3.201 \(\int \sinh (x) \tanh (3 x) \, dx\)

Optimal. Leaf size=19 \[ \sinh (x)-\frac{1}{3} \tan ^{-1}(\sinh (x))-\frac{1}{3} \tan ^{-1}(2 \sinh (x)) \]

[Out]

-ArcTan[Sinh[x]]/3 - ArcTan[2*Sinh[x]]/3 + Sinh[x]

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Rubi [A]  time = 0.0433314, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {1279, 1163, 203} \[ \sinh (x)-\frac{1}{3} \tan ^{-1}(\sinh (x))-\frac{1}{3} \tan ^{-1}(2 \sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]*Tanh[3*x],x]

[Out]

-ArcTan[Sinh[x]]/3 - ArcTan[2*Sinh[x]]/3 + Sinh[x]

Rule 1279

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(e*f
*(f*x)^(m - 1)*(a + b*x^2 + c*x^4)^(p + 1))/(c*(m + 4*p + 3)), x] - Dist[f^2/(c*(m + 4*p + 3)), Int[(f*x)^(m -
2)*(a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[
{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (Inte
gerQ[p] || IntegerQ[m])

Rule 1163

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && GtQ[b^2
- 4*a*c, 0]

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 (3 x) \, dx &=-\operatorname{Subst}\left (\int \frac{x^2 \left (-3-4 x^2\right )}{1+5 x^2+4 x^4} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)+\frac{1}{4} \operatorname{Subst}\left (\int \frac{-4-8 x^2}{1+5 x^2+4 x^4} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{1+4 x^2} \, dx,x,\sinh (x)\right )-\frac{4}{3} \operatorname{Subst}\left (\int \frac{1}{4+4 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{3} \tan ^{-1}(\sinh (x))-\frac{1}{3} \tan ^{-1}(2 \sinh (x))+\sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.028378, size = 19, normalized size = 1. \[ \sinh (x)-\frac{1}{3} \tan ^{-1}(\sinh (x))-\frac{1}{3} \tan ^{-1}(2 \sinh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]*Tanh[3*x],x]

[Out]

-ArcTan[Sinh[x]]/3 - ArcTan[2*Sinh[x]]/3 + Sinh[x]

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Maple [C]  time = 0.066, size = 60, normalized size = 3.2 \begin{align*}{\frac{{{\rm e}^{x}}}{2}}-{\frac{{{\rm e}^{-x}}}{2}}+{\frac{i}{3}}\ln \left ({{\rm e}^{x}}-i \right ) -{\frac{i}{3}}\ln \left ({{\rm e}^{x}}+i \right ) +{\frac{i}{6}}\ln \left ({{\rm e}^{2\,x}}-i{{\rm e}^{x}}-1 \right ) -{\frac{i}{6}}\ln \left ({{\rm e}^{2\,x}}+i{{\rm e}^{x}}-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)*tanh(3*x),x)

[Out]

1/2*exp(x)-1/2*exp(-x)+1/3*I*ln(exp(x)-I)-1/3*I*ln(exp(x)+I)+1/6*I*ln(exp(2*x)-I*exp(x)-1)-1/6*I*ln(exp(2*x)+I
*exp(x)-1)

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Maxima [B]  time = 1.61363, size = 62, normalized size = 3.26 \begin{align*} \frac{1}{3} \, \arctan \left (\sqrt{3} + 2 \, e^{\left (-x\right )}\right ) + \frac{1}{3} \, \arctan \left (-\sqrt{3} + 2 \, e^{\left (-x\right )}\right ) + \frac{2}{3} \, \arctan \left (e^{\left (-x\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(3*x),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.01357, size = 302, normalized size = 15.89 \begin{align*} \frac{2 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (-\frac{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - 6 \,{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 3 \, \cosh \left (x\right )^{2} + 6 \, \cosh \left (x\right ) \sinh \left (x\right ) + 3 \, \sinh \left (x\right )^{2} - 3}{6 \,{\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(3*x),x, algorithm="fricas")

[Out]

1/6*(2*(cosh(x) + sinh(x))*arctan(-(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)/(cosh(x) - sinh(x))) - 6*(cosh(
x) + sinh(x))*arctan(cosh(x) + sinh(x)) + 3*cosh(x)^2 + 6*cosh(x)*sinh(x) + 3*sinh(x)^2 - 3)/(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 (3 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(3*x),x)

[Out]

Integral(sinh(x)*tanh(3*x), x)

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Giac [B]  time = 1.19821, size = 58, normalized size = 3.05 \begin{align*} -\frac{1}{3} \, \pi - \frac{1}{3} \, \arctan \left ({\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - \frac{1}{3} \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\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(3*x),x, algorithm="giac")

[Out]

-1/3*pi - 1/3*arctan((e^(2*x) - 1)*e^(-x)) - 1/3*arctan(1/2*(e^(2*x) - 1)*e^(-x)) - 1/2*e^(-x) + 1/2*e^x