3.13.48 \(\int \frac {(-3+k^2) x+2 k^2 x^3}{\sqrt [4]{(1-x^2) (1-k^2 x^2)} (-1+d+(3-d k^2) x^2-3 x^4+x^6)} \, dx\)

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

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Rubi [F]  time = 6.48, 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 (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-3 + k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (3 - d*k^2)*x^2 - 3*x^4 + x^6)),x]

[Out]

(-4*k^2*(1 - x^2)^(1/4)*(1 - k^2*x^2)^(1/4)*Defer[Subst][Defer[Int][x^2/((1 - k^2 + k^2*x^4)^(1/4)*(d*(1 - k^2
) + d*k^2*x^4 - x^12)), x], x, (1 - x^2)^(1/4)])/((1 - x^2)*(1 - k^2*x^2))^(1/4) - (4*k^2*(1 - x^2)^(1/4)*(1 -
k^2*x^2)^(1/4)*Defer[Subst][Defer[Int][x^6/((1 - k^2 + k^2*x^4)^(1/4)*(-(d*(1 - k^2)) - d*k^2*x^4 + x^12)), x
], x, (1 - x^2)^(1/4)])/((1 - x^2)*(1 - k^2*x^2))^(1/4) + (2*(3 - k^2)*(1 - x^2)^(1/4)*(1 - k^2*x^2)^(1/4)*Def
er[Subst][Defer[Int][x^2/((1 - k^2 + k^2*x^4)^(1/4)*(d - x^12 + d*k^2*(-1 + x^4))), x], x, (1 - x^2)^(1/4)])/(
(1 - x^2)*(1 - k^2*x^2))^(1/4)

Rubi steps

\begin {align*} \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx &=\int \frac {x \left (-3+k^2+2 k^2 x^2\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {-3+k^2+2 k^2 x}{\sqrt [4]{(1-x) \left (1-k^2 x\right )} \left (-1+d+\left (3-d k^2\right ) x-3 x^2+x^3\right )} \, dx,x,x^2\right )\\ &=\frac {\left (\sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {-3+k^2+2 k^2 x}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (-1+d+\left (3-d k^2\right ) x-3 x^2+x^3\right )} \, dx,x,x^2\right )}{2 \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 ) \operatorname {Subst}\left (\int \left (\frac {3 \left (1-\frac {k^2}{3}\right )}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (1-d-\left (3-d k^2\right ) x+3 x^2-x^3\right )}+\frac {2 k^2 x}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (-1+d+\left (3-d k^2\right ) x-3 x^2+x^3\right )}\right ) \, dx,x,x^2\right )}{2 \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (-1+d+\left (3-d k^2\right ) x-3 x^2+x^3\right )} \, dx,x,x^2\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (\left (3-k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-x} \sqrt [4]{1-k^2 x} \left (1-d-\left (3-d k^2\right ) x+3 x^2-x^3\right )} \, dx,x,x^2\right )}{2 \sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2-x^6}{\sqrt [4]{1+k^2 \left (-1+x^4\right )} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (3-k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1+k^2 \left (-1+x^4\right )} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (1-x^4\right )}{\sqrt [4]{1+k^2 \left (-1+x^4\right )} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (3-k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (1-x^4\right )}{\sqrt [4]{1-k^2+k^2 x^4} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (3-k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (d \left (1-k^2\right )+d k^2 x^4-x^{12}\right )}+\frac {x^6}{\sqrt [4]{1-k^2+k^2 x^4} \left (-d \left (1-k^2\right )-d k^2 x^4+x^{12}\right )}\right ) \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (3-k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (d \left (1-k^2\right )+d k^2 x^4-x^{12}\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (4 k^2 \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{1-k^2+k^2 x^4} \left (-d \left (1-k^2\right )-d k^2 x^4+x^{12}\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (3-k^2\right ) \sqrt [4]{1-x^2} \sqrt [4]{1-k^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{1-k^2+k^2 x^4} \left (d-x^{12}+d k^2 \left (-1+x^4\right )\right )} \, dx,x,\sqrt [4]{1-x^2}\right )}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}

