3.3 \(\int \sec (2 a x) \, dx\)

Optimal. Leaf size=13 \[ \frac{\tanh ^{-1}(\sin (2 a x))}{2 a} \]

[Out]

ArcTanh[Sin[2*a*x]]/(2*a)

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Rubi [A]  time = 0.0037081, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 5, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3770} \[ \frac{\tanh ^{-1}(\sin (2 a x))}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sec[2*a*x],x]

[Out]

ArcTanh[Sin[2*a*x]]/(2*a)

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 0.0082175, size = 37, normalized size = 2.85 \[ \frac{\log (\sin (a x)+\cos (a x))}{2 a}-\frac{\log (\cos (a x)-\sin (a x))}{2 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[2*a*x],x]

[Out]

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

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Maple [A]  time = 0.003, size = 18, normalized size = 1.4 \begin{align*}{\frac{\ln \left ( \sec \left ( 2\,ax \right ) +\tan \left ( 2\,ax \right ) \right ) }{2\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(2*a*x),x)

[Out]

1/2/a*ln(sec(2*a*x)+tan(2*a*x))

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Maxima [A]  time = 0.922288, size = 23, normalized size = 1.77 \begin{align*} \frac{\log \left (\sec \left (2 \, a x\right ) + \tan \left (2 \, a x\right )\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*a*x),x, algorithm="maxima")

[Out]

1/2*log(sec(2*a*x) + tan(2*a*x))/a

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Fricas [B]  time = 2.03189, size = 70, normalized size = 5.38 \begin{align*} \frac{\log \left (\sin \left (2 \, a x\right ) + 1\right ) - \log \left (-\sin \left (2 \, a x\right ) + 1\right )}{4 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*a*x),x, algorithm="fricas")

[Out]

1/4*(log(sin(2*a*x) + 1) - log(-sin(2*a*x) + 1))/a

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Sympy [A]  time = 0.099176, size = 29, normalized size = 2.23 \begin{align*} \begin{cases} \frac{- \frac{\log{\left (\sin{\left (2 a x \right )} - 1 \right )}}{2} + \frac{\log{\left (\sin{\left (2 a x \right )} + 1 \right )}}{2}}{2 a} & \text{for}\: 2 a \neq 0 \\x & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*a*x),x)

[Out]

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

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Giac [B]  time = 1.04893, size = 54, normalized size = 4.15 \begin{align*} \frac{\log \left ({\left | \frac{1}{\sin \left (2 \, a x\right )} + \sin \left (2 \, a x\right ) + 2 \right |}\right ) - \log \left ({\left | \frac{1}{\sin \left (2 \, a x\right )} + \sin \left (2 \, a x\right ) - 2 \right |}\right )}{8 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(2*a*x),x, algorithm="giac")

[Out]

1/8*(log(abs(1/sin(2*a*x) + sin(2*a*x) + 2)) - log(abs(1/sin(2*a*x) + sin(2*a*x) - 2)))/a