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

Optimal. Leaf size=383 \[ \frac {\left (a+\sqrt {b}\right ) \log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\sqrt [6]{b} x\right )}{2 b^{2/3}}+\frac {\left (a-\sqrt {b}\right ) \log \left (\sqrt [6]{b} x+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}+\frac {\left (\sqrt {b}-a\right ) \log \left (-\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\left (-a-\sqrt {b}\right ) \log \left (\sqrt [6]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}-\frac {\sqrt {3} \left (a-\sqrt {b}\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{2 b^{2/3}}-\frac {\sqrt {3} \left (a+\sqrt {b}\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{2 b^{2/3}} \]

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Rubi [F]  time = 28.93, 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 (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(9*(1 + k)*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][x^4/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/3
)*(1 - 2*(1 + k)*x^3 + (1 + k*(4 + k))*x^6 - 2*k*(1 + k)*x^9 - b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((1 - x)
*x*(1 - k*x))^(1/3) - (3*(1 + 2*a + 4*k + k^2)*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][x
^7/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/3)*(1 - 2*(1 + k)*x^3 + (1 + k*(4 + k))*x^6 - 2*k*(1 + k)*x^9 - b*(1 - k^2/
b)*x^12)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(1/3) + (3*(1 + k)*(a + k)*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(
1/3)*Defer[Subst][Defer[Int][x^10/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/3)*(1 - 2*(1 + k)*x^3 + (1 + k*(4 + k))*x^6
- 2*k*(1 + k)*x^9 - b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(1/3) + (6*(1 - x)^(1/3)*x^(1
/3)*(1 - k*x)^(1/3)*Defer[Subst][Defer[Int][x/((1 - x^3)^(1/3)*(1 - k*x^3)^(1/3)*(-1 + 2*(1 + k)*x^3 - (1 + k*
(4 + k))*x^6 + 2*k*(1 + k)*x^9 + b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(1/3)

Rubi steps

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

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Mathematica [F]  time = 7.62, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.40, size = 383, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \left (a-\sqrt {b}\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{2/3}}-\frac {\sqrt {3} \left (a+\sqrt {b}\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{2/3}}+\frac {\left (a+\sqrt {b}\right ) \log \left (-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\left (a-\sqrt {b}\right ) \log \left (\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\left (-a+\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\left (-a-\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\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[((-2 + (1 + k)*x)*(1 - (1 + k)*x + (a + k)*x^2))/(((1 - x)*x*(1 - k*x))^(1/3)*(1 - (2 + 2*k
)*x + (1 + 4*k + k^2)*x^2 - 2*(k + k^2)*x^3 + (-b + k^2)*x^4)),x]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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