3.28.49 \(\int \frac {(2-k^2) x-2 x^3+k^2 x^5}{((1-x^2) (1-k^2 x^2))^{2/3} (1-d+(-2+d k^2) x^2+x^4)} \, dx\) [2749]

Optimal. Leaf size=256 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2 \sqrt [3]{d}-2 \sqrt [3]{d} k^2 x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 \sqrt [3]{d}}-\frac {\log \left (-\sqrt [3]{d}+\sqrt [3]{d} k^2 x^2+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}}+\frac {\log \left (d^{2/3}-2 d^{2/3} k^2 x^2+d^{2/3} k^4 x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} k^2 x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+\left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 \sqrt [3]{d}} \]

[Out]

-1/2*3^(1/2)*arctan(3^(1/2)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3)/(2*d^(1/3)-2*d^(1/3)*k^2*x^2+(1+(-k^2-1)*x^2+k^2*x^
4)^(2/3)))/d^(1/3)-1/2*ln(-d^(1/3)+d^(1/3)*k^2*x^2+(1+(-k^2-1)*x^2+k^2*x^4)^(2/3))/d^(1/3)+1/4*ln(d^(2/3)-2*d^
(2/3)*k^2*x^2+d^(2/3)*k^4*x^4+(d^(1/3)-d^(1/3)*k^2*x^2)*(1+(-k^2-1)*x^2+k^2*x^4)^(2/3)+(1+(-k^2-1)*x^2+k^2*x^4
)^(4/3))/d^(1/3)

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Rubi [F]
time = 1.75, 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 (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-((3^(3/4)*Sqrt[2 + Sqrt[3]]*k^(4/3)*Sqrt[(-1 - k^2 + 2*k^2*x^2)^2]*((-1 + k^2)^(2/3) + 2^(2/3)*k^(2/3)*((1 -
x^2)*(1 - k^2*x^2))^(1/3))*Sqrt[((-1 + k^2)^(4/3) - 2^(2/3)*k^(2/3)*(-1 + k^2)^(2/3)*((1 - x^2)*(1 - k^2*x^2))
^(1/3) + 2*2^(1/3)*k^(4/3)*((1 - x^2)*(1 - k^2*x^2))^(2/3))/((1 + Sqrt[3])*(-1 + k^2)^(2/3) + 2^(2/3)*k^(2/3)*
((1 - x^2)*(1 - k^2*x^2))^(1/3))^2]*EllipticF[ArcSin[((1 - Sqrt[3])*(-1 + k^2)^(2/3) + 2^(2/3)*k^(2/3)*((1 - x
^2)*(1 - k^2*x^2))^(1/3))/((1 + Sqrt[3])*(-1 + k^2)^(2/3) + 2^(2/3)*k^(2/3)*((1 - x^2)*(1 - k^2*x^2))^(1/3))],
-7 - 4*Sqrt[3]])/(2^(2/3)*(1 + k^2 - 2*k^2*x^2)*Sqrt[(-1 - k^2*(1 - 2*x^2))^2]*Sqrt[((-1 + k^2)^(2/3)*((-1 +
k^2)^(2/3) + 2^(2/3)*k^(2/3)*((1 - x^2)*(1 - k^2*x^2))^(1/3)))/((1 + Sqrt[3])*(-1 + k^2)^(2/3) + 2^(2/3)*k^(2/
3)*((1 - x^2)*(1 - k^2*x^2))^(1/3))^2])) + Defer[Subst][Defer[Int][(2 - (2 - d)*k^2 + (-2 + 2*k^2 - d*k^4)*x)/
((1 - d + (-2 + d*k^2)*x + x^2)*(1 + (-1 - k^2)*x + k^2*x^2)^(2/3)), x], x, x^2]/2