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Mathematica [F]  time = 2.91, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-3+k^2\right ) x+2 k^2 x^3}{\sqrt [4]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (-1+d+\left (3-d k^2\right ) x^2-3 x^4+x^6\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[((-3 + k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (3 - d*k^2)*x^2 - 3*x^4 + x^6)
),x]

[Out]

Integrate[((-3 + k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (3 - d*k^2)*x^2 - 3*x^4 + x^6)
), x]

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IntegrateAlgebraic [A]  time = 12.95, 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+x^2}\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+x^2}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-3 + k^2)*x + 2*k^2*x^3)/(((1 - x^2)*(1 - k^2*x^2))^(1/4)*(-1 + d + (3 - d*k^2)*x^2 - 3*x
^4 + x^6)),x]

[Out]

ArcTan[(d^(1/4)*(1 + (-1 - k^2)*x^2 + k^2*x^4)^(1/4))/(-1 + x^2)]/d^(3/4) - ArcTanh[(d^(1/4)*(1 + (-1 - k^2)*x
^2 + k^2*x^4)^(1/4))/(-1 + x^2)]/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(((k^2-3)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(-d*k^2+3)*x^2-3*x^4+x^6),x, algorithm="fr
icas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, k^{2} x^{3} + {\left (k^{2} - 3\right )} x}{{\left (x^{6} - 3 \, x^{4} - {\left (d k^{2} - 3\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(((k^2-3)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(-d*k^2+3)*x^2-3*x^4+x^6),x, algorithm="gi
ac")

[Out]

integrate((2*k^2*x^3 + (k^2 - 3)*x)/((x^6 - 3*x^4 - (d*k^2 - 3)*x^2 + d - 1)*((k^2*x^2 - 1)*(x^2 - 1))^(1/4)),
x)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (k^{2}-3\right ) x +2 k^{2} x^{3}}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {1}{4}} \left (-1+d +\left (-d \,k^{2}+3\right ) x^{2}-3 x^{4}+x^{6}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((k^2-3)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(-d*k^2+3)*x^2-3*x^4+x^6),x)

[Out]

int(((k^2-3)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(-d*k^2+3)*x^2-3*x^4+x^6),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, k^{2} x^{3} + {\left (k^{2} - 3\right )} x}{{\left (x^{6} - 3 \, x^{4} - {\left (d k^{2} - 3\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(((k^2-3)*x+2*k^2*x^3)/((-x^2+1)*(-k^2*x^2+1))^(1/4)/(-1+d+(-d*k^2+3)*x^2-3*x^4+x^6),x, algorithm="ma
xima")

[Out]

integrate((2*k^2*x^3 + (k^2 - 3)*x)/((x^6 - 3*x^4 - (d*k^2 - 3)*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 {x\,\left (k^2-3\right )+2\,k^2\,x^3}{{\left (\left (x^2-1\right )\,\left (k^2\,x^2-1\right )\right )}^{1/4}\,\left (-x^6+3\,x^4+\left (d\,k^2-3\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 - 3) + 2*k^2*x^3)/(((x^2 - 1)*(k^2*x^2 - 1))^(1/4)*(x^2*(d*k^2 - 3) - d + 3*x^4 - x^6 + 1)),x)

[Out]

-int((x*(k^2 - 3) + 2*k^2*x^3)/(((x^2 - 1)*(k^2*x^2 - 1))^(1/4)*(x^2*(d*k^2 - 3) - d + 3*x^4 - x^6 + 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(((k**2-3)*x+2*k**2*x**3)/((-x**2+1)*(-k**2*x**2+1))**(1/4)/(-1+d+(-d*k**2+3)*x**2-3*x**4+x**6),x)

[Out]

Timed out

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