3.24.17 \(\int \frac {5 x-4 (1+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} (-1+(1+k) x-k x^2+b x^5)} \, dx\)

Optimal. Leaf size=179 \[ -\frac {\log \left (b^{2/3} x^4+\sqrt [3]{b} x^2 \sqrt [3]{k x^3+(-k-1) x^2+x}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 \sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\sqrt [3]{b} x^2\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k x^3+(-k-1) x^2+x}}{2 \sqrt [3]{b} x^2+\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 13.58, 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 {5 x-4 (1+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(12*(1 + k)*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][x^7/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/
3)*(1 - (1 + k)*x^3 + k*x^6 - b*x^15)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(1/3) + (15*(1 - x)^(1/3)*x^(1/
3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][x^4/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/3)*(-1 + (1 + k)*x^3 - k*x^6 +
b*x^15)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(1/3) + (9*k*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subs
t][Defer[Int][x^10/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/3)*(-1 + (1 + k)*x^3 - k*x^6 + b*x^15)), x], x, x^(1/3)])/(
(1 - x)*x*(1 - k*x))^(1/3)

Rubi steps

\begin {align*} \int \frac {5 x-4 (1+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx &=\int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx\\ &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {x^{2/3} \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (5-4 (1+k) x^3+3 k x^6\right )}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \left (\frac {4 (1+k) x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )}+\frac {5 x^4}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}+\frac {3 k x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {\left (15 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (9 k \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (12 (1+k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ \end {align*}

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Mathematica [F]  time = 3.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 x-4 (1+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.13, size = 179, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b} x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3} x^4+\sqrt [3]{b} x^2 \sqrt [3]{x+(-1-k) x^2+k x^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[(5*x - 4*(1 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (1 + k)*x - k*x^2 + b*x^
5)),x]

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {5 x -4 \left (1+k \right ) x^{2}+3 k \,x^{3}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-1+\left (1+k \right ) x -k \,x^{2}+b \,x^{5}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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