3.247 \(\int \sec (\tanh (a+b x)) \, dx\)

Optimal. Leaf size=65 \[ \frac{1}{2} \text{Unintegrable}\left (\frac{\text{sech}^2(a+b x) \sec (\tanh (a+b x))}{\tanh (a+b x)+1},x\right )-\frac{1}{2} \text{Unintegrable}\left (\frac{\text{sech}^2(a+b x) \sec (\tanh (a+b x))}{\tanh (a+b x)-1},x\right ) \]

[Out]

-Unintegrable[(Sec[Tanh[a + b*x]]*Sech[a + b*x]^2)/(-1 + Tanh[a + b*x]), x]/2 + Unintegrable[(Sec[Tanh[a + b*x
]]*Sech[a + b*x]^2)/(1 + Tanh[a + b*x]), x]/2

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Rubi [A]  time = 0.0771812, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \sec (\tanh (a+b x)) \, dx \]

Verification is Not applicable to the result.

[In]

Int[Sec[Tanh[a + b*x]],x]

[Out]

-Defer[Subst][Defer[Int][Sec[x]/(-1 + x), x], x, Tanh[a + b*x]]/(2*b) + Defer[Subst][Defer[Int][Sec[x]/(1 + x)
, x], x, Tanh[a + b*x]]/(2*b)

Rubi steps

\begin{align*} \int \sec (\tanh (a+b x)) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sec (x)}{1-x^2} \, dx,x,\tanh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{\sec (x)}{2 (-1+x)}+\frac{\sec (x)}{2 (1+x)}\right ) \, dx,x,\tanh (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sec (x)}{-1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\sec (x)}{1+x} \, dx,x,\tanh (a+b x)\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 4.66012, size = 0, normalized size = 0. \[ \int \sec (\tanh (a+b x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[Tanh[a + b*x]],x]

[Out]

Integrate[Sec[Tanh[a + b*x]], x]

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Maple [A]  time = 0.095, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( \tanh \left ( bx+a \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(tanh(b*x+a)),x)

[Out]

int(sec(tanh(b*x+a)),x)

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Maxima [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (\tanh \left (b x + a\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(tanh(b*x+a)),x, algorithm="maxima")

[Out]

integrate(sec(tanh(b*x + a)), x)

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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sec \left (\tanh \left (b x + a\right )\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(tanh(b*x+a)),x, algorithm="fricas")

[Out]

integral(sec(tanh(b*x + a)), x)

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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec{\left (\tanh{\left (a + b x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(tanh(b*x+a)),x)

[Out]

Integral(sec(tanh(a + b*x)), x)

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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec \left (\tanh \left (b x + a\right )\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(tanh(b*x+a)),x, algorithm="giac")

[Out]

integrate(sec(tanh(b*x + a)), x)