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

Optimal. Leaf size=186 \[ \frac {\log \left (b x^2-b^{2/3} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k x^3+(-k-1) x^2+x}}{2 \sqrt [3]{b} x^2+\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{b^{2/3}}-\frac {\log \left (b^{5/3} x^2 \sqrt [3]{k x^3+(-k-1) x^2+x}+b^{4/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+b^2 x^4\right )}{2 b^{2/3}} \]

________________________________________________________________________________________

Rubi [F]  time = 18.27, 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 {x^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(9*k*x*((1 - x)/(1 - k*x))^(2/3)*(1 - k*x)*Hypergeometric2F1[1/3, 2/3, 4/3, ((1 - k)*x)/(1 - k*x)])/(b*((1 - x
)*x*(1 - k*x))^(2/3)) + (9*k*(1 + k)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^3/((1 - x
^3)^(2/3)*(1 - k*x^3)^(2/3)*(1 - (1 + k)*x^3 + k*x^6 - b*x^15)), x], x, x^(1/3)])/(b*((1 - x)*x*(1 - k*x))^(2/
3)) + (12*(1 + k)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^12/((1 - x^3)^(2/3)*(1 - k*x
^3)^(2/3)*(1 - (1 + k)*x^3 + k*x^6 - b*x^15)), 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][1/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(-1 + (1 + k)*x^3 - k*
x^6 + b*x^15)), x], x, x^(1/3)])/(b*((1 - x)*x*(1 - k*x))^(2/3)) + (9*k^2*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3
)*Defer[Subst][Defer[Int][x^6/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(-1 + (1 + k)*x^3 - k*x^6 + b*x^15)), x], x,
x^(1/3)])/(b*((1 - x)*x*(1 - k*x))^(2/3)) + (15*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][
x^9/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(-1 + (1 + k)*x^3 - k*x^6 + b*x^15)), x], x, x^(1/3)])/((1 - x)*x*(1 -
k*x))^(2/3)

Rubi steps

\begin {align*} \int \frac {x^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {x^{7/3} \left (5-4 (1+k) x+3 k x^2\right )}{(1-x)^{2/3} (1-k x)^{2/3} \left (-1+(1+k) x-k x^2+b x^5\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 {x^9 \left (5-4 (1+k) x^3+3 k x^6\right )}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+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 \left (\frac {3 k}{b \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}+\frac {3 k-3 k (1+k) x^3+3 k^2 x^6+5 b x^9-4 b (1+k) x^{12}}{b \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+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 {3 k-3 k (1+k) x^3+3 k^2 x^6+5 b x^9-4 b (1+k) x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((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 {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}\\ &=\frac {9 k x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{b ((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 {3 k (1+k) x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )}+\frac {4 b (1+k) x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )}+\frac {3 k}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}+\frac {3 k^2 x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}+\frac {5 b x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}\\ &=\frac {9 k x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (15 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+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 {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (9 k^2 (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 )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (12 (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (9 k (1+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 )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 1.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 8.71, size = 186, 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} x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (b x^2-b^{2/3} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (b^2 x^4+b^{5/3} x^2 \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{4/3} \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[(x^3*(5 - 4*(1 + k)*x + 3*k*x^2))/(((1 - x)*x*(1 - k*x))^(2/3)*(-1 + (1 + k)*x - k*x^2 + b*
x^5)),x]

[Out]

-((Sqrt[3]*ArcTan[(Sqrt[3]*(x + (-1 - k)*x^2 + k*x^3)^(1/3))/(2*b^(1/3)*x^2 + (x + (-1 - k)*x^2 + k*x^3)^(1/3)
)])/b^(2/3)) + Log[b*x^2 - b^(2/3)*(x + (-1 - k)*x^2 + k*x^3)^(1/3)]/b^(2/3) - Log[b^2*x^4 + b^(5/3)*x^2*(x +
(-1 - k)*x^2 + k*x^3)^(1/3) + b^(4/3)*(x + (-1 - k)*x^2 + k*x^3)^(2/3)]/(2*b^(2/3))

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (3\,k\,x^2-4\,x\,\left (k+1\right )+5\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (b\,x^5-k\,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^3*(3*k*x^2 - 4*x*(k + 1) + 5))/((x*(k*x - 1)*(x - 1))^(2/3)*(b*x^5 + x*(k + 1) - k*x^2 - 1)),x)

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________