### 3.229 $$\int \cosh (x) \tanh (2 x) \, dx$$

Optimal. Leaf size=19 $\cosh (x)-\frac{\tanh ^{-1}\left (\sqrt{2} \cosh (x)\right )}{\sqrt{2}}$

[Out]

-(ArcTanh[Sqrt[2]*Cosh[x]]/Sqrt[2]) + Cosh[x]

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Rubi [A]  time = 0.031566, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 7, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.429, Rules used = {12, 321, 207} $\cosh (x)-\frac{\tanh ^{-1}\left (\sqrt{2} \cosh (x)\right )}{\sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[Cosh[x]*Tanh[2*x],x]

[Out]

-(ArcTanh[Sqrt[2]*Cosh[x]]/Sqrt[2]) + Cosh[x]

Rule 12

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

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

Rubi steps

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

Mathematica [C]  time = 0.17027, size = 164, normalized size = 8.63 $\frac{4 \sqrt{2} \cosh (x)-4 \tanh ^{-1}\left (\sqrt{2}-i \tanh \left (\frac{x}{2}\right )\right )+\log \left (\sqrt{2}-2 \cosh (x)\right )-\log \left (2 \cosh (x)+\sqrt{2}\right )-2 i \tan ^{-1}\left (\frac{\sinh \left (\frac{x}{2}\right )+\cosh \left (\frac{x}{2}\right )}{\left (1+\sqrt{2}\right ) \cosh \left (\frac{x}{2}\right )-\left (\sqrt{2}-1\right ) \sinh \left (\frac{x}{2}\right )}\right )+2 i \tan ^{-1}\left (\frac{\sinh \left (\frac{x}{2}\right )+\cosh \left (\frac{x}{2}\right )}{\left (\sqrt{2}-1\right ) \cosh \left (\frac{x}{2}\right )-\left (1+\sqrt{2}\right ) \sinh \left (\frac{x}{2}\right )}\right )}{4 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cosh[x]*Tanh[2*x],x]

[Out]

((-2*I)*ArcTan[(Cosh[x/2] + Sinh[x/2])/((1 + Sqrt[2])*Cosh[x/2] - (-1 + Sqrt[2])*Sinh[x/2])] + (2*I)*ArcTan[(C
osh[x/2] + Sinh[x/2])/((-1 + Sqrt[2])*Cosh[x/2] - (1 + Sqrt[2])*Sinh[x/2])] - 4*ArcTanh[Sqrt[2] - I*Tanh[x/2]]
+ 4*Sqrt[2]*Cosh[x] + Log[Sqrt[2] - 2*Cosh[x]] - Log[Sqrt[2] + 2*Cosh[x]])/(4*Sqrt[2])

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Maple [A]  time = 0.013, size = 16, normalized size = 0.8 \begin{align*} \cosh \left ( x \right ) -{\frac{{\it Artanh} \left ( \cosh \left ( x \right ) \sqrt{2} \right ) \sqrt{2}}{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*tanh(2*x),x)

[Out]

cosh(x)-1/2*arctanh(cosh(x)*2^(1/2))*2^(1/2)

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Maxima [B]  time = 1.7082, size = 70, normalized size = 3.68 \begin{align*} -\frac{1}{4} \, \sqrt{2} \log \left (\sqrt{2} e^{\left (-x\right )} + e^{\left (-2 \, x\right )} + 1\right ) + \frac{1}{4} \, \sqrt{2} \log \left (-\sqrt{2} e^{\left (-x\right )} + e^{\left (-2 \, x\right )} + 1\right ) + \frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(2*x),x, algorithm="maxima")

[Out]

-1/4*sqrt(2)*log(sqrt(2)*e^(-x) + e^(-2*x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*e^(-x) + e^(-2*x) + 1) + 1/2*e^(-x)
+ 1/2*e^x

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Fricas [B]  time = 2.11186, size = 259, normalized size = 13.63 \begin{align*} \frac{2 \, \cosh \left (x\right )^{2} +{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \log \left (\frac{\cosh \left (x\right )^{2} + \sinh \left (x\right )^{2} - 2 \, \sqrt{2} \cosh \left (x\right ) + 2}{\cosh \left (x\right )^{2} + \sinh \left (x\right )^{2}}\right ) + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} + 2}{4 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(2*x),x, algorithm="fricas")

[Out]

1/4*(2*cosh(x)^2 + (sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*log((cosh(x)^2 + sinh(x)^2 - 2*sqrt(2)*cosh(x) + 2)/(co
sh(x)^2 + sinh(x)^2)) + 4*cosh(x)*sinh(x) + 2*sinh(x)^2 + 2)/(cosh(x) + sinh(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cosh{\left (x \right )} \tanh{\left (2 x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(2*x),x)

[Out]

Integral(cosh(x)*tanh(2*x), x)

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Giac [B]  time = 1.14523, size = 61, normalized size = 3.21 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (-x\right )} - e^{x}}{\sqrt{2} + e^{\left (-x\right )} + e^{x}}\right ) + \frac{1}{2} \, e^{\left (-x\right )} + \frac{1}{2} \, e^{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*tanh(2*x),x, algorithm="giac")

[Out]

1/4*sqrt(2)*log(-(sqrt(2) - e^(-x) - e^x)/(sqrt(2) + e^(-x) + e^x)) + 1/2*e^(-x) + 1/2*e^x