3.27.90 \(\int \frac {(2-k^2) x-2 x^3+k^2 x^5}{((1-x^2) (1-k^2 x^2))^{2/3} (-1+d+(-2 d+k^2) x^2+d x^4)} \, dx\) [2690]

Optimal. Leaf size=243 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2-2 k^2 x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 d^{2/3}}-\frac {\log \left (-1+k^2 x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 d^{2/3}}+\frac {\log \left (1-2 k^2 x^2+k^4 x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} k^2 x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 d^{2/3}} \]

[Out]

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

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Rubi [F]
time = 1.76, 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+d 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 + d*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)*d*(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][(k^2 + 2*d*(1 - k^2) + (-k^4 - 2*d*(1 - k
^2))*x)/((-1 + d + (-2*d + k^2)*x + d*x^2)*(1 + (-1 - k^2)*x + k^2*x^2)^(2/3)), x], x, x^2]/(2*d)

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+d 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+d 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+d 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+d 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+d 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+\sqrt {4 d-4 d k^2+k^4}\right ) \sqrt [3]{1-x}}{\left (-2 d+k^2-\sqrt {4 d-4 d k^2+k^4}+2 d x\right ) \left (1-k^2 x\right )^{2/3}}+\frac {\left (-k^2-\sqrt {4 d-4 d k^2+k^4}\right ) \sqrt [3]{1-x}}{\left (-2 d+k^2+\sqrt {4 d-4 d k^2+k^4}+2 d 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-\sqrt {4 d-4 d k^2+k^4}\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 {4 d-4 d k^2+k^4}+2 d 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+\sqrt {4 d-4 d k^2+k^4}\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 {4 d-4 d k^2+k^4}+2 d 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-\sqrt {4 d-4 d k^2+k^4}\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 {4 d-4 d k^2+k^4}+2 d 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+\sqrt {4 d-4 d k^2+k^4}\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 {4 d-4 d k^2+k^4}+2 d 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 d \left (1-x^2\right )}{k^2-\sqrt {4 d-4 d k^2+k^4}}\right )}{8 \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 d \left (1-x^2\right )}{k^2+\sqrt {4 d-4 d k^2+k^4}}\right )}{8 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}\\ \end {align*}

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Mathematica [F]
time = 26.24, size = 0, normalized size = 0.00 \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+d x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[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 + d*x^4)
),x]

[Out]

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 + d*x^4)
), x]

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

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+(k^2-2*d)*x^2+d*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+(k^2-2*d)*x^2+d*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+(k^2-2*d)*x^2+d*x^4),x, algorithm="ma
xima")

[Out]

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

[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+(k**2-2*d)*x**2+d*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+(k^2-2*d)*x^2+d*x^4),x, algorithm="gi
ac")

[Out]

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

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