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

Optimal. Leaf size=113 \[ 2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x-1}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{x-1}\right )+\frac {4 \left (k x^3-k x^2-x^2+x\right )^{3/4}}{x (k x-1)} \]

________________________________________________________________________________________

Rubi [F]  time = 26.45, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-1 + x)^3*(-1 + 2*(-1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + k*x)*(-1 + (3 + d)*x - (3 +
d*k)*x^2 + x^3)),x]

[Out]

(-4*(1 - x)^(1/4)*(1 - k*x)^(1/4)*AppellF1[-1/4, -11/4, 5/4, 3/4, x, k*x])/((1 - x)*x*(1 - k*x))^(1/4) + (4*(5
+ d - 2*k)*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][(x^2*(1 - x^4)^(11/4))/((1 - k*x^4)^
(5/4)*(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*(
3 + k + d*k)*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][(x^6*(1 - x^4)^(11/4))/((1 - k*x^4)
^(5/4)*(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*
(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][(x^10*(1 - x^4)^(11/4))/((1 - k*x^4)^(5/4)*(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*} \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx &=\frac {\left (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{\sqrt [4]{1-x} x^{5/4} \sqrt [4]{1-k x} (-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 (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {(1-x)^{11/4} \left (-1+2 (-1+k) x+k x^2\right )}{x^{5/4} \sqrt [4]{1-k x} (-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 (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {(1-x)^{11/4} \left (-1+2 (-1+k) x+k x^2\right )}{x^{5/4} (1-k x)^{5/4} \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 ) \operatorname {Subst}\left (\int \frac {\left (1-x^4\right )^{11/4} \left (-1+2 (-1+k) x^4+k x^8\right )}{x^2 \left (1-k x^4\right )^{5/4} \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 ) \operatorname {Subst}\left (\int \left (\frac {\left (1-x^4\right )^{11/4}}{x^2 \left (1-k x^4\right )^{5/4}}+\frac {x^2 \left (1-x^4\right )^{11/4} \left (5+d-2 k-(3+k+d k) x^4+x^8\right )}{\left (1-k x^4\right )^{5/4} \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 ) \operatorname {Subst}\left (\int \frac {\left (1-x^4\right )^{11/4}}{x^2 \left (1-k x^4\right )^{5/4}} \, 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 ) \operatorname {Subst}\left (\int \frac {x^2 \left (1-x^4\right )^{11/4} \left (5+d-2 k-(3+k+d k) x^4+x^8\right )}{\left (1-k x^4\right )^{5/4} \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 {4 \sqrt [4]{1-x} \sqrt [4]{1-k x} F_1\left (-\frac {1}{4};-\frac {11}{4},\frac {5}{4};\frac {3}{4};x,k 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 ) \operatorname {Subst}\left (\int \left (\frac {5 \left (1+\frac {1}{5} (d-2 k)\right ) x^2 \left (1-x^4\right )^{11/4}}{\left (1-k x^4\right )^{5/4} \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 {(-3-k-d k) x^6 \left (1-x^4\right )^{11/4}}{\left (1-k x^4\right )^{5/4} \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 {x^{10} \left (1-x^4\right )^{11/4}}{\left (1-k x^4\right )^{5/4} \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 {4 \sqrt [4]{1-x} \sqrt [4]{1-k x} F_1\left (-\frac {1}{4};-\frac {11}{4},\frac {5}{4};\frac {3}{4};x,k 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 ) \operatorname {Subst}\left (\int \frac {x^{10} \left (1-x^4\right )^{11/4}}{\left (1-k x^4\right )^{5/4} \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 (5+d-2 k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (1-x^4\right )^{11/4}}{\left (1-k x^4\right )^{5/4} \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 (-3-k-d k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6 \left (1-x^4\right )^{11/4}}{\left (1-k x^4\right )^{5/4} \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 [F]  time = 5.96, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+x)^3 \left (-1+2 (-1+k) x+k x^2\right )}{x \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-1+(3+d) x-(3+d k) x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-1 + x)^3*(-1 + 2*(-1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + k*x)*(-1 + (3 + d)*x -
(3 + d*k)*x^2 + x^3)),x]

[Out]

Integrate[((-1 + x)^3*(-1 + 2*(-1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + k*x)*(-1 + (3 + d)*x -
(3 + d*k)*x^2 + x^3)), x]

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 180.03, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((-1 + x)^3*(-1 + 2*(-1 + k)*x + k*x^2))/(x*((1 - x)*x*(1 - k*x))^(1/4)*(-1 + k*x)*(-1 + (3
+ d)*x - (3 + d*k)*x^2 + x^3)),x]

[Out]

$Aborted

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x, al
gorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (k x^{2} + 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )}^{3}}{\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 )} {\left (k x - 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x, al
gorithm="giac")

[Out]

integrate(-(k*x^2 + 2*(k - 1)*x - 1)*(x - 1)^3/(((k*x - 1)*(x - 1)*x)^(1/4)*((d*k + 3)*x^2 - x^3 - (d + 3)*x +
1)*(k*x - 1)*x), x)

________________________________________________________________________________________

maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-1+x \right )^{3} \left (-1+2 \left (-1+k \right ) x +k \,x^{2}\right )}{x \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (k x -1\right ) \left (-1+\left (3+d \right ) x -\left (d k +3\right ) x^{2}+x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x)

[Out]

int((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (k x^{2} + 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )}^{3}}{\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 )} {\left (k x - 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)^3*(-1+2*(-1+k)*x+k*x^2)/x/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-1+(3+d)*x-(d*k+3)*x^2+x^3),x, al
gorithm="maxima")

[Out]

-integrate((k*x^2 + 2*(k - 1)*x - 1)*(x - 1)^3/(((k*x - 1)*(x - 1)*x)^(1/4)*((d*k + 3)*x^2 - x^3 - (d + 3)*x +
1)*(k*x - 1)*x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (x-1\right )}^3\,\left (2\,x\,\left (k-1\right )+k\,x^2-1\right )}{x\,\left (k\,x-1\right )\,{\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x - 1)^3*(2*x*(k - 1) + k*x^2 - 1))/(x*(k*x - 1)*(x*(k*x - 1)*(x - 1))^(1/4)*(x*(d + 3) - x^2*(d*k + 3)
+ x^3 - 1)),x)

[Out]

int(((x - 1)^3*(2*x*(k - 1) + k*x^2 - 1))/(x*(k*x - 1)*(x*(k*x - 1)*(x - 1))^(1/4)*(x*(d + 3) - x^2*(d*k + 3)
+ x^3 - 1)), x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)**3*(-1+2*(-1+k)*x+k*x**2)/x/((1-x)*x*(-k*x+1))**(1/4)/(k*x-1)/(-1+(3+d)*x-(d*k+3)*x**2+x**3),
x)

[Out]

Timed out

________________________________________________________________________________________