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

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

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Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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*}

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Mathematica [B]  time = 0.01, 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]

-1/2*Log[Cos[a*x] - Sin[a*x]]/a + Log[Cos[a*x] + Sin[a*x]]/(2*a)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec (2 a x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [B]  time = 1.16, size = 26, normalized size = 2.00 \[ \frac {\log \left (\sin \left (2 \, a x\right ) + 1\right ) - \log \left (-\sin \left (2 \, a x\right ) + 1\right )}{4 \, a} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (pi/x/2)>(-pi/x/2)
Unable to check sign: (pi/x/2)>(-pi/x/2)2/a*1/2*(-1/8*ln(abs(sin(2*a*x)+1/sin(2*a*x)-2))+1/8*ln(abs(sin(2*a*x)
+1/sin(2*a*x)+2)))

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maple [A]  time = 0.08, size = 18, normalized size = 1.38

method	result	size
derivativedivides	\(\frac {\ln \left (\sec \left (2 a x \right )+\tan \left (2 a x \right )\right )}{2 a}\)	\(18\)
default	\(\frac {\ln \left (\sec \left (2 a x \right )+\tan \left (2 a x \right )\right )}{2 a}\)	\(18\)
norman	\(-\frac {\ln \left (\tan \left (a x \right )-1\right )}{2 a}+\frac {\ln \left (\tan \left (a x \right )+1\right )}{2 a}\)	\(26\)
risch	\(\frac {\ln \left ({\mathrm e}^{2 i a x}+i\right )}{2 a}-\frac {\ln \left ({\mathrm e}^{2 i a x}-i\right )}{2 a}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(2*a*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.43, size = 17, normalized size = 1.31 \[ \frac {\log \left (\sec \left (2 \, a x\right ) + \tan \left (2 \, a x\right )\right )}{2 \, a} \]

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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mupad [B]  time = 0.23, size = 11, normalized size = 0.85 \[ \frac {\mathrm {atanh}\left (\sin \left (2\,a\,x\right )\right )}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(2*a*x),x)

[Out]

atanh(sin(2*a*x))/(2*a)

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sympy [A]  time = 0.11, size = 29, normalized size = 2.23 \[ \begin {cases} \frac {- \frac {\log {\left (\sin {\left (2 a x \right )} - 1 \right )}}{2} + \frac {\log {\left (\sin {\left (2 a x \right )} + 1 \right )}}{2}}{2 a} & \text {for}\: 2 a \neq 0 \\x & \text {otherwise} \end {cases} \]

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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