3.33 \(\int \frac {1}{(1+\cosh (c+d x))^2} \, dx\)

Optimal. Leaf size=47 \[ \frac {\sinh (c+d x)}{3 d (\cosh (c+d x)+1)}+\frac {\sinh (c+d x)}{3 d (\cosh (c+d x)+1)^2} \]

[Out]

1/3*sinh(d*x+c)/d/(1+cosh(d*x+c))^2+1/3*sinh(d*x+c)/d/(1+cosh(d*x+c))

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2650, 2648} \[ \frac {\sinh (c+d x)}{3 d (\cosh (c+d x)+1)}+\frac {\sinh (c+d x)}{3 d (\cosh (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Cosh[c + d*x])^(-2),x]

[Out]

Sinh[c + d*x]/(3*d*(1 + Cosh[c + d*x])^2) + Sinh[c + d*x]/(3*d*(1 + Cosh[c + d*x]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

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

Rubi steps

\begin {align*} \int \frac {1}{(1+\cosh (c+d x))^2} \, dx &=\frac {\sinh (c+d x)}{3 d (1+\cosh (c+d x))^2}+\frac {1}{3} \int \frac {1}{1+\cosh (c+d x)} \, dx\\ &=\frac {\sinh (c+d x)}{3 d (1+\cosh (c+d x))^2}+\frac {\sinh (c+d x)}{3 d (1+\cosh (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 34, normalized size = 0.72 \[ \frac {4 \sinh (c+d x)+\sinh (2 (c+d x))}{6 d (\cosh (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Cosh[c + d*x])^(-2),x]

[Out]

(4*Sinh[c + d*x] + Sinh[2*(c + d*x)])/(6*d*(1 + Cosh[c + d*x])^2)

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fricas [B]  time = 0.45, size = 113, normalized size = 2.40 \[ -\frac {2 \, {\left (3 \, \cosh \left (d x + c\right ) + 3 \, \sinh \left (d x + c\right ) + 1\right )}}{3 \, {\left (d \cosh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (d \cosh \left (d x + c\right ) + d\right )} \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right ) + 3 \, {\left (d \cosh \left (d x + c\right )^{2} + 2 \, d \cosh \left (d x + c\right ) + d\right )} \sinh \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))^2,x, algorithm="fricas")

[Out]

-2/3*(3*cosh(d*x + c) + 3*sinh(d*x + c) + 1)/(d*cosh(d*x + c)^3 + d*sinh(d*x + c)^3 + 3*d*cosh(d*x + c)^2 + 3*
(d*cosh(d*x + c) + d)*sinh(d*x + c)^2 + 3*d*cosh(d*x + c) + 3*(d*cosh(d*x + c)^2 + 2*d*cosh(d*x + c) + d)*sinh
(d*x + c) + d)

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giac [A]  time = 0.13, size = 25, normalized size = 0.53 \[ -\frac {2 \, {\left (3 \, e^{\left (d x + c\right )} + 1\right )}}{3 \, d {\left (e^{\left (d x + c\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))^2,x, algorithm="giac")

[Out]

-2/3*(3*e^(d*x + c) + 1)/(d*(e^(d*x + c) + 1)^3)

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maple [A]  time = 0.06, size = 30, normalized size = 0.64 \[ \frac {-\frac {\left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6}+\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+cosh(d*x+c))^2,x)

[Out]

1/d*(-1/6*tanh(1/2*d*x+1/2*c)^3+1/2*tanh(1/2*d*x+1/2*c))

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maxima [B]  time = 0.31, size = 90, normalized size = 1.91 \[ \frac {2 \, e^{\left (-d x - c\right )}}{d {\left (3 \, e^{\left (-d x - c\right )} + 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-3 \, d x - 3 \, c\right )} + 1\right )}} + \frac {2}{3 \, d {\left (3 \, e^{\left (-d x - c\right )} + 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-3 \, d x - 3 \, c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))^2,x, algorithm="maxima")

[Out]

2*e^(-d*x - c)/(d*(3*e^(-d*x - c) + 3*e^(-2*d*x - 2*c) + e^(-3*d*x - 3*c) + 1)) + 2/3/(d*(3*e^(-d*x - c) + 3*e
^(-2*d*x - 2*c) + e^(-3*d*x - 3*c) + 1))

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mupad [B]  time = 0.06, size = 25, normalized size = 0.53 \[ -\frac {2\,\left (3\,{\mathrm {e}}^{c+d\,x}+1\right )}{3\,d\,{\left ({\mathrm {e}}^{c+d\,x}+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(c + d*x) + 1)^2,x)

[Out]

-(2*(3*exp(c + d*x) + 1))/(3*d*(exp(c + d*x) + 1)^3)

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sympy [A]  time = 1.01, size = 36, normalized size = 0.77 \[ \begin {cases} - \frac {\tanh ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 d} + \frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2 d} & \text {for}\: d \neq 0 \\\frac {x}{\left (\cosh {\relax (c )} + 1\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+cosh(d*x+c))**2,x)

[Out]

Piecewise((-tanh(c/2 + d*x/2)**3/(6*d) + tanh(c/2 + d*x/2)/(2*d), Ne(d, 0)), (x/(cosh(c) + 1)**2, True))

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