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

Optimal. Leaf size=404 \[ -\frac {\log \left (\sqrt [6]{b} k^2 x-\sqrt [6]{b} k-k \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [6]{b} k^2 x-\sqrt [6]{b} k+k \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{\sqrt {3} \sqrt [3]{b}}+\frac {k^2 x^2}{\sqrt {3}}-\frac {2 k x}{\sqrt {3}}+\frac {1}{\sqrt {3}}}{(k x-1)^2}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [3]{b} k^4 x^2-2 \sqrt [3]{b} k^3 x+\sqrt [3]{b} k^2+\sqrt [3]{k x^3+(-k-1) x^2+x} \left (\sqrt [6]{b} k^2-\sqrt [6]{b} k^3 x\right )+k^2 \left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\log \left (\sqrt [3]{b} k^4 x^2-2 \sqrt [3]{b} k^3 x+\sqrt [3]{b} k^2+\sqrt [3]{k x^3+(-k-1) x^2+x} \left (\sqrt [6]{b} k^3 x-\sqrt [6]{b} k^2\right )+k^2 \left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}} \]

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Rubi [F]  time = 11.40, 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+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 6.20, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(1+(-2+k) x) \left (1-2 k x+k^2 x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (b-4 b k x+\left (-1+6 b k^2\right ) x^2+\left (2-4 b k^3\right ) x^3+\left (-1+b k^4\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 6.57, size = 322, normalized size = 0.80 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b}-2 \sqrt {3} \sqrt [3]{b} k x+\sqrt {3} \sqrt [3]{b} k^2 x^2}{\sqrt [3]{b}-2 \sqrt [3]{b} k x+\sqrt [3]{b} k^2 x^2+2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [3]{b} k-2 \sqrt [3]{b} k^2 x+\sqrt [3]{b} k^3 x^2-k \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}}+\frac {\log \left (b^{2/3} k^2-4 b^{2/3} k^3 x+6 b^{2/3} k^4 x^2-4 b^{2/3} k^5 x^3+b^{2/3} k^6 x^4+\left (\sqrt [3]{b} k^2-2 \sqrt [3]{b} k^3 x+\sqrt [3]{b} k^4 x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+k^2 \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{4 b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3) - 2*Sqrt[3]*b^(1/3)*k*x + Sqrt[3]*b^(1/3)*k^2*x^2)/(b^(1/3) - 2*b^(1/3)*
k*x + b^(1/3)*k^2*x^2 + 2*(x + (-1 - k)*x^2 + k*x^3)^(2/3))])/b^(2/3) - Log[b^(1/3)*k - 2*b^(1/3)*k^2*x + b^(1
/3)*k^3*x^2 - k*(x + (-1 - k)*x^2 + k*x^3)^(2/3)]/(2*b^(2/3)) + Log[b^(2/3)*k^2 - 4*b^(2/3)*k^3*x + 6*b^(2/3)*
k^4*x^2 - 4*b^(2/3)*k^5*x^3 + b^(2/3)*k^6*x^4 + (b^(1/3)*k^2 - 2*b^(1/3)*k^3*x + b^(1/3)*k^4*x^2)*(x + (-1 - k
)*x^2 + k*x^3)^(2/3) + k^2*(x + (-1 - k)*x^2 + k*x^3)^(4/3)]/(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+(-2+k)*x)*(k^2*x^2-2*k*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(b-4*b*k*x+(6*b*k^2-1)*x^2+(-4*b*k^3+2)*x^3+
(b*k^4-1)*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 (k^{2} x^{2} - 2 \, k x + 1\right )} {\left ({\left (k - 2\right )} x + 1\right )}}{{\left ({\left (b k^{4} - 1\right )} x^{4} - 2 \, {\left (2 \, b k^{3} - 1\right )} x^{3} - 4 \, b k x + {\left (6 \, b k^{2} - 1\right )} x^{2} + 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+(-2+k)*x)*(k^2*x^2-2*k*x+1)/((1-x)*x*(-k*x+1))^(1/3)/(b-4*b*k*x+(6*b*k^2-1)*x^2+(-4*b*k^3+2)*x^3+
(b*k^4-1)*x^4),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (1+\left (-2+k \right ) x \right ) \left (k^{2} x^{2}-2 k x +1\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (b -4 b k x +\left (6 b \,k^{2}-1\right ) x^{2}+\left (-4 b \,k^{3}+2\right ) x^{3}+\left (b \,k^{4}-1\right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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