Integrand size = 51, antiderivative size = 214 \[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b}-2 \sqrt [3]{b} k x+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b}+\sqrt [3]{b} k x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3}-2 b^{2/3} k x+b^{2/3} k^2 x^2+\left (\sqrt [3]{b}-\sqrt [3]{b} k x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \]
3^(1/2)*arctan(3^(1/2)*(x+(-1-k)*x^2+k*x^3)^(1/3)/(2*b^(1/3)-2*b^(1/3)*k*x +(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(1/3)+ln(-b^(1/3)+b^(1/3)*k*x+(x+(-1-k)*x^ 2+k*x^3)^(1/3))/b^(1/3)-1/2*ln(b^(2/3)-2*b^(2/3)*k*x+b^(2/3)*k^2*x^2+(b^(1 /3)-b^(1/3)*k*x)*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^ (1/3)
Time = 15.45 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.71 \[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{(-1+x) x (-1+k x)}}{\sqrt [3]{b} (2-2 k x)+\sqrt [3]{(-1+x) x (-1+k x)}}\right )+2 \log \left (\sqrt [3]{b} (-1+k x)+\sqrt [3]{(-1+x) x (-1+k x)}\right )-\log \left (b^{2/3} (-1+k x)^2+\sqrt [3]{b} (1-k x) \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{2 \sqrt [3]{b}} \]
Integrate[(-1 + (2 - k)*x)/(((1 - x)*x*(1 - k*x))^(1/3)*(b - (1 + 2*b*k)*x + (1 + b*k^2)*x^2)),x]
(2*Sqrt[3]*ArcTan[(Sqrt[3]*((-1 + x)*x*(-1 + k*x))^(1/3))/(b^(1/3)*(2 - 2* k*x) + ((-1 + x)*x*(-1 + k*x))^(1/3))] + 2*Log[b^(1/3)*(-1 + k*x) + ((-1 + x)*x*(-1 + k*x))^(1/3)] - Log[b^(2/3)*(-1 + k*x)^2 + b^(1/3)*(1 - k*x)*(( -1 + x)*x*(-1 + k*x))^(1/3) + ((-1 + x)*x*(-1 + k*x))^(2/3)])/(2*b^(1/3))
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.
| \(\displaystyle \int \frac {(2-k) x-1}{\sqrt [3]{(1-x) x (1-k x)} \left (x^2 \left (b k^2+1\right )-x (2 b k+1)+b\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {1-(2-k) x}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {1-(2-k) x}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} (1-(2-k) x)}{\sqrt [3]{k x^2-(k+1) x+1} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \left (\frac {(k-2) x^{4/3}}{\sqrt [3]{k x^2-(k+1) x+1} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}+\frac {\sqrt [3]{x}}{\sqrt [3]{k x^2-(k+1) x+1} \left (\left (b k^2+1\right ) x^2-(2 b k+1) x+b\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-\frac {(2-k) \left (2 b k-\sqrt {1-4 b (1-k)}+1\right ) \int \frac {\sqrt [3]{x}}{\left (2 b k-2 \left (b k^2+1\right ) x-\sqrt {4 k b-4 b+1}+1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {1-4 b (1-k)}}+\frac {2 \left (b k^2+1\right ) \int \frac {\sqrt [3]{x}}{\left (2 b k-2 \left (b k^2+1\right ) x-\sqrt {4 k b-4 b+1}+1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {1-4 b (1-k)}}+\frac {(2-k) \left (2 b k+\sqrt {1-4 b (1-k)}+1\right ) \int \frac {\sqrt [3]{x}}{\left (2 b k-2 \left (b k^2+1\right ) x+\sqrt {4 k b-4 b+1}+1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {1-4 b (1-k)}}+\frac {2 \left (b k^2+1\right ) \int \frac {\sqrt [3]{x}}{\left (-2 b k+2 \left (b k^2+1\right ) x-\sqrt {4 k b-4 b+1}-1\right ) \sqrt [3]{k x^2+(-k-1) x+1}}d\sqrt [3]{x}}{\sqrt {1-4 b (1-k)}}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\) |
3.26.42.3.1 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]
\[\int \frac {-1+\left (2-k \right ) x}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (b -\left (2 b k +1\right ) x +\left (b \,k^{2}+1\right ) x^{2}\right )}d x\]
Timed out. \[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\text {Timed out} \]
integrate((-1+(2-k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-(2*b*k+1)*x+(b*k^2+1)*x ^2),x, algorithm="fricas")
Timed out. \[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int { -\frac {{\left (k - 2\right )} x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b k^{2} + 1\right )} x^{2} - {\left (2 \, b k + 1\right )} x + b\right )}} \,d x } \]
integrate((-1+(2-k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-(2*b*k+1)*x+(b*k^2+1)*x ^2),x, algorithm="maxima")
-integrate(((k - 2)*x + 1)/(((k*x - 1)*(x - 1)*x)^(1/3)*((b*k^2 + 1)*x^2 - (2*b*k + 1)*x + b)), x)
\[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int { -\frac {{\left (k - 2\right )} x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b k^{2} + 1\right )} x^{2} - {\left (2 \, b k + 1\right )} x + b\right )}} \,d x } \]
integrate((-1+(2-k)*x)/((1-x)*x*(-k*x+1))^(1/3)/(b-(2*b*k+1)*x+(b*k^2+1)*x ^2),x, algorithm="giac")
integrate(-((k - 2)*x + 1)/(((k*x - 1)*(x - 1)*x)^(1/3)*((b*k^2 + 1)*x^2 - (2*b*k + 1)*x + b)), x)
Timed out. \[ \int \frac {-1+(2-k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int -\frac {x\,\left (k-2\right )+1}{\left (\left (b\,k^2+1\right )\,x^2+\left (-2\,b\,k-1\right )\,x+b\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}} \,d x \]