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

Optimal. Leaf size=236 \[ \frac {\log \left (d^{2/3} k^2 x^2+2 d^{2/3} k x+d^{2/3}+\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (\sqrt [3]{d} k x+\sqrt [3]{d}\right )+\left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d} k x-\sqrt [3]{d}+\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}\right )}{\sqrt [3]{d}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{2 \sqrt [3]{d} k x+2 \sqrt [3]{d}+\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{\sqrt [3]{d}} \]

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Rubi [F]  time = 6.51, 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 {3 k+\left (2-k^2\right ) x-3 k x^2-k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d-(1+2 d) k x+\left (-1-d k^2\right ) x^2+k x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-((k*x*(1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*AppellF1[1/2, 1/3, 1/3, 3/2, x^2, k^2*x^2])/((1 - x^2)*(1 - k^2*x^2
))^(1/3)) + ((4 - d)*k*(1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*Defer[Int][1/((1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*(
1 - d - (1 + 2*d)*k*x - (1 + d*k^2)*x^2 + k*x^3)), x])/((1 - x^2)*(1 - k^2*x^2))^(1/3) + (2*(1 - (1 + d)*k^2)*
(1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*Defer[Int][x/((1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*(1 - d - (1 + 2*d)*k*x -
(1 + d*k^2)*x^2 + k*x^3)), x])/((1 - x^2)*(1 - k^2*x^2))^(1/3) - (k*(4 + d*k^2)*(1 - x^2)^(1/3)*(1 - k^2*x^2)
^(1/3)*Defer[Int][x^2/((1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*(1 - d - (1 + 2*d)*k*x - (1 + d*k^2)*x^2 + k*x^3)),
x])/((1 - x^2)*(1 - k^2*x^2))^(1/3)

Rubi steps

\begin {align*} \int \frac {3 k+\left (2-k^2\right ) x-3 k x^2-k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d-(1+2 d) k x+\left (-1-d k^2\right ) x^2+k x^3\right )} \, dx &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {3 k+\left (2-k^2\right ) x-3 k x^2-k^2 x^3}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x+\left (-1-d k^2\right ) x^2+k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {3 k+\left (2-k^2\right ) x-3 k x^2-k^2 x^3}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (-\frac {k}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}}+\frac {(4-d) k+2 \left (1-(1+d) k^2\right ) x-k \left (4+d k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {(4-d) k+2 \left (1-(1+d) k^2\right ) x-k \left (4+d k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (k \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {k x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {(4-d) k}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )}+\frac {2 \left (1-(1+d) k^2\right ) x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )}+\frac {k \left (-4-d k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {k x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left ((4-d) k \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (k \left (4+d k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (1-(1+d) k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(1+2 d) k x-\left (1+d k^2\right ) x^2+k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}

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Mathematica [F]  time = 1.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 k+\left (2-k^2\right ) x-3 k x^2-k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d-(1+2 d) k x+\left (-1-d k^2\right ) x^2+k x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 5.65, size = 236, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{2 \sqrt [3]{d}+2 \sqrt [3]{d} k x+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{\sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d}-\sqrt [3]{d} k x+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{\sqrt [3]{d}}+\frac {\log \left (d^{2/3}+2 d^{2/3} k x+d^{2/3} k^2 x^2+\left (\sqrt [3]{d}+\sqrt [3]{d} k x\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[3]*ArcTan[(Sqrt[3]*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(1/3))/(2*d^(1/3) + 2*d^(1/3)*k*x + (1 + (-1 - k^2)*
x^2 + k^2*x^4)^(1/3))])/d^(1/3)) - Log[-d^(1/3) - d^(1/3)*k*x + (1 + (-1 - k^2)*x^2 + k^2*x^4)^(1/3)]/d^(1/3)
+ Log[d^(2/3) + 2*d^(2/3)*k*x + d^(2/3)*k^2*x^2 + (d^(1/3) + d^(1/3)*k*x)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(1/3)
+ (1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/3)]/(2*d^(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((3*k+(-k^2+2)*x-3*k*x^2-k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(1-d-(1+2*d)*k*x+(-d*k^2-1)*x^2+k*x^3
),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 {k^{2} x^{3} + 3 \, k x^{2} + {\left (k^{2} - 2\right )} x - 3 \, k}{{\left (k x^{3} - {\left (2 \, d + 1\right )} k x - {\left (d k^{2} + 1\right )} x^{2} - d + 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*k+(-k^2+2)*x-3*k*x^2-k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(1-d-(1+2*d)*k*x+(-d*k^2-1)*x^2+k*x^3
),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*k+(-k^2+2)*x-3*k*x^2-k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(1-d-(1+2*d)*k*x+(-d*k^2-1)*x^2+k*x^3
),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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