6)*k^2+b^(1/6)*k^3*x)*(x+(-1-k)*x^2+k*x ^3)^(1/3)+k^2*(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(2/3)
\[ \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx=\int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx \]
Integrate[((1 + (-2 + k)*x)*(1 - 2*k*x + k^2*x^2))/(((1 - x)*x*(1 - k*x))^ (1/3)*(b - 4*b*k*x + (-1 + 6*b*k^2)*x^2 + (2 - 4*b*k^3)*x^3 + (-1 + b*k^4) *x^4)),x]
Integrate[((1 + (-2 + k)*x)*(1 - 2*k*x + k^2*x^2))/(((1 - x)*x*(1 - k*x))^ (1/3)*(b - 4*b*k*x + (-1 + 6*b*k^2)*x^2 + (2 - 4*b*k^3)*x^3 + (-1 + 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.
| \(\displaystyle \int \frac {((k-2) x+1) \left (k^2 x^2-2 k x+1\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (x^4 \left (b k^4-1\right )+x^3 \left (2-4 b k^3\right )+x^2 \left (6 b k^2-1\right )-4 b k x+b\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {(1-(2-k) x) \left (k^2 x^2-2 k x+1\right )}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (1-b k^4\right ) x^4\right )+2 \left (1-2 b k^3\right ) x^3-\left (1-6 b k^2\right ) x^2-4 b k x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} (1-(2-k) x) \left (k^2 x^2-2 k x+1\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (1-b k^4\right ) x^4\right )+2 \left (1-2 b k^3\right ) x^3-\left (1-6 b k^2\right ) x^2-4 b k x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 1380 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {k^2 \sqrt [3]{x} (1-(2-k) x) (1-k x)^2}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (1-b k^4\right ) x^4\right )+2 \left (1-2 b k^3\right ) x^3-\left (1-6 b k^2\right ) x^2-4 b k x+b\right )}d\sqrt [3]{x}}{k^2 \sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} (1-(2-k) x) (1-k x)^2}{\sqrt [3]{k x^2-(k+1) x+1} \left (-\left (\left (1-b k^4\right ) x^4\right )+2 \left (1-2 b k^3\right ) x^3-\left (1-6 b k^2\right ) x^2-4 b k x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 1395 |
| \(\displaystyle \frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \int \frac {\sqrt [3]{x} (1-(2-k) x) (1-k x)^{5/3}}{\sqrt [3]{1-x} \left (-\left (\left (1-b k^4\right ) x^4\right )+2 \left (1-2 b k^3\right ) x^3-\left (1-6 b k^2\right ) x^2-4 b k x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \int \frac {\sqrt [3]{x} (-((k-2) x)-1) (1-k x)^{5/3}}{\sqrt [3]{1-x} \left ((x-1)^2 x^2-b (k x-1)^4\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \int \left (\frac {(k-2) x^{4/3} (1-k x)^{5/3}}{\sqrt [3]{1-x} \left (-\left (\left (1-b k^4\right ) x^4\right )+2 \left (1-2 b k^3\right ) x^3-\left (1-6 b k^2\right ) x^2-4 b k x+b\right )}+\frac {\sqrt [3]{x} (1-k x)^{5/3}}{\sqrt [3]{1-x} \left (-\left (\left (1-b k^4\right ) x^4\right )+2 \left (1-2 b k^3\right ) x^3-\left (1-6 b k^2\right ) x^2-4 b k x+b\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (\int \frac {\sqrt [3]{x} (1-k x)^{5/3}}{\sqrt [3]{1-x} \left (b (k x-1)^4-(x-1)^2 x^2\right )}d\sqrt [3]{x}-(2-k) \int \frac {x^{4/3} (1-k x)^{5/3}}{\sqrt [3]{1-x} \left (b (k x-1)^4-(x-1)^2 x^2\right )}d\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
