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

Optimal. Leaf size=129 \[ -\frac {\tan ^{-1}\left (\frac {2 \sqrt [4]{2} \sqrt {b x^2+2}+2\ 2^{3/4}}{2 \sqrt {b} x \sqrt [4]{b x^2+2}}\right )}{2\ 2^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {2\ 2^{3/4}-2 \sqrt [4]{2} \sqrt {b x^2+2}}{2 \sqrt {b} x \sqrt [4]{b x^2+2}}\right )}{2\ 2^{3/4} \sqrt {b}} \]

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Rubi [A]  time = 0.02, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {397} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {2 \sqrt [4]{2} \sqrt {b x^2+2}+2\ 2^{3/4}}{2 \sqrt {b} x \sqrt [4]{b x^2+2}}\right )}{2\ 2^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {2\ 2^{3/4}-2 \sqrt [4]{2} \sqrt {b x^2+2}}{2 \sqrt {b} x \sqrt [4]{b x^2+2}}\right )}{2\ 2^{3/4} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[(2*2^(3/4) + 2*2^(1/4)*Sqrt[2 + b*x^2])/(2*Sqrt[b]*x*(2 + b*x^2)^(1/4))]/(2*2^(3/4)*Sqrt[b]) - ArcTanh
[(2*2^(3/4) - 2*2^(1/4)*Sqrt[2 + b*x^2])/(2*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\ 2^{3/4}+2 \sqrt [4]{2} \sqrt {2+b x^2}}{2 \sqrt {b} x \sqrt [4]{2+b x^2}}\right )}{2\ 2^{3/4} \sqrt {b}}-\frac {\tanh ^{-1}\left (\frac {2\ 2^{3/4}-2 \sqrt [4]{2} \sqrt {2+b x^2}}{2 \sqrt {b} x \sqrt [4]{2+b x^2}}\right )}{2\ 2^{3/4} \sqrt {b}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 144, normalized size = 1.12 \begin {gather*} -\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]{b x^2+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 )} \end {gather*}

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

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IntegrateAlgebraic [A]  time = 0.27, size = 137, normalized size = 1.06 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\frac {\sqrt {b} x^2}{2 \sqrt [4]{2}}-\frac {\sqrt [4]{2} \sqrt {b x^2+2}}{\sqrt {b}}}{x \sqrt [4]{b x^2+2}}\right )}{4\ 2^{3/4} \sqrt {b}}+\frac {\tanh ^{-1}\left (\frac {2\ 2^{3/4} \sqrt {b} x \sqrt [4]{b x^2+2}}{\sqrt {2} b x^2+4 \sqrt {b x^2+2}}\right )}{4\ 2^{3/4} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((2 + b*x^2)^(1/4)*(4 + b*x^2)),x]

[Out]

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

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fricas [B]  time = 28.02, size = 755, normalized size = 5.85

result too large to display

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*sqrt(b^(-2))*x^2 -
2*sqrt(b*x^2 + 2))/(b*x^2 + 4)))/(b^2*x^4 - 8*b*x^2 - 16)) + 1/16*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{2} + 4\right )} {\left (b x^{2} + 2\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

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)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \,x^{2}+2\right )^{\frac {1}{4}} \left (b \,x^{2}+4\right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (b x^{2} + 4\right )} {\left (b x^{2} + 2\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

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

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

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 + 2)**(1/4)*(b*x**2 + 4)), x)

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