Integrand size = 53, antiderivative size = 79 \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}\right )}{d^{3/4}} \]
2*arctan(d^(1/4)*(x+(-1-k)*x^2+k*x^3)^(1/4)/(-1+x))/d^(3/4)-2*arctanh(d^(1 /4)*(x+(-1-k)*x^2+k*x^3)^(1/4)/(-1+x))/d^(3/4)
Time = 11.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{(-1+x) x (-1+k x)}}{-1+x}\right )+\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{(-1+x) x (-1+k x)}}{1-x}\right )\right )}{d^{3/4}} \]
Integrate[(-1 + 2*(-1 + k)*x + k*x^2)/(((1 - x)*x*(1 - k*x))^(1/4)*(-1 + ( 3 + d)*x - (3 + d*k)*x^2 + x^3)),x]
(2*(ArcTan[(d^(1/4)*((-1 + x)*x*(-1 + k*x))^(1/4))/(-1 + x)] + ArcTanh[(d^ (1/4)*((-1 + x)*x*(-1 + k*x))^(1/4))/(1 - x)]))/d^(3/4)
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.
| \(\displaystyle \int \frac {k x^2+2 (k-1) x-1}{\sqrt [4]{(1-x) x (1-k x)} \left (-x^2 (d k+3)+(d+3) x+x^3-1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {-k x^2+2 (1-k) x+1}{\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+(d k+3) x^2-(d+3) x+1\right )}dx}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {\sqrt {x} \left (-k x^2+2 (1-k) x+1\right )}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+(d k+3) x^2-(d+3) x+1\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \left (\frac {k x^{5/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (x^3-3 \left (\frac {d k}{3}+1\right ) x^2+3 \left (\frac {d}{3}+1\right ) x-1\right )}+\frac {2 (1-k) x^{3/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}+\frac {\sqrt {x}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}\right )d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \left (\int \frac {\sqrt {x}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}d\sqrt [4]{x}+2 (1-k) \int \frac {x^{3/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+3 \left (\frac {d k}{3}+1\right ) x^2-3 \left (\frac {d}{3}+1\right ) x+1\right )}d\sqrt [4]{x}+k \int \frac {x^{5/2}}{\sqrt [4]{k x^2-(k+1) x+1} \left (x^3-3 \left (\frac {d k}{3}+1\right ) x^2+3 \left (\frac {d}{3}+1\right ) x-1\right )}d\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\) |
Int[(-1 + 2*(-1 + k)*x + k*x^2)/(((1 - x)*x*(1 - k*x))^(1/4)*(-1 + (3 + d) *x - (3 + d*k)*x^2 + x^3)),x]
3.11.46.3.1 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]
\[\int \frac {-1+2 \left (-1+k \right ) x +k \,x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (-1+\left (3+d \right ) x -\left (d k +3\right ) x^{2}+x^{3}\right )}d x\]
Timed out. \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\text {Timed out} \]
integrate((-1+2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-1+(3+d)*x-(d*k+ 3)*x^2+x^3),x, algorithm="fricas")
Timed out. \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\text {Timed out} \]
\[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int { -\frac {k x^{2} + 2 \, {\left (k - 1\right )} x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )}} \,d x } \]
integrate((-1+2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-1+(3+d)*x-(d*k+ 3)*x^2+x^3),x, algorithm="maxima")
-integrate((k*x^2 + 2*(k - 1)*x - 1)/(((k*x - 1)*(x - 1)*x)^(1/4)*((d*k + 3)*x^2 - x^3 - (d + 3)*x + 1)), x)
\[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int { -\frac {k x^{2} + 2 \, {\left (k - 1\right )} x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )}} \,d x } \]
integrate((-1+2*(-1+k)*x+k*x^2)/((1-x)*x*(-k*x+1))^(1/4)/(-1+(3+d)*x-(d*k+ 3)*x^2+x^3),x, algorithm="giac")
integrate(-(k*x^2 + 2*(k - 1)*x - 1)/(((k*x - 1)*(x - 1)*x)^(1/4)*((d*k + 3)*x^2 - x^3 - (d + 3)*x + 1)), x)
Timed out. \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int \frac {2\,x\,\left (k-1\right )+k\,x^2-1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (x^3+\left (-d\,k-3\right )\,x^2+\left (d+3\right )\,x-1\right )} \,d x \]
int((2*x*(k - 1) + k*x^2 - 1)/((x*(k*x - 1)*(x - 1))^(1/4)*(x*(d + 3) - x^ 2*(d*k + 3) + x^3 - 1)),x)