3.77 \(\int \frac{1}{a-a \text{sech}(c+d x)} \, dx\)

Optimal. Leaf size=30 \[ \frac{x}{a}-\frac{\tanh (c+d x)}{d (a-a \text{sech}(c+d x))} \]

[Out]

x/a - Tanh[c + d*x]/(d*(a - a*Sech[c + d*x]))

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Rubi [A]  time = 0.0154844, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3777, 8} \[ \frac{x}{a}-\frac{\tanh (c+d x)}{d (a-a \text{sech}(c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sech[c + d*x])^(-1),x]

[Out]

x/a - Tanh[c + d*x]/(d*(a - a*Sech[c + d*x]))

Rule 3777

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(Cot[c + d*x]*(a + b*Csc[c + d*x])^n)/(d*(
2*n + 1)), x] + Dist[1/(a^2*(2*n + 1)), Int[(a + b*Csc[c + d*x])^(n + 1)*(a*(2*n + 1) - b*(n + 1)*Csc[c + d*x]
), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{a-a \text{sech}(c+d x)} \, dx &=-\frac{\tanh (c+d x)}{d (a-a \text{sech}(c+d x))}+\frac{\int a \, dx}{a^2}\\ &=\frac{x}{a}-\frac{\tanh (c+d x)}{d (a-a \text{sech}(c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.140193, size = 59, normalized size = 1.97 \[ \frac{\text{csch}\left (\frac{c}{2}\right ) \text{csch}\left (\frac{1}{2} (c+d x)\right ) \left (d x \cosh \left (c+\frac{d x}{2}\right )+2 \sinh \left (\frac{d x}{2}\right )-d x \cosh \left (\frac{d x}{2}\right )\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sech[c + d*x])^(-1),x]

[Out]

(Csch[c/2]*Csch[(c + d*x)/2]*(-(d*x*Cosh[(d*x)/2]) + d*x*Cosh[c + (d*x)/2] + 2*Sinh[(d*x)/2]))/(2*a*d)

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Maple [A]  time = 0.032, size = 60, normalized size = 2. \begin{align*}{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{da} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}}-{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-a*sech(d*x+c)),x)

[Out]

1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/a/tanh(1/2*d*x+1/2*c)-1/d/a*ln(tanh(1/2*d*x+1/2*c)-1)

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Maxima [A]  time = 1.11493, size = 47, normalized size = 1.57 \begin{align*} \frac{d x + c}{a d} + \frac{2}{{\left (a e^{\left (-d x - c\right )} - a\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sech(d*x+c)),x, algorithm="maxima")

[Out]

(d*x + c)/(a*d) + 2/((a*e^(-d*x - c) - a)*d)

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Fricas [A]  time = 2.42036, size = 131, normalized size = 4.37 \begin{align*} \frac{d x \cosh \left (d x + c\right ) + d x \sinh \left (d x + c\right ) - d x - 2}{a d \cosh \left (d x + c\right ) + a d \sinh \left (d x + c\right ) - a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sech(d*x+c)),x, algorithm="fricas")

[Out]

(d*x*cosh(d*x + c) + d*x*sinh(d*x + c) - d*x - 2)/(a*d*cosh(d*x + c) + a*d*sinh(d*x + c) - a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{\operatorname{sech}{\left (c + d x \right )} - 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sech(d*x+c)),x)

[Out]

-Integral(1/(sech(c + d*x) - 1), x)/a

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Giac [A]  time = 1.16877, size = 42, normalized size = 1.4 \begin{align*} \frac{d x + c}{a d} - \frac{2}{a d{\left (e^{\left (d x + c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sech(d*x+c)),x, algorithm="giac")

[Out]

(d*x + c)/(a*d) - 2/(a*d*(e^(d*x + c) - 1))