3.1 \(\int \frac {1}{\sqrt {2}+\cos (z)+\sin (z)} \, dz\)

Optimal. Leaf size=22 \[ -\frac {1-\sqrt {2} \sin (z)}{\cos (z)-\sin (z)} \]

[Out]

(-1+sin(z)*2^(1/2))/(cos(z)-sin(z))

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Rubi [A]  time = 0.14, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3114} \[ -\frac {1-\sqrt {2} \sin (z)}{\cos (z)-\sin (z)} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[2] + Cos[z] + Sin[z])^(-1),z]

[Out]

-((1 - Sqrt[2]*Sin[z])/(Cos[z] - Sin[z]))

Rule 3114

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> -Simp[(c - a*Sin
[d + e*x])/(c*e*(c*Cos[d + e*x] - b*Sin[d + e*x])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[a^2 - b^2 - c^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {2}+\cos (z)+\sin (z)} \, dz &=-\frac {1-\sqrt {2} \sin (z)}{\cos (z)-\sin (z)}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 77, normalized size = 3.50 \[ \frac {\left ((1+i)-i \sqrt {2}\right ) \sin \left (\frac {z}{2}\right )-\left (\sqrt {2}+(1+3 i)\right ) \cos \left (\frac {z}{2}\right )}{i \left (\sqrt {2}+(-1-i)\right ) \sin \left (\frac {z}{2}\right )+\left (\sqrt {2}+(1+i)\right ) \cos \left (\frac {z}{2}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[2] + Cos[z] + Sin[z])^(-1),z]

[Out]

(-(((1 + 3*I) + Sqrt[2])*Cos[z/2]) + ((1 + I) - I*Sqrt[2])*Sin[z/2])/(((1 + I) + Sqrt[2])*Cos[z/2] + I*((-1 -
I) + Sqrt[2])*Sin[z/2])

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fricas [A]  time = 0.41, size = 25, normalized size = 1.14 \[ \frac {\sqrt {2} \cos \relax (z) + \sqrt {2} \sin \relax (z) - 2}{2 \, {\left (\cos \relax (z) - \sin \relax (z)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cos(z)+sin(z)+2^(1/2)),z, algorithm="fricas")

[Out]

1/2*(sqrt(2)*cos(z) + sqrt(2)*sin(z) - 2)/(cos(z) - sin(z))

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giac [A]  time = 0.18, size = 18, normalized size = 0.82 \[ -\frac {2 \, {\left (\sqrt {2} + 1\right )}}{\sqrt {2} + \tan \left (\frac {1}{2} \, z\right ) + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cos(z)+sin(z)+2^(1/2)),z, algorithm="giac")

[Out]

-2*(sqrt(2) + 1)/(sqrt(2) + tan(1/2*z) + 1)

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maple [A]  time = 0.08, size = 21, normalized size = 0.95 \[ -\frac {2}{\left (\sqrt {2}-1\right ) \left (\tan \left (\frac {z}{2}\right )+\sqrt {2}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(z)+sin(z)+2^(1/2)),z)

[Out]

-2/(2^(1/2)-1)/(tan(1/2*z)+2^(1/2)+1)

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maxima [A]  time = 1.12, size = 20, normalized size = 0.91 \[ -\frac {2}{\frac {{\left (\sqrt {2} - 1\right )} \sin \relax (z)}{\cos \relax (z) + 1} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cos(z)+sin(z)+2^(1/2)),z, algorithm="maxima")

[Out]

-2/((sqrt(2) - 1)*sin(z)/(cos(z) + 1) + 1)

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mupad [B]  time = 0.14, size = 16, normalized size = 0.73 \[ -\frac {2}{\mathrm {tan}\left (\frac {z}{2}\right )\,\left (\sqrt {2}-1\right )+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(z) + sin(z) + 2^(1/2)),z)

[Out]

-2/(tan(z/2)*(2^(1/2) - 1) + 1)

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sympy [B]  time = 6.73, size = 61, normalized size = 2.77 \[ - \frac {198}{- 239 \tan {\left (\frac {z}{2} \right )} + 169 \sqrt {2} \tan {\left (\frac {z}{2} \right )} - 70 \sqrt {2} + 99} + \frac {140 \sqrt {2}}{- 239 \tan {\left (\frac {z}{2} \right )} + 169 \sqrt {2} \tan {\left (\frac {z}{2} \right )} - 70 \sqrt {2} + 99} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(cos(z)+sin(z)+2**(1/2)),z)

[Out]

-198/(-239*tan(z/2) + 169*sqrt(2)*tan(z/2) - 70*sqrt(2) + 99) + 140*sqrt(2)/(-239*tan(z/2) + 169*sqrt(2)*tan(z
/2) - 70*sqrt(2) + 99)

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