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

Optimal. Leaf size=67 \[ \frac {1}{2} \text {Int}\left (\frac {\text {sech}^2(a+b x) \sec (\tanh (a+b x))}{\tanh (a+b x)+1},x\right )-\frac {1}{2} \text {Int}\left (\frac {\text {sech}^2(a+b x) \sec (\tanh (a+b x))}{\tanh (a+b x)-1},x\right ) \]

[Out]

-1/2*Unintegrable(sec(tanh(b*x+a))*sech(b*x+a)^2/(-1+tanh(b*x+a)),x)+1/2*Unintegrable(sec(tanh(b*x+a))*sech(b*
x+a)^2/(1+tanh(b*x+a)),x)

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Rubi [A]  time = 0.08, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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*}

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Mathematica [A]  time = 5.69, size = 0, normalized size = 0.00 \[ \int \sec (\tanh (a+b x)) \, dx \]

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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fricas [A]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sec \left (\tanh \left (b x + a\right )\right ), x\right ) \]

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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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec \left (\tanh \left (b x + a\right )\right )\,{d x} \]

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)

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maple [A]  time = 0.50, size = 0, normalized size = 0.00 \[ \int \sec \left (\tanh \left (b x +a \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \sec \left (\tanh \left (b x + a\right )\right )\,{d x} \]

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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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\cos \left (\mathrm {tanh}\left (a+b\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cos(tanh(a + b*x)),x)

[Out]

int(1/cos(tanh(a + b*x)), x)

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sec {\left (\tanh {\left (a + b x \right )} \right )}\, dx \]

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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