3.11.70 \(\int \frac {1}{(2-3 x^2)^{3/4} (4-3 x^2)} \, dx\) [1070]

Optimal. Leaf size=148 \[ \frac {\tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{4 \sqrt [4]{2} \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{4 \sqrt [4]{2} \sqrt {3}}+\frac {F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2 \sqrt [4]{2} \sqrt {3}} \]

[Out]

1/24*2^(3/4)*arctan(1/3*(2^(3/4)-2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))*3^(1/2)-1/24*2^(3/4)*ar
ctanh(1/3*(2^(3/4)+2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))*3^(1/2)+1/12*2^(3/4)*(cos(1/2*arcsin(
1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcsin(1/2*x*6^(1/2)))*EllipticF(sin(1/2*arcsin(1/2*x*6^(1/2))),2^(1/2))*3^(1
/2)

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Rubi [A]
time = 0.03, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {409, 238, 452} \begin {gather*} \frac {F\left (\left .\frac {1}{2} \text {ArcSin}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2 \sqrt [4]{2} \sqrt {3}}+\frac {\text {ArcTan}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{4 \sqrt [4]{2} \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{4 \sqrt [4]{2} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))]/(4*2^(1/4)*Sqrt[3]) - ArcTanh[(2^(3/
4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))]/(4*2^(1/4)*Sqrt[3]) + EllipticF[ArcSin[Sqrt[3/2]*
x]/2, 2]/(2*2^(1/4)*Sqrt[3])

Rule 238

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2]))*EllipticF[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 409

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[1/c, Int[1/(a + b*x^2)^(3/4), x],
x] - Dist[d/c, Int[x^2/((a + b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d,
0]

Rule 452

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-b/(a*d*Rt[b^2/a, 4]^3))*Ar
cTan[(b + Rt[b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))], x] + Simp[(b/(a*d*Rt[b^2/a, 4
]^3))*ArcTanh[(b - Rt[b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c
, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[b^2/a]

Rubi steps

\begin {align*} \int \frac {1}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx &=\frac {1}{4} \int \frac {1}{\left (2-3 x^2\right )^{3/4}} \, dx+\frac {3}{4} \int \frac {x^2}{\left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx\\ &=\frac {\tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{4 \sqrt [4]{2} \sqrt {3}}-\frac {\tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{4 \sqrt [4]{2} \sqrt {3}}+\frac {F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{2 \sqrt [4]{2} \sqrt {3}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 6.43, size = 63, normalized size = 0.43 \begin {gather*} -\frac {\sqrt {x^2} \left (\Pi \left (-i;\left .\sin ^{-1}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )\right |-1\right )+\Pi \left (i;\left .\sin ^{-1}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )\right |-1\right )\right )}{2 \sqrt [4]{2} \sqrt {3} x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[x^2]*(EllipticPi[-I, ArcSin[(1 - (3*x^2)/2)^(1/4)], -1] + EllipticPi[I, ArcSin[(1 - (3*x^2)/2)^(1/4
)], -1]))/(2^(1/4)*Sqrt[3]*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-3*x^2 + 2)^(1/4)/(9*x^4 - 18*x^2 + 8), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(3*x**2*(2 - 3*x**2)**(3/4) - 4*(2 - 3*x**2)**(3/4)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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