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

Optimal. Leaf size=209 \[ \frac {(-a-b) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 \sqrt [3]{b}}+\frac {(a+b) \log \left (\sqrt [3]{k x^3+(-k-1) x^2+x}-\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac {\left (-\sqrt {3} a-\sqrt {3} b\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{\sqrt [3]{b}}+\frac {3 \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{2 x^2} \]

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Rubi [F]  time = 23.85, 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 )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\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))/(x^2*((1 - x)*x*(1 - k*x))^(1/3)*(1 - (1 + k)*x + (-b + k
)*x^2)),x]

[Out]

(3*(2*k^2 - a*(1 + 2*b + k^2) - b*(1 + 4*k + k^2))*(1 - x)*(1 - k*x))/(4*(b - k)^2*x*((1 - x)*x*(1 - k*x))^(1/
3)) + (3*(1 + k)*(a + k)*(1 - x)*(((1 - k)*x)/(1 - k*x))^(4/3)*(1 - k*x)*Hypergeometric2F1[2/3, 4/3, 5/3, (1 -
x)/(1 - k*x)])/(2*(1 - k)*(b - k)*x*((1 - x)*x*(1 - k*x))^(1/3)) + (3*(1 + k)*(2*k^2 - a*(1 + 2*b + k^2) - b*
(1 + 4*k + k^2))*(1 - x)*(((1 - k)*x)/(1 - k*x))^(4/3)*(1 - k*x)*Hypergeometric2F1[2/3, 4/3, 5/3, (1 - x)/(1 -
k*x)])/(4*(1 - k)*(b - k)^2*x*((1 - x)*x*(1 - k*x))^(1/3)) + ((a + b)*(1 + k^3 + 3*b*(1 + k) + (4*b^2 + (1 -
k)^2*(1 + k + k^2) + b*(5 + 2*k + 5*k^2))/Sqrt[4*b + (-1 + k)^2])*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[
Int][1/((1 - x)^(1/3)*x^(7/3)*(1 - k*x)^(1/3)*(-1 - k - Sqrt[1 + 4*b - 2*k + k^2] + 2*(-b + k)*x)), x])/((b -
k)^2*((1 - x)*x*(1 - k*x))^(1/3)) + ((a + b)*(1 + k^3 + 3*b*(1 + k) - (1 + 4*b^2 - k - k^3 + k^4 + b*(5 + 2*k
+ 5*k^2))/Sqrt[4*b + (-1 + k)^2])*(1 - x)^(1/3)*x^(1/3)*(1 - k*x)^(1/3)*Defer[Int][1/((1 - x)^(1/3)*x^(7/3)*(1
- k*x)^(1/3)*(-1 - k + Sqrt[1 + 4*b - 2*k + k^2] + 2*(-b + k)*x)), x])/((b - k)^2*((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 )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-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 (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (1-(1+k) x+(-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 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )}{(b-k)^2 \sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x}}-\frac {(1+k) (a+k)}{(b-k) \sqrt [3]{1-x} x^{4/3} \sqrt [3]{1-k x}}-\frac {(a+b) \left (1+2 b+k^2\right )-(a+b) (1+k) \left (1+3 b-k+k^2\right ) x}{(b-k)^2 \sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (1+(-1-k) x+(-b+k) x^2\right )}\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 {(a+b) \left (1+2 b+k^2\right )-(a+b) (1+k) \left (1+3 b-k+k^2\right ) x}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (1+(-1-k) x+(-b+k) x^2\right )} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left ((1+k) (a+k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{4/3} \sqrt [3]{1-k x}} \, dx}{(b-k) \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (\left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x}} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {3 \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) (1-k x)}{4 (b-k)^2 x \sqrt [3]{(1-x) x (1-k x)}}+\frac {3 (1+k) (a+k) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{4/3} (1-k x) \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};\frac {1-x}{1-k x}\right )}{2 (1-k) (b-k) x \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 {-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )-\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k-\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )}+\frac {-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )+\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k+\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )}\right ) \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left ((1+k) \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{4/3} \sqrt [3]{1-k x}} \, dx}{2 (b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {3 \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) (1-k x)}{4 (b-k)^2 x \sqrt [3]{(1-x) x (1-k x)}}+\frac {3 (1+k) (a+k) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{4/3} (1-k x) \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};\frac {1-x}{1-k x}\right )}{2 (1-k) (b-k) x \sqrt [3]{(1-x) x (1-k x)}}+\frac {3 (1+k) \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{4/3} (1-k x) \, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};\frac {1-x}{1-k x}\right )}{4 (1-k) (b-k)^2 x \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (\left (-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )+\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k+\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}+\frac {\left ((a+b) \left (1+k^3+3 b (1+k)+\frac {4 b^2+(1-k)^2 \left (1+k+k^2\right )+b \left (5+2 k+5 k^2\right )}{\sqrt {4 b+(-1+k)^2}}\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k-\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}\\ \end {align*}

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Mathematica [F]  time = 12.81, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\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))/(x^2*((1 - x)*x*(1 - k*x))^(1/3)*(1 - (1 + k)*x + (
-b + k)*x^2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.39, size = 209, normalized size = 1.00 \begin {gather*} \frac {3 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 x^2}+\frac {\left (-\sqrt {3} a-\sqrt {3} b\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {(a+b) \log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}+\frac {(-a-b) \log \left (b^{2/3} x^2+\sqrt [3]{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 )}{2 \sqrt [3]{b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((-2+(1+k)*x)*(1-(1+k)*x+(a+k)*x^2)/x^2/((1-x)*x*(-k*x+1))^(1/3)/(1-(1+k)*x+(-b+k)*x^2),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 x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )} x^{2}}\,{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)/x^2/((1-x)*x*(-k*x+1))^(1/3)/(1-(1+k)*x+(-b+k)*x^2),x, algorithm=
"maxima")

[Out]

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

[Out]

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

[Out]

Timed out

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