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

Optimal. Leaf size=223 \[ -\frac {\log \left (\sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}+k x-1\right )}{d^{2/3}}+\frac {\log \left (d^{2/3} \left (k^2 x^4+\left (-k^2-1\right ) x^2+1\right )^{2/3}+\sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1} \left (\sqrt [3]{d}-\sqrt [3]{d} k x\right )+k^2 x^2-2 k x+1\right )}{2 d^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}{\sqrt [3]{d} \sqrt [3]{k^2 x^4+\left (-k^2-1\right ) x^2+1}-2 k x+2}\right )}{d^{2/3}} \]

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Rubi [F]  time = 7.20, 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 {3 k+\left (-2+k^2\right ) x-3 k x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(k*x*(1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*AppellF1[1/2, 1/3, 1/3, 3/2, x^2, k^2*x^2])/(d*((1 - x^2)*(1 - k^2*x^
2))^(1/3)) - ((1 - 4*d)*k*(1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*Defer[Int][1/((1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3
)*(1 - d - (2 + d)*k*x + (d + k^2)*x^2 + d*k*x^3)), x])/(d*((1 - x^2)*(1 - k^2*x^2))^(1/3)) + (2*(k^2 - d*(1 -
k^2))*(1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*Defer[Int][x/((1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*(1 - d - (2 + d)*
k*x + (d + k^2)*x^2 + d*k*x^3)), x])/(d*((1 - x^2)*(1 - k^2*x^2))^(1/3)) - (k*(4*d + k^2)*(1 - x^2)^(1/3)*(1 -
k^2*x^2)^(1/3)*Defer[Int][x^2/((1 - x^2)^(1/3)*(1 - k^2*x^2)^(1/3)*(1 - d - (2 + d)*k*x + (d + k^2)*x^2 + d*k
*x^3)), x])/(d*((1 - x^2)*(1 - k^2*x^2))^(1/3))

Rubi steps

\begin {align*} \int \frac {3 k+\left (-2+k^2\right ) x-3 k x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {3 k+\left (-2+k^2\right ) x-3 k x^2+k^2 x^3}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {3 k-\left (2-k^2\right ) x-3 k x^2+k^2 x^3}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {k}{d \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}}-\frac {k-4 d k-2 \left (k^2-d \left (1-k^2\right )\right ) x+k \left (4 d+k^2\right ) x^2}{d \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )}\right ) \, dx}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=-\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {k-4 d k-2 \left (k^2-d \left (1-k^2\right )\right ) x+k \left (4 d+k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (k \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {k x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {(1-4 d) k-2 \left (k^2-d \left (1-k^2\right )\right ) x+k \left (4 d+k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {k x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \left (\frac {k-4 d k}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )}+\frac {2 \left (-k^2+d \left (1-k^2\right )\right ) x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )}+\frac {k \left (4 d+k^2\right ) x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )}\right ) \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ &=\frac {k x \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} F_1\left (\frac {1}{2};\frac {1}{3},\frac {1}{3};\frac {3}{2};x^2,k^2 x^2\right )}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left ((1-4 d) k \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {1}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}-\frac {\left (k \left (4 d+k^2\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x^2}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}+\frac {\left (2 \left (k^2-d \left (1-k^2\right )\right ) \sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2}\right ) \int \frac {x}{\sqrt [3]{1-x^2} \sqrt [3]{1-k^2 x^2} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx}{d \sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )}}\\ \end {align*}

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Mathematica [F]  time = 1.54, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 k+\left (-2+k^2\right ) x-3 k x^2+k^2 x^3}{\sqrt [3]{\left (1-x^2\right ) \left (1-k^2 x^2\right )} \left (1-d-(2+d) k x+\left (d+k^2\right ) x^2+d k x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 5.65, size = 223, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}{2-2 k x+\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{2/3}}-\frac {\log \left (-1+k x+\sqrt [3]{d} \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}\right )}{d^{2/3}}+\frac {\log \left (1-2 k x+k^2 x^2+\left (\sqrt [3]{d}-\sqrt [3]{d} k x\right ) \sqrt [3]{1+\left (-1-k^2\right ) x^2+k^2 x^4}+d^{2/3} \left (1+\left (-1-k^2\right ) x^2+k^2 x^4\right )^{2/3}\right )}{2 d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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