3.1.3 \(\int \frac {\text {sech}^2(2+3 x)}{1+2 \tanh ^2(2+3 x)} \, dx\) [3]

Optimal. Leaf size=22 \[ \frac {\text {ArcTan}\left (\sqrt {2} \tanh (2+3 x)\right )}{3 \sqrt {2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3756, 209} \begin {gather*} \frac {\text {ArcTan}\left (\sqrt {2} \tanh (3 x+2)\right )}{3 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sech[2 + 3*x]^2/(1 + 2*Tanh[2 + 3*x]^2),x]

[Out]

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

Rule 209

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

Rule 3756

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {\text {sech}^2(2+3 x)}{1+2 \tanh ^2(2+3 x)} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\tanh (2+3 x)\right )\\ &=\frac {\tan ^{-1}\left (\sqrt {2} \tanh (2+3 x)\right )}{3 \sqrt {2}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(47\) vs. \(2(22)=44\).
time = 0.06, size = 47, normalized size = 2.14 \begin {gather*} \frac {\text {ArcTan}\left (\frac {3 \left (-1+e^8\right )+\left (3+2 e^4+3 e^8\right ) \tanh (3 x)}{4 \sqrt {2} e^4}\right )}{3 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sech[2 + 3*x]^2/(1 + 2*Tanh[2 + 3*x]^2),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 [B] Leaf count of result is larger than twice the leaf count of optimal. \(91\) vs. \(2(16)=32\).
time = 3.69, size = 92, normalized size = 4.18

method	result	size
risch	\(\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{4+6 x}-\frac {1}{3}+\frac {2 i \sqrt {2}}{3}\right )}{12}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{4+6 x}-\frac {1}{3}-\frac {2 i \sqrt {2}}{3}\right )}{12}\)	\(44\)
derivativedivides	\(-\frac {\left (3+\sqrt {6}\right ) \sqrt {6}\, \arctan \left (\frac {2 \tanh \left (1+\frac {3 x}{2}\right )}{2 \sqrt {3}+2 \sqrt {2}}\right )}{9 \left (2 \sqrt {3}+2 \sqrt {2}\right )}-\frac {\sqrt {6}\, \left (-3+\sqrt {6}\right ) \arctan \left (\frac {2 \tanh \left (1+\frac {3 x}{2}\right )}{2 \sqrt {3}-2 \sqrt {2}}\right )}{9 \left (2 \sqrt {3}-2 \sqrt {2}\right )}\)	\(92\)
default	\(-\frac {\left (3+\sqrt {6}\right ) \sqrt {6}\, \arctan \left (\frac {2 \tanh \left (1+\frac {3 x}{2}\right )}{2 \sqrt {3}+2 \sqrt {2}}\right )}{9 \left (2 \sqrt {3}+2 \sqrt {2}\right )}-\frac {\sqrt {6}\, \left (-3+\sqrt {6}\right ) \arctan \left (\frac {2 \tanh \left (1+\frac {3 x}{2}\right )}{2 \sqrt {3}-2 \sqrt {2}}\right )}{9 \left (2 \sqrt {3}-2 \sqrt {2}\right )}\)	\(92\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(2+3*x)^2/(1+2*tanh(2+3*x)^2),x,method=_RETURNVERBOSE)

[Out]

-1/9*(3+6^(1/2))*6^(1/2)/(2*3^(1/2)+2*2^(1/2))*arctan(2*tanh(1+3/2*x)/(2*3^(1/2)+2*2^(1/2)))-1/9*6^(1/2)*(-3+6
^(1/2))/(2*3^(1/2)-2*2^(1/2))*arctan(2*tanh(1+3/2*x)/(2*3^(1/2)-2*2^(1/2)))

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Maxima [A]
time = 0.50, size = 21, normalized size = 0.95 \begin {gather*} -\frac {1}{6} \, \sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} {\left (3 \, e^{\left (-6 \, x - 4\right )} - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2+3*x)^2/(1+2*tanh(2+3*x)^2),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] Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (16) = 32\).
time = 0.41, size = 47, normalized size = 2.14 \begin {gather*} -\frac {1}{6} \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} \cosh \left (3 \, x + 2\right ) + 2 \, \sqrt {2} \sinh \left (3 \, x + 2\right )}{2 \, {\left (\cosh \left (3 \, x + 2\right ) - \sinh \left (3 \, x + 2\right )\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2+3*x)^2/(1+2*tanh(2+3*x)^2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {sech}^{2}{\left (3 x + 2 \right )}}{2 \tanh ^{2}{\left (3 x + 2 \right )} + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2+3*x)**2/(1+2*tanh(2+3*x)**2),x)

[Out]

Integral(sech(3*x + 2)**2/(2*tanh(3*x + 2)**2 + 1), x)

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Giac [A]
time = 0.39, size = 21, normalized size = 0.95 \begin {gather*} \frac {1}{6} \, \sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} {\left (3 \, e^{\left (6 \, x + 4\right )} - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(2+3*x)^2/(1+2*tanh(2+3*x)^2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.49, size = 21, normalized size = 0.95 \begin {gather*} \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (3\,{\mathrm {e}}^{6\,x+4}-1\right )}{4}\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(3*x + 2)^2*(2*tanh(3*x + 2)^2 + 1)),x)

[Out]

(2^(1/2)*atan((2^(1/2)*(3*exp(6*x + 4) - 1))/4))/6

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