3.26.78 \(\int \frac {1-(-3+2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} (-1+(5+b) x-(10+b k) x^2+10 x^3-5 x^4+x^5)} \, dx\)

Optimal. Leaf size=222 \[ \frac {\log \left (-\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x^2-2 x+1\right )}{b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+2 x^2-4 x+2}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\left (\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+x^4-4 x^3+6 x^2-4 x+1\right )}{2 b^{2/3}} \]

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Rubi [F]  time = 13.14, 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-(-3+2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(5+b) x-(10+b k) x^2+10 x^3-5 x^4+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 - (-3 + 2*k)*x - (4 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (5 + b)*x - (10 + b*k)*x^2 +
10*x^3 - 5*x^4 + x^5)),x]

[Out]

(-6*(2 - k)*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][(x^4*(1 - x^3)^(2/3))/((1 - k*x^3)^(
1/3)*(1 - 5*(1 + b/5)*x^3 + 10*(1 + (b*k)/10)*x^6 - 10*x^9 + 5*x^12 - 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[Subst][Defer[Int][(x^7*(1 - x^3)^(2/3))/((1 -
k*x^3)^(1/3)*(1 - 5*(1 + b/5)*x^3 + 10*(1 + (b*k)/10)*x^6 - 10*x^9 + 5*x^12 - x^15)), x], x, x^(1/3)])/((1 - x
)*x*(1 - k*x))^(1/3) + (3*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][(x*(1 - x^3)^(2/3))/((
1 - k*x^3)^(1/3)*(-1 + 5*(1 + b/5)*x^3 - 10*(1 + (b*k)/10)*x^6 + 10*x^9 - 5*x^12 + x^15)), x], x, x^(1/3)])/((
1 - x)*x*(1 - k*x))^(1/3)

Rubi steps

\begin {align*} \int \frac {1-(-3+2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(5+b) x-(10+b k) x^2+10 x^3-5 x^4+x^5\right )} \, dx &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1-(-3+2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (-1+(5+b) x-(10+b k) x^2+10 x^3-5 x^4+x^5\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(1-x)^{2/3} \left (1+(4-2 k) x-3 k x^2\right )}{\sqrt [3]{x} \sqrt [3]{1-k x} \left (-1+(5+b) x-(10+b k) x^2+10 x^3-5 x^4+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 \left (1-x^3\right )^{2/3} \left (1+(4-2 k) x^3-3 k x^6\right )}{\sqrt [3]{1-k x^3} \left (-1+(5+b) x^3-(10+b k) x^6+10 x^9-5 x^{12}+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 {2 (-2+k) x^4 \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-5 \left (1+\frac {b}{5}\right ) x^3+10 \left (1+\frac {b k}{10}\right ) x^6-10 x^9+5 x^{12}-x^{15}\right )}+\frac {3 k x^7 \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-5 \left (1+\frac {b}{5}\right ) x^3+10 \left (1+\frac {b k}{10}\right ) x^6-10 x^9+5 x^{12}-x^{15}\right )}+\frac {x \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (-1+5 \left (1+\frac {b}{5}\right ) x^3-10 \left (1+\frac {b k}{10}\right ) x^6+10 x^9-5 x^{12}+x^{15}\right )}\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 \frac {x \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (-1+5 \left (1+\frac {b}{5}\right ) x^3-10 \left (1+\frac {b k}{10}\right ) x^6+10 x^9-5 x^{12}+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (6 (2-k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-5 \left (1+\frac {b}{5}\right ) x^3+10 \left (1+\frac {b k}{10}\right ) x^6-10 x^9+5 x^{12}-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^7 \left (1-x^3\right )^{2/3}}{\sqrt [3]{1-k x^3} \left (1-5 \left (1+\frac {b}{5}\right ) x^3+10 \left (1+\frac {b k}{10}\right ) x^6-10 x^9+5 x^{12}-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 = 4.36, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-(-3+2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-1+(5+b) x-(10+b k) x^2+10 x^3-5 x^4+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(1 - (-3 + 2*k)*x - (4 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (5 + b)*x - (10 + b*k)
*x^2 + 10*x^3 - 5*x^4 + x^5)),x]

[Out]

Integrate[(1 - (-3 + 2*k)*x - (4 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (5 + b)*x - (10 + b*k)
*x^2 + 10*x^3 - 5*x^4 + x^5)), x]

