3.27.8 \(\int \frac {(-1+2 k^2) x-2 k^4 x^3+k^4 x^5}{((1-x^2) (1-k^2 x^2))^{2/3} (-1+d+(1-2 d k^2) x^2+d k^4 x^4)} \, dx\) [2608]

Optimal. Leaf size=226 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}{2-2 x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}}\right )}{2 d^{2/3}}-\frac {\log \left (-1+x^2+\sqrt [3]{d} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 d^{2/3}}+\frac {\log \left (1-2 x^2+x^4+\left (\sqrt [3]{d}-\sqrt [3]{d} x^2\right ) \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{4/3}\right )}{4 d^{2/3}} \]

[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*x^2+d^(1/3)*(1+(-k^2-1)*x^2+k^2*x^4)^(
2/3)))/d^(2/3)-1/2*ln(-1+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*x^2+x^4+(d^(1/3)-d^(1/
3)*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 = 1.74, 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 (-1+2 k^2\right ) x-2 k^4 x^3+k^4 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x^2+d k^4 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {\left (-1+2 k^2\right ) x-2 k^4 x^3+k^4 x^5}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x^2+d k^4 x^4\right )} \, dx &=\int \frac {x \left (-1+2 k^2-2 k^4 x^2+k^4 x^4\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x^2+d k^4 x^4\right )} \, dx\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {-1+2 k^2-2 k^4 x+k^4 x^2}{\left ((1-x) \left (1-k^2 x\right )\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x+d k^4 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 {-1+2 k^2-2 k^4 x+k^4 x^2}{(1-x)^{2/3} \left (1-k^2 x\right )^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x+d k^4 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-k^2 x} \left (-1+2 k^2-k^2 x\right )}{(1-x)^{2/3} \left (-1+d+\left (1-2 d k^2\right ) x+d k^4 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+k^2 \sqrt {1-4 d k^2+4 d k^4}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2-\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )}+\frac {\left (-k^2-k^2 \sqrt {1-4 d k^2+4 d k^4}\right ) \sqrt [3]{1-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2+\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )}\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 (k^2 \left (1-\sqrt {1-4 d k^2 \left (1-k^2\right )}\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-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2-\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\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 (k^2 \left (1+\sqrt {1-4 d k^2 \left (1-k^2\right )}\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-k^2 x}}{(1-x)^{2/3} \left (1-2 d k^2+\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\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 (k^2 \left (1-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (1-2 d k^2-\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}-\frac {\left (k^2 \left (1+\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right )^{2/3} \left (1-k^2 x^2\right )\right ) \text {Subst}\left (\int \frac {\sqrt [3]{-\frac {1}{-1+k^2}+\frac {k^2 x}{-1+k^2}}}{(1-x)^{2/3} \left (1-2 d k^2+\sqrt {1-4 d k^2+4 d k^4}+2 d k^4 x\right )} \, dx,x,x^2\right )}{2 \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3} \sqrt [3]{\frac {-1+k^2 x^2}{-1+k^2}}}\\ &=\frac {3 k^2 \left (1-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d k^4 \left (1-x^2\right )}{1-2 d k^2 \left (1-k^2\right )-\sqrt {1-4 d k^2 \left (1-k^2\right )}}\right )}{2 \left (1-2 d k^2 \left (1-k^2\right )-\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{2/3}}+\frac {3 k^2 \left (1+\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \left (1-x^2\right ) \left (1-k^2 x^2\right ) F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {k^2 \left (1-x^2\right )}{1-k^2},\frac {2 d k^4 \left (1-x^2\right )}{1-2 d k^2 \left (1-k^2\right )+\sqrt {1-4 d k^2 \left (1-k^2\right )}}\right )}{2 \left (1-2 d k^2 \left (1-k^2\right )+\sqrt {1-4 d k^2 \left (1-k^2\right )}\right ) \sqrt [3]{\frac {1-k^2 x^2}{1-k^2}} \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 = 14.66, size = 208, normalized size = 0.92 \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} \sqrt [3]{d} \left (-1+k^2 x^2\right )^{2/3}}{2 \sqrt [3]{-1+x^2}-\sqrt [3]{d} \left (-1+k^2 x^2\right )^{2/3}}\right )-2 \log \left (\sqrt [3]{-1+x^2}+\sqrt [3]{d} \left (-1+k^2 x^2\right )^{2/3}\right )+\log \left (\left (-1+x^2\right )^{2/3}-\sqrt [3]{d} \sqrt [3]{-1+x^2} \left (-1+k^2 x^2\right )^{2/3}+d^{2/3} \left (-1+k^2 x^2\right )^{4/3}\right )\right )}{4 d^{2/3} \left (\left (-1+x^2\right ) \left (-1+k^2 x^2\right )\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1 + x^2)^(2/3)*(-1 + k^2*x^2)^(2/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*d^(1/3)*(-1 + k^2*x^2)^(2/3))/(2*(-1 + x^2)^
(1/3) - d^(1/3)*(-1 + k^2*x^2)^(2/3))] - 2*Log[(-1 + x^2)^(1/3) + d^(1/3)*(-1 + k^2*x^2)^(2/3)] + Log[(-1 + x^
2)^(2/3) - d^(1/3)*(-1 + x^2)^(1/3)*(-1 + k^2*x^2)^(2/3) + d^(2/3)*(-1 + k^2*x^2)^(4/3)]))/(4*d^(2/3)*((-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 \[\int \frac {\left (2 k^{2}-1\right ) x -2 k^{4} x^{3}+k^{4} x^{5}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {2}{3}} \left (-1+d +\left (-2 d \,k^{2}+1\right ) x^{2}+d \,k^{4} x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(-2*d*k^2+1)*x^2+d*k^4*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(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(-2*d*k^2+1)*x^2+d*k^4*x^4),x, a
lgorithm="maxima")

[Out]

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

[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(((2*k**2-1)*x-2*k**4*x**3+k**4*x**5)/((-x**2+1)*(-k**2*x**2+1))**(2/3)/(-1+d+(-2*d*k**2+1)*x**2+d*k*
*4*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(((2*k^2-1)*x-2*k^4*x^3+k^4*x^5)/((-x^2+1)*(-k^2*x^2+1))^(2/3)/(-1+d+(-2*d*k^2+1)*x^2+d*k^4*x^4),x, a
lgorithm="giac")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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