3.1 \(\int \frac{2}{-1+3 \cosh (4+6 x)} \, dx\)

Optimal. Leaf size=22 \[ \frac{\tan ^{-1}\left (\sqrt{2} \tanh (3 x+2)\right )}{3 \sqrt{2}} \]

[Out]

ArcTan[Sqrt[2]*Tanh[2 + 3*x]]/(3*Sqrt[2])

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Rubi [A]  time = 0.0155441, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {12, 2659, 206} \[ \frac{\tan ^{-1}\left (\sqrt{2} \tanh (3 x+2)\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[2/(-1 + 3*Cosh[4 + 6*x]),x]

[Out]

ArcTan[Sqrt[2]*Tanh[2 + 3*x]]/(3*Sqrt[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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{2}{-1+3 \cosh (4+6 x)} \, dx &=2 \int \frac{1}{-1+3 \cosh (4+6 x)} \, dx\\ &=-\left (\frac{2}{3} i \operatorname{Subst}\left (\int \frac{1}{2-4 x^2} \, dx,x,\tan \left (\frac{1}{2} (4 i+6 i x)\right )\right )\right )\\ &=\frac{\tan ^{-1}\left (\sqrt{2} \tanh (2+3 x)\right )}{3 \sqrt{2}}\\ \end{align*}

Mathematica [B]  time = 0.0704974, size = 47, normalized size = 2.14 \[ \frac{\tan ^{-1}\left (\frac{\left (3+2 e^4+3 e^8\right ) \tanh (3 x)+3 \left (e^8-1\right )}{4 \sqrt{2} e^4}\right )}{3 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Integrate[2/(-1 + 3*Cosh[4 + 6*x]),x]

[Out]

ArcTan[(3*(-1 + E^8) + (3 + 2*E^4 + 3*E^8)*Tanh[3*x])/(4*Sqrt[2]*E^4)]/(3*Sqrt[2])

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Maple [A]  time = 0.014, size = 17, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( \sqrt{2}\tanh \left ( 2+3\,x \right ) \right ) \sqrt{2}}{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(-1+3*cosh(4+6*x)),x)

[Out]

1/6*arctan(2^(1/2)*tanh(2+3*x))*2^(1/2)

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Maxima [A]  time = 1.48378, size = 28, normalized size = 1.27 \begin{align*} -\frac{1}{6} \, \sqrt{2} \arctan \left (\frac{1}{4} \, \sqrt{2}{\left (3 \, e^{\left (-6 \, x - 4\right )} - 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(-1+3*cosh(4+6*x)),x, algorithm="maxima")

[Out]

-1/6*sqrt(2)*arctan(1/4*sqrt(2)*(3*e^(-6*x - 4) - 1))

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Fricas [B]  time = 2.04042, size = 120, normalized size = 5.45 \begin{align*} \frac{1}{6} \, \sqrt{2} \arctan \left (\frac{3}{4} \, \sqrt{2} \cosh \left (6 \, x + 4\right ) + \frac{3}{4} \, \sqrt{2} \sinh \left (6 \, x + 4\right ) - \frac{1}{4} \, \sqrt{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(-1+3*cosh(4+6*x)),x, algorithm="fricas")

[Out]

1/6*sqrt(2)*arctan(3/4*sqrt(2)*cosh(6*x + 4) + 3/4*sqrt(2)*sinh(6*x + 4) - 1/4*sqrt(2))

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Sympy [A]  time = 0.322538, size = 19, normalized size = 0.86 \begin{align*} \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} \tanh{\left (3 x + 2 \right )} \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(-1+3*cosh(4+6*x)),x)

[Out]

sqrt(2)*atan(sqrt(2)*tanh(3*x + 2))/6

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Giac [A]  time = 1.186, size = 28, normalized size = 1.27 \begin{align*} \frac{1}{6} \, \sqrt{2} \arctan \left (\frac{1}{4} \, \sqrt{2}{\left (3 \, e^{\left (6 \, x + 4\right )} - 1\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(-1+3*cosh(4+6*x)),x, algorithm="giac")

[Out]

1/6*sqrt(2)*arctan(1/4*sqrt(2)*(3*e^(6*x + 4) - 1))