3.188 \(\int \frac {1}{\sec (x)+\tan (x)} \, dx\)

Optimal. Leaf size=5 \[ \log (\sin (x)+1) \]

[Out]

ln(1+sin(x))

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Rubi [A]  time = 0.02, antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3159, 2667, 31} \[ \log (\sin (x)+1) \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x] + Tan[x])^(-1),x]

[Out]

Log[1 + Sin[x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 3159

Int[((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + (c_.)*tan[(d_.) + (e_.)*(x_)])^(-1), x_Symbol] :> Int[Cos[d + e*x
]/(b + a*Cos[d + e*x] + c*Sin[d + e*x]), x] /; FreeQ[{a, b, c, d, e}, x]

Rubi steps

\begin {align*} \int \frac {1}{\sec (x)+\tan (x)} \, dx &=\int \frac {\cos (x)}{1+\sin (x)} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\sin (x)\right )\\ &=\log (1+\sin (x))\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 16, normalized size = 3.20 \[ 2 \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x] + Tan[x])^(-1),x]

[Out]

2*Log[Cos[x/2] + Sin[x/2]]

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fricas [A]  time = 1.05, size = 5, normalized size = 1.00 \[ \log \left (\sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x)),x, algorithm="fricas")

[Out]

log(sin(x) + 1)

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giac [B]  time = 5.74, size = 22, normalized size = 4.40 \[ -\log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right ) + 2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x)),x, algorithm="giac")

[Out]

-log(tan(1/2*x)^2 + 1) + 2*log(abs(tan(1/2*x) + 1))

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maple [A]  time = 0.08, size = 6, normalized size = 1.20 \[ \ln \left (1+\sin \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sec(x)+tan(x)),x)

[Out]

ln(1+sin(x))

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maxima [B]  time = 0.32, size = 31, normalized size = 6.20 \[ 2 \, \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right ) - \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x)),x, algorithm="maxima")

[Out]

2*log(sin(x)/(cos(x) + 1) + 1) - log(sin(x)^2/(cos(x) + 1)^2 + 1)

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mupad [B]  time = 1.11, size = 21, normalized size = 4.20 \[ 2\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )+1\right )-\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(tan(x) + 1/cos(x)),x)

[Out]

2*log(tan(x/2) + 1) - log(tan(x/2)^2 + 1)

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sympy [B]  time = 0.13, size = 17, normalized size = 3.40 \[ \log {\left (\tan {\relax (x )} + \sec {\relax (x )} \right )} - \frac {\log {\left (\tan ^{2}{\relax (x )} + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x)),x)

[Out]

log(tan(x) + sec(x)) - log(tan(x)**2 + 1)/2

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