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

Optimal. Leaf size=212 \[ \frac {(a+b) \log \left (x-\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{b^{2/3}}+\frac {(-a-b) \log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+x^2\right )}{2 b^{2/3}}+\frac {\left (-\sqrt {3} a-\sqrt {3} b\right ) \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x}\right )}{b^{2/3}}+\frac {3 \left (k x^3-k x^2-x^2+x\right )^{2/3}}{(x-1) (k x-1)} \]

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Rubi [F]  time = 55.13, 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 {(-2+(1+k) x) \left (a-a (1+k) x+(1+a k) x^2\right )}{(-1+x) \sqrt [3]{(1-x) x (1-k x)} (-1+k x) \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(-3*(1 + k)*(1 + a*k)*x)/((1 - k)*(1 - b*k)*((1 - x)*x*(1 - k*x))^(1/3)) + (3*(2 + a + b + 4*a*k + a*(1 - 2*b)
*k^2 + b*k^2)*x)/((1 - k)*(1 - b*k)^2*((1 - x)*x*(1 - k*x))^(1/3)) - (3*(1 + k)*(1 + a*k)*(1 - x)*(((1 - k)*x)
/(1 - k*x))^(1/3)*Hypergeometric2F1[1/3, 2/3, 5/3, (1 - x)/(1 - k*x)])/((1 - k)^2*(1 - b*k)*((1 - x)*x*(1 - k*
x))^(1/3)) + (3*(1 + k)*(2 + a + b + 4*a*k + a*(1 - 2*b)*k^2 + b*k^2)*(1 - x)*(((1 - k)*x)/(1 - k*x))^(1/3)*Hy
pergeometric2F1[1/3, 2/3, 5/3, (1 - x)/(1 - k*x)])/(2*(1 - k)^2*(1 - b*k)^2*((1 - x)*x*(1 - k*x))^(1/3)) + ((a
+ b)*(3 + b + 3*k + b*k^3 + (4 + b^2*(1 - k)^2*(1 + k + k^2) + b*(5 + 2*k + 5*k^2))/(Sqrt[b]*Sqrt[4 + b*(1 -
k)^2]))*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Int][1/((1 - x)^(4/3)*x^(1/3)*(1 - k*x)^(4/3)*(-(b*(1 + k)
) - Sqrt[b]*Sqrt[4 + b - 2*b*k + b*k^2] + 2*(-1 + b*k)*x)), x])/((1 - b*k)^2*((1 - x)*x*(1 - k*x))^(1/3)) + ((
a + b)*(3*(1 + k) + b*(1 + k^3) - (4 + b*(5 + 2*k + 5*k^2) + b^2*(1 - k - k^3 + k^4))/(Sqrt[b]*Sqrt[4 + b*(1 -
k)^2]))*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Int][1/((1 - x)^(4/3)*x^(1/3)*(1 - k*x)^(4/3)*(-(b*(1 + k
)) + Sqrt[b]*Sqrt[4 + b - 2*b*k + b*k^2] + 2*(-1 + b*k)*x)), x])/((1 - b*k)^2*((1 - x)*x*(1 - k*x))^(1/3))

Rubi steps

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

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Mathematica [F]  time = 9.44, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-2+(1+k) x) \left (a-a (1+k) x+(1+a k) x^2\right )}{(-1+x) \sqrt [3]{(1-x) x (1-k x)} (-1+k x) \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.48, size = 212, normalized size = 1.00 \begin {gather*} \frac {3 \left (x-x^2-k x^2+k x^3\right )^{2/3}}{(-1+x) (-1+k x)}+\frac {\left (-\sqrt {3} a-\sqrt {3} b\right ) \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {(a+b) \log \left (x-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}+\frac {(-a-b) \log \left (x^2+\sqrt [3]{b} x \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[((-2 + (1 + k)*x)*(a - a*(1 + k)*x + (1 + a*k)*x^2))/((-1 + x)*((1 - x)*x*(1 - k*x))^(1/3)*
(-1 + k*x)*(b - b*(1 + k)*x + (-1 + b*k)*x^2)),x]

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.22, size = 0, normalized size = 0.00 \[\int \frac {\left (-2+\left (1+k \right ) x \right ) \left (a -a \left (1+k \right ) x +\left (a k +1\right ) x^{2}\right )}{\left (-1+x \right ) \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (k x -1\right ) \left (b -b \left (1+k \right ) x +\left (b k -1\right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out

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