3.28.48 \(\int \frac {(1-2 k^2) x+k^2 x^3}{\sqrt [3]{(1-x^2) (1-k^2 x^2)} (-1+d+(1-2 d k^2) x^2+d k^4 x^4)} \, dx\)

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

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Rubi [F]  time = 0.87, 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+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \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 + k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/3)*(-1 + d + (1 - 2*d*k^2)*x^2 + d*k^4*x^4)),x]

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 14.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1-2 k^2\right ) x+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \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]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 4.79, size = 256, 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^2 x^2+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{2 \sqrt [3]{d}}+\frac {\log \left (-\sqrt [3]{d}+\sqrt [3]{d} k^2 x^2+\sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\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 ) \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 )}{4 \sqrt [3]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 - 2*k^2)*x + k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/3)*(-1 + d + (1 - 2*d*k^2)*x^2 + d*
k^4*x^4)),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^2*x^2 + (1 + (-1 - k^2
)*x^2 + k^2*x^4)^(1/3))])/(2*d^(1/3)) + Log[-d^(1/3) + d^(1/3)*k^2*x^2 + (1 + (-1 - k^2)*x^2 + k^2*x^4)^(1/3)]
/(2*d^(1/3)) - Log[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 + (-1 - k^2)
*x^2 + k^2*x^4)^(1/3) + (1 + (-1 - k^2)*x^2 + k^2*x^4)^(2/3)]/(4*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(((-2*k^2+1)*x+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d+(-2*d*k^2+1)*x^2+d*k^4*x^4),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} - {\left (2 \, k^{2} - 1\right )} x}{{\left (d k^{4} x^{4} - {\left (2 \, 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(((-2*k^2+1)*x+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d+(-2*d*k^2+1)*x^2+d*k^4*x^4),x, algorithm=
"giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((-2*k^2+1)*x+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-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*} \int \frac {k^{2} x^{3} - {\left (2 \, k^{2} - 1\right )} x}{{\left (d k^{4} x^{4} - {\left (2 \, 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(((-2*k^2+1)*x+k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/3)/(-1+d+(-2*d*k^2+1)*x^2+d*k^4*x^4),x, algorithm=
"maxima")

[Out]

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

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

[Out]

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

[Out]

Timed out

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