\(\int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} (-1+(3+d) x-(3+d k) x^2+x^3)} \, dx\) [1046]

Optimal result

Integrand size = 53, antiderivative size = 79 \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}\right )}{d^{3/4}} \]

[Out]

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)

Rubi [F]

\[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx \]

[In]

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]

[Out]

(4*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^2/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*(1 - 3
*(1 + d/3)*x^4 + 3*(1 + (d*k)/3)*x^8 - x^12)), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) + (8*(1 - k)*(1 -
x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^6/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*(1 - 3*(1 + d/
3)*x^4 + 3*(1 + (d*k)/3)*x^8 - x^12)), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) + (4*k*(1 - x)^(1/4)*x^(1/
4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][x^10/((1 - x^4)^(1/4)*(1 - k*x^4)^(1/4)*(-1 + 3*(1 + d/3)*x^4 - 3*(
1 + (d*k)/3)*x^8 + x^12)), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4)

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx}{\sqrt [4]{(1-x) x (1-k x)}} \\ & = \frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \text {Subst}\left (\int \frac {x^2 \left (-1+2 (-1+k) x^4+k x^8\right )}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+(3+d) x^4-(3+d k) x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}} \\ & = \frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \text {Subst}\left (\int \left (\frac {x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) x^8-x^{12}\right )}+\frac {2 (1-k) x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) x^8-x^{12}\right )}+\frac {k x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+3 \left (1+\frac {d}{3}\right ) x^4-3 \left (1+\frac {d k}{3}\right ) x^8+x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}} \\ & = \frac {\left (4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) x^8-x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (8 (1-k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \text {Subst}\left (\int \frac {x^6}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (1-3 \left (1+\frac {d}{3}\right ) x^4+3 \left (1+\frac {d k}{3}\right ) x^8-x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 k \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \text {Subst}\left (\int \frac {x^{10}}{\sqrt [4]{1-x^4} \sqrt [4]{1-k x^4} \left (-1+3 \left (1+\frac {d}{3}\right ) x^4-3 \left (1+\frac {d k}{3}\right ) x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 11.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{(-1+x) x (-1+k x)}}{-1+x}\right )+\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{(-1+x) x (-1+k x)}}{1-x}\right )\right )}{d^{3/4}} \]

[In]

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]

[Out]

(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)

Maple [F]

\[\int \frac {-1+2 \left (-1+k \right ) x +k \,x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (-1+\left (3+d \right ) x -\left (d k +3\right ) x^{2}+x^{3}\right )}d x\]

[In]

int((-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)

[Out]

int((-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)

Fricas [F(-1)]

Timed out. \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\text {Timed out} \]

[In]

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")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\text {Timed out} \]

[In]

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)

[Out]

Timed out

Maxima [F]

\[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int { -\frac {k x^{2} + 2 \, {\left (k - 1\right )} x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )}} \,d x } \]

[In]

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")

[Out]

-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)

Giac [F]

\[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int { -\frac {k x^{2} + 2 \, {\left (k - 1\right )} x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left ({\left (d k + 3\right )} x^{2} - x^{3} - {\left (d + 3\right )} x + 1\right )}} \,d x } \]

[In]

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")

[Out]

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)

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+2 (-1+k) x+k x^2}{\sqrt [4]{(1-x) x (1-k x)} \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx=\int \frac {2\,x\,\left (k-1\right )+k\,x^2-1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (x^3+\left (-d\,k-3\right )\,x^2+\left (d+3\right )\,x-1\right )} \,d x \]

[In]

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)

[Out]

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)