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

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

[Out]

Log[1 + Sin[x]]

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Rubi [A]  time = 0.0244933, antiderivative size = 5, normalized size of antiderivative = 1., 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 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]

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 31

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

Mathematica [B]  time = 0.0138217, size = 16, normalized size = 3.2 \[ 2 \log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right ) \]

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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Maple [A]  time = 0.051, size = 6, normalized size = 1.2 \begin{align*} \ln \left ( 1+\sin \left ( x \right ) \right ) \end{align*}

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 = 0.979022, size = 42, normalized size = 8.4 \begin{align*} 2 \, \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right ) - \log \left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right ) \end{align*}

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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Fricas [A]  time = 2.1763, size = 23, normalized size = 4.6 \begin{align*} \log \left (\sin \left (x\right ) + 1\right ) \end{align*}

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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Sympy [B]  time = 0.165543, size = 17, normalized size = 3.4 \begin{align*} \log{\left (\tan{\left (x \right )} + \sec{\left (x \right )} \right )} - \frac{\log{\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{2} \end{align*}

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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Giac [B]  time = 1.12336, size = 30, normalized size = 6. \begin{align*} -\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 ) \end{align*}

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))