3.112 \(\int \sec (a+b x) \, dx\)

Optimal. Leaf size=11 \[ \frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

[Out]

ArcTanh[Sin[a + b*x]]/b

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Rubi [A]  time = 0.0039425, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3770} \[ \frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b

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 (a+b x) \, dx &=\frac{\tanh ^{-1}(\sin (a+b x))}{b}\\ \end{align*}

Mathematica [A]  time = 0.0015468, size = 11, normalized size = 1. \[ \frac{\tanh ^{-1}(\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b

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Maple [A]  time = 0.006, size = 19, normalized size = 1.7 \begin{align*}{\frac{\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(b*x+a),x)

[Out]

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

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Maxima [B]  time = 0.939586, size = 35, normalized size = 3.18 \begin{align*} \frac{\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (\sin \left (b x + a\right ) - 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.80147, size = 76, normalized size = 6.91 \begin{align*} \frac{\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.613017, size = 34, normalized size = 3.09 \begin{align*} \begin{cases} - \frac{\log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} - 1 \right )}}{b} + \frac{\log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 1 \right )}}{b} & \text{for}\: b \neq 0 \\\frac{x}{\cos{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(b*x+a),x)

[Out]

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

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Giac [B]  time = 1.11236, size = 38, normalized size = 3.45 \begin{align*} \frac{\log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) - \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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