Int[((1 + (-2 + k)*x)*(1 - 2*k*x + k^2*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)* (b - 4*b*k*x + (-1 + 6*b*k^2)*x^2 + (2 - 4*b*k^3)*x^3 + (-1 + b*k^4)*x^4)) ,x]
3.31.5.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
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[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
\[\int \frac {\left (1+\left (-2+k \right ) x \right ) \left (k^{2} x^{2}-2 k x +1\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (b -4 b k x +\left (6 b \,k^{2}-1\right ) x^{2}+\left (-4 b \,k^{3}+2\right ) x^{3}+\left (b \,k^{4}-1\right ) x^{4}\right )}d x\]
int((1+(-2+k)*x)*(k^2*x^2-2*k*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(b-4*b*k*x+(6* b*k^2-1)*x^2+(-4*b*k^3+2)*x^3+(b*k^4-1)*x^4),x)
int((1+(-2+k)*x)*(k^2*x^2-2*k*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(b-4*b*k*x+(6* b*k^2-1)*x^2+(-4*b*k^3+2)*x^3+(b*k^4-1)*x^4),x)
Timed out. \[ \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate((1+(-2+k)*x)*(k^2*x^2-2*k*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(b-4*b*k *x+(6*b*k^2-1)*x^2+(-4*b*k^3+2)*x^3+(b*k^4-1)*x^4),x, algorithm="fricas")
Timed out. \[ \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx=\text {Timed out} \]
integrate((1+(-2+k)*x)*(k**2*x**2-2*k*x+1)/((1-x)*x*(-k*x+1))**(1/3)/(b-4* b*k*x+(6*b*k**2-1)*x**2+(-4*b*k**3+2)*x**3+(b*k**4-1)*x**4),x)
\[ \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx=\int { \frac {{\left (k^{2} x^{2} - 2 \, k x + 1\right )} {\left ({\left (k - 2\right )} x + 1\right )}}{{\left ({\left (b k^{4} - 1\right )} x^{4} - 2 \, {\left (2 \, b k^{3} - 1\right )} x^{3} - 4 \, b k x + {\left (6 \, b k^{2} - 1\right )} x^{2} + b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((1+(-2+k)*x)*(k^2*x^2-2*k*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(b-4*b*k *x+(6*b*k^2-1)*x^2+(-4*b*k^3+2)*x^3+(b*k^4-1)*x^4),x, algorithm="maxima")
integrate((k^2*x^2 - 2*k*x + 1)*((k - 2)*x + 1)/(((b*k^4 - 1)*x^4 - 2*(2*b *k^3 - 1)*x^3 - 4*b*k*x + (6*b*k^2 - 1)*x^2 + b)*((k*x - 1)*(x - 1)*x)^(1/ 3)), x)
\[ \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx=\int { \frac {{\left (k^{2} x^{2} - 2 \, k x + 1\right )} {\left ({\left (k - 2\right )} x + 1\right )}}{{\left ({\left (b k^{4} - 1\right )} x^{4} - 2 \, {\left (2 \, b k^{3} - 1\right )} x^{3} - 4 \, b k x + {\left (6 \, b k^{2} - 1\right )} x^{2} + b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
integrate((1+(-2+k)*x)*(k^2*x^2-2*k*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(b-4*b*k *x+(6*b*k^2-1)*x^2+(-4*b*k^3+2)*x^3+(b*k^4-1)*x^4),x, algorithm="giac")
integrate((k^2*x^2 - 2*k*x + 1)*((k - 2)*x + 1)/(((b*k^4 - 1)*x^4 - 2*(2*b *k^3 - 1)*x^3 - 4*b*k*x + (6*b*k^2 - 1)*x^2 + b)*((k*x - 1)*(x - 1)*x)^(1/ 3)), x)
Timed out. \[ \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx=\int \frac {\left (x\,\left (k-2\right )+1\right )\,\left (k^2\,x^2-2\,k\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b\,k^4-1\right )\,x^4+\left (2-4\,b\,k^3\right )\,x^3+\left (6\,b\,k^2-1\right )\,x^2-4\,b\,k\,x+b\right )} \,d x \]
int(((x*(k - 2) + 1)*(k^2*x^2 - 2*k*x + 1))/((x*(k*x - 1)*(x - 1))^(1/3)*( b + x^4*(b*k^4 - 1) + x^2*(6*b*k^2 - 1) - x^3*(4*b*k^3 - 2) - 4*b*k*x)),x)