3.28.94 \(\int \frac {(-1+2 (-2+k) x+3 k x^2) (-1+3 x-3 x^2+x^3)}{((1-x) x (1-k x))^{2/3} (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5)} \, dx\)

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

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Rubi [F]  time = 10.26, 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 {\left (-1+2 (-2+k) x+3 k x^2\right ) \left (-1+3 x-3 x^2+x^3\right )}{((1-x) x (1-k x))^{2/3} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(6*(2 - k)*(-1 + x)^(2/3)*x^(2/3)*(-1 + k*x)^(2/3)*Defer[Subst][Defer[Int][(x^3*(-1 + x^3)^(7/3))/((-1 + k*x^3
)^(2/3)*(-x^3 + k*x^6 - b*(-1 + x^3)^5)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(2/3) - (9*k*(-1 + x)^(2/3)*x
^(2/3)*(-1 + k*x)^(2/3)*Defer[Subst][Defer[Int][(x^6*(-1 + x^3)^(7/3))/((-1 + k*x^3)^(2/3)*(-x^3 + k*x^6 - b*(
-1 + x^3)^5)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(2/3) - (3*(-1 + x)^(2/3)*x^(2/3)*(-1 + k*x)^(2/3)*Defer
[Subst][Defer[Int][(-1 + x^3)^(7/3)/((-1 + k*x^3)^(2/3)*(x^3 - k*x^6 + b*(-1 + x^3)^5)), x], x, x^(1/3)])/((1
- x)*x*(1 - k*x))^(2/3)

Rubi steps

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

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Mathematica [F]  time = 1.55, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+2 (-2+k) x+3 k x^2\right ) \left (-1+3 x-3 x^2+x^3\right )}{((1-x) x (1-k x))^{2/3} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 6.64, size = 270, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b}-2 \sqrt {3} \sqrt [3]{b} x+\sqrt {3} \sqrt [3]{b} x^2}{\sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{b} x^2+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (-\sqrt [3]{b}+2 \sqrt [3]{b} x-\sqrt [3]{b} x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3}-4 b^{2/3} x+6 b^{2/3} x^2-4 b^{2/3} x^3+b^{2/3} x^4+\left (\sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{b} 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 b^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*b^(1/3) - 2*Sqrt[3]*b^(1/3)*x + Sqrt[3]*b^(1/3)*x^2)/(b^(1/3) - 2*b^(1/3)*x + b^(1/3)
*x^2 + 2*(x + (-1 - k)*x^2 + k*x^3)^(1/3))])/b^(2/3) + Log[-b^(1/3) + 2*b^(1/3)*x - b^(1/3)*x^2 + (x + (-1 - k
)*x^2 + k*x^3)^(1/3)]/b^(2/3) - Log[b^(2/3) - 4*b^(2/3)*x + 6*b^(2/3)*x^2 - 4*b^(2/3)*x^3 + b^(2/3)*x^4 + (b^(
1/3) - 2*b^(1/3)*x + b^(1/3)*x^2)*(x + (-1 - k)*x^2 + k*x^3)^(1/3) + (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((-1+2*(-2+k)*x+3*k*x^2)*(x^3-3*x^2+3*x-1)/((1-x)*x*(-k*x+1))^(2/3)/(-b+(1+5*b)*x-(10*b+k)*x^2+10*b*x
^3-5*b*x^4+b*x^5),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 (3 \, k x^{2} + 2 \, {\left (k - 2\right )} x - 1\right )} {\left (x^{3} - 3 \, x^{2} + 3 \, x - 1\right )}}{{\left (b x^{5} - 5 \, b x^{4} + 10 \, b x^{3} - {\left (10 \, b + k\right )} x^{2} + {\left (5 \, b + 1\right )} x - 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+2*(-2+k)*x+3*k*x^2)*(x^3-3*x^2+3*x-1)/((1-x)*x*(-k*x+1))^(2/3)/(-b+(1+5*b)*x-(10*b+k)*x^2+10*b*x
^3-5*b*x^4+b*x^5),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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