3.14.30 \(\int \frac {(1-2 x+x^2) (-2+(-1+k) (1+k) x+2 k^2 x^2)}{((1-x^2) (1-k^2 x^2))^{3/4} (-1+d-(1+3 d) x+(3 d+k^2) x^2+(-d+k^2) x^3)} \, dx\)

Optimal. Leaf size=96 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt [4]{d} x-\sqrt [4]{d}}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{d^{3/4}}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{d} x-\sqrt [4]{d}}{\sqrt [4]{k^2 x^4+\left (-k^2-1\right ) x^2+1}}\right )}{d^{3/4}} \]

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Rubi [F]  time = 180.01, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

Rubi steps

\begin {align*} \int \frac {\left (1-2 x+x^2\right ) \left (-2+(-1+k) (1+k) x+2 k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d-(1+3 d) x+\left (3 d+k^2\right ) x^2+\left (-d+k^2\right ) x^3\right )} \, dx &=\int \frac {(-1+x)^2 \left (-2+(-1+k) (1+k) x+2 k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d-(1+3 d) x+\left (3 d+k^2\right ) x^2+\left (-d+k^2\right ) x^3\right )} \, dx\\ &=\frac {\left (\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {(-1+x)^2 \left (-2+(-1+k) (1+k) x+2 k^2 x^2\right )}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (-1+d-(1+3 d) x+\left (3 d+k^2\right ) x^2+\left (-d+k^2\right ) x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=\frac {\left (\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4}\right ) \int \frac {(1-x)^2 \left (2+(1-k) (1+k) x-2 k^2 x^2\right )}{\left (1-x^2\right )^{3/4} \left (1-k^2 x^2\right )^{3/4} \left (1-d+(1+3 d) x-\left (3 d+k^2\right ) x^2+\left (d-k^2\right ) x^3\right )} \, dx}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4}}\\ &=\text {rest of steps removed due to Latex formating problem} \end {align*}

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Mathematica [F]  time = 2.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1-2 x+x^2\right ) \left (-2+(-1+k) (1+k) x+2 k^2 x^2\right )}{\left (\left (1-x^2\right ) \left (1-k^2 x^2\right )\right )^{3/4} \left (-1+d-(1+3 d) x+\left (3 d+k^2\right ) x^2+\left (-d+k^2\right ) x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 15.81, size = 96, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{3/4}}+\frac {\tanh ^{-1}\left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{1+\left (-1-k^2\right ) x^2+k^2 x^4}}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (x^{2}-2 x +1\right ) \left (-2+\left (-1+k \right ) \left (1+k \right ) x +2 k^{2} x^{2}\right )}{\left (\left (-x^{2}+1\right ) \left (-k^{2} x^{2}+1\right )\right )^{\frac {3}{4}} \left (-1+d -\left (1+3 d \right ) x +\left (k^{2}+3 d \right ) x^{2}+\left (k^{2}-d \right ) x^{3}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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