3.307 \(\int \frac{1}{\sqrt [4]{2-b x^2} (4-b x^2)} \, dx\)

Optimal. Leaf size=124 \[ \frac{\tan ^{-1}\left (\frac{2-\sqrt{2} \sqrt{2-b x^2}}{\sqrt [4]{2} \sqrt{b} x \sqrt [4]{2-b x^2}}\right )}{2\ 2^{3/4} \sqrt{b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{2-b x^2}+2}{\sqrt [4]{2} \sqrt{b} x \sqrt [4]{2-b x^2}}\right )}{2\ 2^{3/4} \sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0204322, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.043, Rules used = {397} \[ \frac{\tan ^{-1}\left (\frac{2-\sqrt{2} \sqrt{2-b x^2}}{\sqrt [4]{2} \sqrt{b} x \sqrt [4]{2-b x^2}}\right )}{2\ 2^{3/4} \sqrt{b}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{2-b x^2}+2}{\sqrt [4]{2} \sqrt{b} x \sqrt [4]{2-b x^2}}\right )}{2\ 2^{3/4} \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 397

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b^2/a, 4]}, -Simp[(b*ArcT
an[(b + q^2*Sqrt[a + b*x^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), x] - Simp[(b*ArcTanh[(b - q^2*Sqrt[a + b*x
^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), 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}{\sqrt [4]{2-b x^2} \left (4-b x^2\right )} \, dx &=\frac{\tan ^{-1}\left (\frac{2-\sqrt{2} \sqrt{2-b x^2}}{\sqrt [4]{2} \sqrt{b} x \sqrt [4]{2-b x^2}}\right )}{2\ 2^{3/4} \sqrt{b}}+\frac{\tanh ^{-1}\left (\frac{2+\sqrt{2} \sqrt{2-b x^2}}{\sqrt [4]{2} \sqrt{b} x \sqrt [4]{2-b x^2}}\right )}{2\ 2^{3/4} \sqrt{b}}\\ \end{align*}

Mathematica [C]  time = 0.131649, size = 145, normalized size = 1.17 \[ -\frac{12 x F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};\frac{b x^2}{2},\frac{b x^2}{4}\right )}{\sqrt [4]{2-b x^2} \left (b x^2-4\right ) \left (b x^2 \left (2 F_1\left (\frac{3}{2};\frac{1}{4},2;\frac{5}{2};\frac{b x^2}{2},\frac{b x^2}{4}\right )+F_1\left (\frac{3}{2};\frac{5}{4},1;\frac{5}{2};\frac{b x^2}{2},\frac{b x^2}{4}\right )\right )+12 F_1\left (\frac{1}{2};\frac{1}{4},1;\frac{3}{2};\frac{b x^2}{2},\frac{b x^2}{4}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-12*x*AppellF1[1/2, 1/4, 1, 3/2, (b*x^2)/2, (b*x^2)/4])/((2 - b*x^2)^(1/4)*(-4 + b*x^2)*(12*AppellF1[1/2, 1/4
, 1, 3/2, (b*x^2)/2, (b*x^2)/4] + b*x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, (b*x^2)/2, (b*x^2)/4] + AppellF1[3/2, 5/
4, 1, 5/2, (b*x^2)/2, (b*x^2)/4])))

