3.219 \(\int \text{csch}(5 x) \sinh (x) \, dx\)

Optimal. Leaf size=75 \[ \frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{5-2 \sqrt{5}}}\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{5+2 \sqrt{5}}}\right ) \]

[Out]

(Sqrt[(5 - Sqrt[5])/2]*ArcTan[Tanh[x]/Sqrt[5 - 2*Sqrt[5]]])/5 - (Sqrt[(5 + Sqrt[5])/2]*ArcTan[Tanh[x]/Sqrt[5 +
2*Sqrt[5]]])/5

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Rubi [A]  time = 0.108097, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {1166, 203} \[ \frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{5-2 \sqrt{5}}}\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{5+2 \sqrt{5}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Csch[5*x]*Sinh[x],x]

[Out]

(Sqrt[(5 - Sqrt[5])/2]*ArcTan[Tanh[x]/Sqrt[5 - 2*Sqrt[5]]])/5 - (Sqrt[(5 + Sqrt[5])/2]*ArcTan[Tanh[x]/Sqrt[5 +
2*Sqrt[5]]])/5

Rule 1166

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] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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 \text{csch}(5 x) \sinh (x) \, dx &=\operatorname{Subst}\left (\int \frac{1-x^2}{5+10 x^2+x^4} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{10} \left (-5+3 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{5-2 \sqrt{5}+x^2} \, dx,x,\tanh (x)\right )-\frac{1}{10} \left (5+3 \sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{5+2 \sqrt{5}+x^2} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{5} \sqrt{\frac{1}{2} \left (5-\sqrt{5}\right )} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{5-2 \sqrt{5}}}\right )-\frac{1}{5} \sqrt{\frac{1}{2} \left (5+\sqrt{5}\right )} \tan ^{-1}\left (\frac{\tanh (x)}{\sqrt{5+2 \sqrt{5}}}\right )\\ \end{align*}

Mathematica [A]  time = 0.103277, size = 84, normalized size = 1.12 \[ \frac{\sqrt{5+\sqrt{5}} \tan ^{-1}\left (\frac{\left (\sqrt{5}-3\right ) \tanh (x)}{\sqrt{10-2 \sqrt{5}}}\right )+\sqrt{5-\sqrt{5}} \tan ^{-1}\left (\frac{\left (3+\sqrt{5}\right ) \tanh (x)}{\sqrt{2 \left (5+\sqrt{5}\right )}}\right )}{5 \sqrt{2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csch[5*x]*Sinh[x],x]

[Out]

(Sqrt[5 + Sqrt[5]]*ArcTan[((-3 + Sqrt[5])*Tanh[x])/Sqrt[10 - 2*Sqrt[5]]] + Sqrt[5 - Sqrt[5]]*ArcTan[((3 + Sqrt
[5])*Tanh[x])/Sqrt[2*(5 + Sqrt[5])]])/(5*Sqrt[2])

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Maple [C]  time = 0.038, size = 41, normalized size = 0.6 \begin{align*} 2\,\sum _{{\it \_R}={\it RootOf} \left ( 32000\,{{\it \_Z}}^{4}+400\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( 4000\,{{\it \_R}}^{3}-200\,{{\it \_R}}^{2}+{{\rm e}^{2\,x}}+30\,{\it \_R}-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(5*x)*sinh(x),x)

[Out]

2*sum(_R*ln(4000*_R^3-200*_R^2+exp(2*x)+30*_R-1),_R=RootOf(32000*_Z^4+400*_Z^2+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(5*x)*sinh(x),x, algorithm="maxima")

[Out]

1/10*(-1)^(3/5)*log((-1)^(1/5) + e^(-2*x)) + 1/10*sqrt(5)*(-1)^(3/5)*log((sqrt(5)*(-1)^(1/5) + (-1)^(1/5)*sqrt
(2*sqrt(5) - 10) + (-1)^(1/5) - 4*e^(-2*x))/(sqrt(5)*(-1)^(1/5) - (-1)^(1/5)*sqrt(2*sqrt(5) - 10) + (-1)^(1/5)
- 4*e^(-2*x)))/sqrt(2*sqrt(5) - 10) - 1/10*sqrt(5)*(-1)^(3/5)*log((sqrt(5)*(-1)^(1/5) - (-1)^(1/5)*sqrt(-2*sq
rt(5) - 10) - (-1)^(1/5) + 4*e^(-2*x))/(sqrt(5)*(-1)^(1/5) + (-1)^(1/5)*sqrt(-2*sqrt(5) - 10) - (-1)^(1/5) + 4
*e^(-2*x)))/sqrt(-2*sqrt(5) - 10) - 1/10*log(-(sqrt(5)*(-1)^(1/5) + (-1)^(1/5))*e^(-2*x) + 2*(-1)^(2/5) + 2*e^
(-4*x))/(sqrt(5)*(-1)^(2/5) + (-1)^(2/5)) + 1/10*log((sqrt(5)*(-1)^(1/5) - (-1)^(1/5))*e^(-2*x) + 2*(-1)^(2/5)
+ 2*e^(-4*x))/(sqrt(5)*(-1)^(2/5) - (-1)^(2/5)) - 1/10*integrate((e^(3*x) + 2*e^(2*x) + 3*e^x + 4)*e^x/(e^(4*
x) + e^(3*x) + e^(2*x) + e^x + 1), x) - 1/10*integrate((e^(3*x) - 2*e^(2*x) + 3*e^x - 4)*e^x/(e^(4*x) - e^(3*x
) + e^(2*x) - e^x + 1), x) + 1/10*log(e^x + 1) + 1/10*log(e^x - 1)

