3.268 \(\int \frac{1}{a \sec (x)+b \tan (x)} \, dx\)

Optimal. Leaf size=11 \[ \frac{\log (a+b \sin (x))}{b} \]

[Out]

Log[a + b*Sin[x]]/b

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Rubi [A]  time = 0.0342235, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3159, 2668, 31} \[ \frac{\log (a+b \sin (x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[(a*Sec[x] + b*Tan[x])^(-1),x]

[Out]

Log[a + b*Sin[x]]/b

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 2668

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [A]  time = 0.0064317, size = 11, normalized size = 1. \[ \frac{\log (a+b \sin (x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Sec[x] + b*Tan[x])^(-1),x]

[Out]

Log[a + b*Sin[x]]/b

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Maple [A]  time = 0.048, size = 12, normalized size = 1.1 \begin{align*}{\frac{\ln \left ( a+b\sin \left ( x \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*sec(x)+b*tan(x)),x)

[Out]

ln(a+b*sin(x))/b

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Maxima [B]  time = 1.53328, size = 68, normalized size = 6.18 \begin{align*} \frac{\log \left (a + \frac{2 \, b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{a \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}}\right )}{b} - \frac{\log \left (\frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.28446, size = 28, normalized size = 2.55 \begin{align*} \frac{\log \left (b \sin \left (x\right ) + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(b*sin(x) + a)/b

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Sympy [A]  time = 0.59975, size = 32, normalized size = 2.91 \begin{align*} \begin{cases} \frac{\log{\left (\frac{a \sec{\left (x \right )}}{b} + \tan{\left (x \right )} \right )}}{b} - \frac{\log{\left (\tan ^{2}{\left (x \right )} + 1 \right )}}{2 b} & \text{for}\: b \neq 0 \\\frac{\tan{\left (x \right )}}{a \sec{\left (x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*sec(x)+b*tan(x)),x)

[Out]

Piecewise((log(a*sec(x)/b + tan(x))/b - log(tan(x)**2 + 1)/(2*b), Ne(b, 0)), (tan(x)/(a*sec(x)), True))

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Giac [A]  time = 1.13482, size = 16, normalized size = 1.45 \begin{align*} \frac{\log \left ({\left | b \sin \left (x\right ) + a \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(abs(b*sin(x) + a))/b