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

Optimal. Leaf size=270 \[ \frac {\log \left (x-\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [6]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x\right )}{2 b^{2/3}}-\frac {\log \left (-\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2\right )}{4 b^{2/3}}-\frac {\log \left (\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2\right )}{4 b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{2 \sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+x^2}\right )}{2 b^{2/3}} \]

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Rubi [F]  time = 21.23, 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) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 3.90, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+x) (-1+k x) \left (-2 x+(1+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (b-2 b (1+k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.55, size = 270, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{2/3}}+\frac {\log \left (x-\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\log \left (x+\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}-\frac {\log \left (x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}-\frac {\log \left (x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\sqrt [3]{b} \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 + x)*(-1 + k*x)*(-2*x + (1 + k)*x^2))/(((1 - x)*x*(1 - k*x))^(2/3)*(b - 2*b*(1 + k)*x
+ (b + 4*b*k + b*k^2)*x^2 - 2*b*k*(1 + k)*x^3 + (-1 + b*k^2)*x^4)),x]

[Out]

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

[Out]

Timed out

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giac [A]  time = 2.63, size = 289, normalized size = 1.07 \begin {gather*} -\frac {{\left | b \right |} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{2 \, \left (-b^{5}\right )^{\frac {1}{3}}} + \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{4}} - \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {2}{3}} \arctan \left (-\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} - 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{4}} - \frac {\left (-b^{5}\right )^{\frac {2}{3}} \log \left (\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{4}} - \frac {\left (-b^{5}\right )^{\frac {2}{3}} \log \left (-\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*abs(b)*log((k - k/x - 1/x + 1/x^2)^(2/3) + (-1/b)^(1/3))/(-b^5)^(1/3) + 1/2*sqrt(3)*(-b^5)^(2/3)*arctan((
sqrt(3)*(-1/b)^(1/6) + 2*(k - k/x - 1/x + 1/x^2)^(1/3))/(-1/b)^(1/6))/b^4 - 1/2*sqrt(3)*(-b^5)^(2/3)*arctan(-(
sqrt(3)*(-1/b)^(1/6) - 2*(k - k/x - 1/x + 1/x^2)^(1/3))/(-1/b)^(1/6))/b^4 - 1/4*(-b^5)^(2/3)*log(sqrt(3)*(k -
k/x - 1/x + 1/x^2)^(1/3)*(-1/b)^(1/6) + (k - k/x - 1/x + 1/x^2)^(2/3) + (-1/b)^(1/3))/b^4 - 1/4*(-b^5)^(2/3)*l
og(-sqrt(3)*(k - k/x - 1/x + 1/x^2)^(1/3)*(-1/b)^(1/6) + (k - k/x - 1/x + 1/x^2)^(2/3) + (-1/b)^(1/3))/b^4

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[\int \frac {\left (-1+x \right ) \left (k x -1\right ) \left (-2 x +\left (1+k \right ) x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (b -2 b \left (1+k \right ) x +\left (b \,k^{2}+4 b k +b \right ) x^{2}-2 b k \left (1+k \right ) x^{3}+\left (b \,k^{2}-1\right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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