3.71 \(\int \frac{1}{3+5 \cosh (c+d x)} \, dx\)

Optimal. Leaf size=22 \[ \frac{\tan ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 d} \]

[Out]

ArcTan[Tanh[(c + d*x)/2]/2]/(2*d)

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Rubi [A]  time = 0.0147783, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2659, 206} \[ \frac{\tan ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Cosh[c + d*x])^(-1),x]

[Out]

ArcTan[Tanh[(c + d*x)/2]/2]/(2*d)

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{3+5 \cosh (c+d x)} \, dx &=-\frac{(2 i) \operatorname{Subst}\left (\int \frac{1}{8-2 x^2} \, dx,x,\tan \left (\frac{1}{2} (i c+i d x)\right )\right )}{d}\\ &=\frac{\tan ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0344725, size = 23, normalized size = 1.05 \[ -\frac{\tan ^{-1}\left (2 \coth \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*Cosh[c + d*x])^(-1),x]

[Out]

-ArcTan[2*Coth[c/2 + (d*x)/2]]/(2*d)

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Maple [A]  time = 0.01, size = 18, normalized size = 0.8 \begin{align*}{\frac{1}{2\,d}\arctan \left ({\frac{1}{2}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3+5*cosh(d*x+c)),x)

[Out]

1/2*arctan(1/2*tanh(1/2*d*x+1/2*c))/d

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Maxima [A]  time = 1.56405, size = 26, normalized size = 1.18 \begin{align*} -\frac{\arctan \left (\frac{5}{4} \, e^{\left (-d x - c\right )} + \frac{3}{4}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*arctan(5/4*e^(-d*x - c) + 3/4)/d

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Fricas [A]  time = 2.28335, size = 80, normalized size = 3.64 \begin{align*} \frac{\arctan \left (\frac{5}{4} \, \cosh \left (d x + c\right ) + \frac{5}{4} \, \sinh \left (d x + c\right ) + \frac{3}{4}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c)),x, algorithm="fricas")

[Out]

1/2*arctan(5/4*cosh(d*x + c) + 5/4*sinh(d*x + c) + 3/4)/d

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Sympy [A]  time = 0.863382, size = 24, normalized size = 1.09 \begin{align*} \begin{cases} \frac{\operatorname{atan}{\left (\frac{\tanh{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2} \right )}}{2 d} & \text{for}\: d \neq 0 \\\frac{x}{5 \cosh{\left (c \right )} + 3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c)),x)

[Out]

Piecewise((atan(tanh(c/2 + d*x/2)/2)/(2*d), Ne(d, 0)), (x/(5*cosh(c) + 3), True))

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Giac [A]  time = 1.19863, size = 22, normalized size = 1. \begin{align*} \frac{\arctan \left (\frac{5}{4} \, e^{\left (d x + c\right )} + \frac{3}{4}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c)),x, algorithm="giac")

[Out]

1/2*arctan(5/4*e^(d*x + c) + 3/4)/d