3.200 \(\int \sinh (x) \tanh (2 x) \, dx\)

Optimal. Leaf size=19 \[ \sinh (x)-\frac{\tan ^{-1}\left (\sqrt{2} \sinh (x)\right )}{\sqrt{2}} \]

[Out]

-(ArcTan[Sqrt[2]*Sinh[x]]/Sqrt[2]) + Sinh[x]

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Rubi [A]  time = 0.025886, 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, 203} \[ \sinh (x)-\frac{\tan ^{-1}\left (\sqrt{2} \sinh (x)\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[Sqrt[2]*Sinh[x]]/Sqrt[2]) + Sinh[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 203

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

Rubi steps

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

Mathematica [A]  time = 0.0151152, size = 19, normalized size = 1. \[ \sinh (x)-\frac{\tan ^{-1}\left (\sqrt{2} \sinh (x)\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[Sqrt[2]*Sinh[x]]/Sqrt[2]) + Sinh[x]

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Maple [C]  time = 0.048, size = 54, normalized size = 2.8 \begin{align*}{\frac{{{\rm e}^{x}}}{2}}-{\frac{{{\rm e}^{-x}}}{2}}+{\frac{i}{4}}\sqrt{2}\ln \left ({{\rm e}^{2\,x}}-i\sqrt{2}{{\rm e}^{x}}-1 \right ) -{\frac{i}{4}}\sqrt{2}\ln \left ({{\rm e}^{2\,x}}+i\sqrt{2}{{\rm e}^{x}}-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*exp(x)-1/2*exp(-x)+1/4*I*2^(1/2)*ln(exp(2*x)-I*2^(1/2)*exp(x)-1)-1/4*I*2^(1/2)*ln(exp(2*x)+I*2^(1/2)*exp(x
)-1)

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Maxima [B]  time = 1.54249, size = 72, normalized size = 3.79 \begin{align*} \frac{1}{2} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, e^{\left (-x\right )}\right )}\right ) + \frac{1}{2} \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, e^{\left (-x\right )}\right )}\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(sinh(x)*tanh(2*x),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.13494, size = 420, normalized size = 22.11 \begin{align*} -\frac{{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \arctan \left (\frac{1}{2} \, \sqrt{2} \cosh \left (x\right ) + \frac{1}{2} \, \sqrt{2} \sinh \left (x\right )\right ) -{\left (\sqrt{2} \cosh \left (x\right ) + \sqrt{2} \sinh \left (x\right )\right )} \arctan \left (-\frac{\sqrt{2} \cosh \left (x\right )^{2} + 2 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{2} \sinh \left (x\right )^{2} + \sqrt{2}}{2 \,{\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) - \cosh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) \sinh \left (x\right ) - \sinh \left (x\right )^{2} + 1}{2 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(x)*tanh(2*x), x)

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Giac [B]  time = 1.19434, size = 49, normalized size = 2.58 \begin{align*} -\frac{1}{4} \, \sqrt{2}{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\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(sinh(x)*tanh(2*x),x, algorithm="giac")

[Out]

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