3.279 \(\int \frac{1}{(\sec (x)+\tan (x))^2} \, dx\)

Optimal. Leaf size=14 \[ -x-\frac{2 \cos (x)}{\sin (x)+1} \]

[Out]

-x - (2*Cos[x])/(1 + Sin[x])

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Rubi [A]  time = 0.0414318, antiderivative size = 14, 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 = {4391, 2680, 8} \[ -x-\frac{2 \cos (x)}{\sin (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[x] + Tan[x])^(-2),x]

[Out]

-x - (2*Cos[x])/(1 + Sin[x])

Rule 4391

Int[(u_.)*((b_.)*sec[(c_.) + (d_.)*(x_)]^(n_.) + (a_.)*tan[(c_.) + (d_.)*(x_)]^(n_.))^(p_), x_Symbol] :> Int[A
ctivateTrig[u]*Sec[c + d*x]^(n*p)*(b + a*Sin[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{(\sec (x)+\tan (x))^2} \, dx &=\int \frac{\cos ^2(x)}{(1+\sin (x))^2} \, dx\\ &=-\frac{2 \cos (x)}{1+\sin (x)}-\int 1 \, dx\\ &=-x-\frac{2 \cos (x)}{1+\sin (x)}\\ \end{align*}

Mathematica [A]  time = 0.0255073, size = 27, normalized size = 1.93 \[ \frac{4 \sin \left (\frac{x}{2}\right )}{\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )}-x \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[x] + Tan[x])^(-2),x]

[Out]

-x + (4*Sin[x/2])/(Cos[x/2] + Sin[x/2])

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Maple [A]  time = 0.059, size = 15, normalized size = 1.1 \begin{align*} -4\, \left ( 1+\tan \left ( x/2 \right ) \right ) ^{-1}-x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sec(x)+tan(x))^2,x)

[Out]

-4/(1+tan(1/2*x))-x

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Maxima [A]  time = 1.49414, size = 38, normalized size = 2.71 \begin{align*} -\frac{4}{\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1} - 2 \, \arctan \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x))^2,x, algorithm="maxima")

[Out]

-4/(sin(x)/(cos(x) + 1) + 1) - 2*arctan(sin(x)/(cos(x) + 1))

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Fricas [A]  time = 2.06679, size = 89, normalized size = 6.36 \begin{align*} -\frac{{\left (x + 2\right )} \cos \left (x\right ) +{\left (x - 2\right )} \sin \left (x\right ) + x + 2}{\cos \left (x\right ) + \sin \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\tan{\left (x \right )} + \sec{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x))**2,x)

[Out]

Integral((tan(x) + sec(x))**(-2), x)

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Giac [A]  time = 1.09812, size = 19, normalized size = 1.36 \begin{align*} -x - \frac{4}{\tan \left (\frac{1}{2} \, x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(sec(x)+tan(x))^2,x, algorithm="giac")

[Out]

-x - 4/(tan(1/2*x) + 1)