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

Optimal. Leaf size=279 \[ -\frac {\log \left (b^{2/3} k^4 x^4-4 b^{2/3} k^3 x^3+6 b^{2/3} k^2 x^2-4 b^{2/3} k x+b^{2/3}+\sqrt [3]{k x^3+(-k-1) x^2+x} \left (\sqrt [3]{b} k^2 x^2-2 \sqrt [3]{b} k x+\sqrt [3]{b}\right )+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 \sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} k^2 x^2+2 \sqrt [3]{b} k x-\sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k x^3+(-k-1) x^2+x}}{2 \sqrt [3]{b} k^2 x^2-4 \sqrt [3]{b} k x+2 \sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{\sqrt [3]{b}} \]

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Rubi [F]  time = 11.02, 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+3 k) x-\left (k+4 k^2\right ) x^2+3 k^2 x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b k) x-\left (1+10 b k^2\right ) x^2+10 b k^3 x^3-5 b k^4 x^4+b k^5 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 5.50, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+(-2+3 k) x-\left (k+4 k^2\right ) x^2+3 k^2 x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b k) x-\left (1+10 b k^2\right ) x^2+10 b k^3 x^3-5 b k^4 x^4+b k^5 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 2.74, size = 279, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b}-4 \sqrt [3]{b} k x+2 \sqrt [3]{b} k^2 x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b}+2 \sqrt [3]{b} k x-\sqrt [3]{b} k^2 x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3}-4 b^{2/3} k x+6 b^{2/3} k^2 x^2-4 b^{2/3} k^3 x^3+b^{2/3} k^4 x^4+\left (\sqrt [3]{b}-2 \sqrt [3]{b} k x+\sqrt [3]{b} k^2 x^2\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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+3*k)*x-(4*k^2+k)*x^2+3*k^2*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(5*b*k+1)*x-(10*b*k^2+1)*x^2+10*
b*k^3*x^3-5*b*k^4*x^4+b*k^5*x^5),x, algorithm="fricas")

[Out]

Timed out

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giac [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+3*k)*x-(4*k^2+k)*x^2+3*k^2*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(5*b*k+1)*x-(10*b*k^2+1)*x^2+10*
b*k^3*x^3-5*b*k^4*x^4+b*k^5*x^5),x, algorithm="giac")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((3*k^2*x^3 - (4*k^2 + k)*x^2 + (3*k - 2)*x + 1)/((b*k^5*x^5 - 5*b*k^4*x^4 + 10*b*k^3*x^3 - (10*b*k^2
+ 1)*x^2 + (5*b*k + 1)*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 {x\,\left (3\,k-2\right )+3\,k^2\,x^3-x^2\,\left (4\,k^2+k\right )+1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (b+x^2\,\left (10\,b\,k^2+1\right )-x\,\left (5\,b\,k+1\right )-10\,b\,k^3\,x^3+5\,b\,k^4\,x^4-b\,k^5\,x^5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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