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Fricas [B]  time = 2.35452, size = 558, normalized size = 7.44 \begin{align*} -\frac{1}{5} \, \sqrt{2} \sqrt{-\sqrt{5} + 5} \arctan \left (\frac{1}{40} \, \sqrt{5} \sqrt{2} \sqrt{-32 \,{\left (\sqrt{5} - 1\right )} e^{\left (2 \, x\right )} + 64 \, e^{\left (4 \, x\right )} + 64} \sqrt{-\sqrt{5} + 5} - \frac{1}{20} \,{\left (4 \, \sqrt{5} \sqrt{2} e^{\left (2 \, x\right )} + \sqrt{5} \sqrt{2} - 5 \, \sqrt{2}\right )} \sqrt{-\sqrt{5} + 5}\right ) + \frac{1}{5} \, \sqrt{2} \sqrt{\sqrt{5} + 5} \arctan \left (-\frac{1}{20} \,{\left (4 \, \sqrt{5} \sqrt{2} e^{\left (2 \, x\right )} + \sqrt{5} \sqrt{2} + 5 \, \sqrt{2}\right )} \sqrt{\sqrt{5} + 5} + \frac{1}{5} \, \sqrt{5} \sqrt{{\left (\sqrt{5} + 1\right )} e^{\left (2 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 2} \sqrt{\sqrt{5} + 5}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(5*x)*sinh(x),x, algorithm="fricas")

[Out]

-1/5*sqrt(2)*sqrt(-sqrt(5) + 5)*arctan(1/40*sqrt(5)*sqrt(2)*sqrt(-32*(sqrt(5) - 1)*e^(2*x) + 64*e^(4*x) + 64)*
sqrt(-sqrt(5) + 5) - 1/20*(4*sqrt(5)*sqrt(2)*e^(2*x) + sqrt(5)*sqrt(2) - 5*sqrt(2))*sqrt(-sqrt(5) + 5)) + 1/5*
sqrt(2)*sqrt(sqrt(5) + 5)*arctan(-1/20*(4*sqrt(5)*sqrt(2)*e^(2*x) + sqrt(5)*sqrt(2) + 5*sqrt(2))*sqrt(sqrt(5)
+ 5) + 1/5*sqrt(5)*sqrt((sqrt(5) + 1)*e^(2*x) + 2*e^(4*x) + 2)*sqrt(sqrt(5) + 5))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (x \right )} \operatorname{csch}{\left (5 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(5*x)*sinh(x),x)

[Out]

Integral(sinh(x)*csch(5*x), x)

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Giac [A]  time = 1.18576, size = 92, normalized size = 1.23 \begin{align*} \frac{1}{10} \, \sqrt{-2 \, \sqrt{5} + 10} \arctan \left (-\frac{\sqrt{5} - 4 \, e^{\left (2 \, x\right )} - 1}{\sqrt{2 \, \sqrt{5} + 10}}\right ) - \frac{1}{10} \, \sqrt{2 \, \sqrt{5} + 10} \arctan \left (\frac{\sqrt{5} + 4 \, e^{\left (2 \, x\right )} + 1}{\sqrt{-2 \, \sqrt{5} + 10}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(5*x)*sinh(x),x, algorithm="giac")

[Out]

1/10*sqrt(-2*sqrt(5) + 10)*arctan(-(sqrt(5) - 4*e^(2*x) - 1)/sqrt(2*sqrt(5) + 10)) - 1/10*sqrt(2*sqrt(5) + 10)
*arctan((sqrt(5) + 4*e^(2*x) + 1)/sqrt(-2*sqrt(5) + 10))