Rubi steps

\begin {align*} \int \frac {\left (2-k^2\right ) x-2 x^3+k^2 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx &=\int \frac {x \left (2-k^2-2 x^2+k^2 x^4\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x^2+x^4\right )} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {2-k^2-2 x+k^2 x^2}{\left ((1-x) \left (1-k^2 x\right )\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x+x^2\right )} \, dx,x,x^2\right )\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {2-k^2-2 x+k^2 x^2}{(1-x)^{2/3} \left (1-k^2 x\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x+x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-x} \left (2-k^2-k^2 x\right )}{\left (1-k^2 x\right )^{2/3} \left (1-d+\left (-2+d k^2\right ) x+x^2\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \left (\frac {\left (-k^2+\frac {\sqrt {4-4 k^2+d k^4}}{\sqrt {d}}\right ) \sqrt [3]{1-x}}{\left (-2+d k^2-\sqrt {d} \sqrt {4-4 k^2+d k^4}+2 x\right ) \left (1-k^2 x\right )^{2/3}}+\frac {\left (-k^2-\frac {\sqrt {4-4 k^2+d k^4}}{\sqrt {d}}\right ) \sqrt [3]{1-x}}{\left (-2+d k^2+\sqrt {d} \sqrt {4-4 k^2+d k^4}+2 x\right ) \left (1-k^2 x\right )^{2/3}}\right ) \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (-k^2-\frac {\sqrt {4-4 k^2+d k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2+d k^2+\sqrt {d} \sqrt {4-4 k^2+d k^4}+2 x\right ) \left (1-k^2 x\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-k^2+\frac {\sqrt {4-4 k^2+d k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2+d k^2-\sqrt {d} \sqrt {4-4 k^2+d k^4}+2 x\right ) \left (1-k^2 x\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {\left (\left (-k^2-\frac {\sqrt {4-4 k^2+d k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (\frac {-1+k^2 x^2}{-1+k^2}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2+d k^2+\sqrt {d} \sqrt {4-4 k^2+d k^4}+2 x\right ) \left (-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {\left (\left (-k^2+\frac {\sqrt {4-4 k^2+d k^4}}{\sqrt {d}}\right ) \left (1-x^2\right )^{2/3} \left (\frac {-1+k^2 x^2}{-1+k^2}\right )^{2/3}\right ) \text {Subst}\left (\int \frac {\sqrt [3]{1-x}}{\left (-2+d k^2-\sqrt {d} \sqrt {4-4 k^2+d k^4}+2 x\right ) \left (-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}\right )^{2/3}} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ &=\frac {3 \left (1-x^2\right )^2 \left (\frac {1-k^2 x^2}{1-k^2}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 \left (1-x^2\right )}{\sqrt {d} \left (\sqrt {d} k^2-\sqrt {4-4 k^2+d k^4}\right )}\right )}{8 d \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 \left (1-x^2\right )^2 \left (\frac {1-k^2 x^2}{1-k^2}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 \left (1-x^2\right )}{\sqrt {d} \left (\sqrt {d} k^2+\sqrt {4-4 k^2+d k^4}\right )}\right )}{8 d \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 12.30, size = 197, normalized size = 0.77 \begin {gather*} \frac {\left (-1+x^2\right )^{2/3} \left (-1+k^2 x^2\right )^{2/3} \left (-2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \left (-1+x^2\right )^{2/3}}{\left (-1+x^2\right )^{2/3}-2 \sqrt [3]{d} \sqrt [3]{-1+k^2 x^2}}\right )-2 \log \left (\left (-1+x^2\right )^{2/3}+\sqrt [3]{d} \sqrt [3]{-1+k^2 x^2}\right )+\log \left (\left (-1+x^2\right )^{4/3}-\sqrt [3]{d} \left (-1+x^2\right )^{2/3} \sqrt [3]{-1+k^2 x^2}+d^{2/3} \left (-1+k^2 x^2\right )^{2/3}\right )\right )}{4 \sqrt [3]{d} \left (\left (-1+x^2\right ) \left (-1+k^2 x^2\right )\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(d*k^2-2)*x^2+x^4),x, algorithm="maxim
a")

[Out]

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

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Fricas [F(-1)] Timed out
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(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(d*k^2-2)*x^2+x^4),x, algorithm="frica
s")

[Out]

Timed out

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Sympy [F(-1)] Timed out
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(((-k**2+2)*x-2*x**3+k**2*x**5)/((-x**2+1)*(-k**2*x**2+1))**(2/3)/(1-d+(d*k**2-2)*x**2+x**4),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-k^2+2)*x-2*x^3+k^2*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(1-d+(d*k^2-2)*x^2+x^4),x, algorithm="giac"
)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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