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

Optimal. Leaf size=115 \[ 2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{k x-1}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k x^3+(-k-1) x^2+x}}{k x-1}\right )+\frac {4 \left (k x^3-k x^2-x^2+x\right )^{3/4}}{(x-1) x} \]

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Rubi [F]  time = 38.09, 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 {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(-4*(1 - x)^(1/4)*(1 - k*x)^(1/4)*AppellF1[-1/4, 5/4, -11/4, 3/4, x, k*x])/((1 - x)*x*(1 - k*x))^(1/4) - (4*(2
- d - 5*k)*(1 - x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][(x^2*(1 - k*x^4)^(11/4))/((1 - x^4)^
(5/4)*(-(-1 + k*x^4)^3 - d*(x^4 - x^8))), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) - (4*(d + k + 3*k^2)*(1
- x)^(1/4)*x^(1/4)*(1 - k*x)^(1/4)*Defer[Subst][Defer[Int][(x^6*(1 - k*x^4)^(11/4))/((1 - x^4)^(5/4)*(-(-1 +
k*x^4)^3 - d*(x^4 - x^8))), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4) + (4*k^3*(1 - x)^(1/4)*x^(1/4)*(1 - k
*x)^(1/4)*Defer[Subst][Defer[Int][(x^10*(1 - k*x^4)^(11/4))/((1 - x^4)^(5/4)*(-(-1 + k*x^4)^3 - d*(x^4 - x^8))
), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(1/4)

Rubi steps

\begin {align*} \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx &=\frac {\left (\sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{\sqrt [4]{1-x} (-1+x) x^{5/4} \sqrt [4]{1-k x} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 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 {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(1-x)^{5/4} x^{5/4} \sqrt [4]{1-k x} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 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-k x)^{3/4} \left (-1-2 (-1+k) x+k x^2\right ) \left (-1+2 k x-k^2 x^2\right )}{(1-x)^{5/4} x^{5/4} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 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-k x)^{3/4} (-1+k x)^2 \left (-1-2 (-1+k) x+k x^2\right )}{(1-x)^{5/4} x^{5/4} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 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-k x)^{11/4} \left (-1-2 (-1+k) x+k x^2\right )}{(1-x)^{5/4} x^{5/4} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 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-k x^4\right )^{11/4} \left (-1-2 (-1+k) x^4+k x^8\right )}{x^2 \left (1-x^4\right )^{5/4} \left (-1+(d+3 k) x^4-\left (d+3 k^2\right ) x^8+k^3 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 \frac {\left (1-k x^4\right )^{11/4} \left (-1-2 (-1+k) x^4+k x^8\right )}{x^2 \left (1-x^4\right )^{5/4} \left (\left (-1+k x^4\right )^3+d \left (x^4-x^8\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 \left (\frac {\left (1-k x^4\right )^{11/4}}{x^2 \left (1-x^4\right )^{5/4}}+\frac {x^2 \left (1-k x^4\right )^{11/4} \left (-2+d+5 k-\left (d+k+3 k^2\right ) x^4+k^3 x^8\right )}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 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-k x^4\right )^{11/4}}{x^2 \left (1-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-k x^4\right )^{11/4} \left (-2+d+5 k-\left (d+k+3 k^2\right ) x^4+k^3 x^8\right )}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 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 {5}{4},-\frac {11}{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 {d \left (1+\frac {-2+5 k}{d}\right ) x^2 \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )}+\frac {\left (-d-k-3 k^2\right ) x^6 \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )}+\frac {k^3 x^{10} \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 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 {5}{4},-\frac {11}{4};\frac {3}{4};x,k x\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 k^3 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10} \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 (-2+d+5 k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 \left (-d-k-3 k^2\right ) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6 \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (1-d \left (1+\frac {3 k}{d}\right ) x^4+d \left (1+\frac {3 k^2}{d}\right ) x^8-k^3 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 {5}{4},-\frac {11}{4};\frac {3}{4};x,k x\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 k^3 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10} \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (-\left (-1+k x^4\right )^3-d \left (x^4-x^8\right )\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 (-2+d+5 k) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (-\left (-1+k x^4\right )^3-d \left (x^4-x^8\right )\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}+\frac {\left (4 \left (-d-k-3 k^2\right ) \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^6 \left (1-k x^4\right )^{11/4}}{\left (1-x^4\right )^{5/4} \left (-\left (-1+k x^4\right )^3-d \left (x^4-x^8\right )\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(1-x) x (1-k x)}}\\ \end {align*}

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Mathematica [F]  time = 7.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1-2 (-1+k) x+k x^2\right ) \left (-1+3 k x-3 k^2 x^2+k^3 x^3\right )}{(-1+x) x \sqrt [4]{(1-x) x (1-k x)} \left (-1+(d+3 k) x-\left (d+3 k^2\right ) x^2+k^3 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.90, size = 115, normalized size = 1.00 \begin {gather*} \frac {4 \left (x-x^2-k x^2+k x^3\right )^{3/4}}{(-1+x) x}+2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+k x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (k^{3} x^{3} - 3 \, k^{2} x^{2} + 3 \, k x - 1\right )} {\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )}}{{\left (k^{3} x^{3} - {\left (3 \, k^{2} + d\right )} x^{2} + {\left (d + 3 \, k\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left (x - 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (k^{3} x^{3} - 3 \, k^{2} x^{2} + 3 \, k x - 1\right )} {\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )}}{{\left (k^{3} x^{3} - {\left (3 \, k^{2} + d\right )} x^{2} + {\left (d + 3 \, k\right )} x - 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{4}} {\left (x - 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (2\,x\,\left (k-1\right )-k\,x^2+1\right )\,\left (-k^3\,x^3+3\,k^2\,x^2-3\,k\,x+1\right )}{x\,\left (x-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (k^3\,x^3-x^2\,\left (3\,k^2+d\right )+x\,\left (d+3\,k\right )-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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