3.22.82 \(\int \frac {x^3 (5-4 (1+k) x+3 k x^2)}{((1-x) x (1-k x))^{2/3} (-b+b (1+k) x-b k x^2+x^5)} \, dx\)

Optimal. Leaf size=162 \[ -\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} x^2 \sqrt [3]{k x^3+(-k-1) x^2+x}+x^4\right )}{2 \sqrt [3]{b}}+\frac {\log \left (x^2-\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{2 \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x^2}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 23.78, 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^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-b+b (1+k) x-b k x^2+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(x^3*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(2/3)*(-b + b*(1 + k)*x - b*k*x^2 + x^5)),x]

[Out]

(9*k*x*((1 - x)/(1 - k*x))^(2/3)*(1 - k*x)*Hypergeometric2F1[1/3, 2/3, 4/3, ((1 - k)*x)/(1 - k*x)])/((1 - x)*x
*(1 - k*x))^(2/3) + (9*b*k*(1 + k)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^3/((1 - x^3
)^(2/3)*(1 - k*x^3)^(2/3)*(b - b*(1 + k)*x^3 + b*k*x^6 - x^15)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(2/3)
+ (12*(1 + k)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^12/((1 - x^3)^(2/3)*(1 - k*x^3)^
(2/3)*(b - b*(1 + k)*x^3 + b*k*x^6 - x^15)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(2/3) + (9*b*k*(1 - x)^(2/
3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][1/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(-b + b*(1 + k)*x^3 -
b*k*x^6 + x^15)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(2/3) + (9*b*k^2*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3
)*Defer[Subst][Defer[Int][x^6/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(-b + b*(1 + k)*x^3 - b*k*x^6 + x^15)), x], x
, x^(1/3)])/((1 - x)*x*(1 - k*x))^(2/3) + (15*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^
9/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(-b + b*(1 + k)*x^3 - b*k*x^6 + x^15)), x], x, x^(1/3)])/((1 - x)*x*(1 -
k*x))^(2/3)

Rubi steps

\begin {align*} \int \frac {x^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-b+b (1+k) x-b k x^2+x^5\right )} \, dx &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {x^{7/3} \left (5-4 (1+k) x+3 k x^2\right )}{(1-x)^{2/3} (1-k x)^{2/3} \left (-b+b (1+k) x-b k x^2+x^5\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^9 \left (5-4 (1+k) x^3+3 k x^6\right )}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {3 k}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}+\frac {3 b k-3 b k (1+k) x^3+3 b k^2 x^6+5 x^9-4 (1+k) x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {3 b k-3 b k (1+k) x^3+3 b k^2 x^6+5 x^9-4 (1+k) x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (9 k (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {9 k x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {3 b k (1+k) x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (b-b (1+k) x^3+b k x^6-x^{15}\right )}+\frac {4 (1+k) x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (b-b (1+k) x^3+b k x^6-x^{15}\right )}+\frac {3 b k}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )}+\frac {3 b k^2 x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )}+\frac {5 x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {9 k x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (15 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (9 b k (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (9 b k^2 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (12 (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (b-b (1+k) x^3+b k x^6-x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (9 b k (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (b-b (1+k) x^3+b k x^6-x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ \end {align*}

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Mathematica [F]  time = 1.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-b+b (1+k) x-b k x^2+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(x^3*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(2/3)*(-b + b*(1 + k)*x - b*k*x^2 + x^5)),x
]

[Out]

Integrate[(x^3*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(2/3)*(-b + b*(1 + k)*x - b*k*x^2 + x^5)),
x]

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IntegrateAlgebraic [A]  time = 6.54, size = 162, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (x^2-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (x^4+\sqrt [3]{b} x^2 \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^3*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(2/3)*(-b + b*(1 + k)*x - b*k*x^2
+ x^5)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*x^2)/(x^2 + 2*b^(1/3)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/b^(1/3) + Log[x^2 - b^(1/3)
*(x + (-1 - k)*x^2 + k*x^3)^(1/3)]/b^(1/3) - Log[x^4 + b^(1/3)*x^2*(x + (-1 - k)*x^2 + k*x^3)^(1/3) + b^(2/3)*
(x + (-1 - k)*x^2 + k*x^3)^(2/3)]/(2*b^(1/3))

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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^3*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(-b+b*(1+k)*x-b*k*x^2+x^5),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 (3 \, k x^{2} - 4 \, {\left (k + 1\right )} x + 5\right )} x^{3}}{{\left (x^{5} - b k x^{2} + b {\left (k + 1\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(-b+b*(1+k)*x-b*k*x^2+x^5),x, algorithm="giac")

[Out]

integrate((3*k*x^2 - 4*(k + 1)*x + 5)*x^3/((x^5 - b*k*x^2 + b*(k + 1)*x - b)*((k*x - 1)*(x - 1)*x)^(2/3)), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (5-4 \left (1+k \right ) x +3 k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (-b +b \left (1+k \right ) x -b k \,x^{2}+x^{5}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(-b+b*(1+k)*x-b*k*x^2+x^5),x)

[Out]

int(x^3*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(-b+b*(1+k)*x-b*k*x^2+x^5),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (3 \, k x^{2} - 4 \, {\left (k + 1\right )} x + 5\right )} x^{3}}{{\left (x^{5} - b k x^{2} + b {\left (k + 1\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(5-4*(1+k)*x+3*k*x^2)/((1-x)*x*(-k*x+1))^(2/3)/(-b+b*(1+k)*x-b*k*x^2+x^5),x, algorithm="maxima")

[Out]

integrate((3*k*x^2 - 4*(k + 1)*x + 5)*x^3/((x^5 - b*k*x^2 + b*(k + 1)*x - b)*((k*x - 1)*(x - 1)*x)^(2/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} -\int \frac {x^3\,\left (3\,k\,x^2-4\,x\,\left (k+1\right )+5\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (-x^5+b\,k\,x^2-b\,\left (k+1\right )\,x+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3*(3*k*x^2 - 4*x*(k + 1) + 5))/((x*(k*x - 1)*(x - 1))^(2/3)*(b - x^5 - b*x*(k + 1) + b*k*x^2)),x)

[Out]

-int((x^3*(3*k*x^2 - 4*x*(k + 1) + 5))/((x*(k*x - 1)*(x - 1))^(2/3)*(b - x^5 - b*x*(k + 1) + b*k*x^2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (3 k x^{2} - 4 k x - 4 x + 5\right )}{\left (x \left (x - 1\right ) \left (k x - 1\right )\right )^{\frac {2}{3}} \left (- b k x^{2} + b k x + b x - b + x^{5}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(5-4*(1+k)*x+3*k*x**2)/((1-x)*x*(-k*x+1))**(2/3)/(-b+b*(1+k)*x-b*k*x**2+x**5),x)

[Out]

Integral(x**3*(3*k*x**2 - 4*k*x - 4*x + 5)/((x*(x - 1)*(k*x - 1))**(2/3)*(-b*k*x**2 + b*k*x + b*x - b + x**5))
, x)

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