3.30.34 \(\int \frac {(-1+(-1+2 k) x) (1-2 x+x^2)}{\sqrt [3]{(1-x) x (1-k x)} (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+(-b+k^2) x^4)} \, dx\)

Optimal. Leaf size=343 \[ -\frac {\log \left (-\sqrt [6]{b} x+\sqrt [6]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [6]{b} x-\sqrt [6]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}+\frac {\log \left (\left (\sqrt [6]{b}-\sqrt [6]{b} x\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\log \left (\left (\sqrt [6]{b} x-\sqrt [6]{b}\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{\sqrt {3} \sqrt [3]{b}}+\frac {x^2}{\sqrt {3}}-\frac {2 x}{\sqrt {3}}+\frac {1}{\sqrt {3}}}{(x-1)^2}\right )}{2 b^{2/3}} \]

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Rubi [F]  time = 10.66, 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+(-1+2 k) x) \left (1-2 x+x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-1 + (-1 + 2*k)*x)*(1 - 2*x + x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-b + 4*b*x + (1 - 6*b)*x^2 + (4*b - 2*
k)*x^3 + (-b + k^2)*x^4)),x]

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 4.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+(-1+2 k) x) \left (1-2 x+x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-1 + (-1 + 2*k)*x)*(1 - 2*x + x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-b + 4*b*x + (1 - 6*b)*x^2 + (4*
b - 2*k)*x^3 + (-b + k^2)*x^4)),x]

[Out]

Integrate[((-1 + (-1 + 2*k)*x)*(1 - 2*x + x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-b + 4*b*x + (1 - 6*b)*x^2 + (4*
b - 2*k)*x^3 + (-b + k^2)*x^4)), x]

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IntegrateAlgebraic [A]  time = 3.48, size = 343, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {1}{\sqrt {3}}-\frac {2 x}{\sqrt {3}}+\frac {x^2}{\sqrt {3}}+\frac {2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{\sqrt {3} \sqrt [3]{b}}}{(-1+x)^2}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [6]{b}-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}-\frac {\log \left (-\sqrt [6]{b}+\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{b} x^2+\left (\sqrt [6]{b}-\sqrt [6]{b} x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\log \left (\sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{b} x^2+\left (-\sqrt [6]{b}+\sqrt [6]{b} x\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + (-1 + 2*k)*x)*(1 - 2*x + x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(-b + 4*b*x + (1 - 6*b)*
x^2 + (4*b - 2*k)*x^3 + (-b + k^2)*x^4)),x]

[Out]

(Sqrt[3]*ArcTan[(1/Sqrt[3] - (2*x)/Sqrt[3] + x^2/Sqrt[3] + (2*(x + (-1 - k)*x^2 + k*x^3)^(2/3))/(Sqrt[3]*b^(1/
3)))/(-1 + x)^2])/(2*b^(2/3)) - Log[b^(1/6) - b^(1/6)*x + (x + (-1 - k)*x^2 + k*x^3)^(1/3)]/(2*b^(2/3)) - Log[
-b^(1/6) + b^(1/6)*x + (x + (-1 - k)*x^2 + k*x^3)^(1/3)]/(2*b^(2/3)) + Log[b^(1/3) - 2*b^(1/3)*x + b^(1/3)*x^2
+ (b^(1/6) - b^(1/6)*x)*(x + (-1 - k)*x^2 + k*x^3)^(1/3) + (x + (-1 - k)*x^2 + k*x^3)^(2/3)]/(4*b^(2/3)) + Lo
g[b^(1/3) - 2*b^(1/3)*x + b^(1/3)*x^2 + (-b^(1/6) + b^(1/6)*x)*(x + (-1 - k)*x^2 + k*x^3)^(1/3) + (x + (-1 - k
)*x^2 + k*x^3)^(2/3)]/(4*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+(-1+2*k)*x)*(x^2-2*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(-b+4*b*x+(1-6*b)*x^2+(4*b-2*k)*x^3+(k^2-b)*x^4
),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 ({\left (2 \, k - 1\right )} x - 1\right )} {\left (x^{2} - 2 \, x + 1\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} + 2 \, {\left (2 \, b - k\right )} x^{3} - {\left (6 \, b - 1\right )} x^{2} + 4 \, b x - b\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+(-1+2*k)*x)*(x^2-2*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(-b+4*b*x+(1-6*b)*x^2+(4*b-2*k)*x^3+(k^2-b)*x^4
),x, algorithm="giac")

[Out]

integrate(((2*k - 1)*x - 1)*(x^2 - 2*x + 1)/(((k^2 - b)*x^4 + 2*(2*b - k)*x^3 - (6*b - 1)*x^2 + 4*b*x - b)*((k
*x - 1)*(x - 1)*x)^(1/3)), x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (-1+\left (-1+2 k \right ) x \right ) \left (x^{2}-2 x +1\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-b +4 b x +\left (1-6 b \right ) x^{2}+\left (4 b -2 k \right ) x^{3}+\left (k^{2}-b \right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+(-1+2*k)*x)*(x^2-2*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(-b+4*b*x+(1-6*b)*x^2+(4*b-2*k)*x^3+(k^2-b)*x^4),x)

[Out]

int((-1+(-1+2*k)*x)*(x^2-2*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(-b+4*b*x+(1-6*b)*x^2+(4*b-2*k)*x^3+(k^2-b)*x^4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (2 \, k - 1\right )} x - 1\right )} {\left (x^{2} - 2 \, x + 1\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} + 2 \, {\left (2 \, b - k\right )} x^{3} - {\left (6 \, b - 1\right )} x^{2} + 4 \, b x - b\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+(-1+2*k)*x)*(x^2-2*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(-b+4*b*x+(1-6*b)*x^2+(4*b-2*k)*x^3+(k^2-b)*x^4
),x, algorithm="maxima")

[Out]

integrate(((2*k - 1)*x - 1)*(x^2 - 2*x + 1)/(((k^2 - b)*x^4 + 2*(2*b - k)*x^3 - (6*b - 1)*x^2 + 4*b*x - b)*((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 {\left (x\,\left (2\,k-1\right )-1\right )\,\left (x^2-2\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b-k^2\right )\,x^4+\left (2\,k-4\,b\right )\,x^3+\left (6\,b-1\right )\,x^2-4\,b\,x+b\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((x*(2*k - 1) - 1)*(x^2 - 2*x + 1))/((x*(k*x - 1)*(x - 1))^(1/3)*(b - x^3*(4*b - 2*k) + x^4*(b - k^2) - 4
*b*x + x^2*(6*b - 1))),x)

[Out]

int(-((x*(2*k - 1) - 1)*(x^2 - 2*x + 1))/((x*(k*x - 1)*(x - 1))^(1/3)*(b - x^3*(4*b - 2*k) + x^4*(b - k^2) - 4
*b*x + x^2*(6*b - 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((-1+(-1+2*k)*x)*(x**2-2*x+1)/((1-x)*x*(-k*x+1))**(1/3)/(-b+4*b*x+(1-6*b)*x**2+(4*b-2*k)*x**3+(k**2-b
)*x**4),x)

[Out]

Timed out

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