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IntegrateAlgebraic [A]  time = 2.98, size = 222, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2-4 x+2 x^2+\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (1-2 x+x^2-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (1-4 x+6 x^2-4 x^3+x^4+\left (\sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{b} x^2\right ) \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 b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 - (-3 + 2*k)*x - (4 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x))^(1/3)*(-1 + (5 + b)*x - (
10 + b*k)*x^2 + 10*x^3 - 5*x^4 + x^5)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3)*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(2 - 4*x + 2*x^2 + b^(1/3)*(x + (-1 - k)*x^
2 + k*x^3)^(1/3))])/b^(2/3) + Log[1 - 2*x + x^2 - b^(1/3)*(x + (-1 - k)*x^2 + k*x^3)^(1/3)]/b^(2/3) - Log[1 -
4*x + 6*x^2 - 4*x^3 + x^4 + (b^(1/3) - 2*b^(1/3)*x + 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^(2/3))

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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((1-(-3+2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(5+b)*x-(b*k+10)*x^2+10*x^3-5*x^4+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} - {\left (k + 4\right )} x^{2} - {\left (2 \, k - 3\right )} x + 1}{{\left (x^{5} - 5 \, x^{4} - {\left (b k + 10\right )} x^{2} + 10 \, x^{3} + {\left (b + 5\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((1-(-3+2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(5+b)*x-(b*k+10)*x^2+10*x^3-5*x^4+x^5)
,x, algorithm="giac")

[Out]

integrate((3*k*x^3 - (k + 4)*x^2 - (2*k - 3)*x + 1)/((x^5 - 5*x^4 - (b*k + 10)*x^2 + 10*x^3 + (b + 5)*x - 1)*(
(k*x - 1)*(x - 1)*x)^(1/3)), x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1-\left (-3+2 k \right ) x -\left (4+k \right ) x^{2}+3 k \,x^{3}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-1+\left (5+b \right ) x -\left (b k +10\right ) x^{2}+10 x^{3}-5 x^{4}+x^{5}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-(-3+2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(5+b)*x-(b*k+10)*x^2+10*x^3-5*x^4+x^5),x)

[Out]

int((1-(-3+2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(5+b)*x-(b*k+10)*x^2+10*x^3-5*x^4+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} - {\left (k + 4\right )} x^{2} - {\left (2 \, k - 3\right )} x + 1}{{\left (x^{5} - 5 \, x^{4} - {\left (b k + 10\right )} x^{2} + 10 \, x^{3} + {\left (b + 5\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((1-(-3+2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-1+(5+b)*x-(b*k+10)*x^2+10*x^3-5*x^4+x^5)
,x, algorithm="maxima")

[Out]

integrate((3*k*x^3 - (k + 4)*x^2 - (2*k - 3)*x + 1)/((x^5 - 5*x^4 - (b*k + 10)*x^2 + 10*x^3 + (b + 5)*x - 1)*(
(k*x - 1)*(x - 1)*x)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {-3\,k\,x^3+\left (k+4\right )\,x^2+\left (2\,k-3\right )\,x-1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (x^5-5\,x^4+10\,x^3+\left (-b\,k-10\right )\,x^2+\left (b+5\right )\,x-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*(2*k - 3) + x^2*(k + 4) - 3*k*x^3 - 1)/((x*(k*x - 1)*(x - 1))^(1/3)*(x*(b + 5) - x^2*(b*k + 10) + 10*x
^3 - 5*x^4 + x^5 - 1)),x)

[Out]

int(-(x*(2*k - 3) + x^2*(k + 4) - 3*k*x^3 - 1)/((x*(k*x - 1)*(x - 1))^(1/3)*(x*(b + 5) - x^2*(b*k + 10) + 10*x
^3 - 5*x^4 + x^5 - 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x - 1\right ) \left (3 k x^{2} + 2 k x - 4 x - 1\right )}{\sqrt [3]{x \left (x - 1\right ) \left (k x - 1\right )} \left (- b k x^{2} + b x + x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-(-3+2*k)*x-(4+k)*x**2+3*k*x**3)/((1-x)*x*(-k*x+1))**(1/3)/(-1+(5+b)*x-(b*k+10)*x**2+10*x**3-5*x**
4+x**5),x)

[Out]

Integral((x - 1)*(3*k*x**2 + 2*k*x - 4*x - 1)/((x*(x - 1)*(k*x - 1))**(1/3)*(-b*k*x**2 + b*x + x**5 - 5*x**4 +
10*x**3 - 10*x**2 + 5*x - 1)), x)

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