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{-b{x}^{2}+4}{\frac{1}{\sqrt [4]{-b{x}^{2}+2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{{\left (b x^{2} - 4\right )}{\left (-b x^{2} + 2\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 73.0235, size = 2229, normalized size = 17.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*(1/2)^(1/4)*(b^(-2))^(1/4)*arctan(-(2*sqrt(2)*(1/2)^(1/4)*(-b*x^2 + 2)^(1/4)*b^2*(b^(-2))^(1/4)*x^
3 + b^2*x^4 + 8*sqrt(2)*(1/2)^(3/4)*(-b*x^2 + 2)^(3/4)*b^2*(b^(-2))^(3/4)*x - 4*b*x^2 + 4*sqrt(1/2)*(b^2*x^2 -
4*b)*sqrt(-b*x^2 + 2)*sqrt(b^(-2)) - 2*sqrt(1/2)*(4*(-b*x^2 + 2)^(1/4)*b*x^2 + 2*sqrt(2)*(1/2)^(3/4)*(b^3*x^3
- 4*b^2*x)*(b^(-2))^(3/4) + 16*sqrt(1/2)*(-b*x^2 + 2)^(3/4)*b*sqrt(b^(-2)) - sqrt(2)*(1/2)^(1/4)*(b^2*x^3 + 4
*b*x)*sqrt(-b*x^2 + 2)*(b^(-2))^(1/4))*sqrt(-(2*sqrt(2)*(1/2)^(3/4)*(-b*x^2 + 2)^(1/4)*b^2*(b^(-2))^(3/4)*x +
sqrt(1/2)*b^2*sqrt(b^(-2))*x^2 + 2*sqrt(-b*x^2 + 2))/(b*x^2 - 4)))/(b^2*x^4 + 8*b*x^2 - 16)) + 1/4*sqrt(2)*(1/
2)^(1/4)*(b^(-2))^(1/4)*arctan(-(2*sqrt(2)*(1/2)^(1/4)*(-b*x^2 + 2)^(1/4)*b^2*(b^(-2))^(1/4)*x^3 - b^2*x^4 + 8
*sqrt(2)*(1/2)^(3/4)*(-b*x^2 + 2)^(3/4)*b^2*(b^(-2))^(3/4)*x + 4*b*x^2 - 4*sqrt(1/2)*(b^2*x^2 - 4*b)*sqrt(-b*x
^2 + 2)*sqrt(b^(-2)) + 2*sqrt(1/2)*(4*(-b*x^2 + 2)^(1/4)*b*x^2 - 2*sqrt(2)*(1/2)^(3/4)*(b^3*x^3 - 4*b^2*x)*(b^
(-2))^(3/4) + 16*sqrt(1/2)*(-b*x^2 + 2)^(3/4)*b*sqrt(b^(-2)) + sqrt(2)*(1/2)^(1/4)*(b^2*x^3 + 4*b*x)*sqrt(-b*x
^2 + 2)*(b^(-2))^(1/4))*sqrt((2*sqrt(2)*(1/2)^(3/4)*(-b*x^2 + 2)^(1/4)*b^2*(b^(-2))^(3/4)*x - sqrt(1/2)*b^2*sq
rt(b^(-2))*x^2 - 2*sqrt(-b*x^2 + 2))/(b*x^2 - 4)))/(b^2*x^4 + 8*b*x^2 - 16)) + 1/16*sqrt(2)*(1/2)^(1/4)*(b^(-2
))^(1/4)*log(-1/2*(2*sqrt(2)*(1/2)^(3/4)*(-b*x^2 + 2)^(1/4)*b^2*(b^(-2))^(3/4)*x + sqrt(1/2)*b^2*sqrt(b^(-2))*
x^2 + 2*sqrt(-b*x^2 + 2))/(b*x^2 - 4)) - 1/16*sqrt(2)*(1/2)^(1/4)*(b^(-2))^(1/4)*log(1/2*(2*sqrt(2)*(1/2)^(3/4
)*(-b*x^2 + 2)^(1/4)*b^2*(b^(-2))^(3/4)*x - sqrt(1/2)*b^2*sqrt(b^(-2))*x^2 - 2*sqrt(-b*x^2 + 2))/(b*x^2 - 4))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{1}{b x^{2} \sqrt [4]{- b x^{2} + 2} - 4 \sqrt [4]{- b x^{2} + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(b*x**2*(-b*x**2 + 2)**(1/4) - 4*(-b*x**2 + 2)**(1/4)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{1}{{\left (b x^{2} - 4\right )}{\left (-b x^{2} + 2\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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