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

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

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Rubi [F]  time = 21.08, 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) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 3.98, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 4.30, size = 116, normalized size = 1.00 \begin {gather*} \frac {4 \left (-x+x^2\right )}{\left (x-x^2-k x^2+k x^3\right )^{3/4}}+2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) x}\right )-2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (-1+x \right ) x \left (-1-2 \left (-1+k \right ) x +k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (k x -1\right ) \left (-d +\left (3 d k +1\right ) x -\left (3 d \,k^{2}+1\right ) x^{2}+d \,k^{3} x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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