3.18.15 \(\int \frac {x (-1+k x) (-1+2 (-1+k) x+k x^2)}{(-1+x) ((1-x) x (1-k x))^{3/4} (-d+(1+3 d) x-(3 d+k) x^2+d x^3)} \, dx\)

Optimal. Leaf size=116 \[ 2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} x-\sqrt [4]{d}}{\sqrt [4]{k x^3+(-k-1) x^2+x}}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} x-\sqrt [4]{d}}{\sqrt [4]{k x^3+(-k-1) x^2+x}}\right )+\frac {4 \sqrt [4]{k x^3+(-k-1) x^2+x}}{x-1} \]

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Rubi [F]  time = 20.57, 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 {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(4*k*(1 - x)^(3/4)*x*(1 - k*x)^(3/4)*AppellF1[1/4, 7/4, -1/4, 5/4, x, k*x])/(d*((1 - x)*x*(1 - k*x))^(3/4)) +
(4*(d + k + 3*d*k)*(1 - x)^(3/4)*x^(3/4)*(1 - k*x)^(3/4)*Defer[Subst][Defer[Int][(x^4*(1 - k*x^4)^(1/4))/((1 -
x^4)^(7/4)*(-x^4 + k*x^8 - d*(-1 + x^4)^3)), x], x, x^(1/4)])/(d*((1 - x)*x*(1 - k*x))^(3/4)) + (4*(d*(2 - 5*
k) - k^2)*(1 - x)^(3/4)*x^(3/4)*(1 - k*x)^(3/4)*Defer[Subst][Defer[Int][(x^8*(1 - k*x^4)^(1/4))/((1 - x^4)^(7/
4)*(-x^4 + k*x^8 - d*(-1 + x^4)^3)), x], x, x^(1/4)])/(d*((1 - x)*x*(1 - k*x))^(3/4)) + (4*k*(1 - x)^(3/4)*x^(
3/4)*(1 - k*x)^(3/4)*Defer[Subst][Defer[Int][(1 - k*x^4)^(1/4)/((1 - x^4)^(7/4)*(x^4 - k*x^8 + d*(-1 + x^4)^3)
), x], x, x^(1/4)])/((1 - x)*x*(1 - k*x))^(3/4)

Rubi steps

\begin {align*} \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx &=\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {\sqrt [4]{x} (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(1-x)^{3/4} (-1+x) (1-k x)^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=-\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {\sqrt [4]{x} (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(1-x)^{7/4} (1-k x)^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {\sqrt [4]{x} \sqrt [4]{1-k x} \left (-1+2 (-1+k) x+k x^2\right )}{(1-x)^{7/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4 \sqrt [4]{1-k x^4} \left (-1+2 (-1+k) x^4+k x^8\right )}{\left (1-x^4\right )^{7/4} \left (-d+(1+3 d) x^4-(3 d+k) x^8+d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4 \sqrt [4]{1-k x^4} \left (-1+2 (-1+k) x^4+k x^8\right )}{\left (1-x^4\right )^{7/4} \left (x^4-k x^8+d \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {k \sqrt [4]{1-k x^4}}{d \left (1-x^4\right )^{7/4}}+\frac {\sqrt [4]{1-k x^4} \left (d k-(d+k+3 d k) x^4-\left (d (2-5 k)-k^2\right ) x^8\right )}{d \left (1-x^4\right )^{7/4} \left (x^4-k x^8+d \left (-1+x^4\right )^3\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}\\ &=\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-k x^4} \left (d k-(d+k+3 d k) x^4-\left (d (2-5 k)-k^2\right ) x^8\right )}{\left (1-x^4\right )^{7/4} \left (x^4-k x^8+d \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 k (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4}} \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}\\ &=\frac {4 k (1-x)^{3/4} x (1-k x)^{3/4} F_1\left (\frac {1}{4};\frac {7}{4},-\frac {1}{4};\frac {5}{4};x,k x\right )}{d ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {(d+k+3 d k) x^4 \sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (d-(1+3 d) x^4+3 d \left (1+\frac {k}{3 d}\right ) x^8-d x^{12}\right )}+\frac {\left (d (2-5 k)-k^2\right ) x^8 \sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (d-(1+3 d) x^4+3 d \left (1+\frac {k}{3 d}\right ) x^8-d x^{12}\right )}+\frac {d k \sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (-d+(1+3 d) x^4-3 d \left (1+\frac {k}{3 d}\right ) x^8+d x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}\\ &=\frac {4 k (1-x)^{3/4} x (1-k x)^{3/4} F_1\left (\frac {1}{4};\frac {7}{4},-\frac {1}{4};\frac {5}{4};x,k x\right )}{d ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 k (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (-d+(1+3 d) x^4-3 d \left (1+\frac {k}{3 d}\right ) x^8+d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (d+k+3 d k) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4 \sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (d-(1+3 d) x^4+3 d \left (1+\frac {k}{3 d}\right ) x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 \left (d (2-5 k)-k^2\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8 \sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (d-(1+3 d) x^4+3 d \left (1+\frac {k}{3 d}\right ) x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}\\ &=\frac {4 k (1-x)^{3/4} x (1-k x)^{3/4} F_1\left (\frac {1}{4};\frac {7}{4},-\frac {1}{4};\frac {5}{4};x,k x\right )}{d ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 k (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (x^4-k x^8+d \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (d+k+3 d k) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^4 \sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (-x^4+k x^8-d \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 \left (d (2-5 k)-k^2\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8 \sqrt [4]{1-k x^4}}{\left (1-x^4\right )^{7/4} \left (-x^4+k x^8-d \left (-1+x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{d ((1-x) x (1-k x))^{3/4}}\\ \end {align*}

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Mathematica [F]  time = 3.88, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x (-1+k x) \left (-1+2 (-1+k) x+k x^2\right )}{(-1+x) ((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 4.23, size = 116, normalized size = 1.00 \begin {gather*} \frac {4 \sqrt [4]{x+(-1-k) x^2+k x^3}}{-1+x}+2 \sqrt [4]{d} \tan ^{-1}\left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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(x*(k*x-1)*(-1+2*(-1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(3/4)/(-d+(1+3*d)*x-(3*d+k)*x^2+d*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 x^{2} + 2 \, {\left (k - 1\right )} x - 1\right )} {\left (k x - 1\right )} x}{{\left (d x^{3} - {\left (3 \, d + k\right )} x^{2} + {\left (3 \, d + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (x - 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x \left (k x -1\right ) \left (-1+2 \left (-1+k \right ) x +k \,x^{2}\right )}{\left (-1+x \right ) \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (-d +\left (1+3 d \right ) x -\left (3 d +k \right ) x^{2}+d \,x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 (k x - 1\right )} x}{{\left (d x^{3} - {\left (3 \, d + k\right )} x^{2} + {\left (3 \, d + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (x - 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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