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

Optimal. Leaf size=90 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{k x-1}\right )}{d^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{k x-1}\right )}{d^{3/4}} \]

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Rubi [F]  time = 5.85, 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 {-2 k-(-1+k) (1+k) x+2 k x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+(3+d) k x-\left (d+3 k^2\right ) x^2+k \left (-d+k^2\right ) x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 1.51, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2 k-(-1+k) (1+k) x+2 k x^2}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+(3+d) k x-\left (d+3 k^2\right ) x^2+k \left (-d+k^2\right ) x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 12.99, size = 90, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+k x}\right )}{d^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{-1+k x}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {-2 k -\left (-1+k \right ) \left (1+k \right ) x +2 k \,x^{2}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {1}{4}} \left (-1+d +\left (3+d \right ) k x -\left (3 k^{2}+d \right ) x^{2}+k \left (k^{2}-d \right ) x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (k + 1\right )} {\left (k - 1\right )} x - 2 \, k x^{2} + 2 \, k}{{\left ({\left (k^{2} - d\right )} k x^{3} + {\left (d + 3\right )} k x - {\left (3 \, k^{2} + d\right )} x^{2} + d - 1\right )} \left ({\left (k^{2} x^{2} - 1\right )} {\left (x^{2} - 1\right )}\right )^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {-2\,k\,x^2+\left (k-1\right )\,\left (k+1\right )\,x+2\,k}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/4}\,\left (k\,\left (d-k^2\right )\,x^3+\left (3\,k^2+d\right )\,x^2-k\,\left (d+3\right )\,x-d+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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