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]

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

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.04, 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]

-1/2*ArcTan[2*Coth[c/2 + (d*x)/2]]/d

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fricas [A]  time = 0.53, size = 24, normalized size = 1.09 \[ \frac {\arctan \left (\frac {5}{4} \, \cosh \left (d x + c\right ) + \frac {5}{4} \, \sinh \left (d x + c\right ) + \frac {3}{4}\right )}{2 \, d} \]

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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giac [A]  time = 0.14, size = 16, normalized size = 0.73 \[ \frac {\arctan \left (\frac {5}{4} \, e^{\left (d x + c\right )} + \frac {3}{4}\right )}{2 \, d} \]

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

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maple [A]  time = 0.06, size = 18, normalized size = 0.82 \[ \frac {\arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{2 d} \]

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 = 0.41, size = 19, normalized size = 0.86 \[ -\frac {\arctan \left (\frac {5}{4} \, e^{\left (-d x - c\right )} + \frac {3}{4}\right )}{2 \, d} \]

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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mupad [B]  time = 0.13, size = 34, normalized size = 1.55 \[ \frac {\mathrm {atan}\left (\frac {3\,\sqrt {d^2}+5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {d^2}}{4\,d}\right )}{2\,\sqrt {d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*cosh(c + d*x) + 3),x)

[Out]

atan((3*(d^2)^(1/2) + 5*exp(d*x)*exp(c)*(d^2)^(1/2))/(4*d))/(2*(d^2)^(1/2))

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sympy [A]  time = 0.78, size = 24, normalized size = 1.09 \[ \begin {cases} \frac {\operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{2 d} & \text {for}\: d \neq 0 \\\frac {x}{5 \cosh {\relax (c )} + 3} & \text {otherwise} \end {cases} \]

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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