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]

1/10*arctan(tanh(x)/(5-2*5^(1/2))^(1/2))*(10-2*5^(1/2))^(1/2)-1/10*arctan(tanh(x)/(5+2*5^(1/2))^(1/2))*(10+2*5
^(1/2))^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [B]  time = 0.48, size = 171, normalized size = 2.28 \[ -\frac {1}{5} \, \sqrt {2} \sqrt {-\sqrt {5} + 5} \arctan \left (\frac {1}{40} \, \sqrt {5} \sqrt {2} \sqrt {-32 \, {\left (\sqrt {5} - 1\right )} e^{\left (2 \, x\right )} + 64 \, e^{\left (4 \, x\right )} + 64} \sqrt {-\sqrt {5} + 5} - \frac {1}{20} \, {\left (4 \, \sqrt {5} \sqrt {2} e^{\left (2 \, x\right )} + \sqrt {5} \sqrt {2} - 5 \, \sqrt {2}\right )} \sqrt {-\sqrt {5} + 5}\right ) + \frac {1}{5} \, \sqrt {2} \sqrt {\sqrt {5} + 5} \arctan \left (-\frac {1}{20} \, {\left (4 \, \sqrt {5} \sqrt {2} e^{\left (2 \, x\right )} + \sqrt {5} \sqrt {2} + 5 \, \sqrt {2}\right )} \sqrt {\sqrt {5} + 5} + \frac {1}{5} \, \sqrt {5} \sqrt {{\left (\sqrt {5} + 1\right )} e^{\left (2 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 2} \sqrt {\sqrt {5} + 5}\right ) \]

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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giac [A]  time = 0.15, size = 68, normalized size = 0.91 \[ \frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (-\frac {\sqrt {5} - 4 \, e^{\left (2 \, x\right )} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) - \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {\sqrt {5} + 4 \, e^{\left (2 \, x\right )} + 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) \]

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))

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maple [C]  time = 0.25, size = 41, normalized size = 0.55 \[ 2 \left (\munderset {\textit {\_R} =\RootOf \left (32000 \textit {\_Z}^{4}+400 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (4000 \textit {\_R}^{3}-200 \textit {\_R}^{2}+{\mathrm e}^{2 x}+30 \textit {\_R} -1\right )\right ) \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{10} \, \left (-1\right )^{\frac {3}{5}} \log \left (\left (-1\right )^{\frac {1}{5}} + e^{\left (-2 \, x\right )}\right ) + \frac {\sqrt {5} \left (-1\right )^{\frac {3}{5}} \log \left (\frac {\sqrt {5} \left (-1\right )^{\frac {1}{5}} + \left (-1\right )^{\frac {1}{5}} \sqrt {2 \, \sqrt {5} - 10} + \left (-1\right )^{\frac {1}{5}} - 4 \, e^{\left (-2 \, x\right )}}{\sqrt {5} \left (-1\right )^{\frac {1}{5}} - \left (-1\right )^{\frac {1}{5}} \sqrt {2 \, \sqrt {5} - 10} + \left (-1\right )^{\frac {1}{5}} - 4 \, e^{\left (-2 \, x\right )}}\right )}{10 \, \sqrt {2 \, \sqrt {5} - 10}} - \frac {\sqrt {5} \left (-1\right )^{\frac {3}{5}} \log \left (\frac {\sqrt {5} \left (-1\right )^{\frac {1}{5}} - \left (-1\right )^{\frac {1}{5}} \sqrt {-2 \, \sqrt {5} - 10} - \left (-1\right )^{\frac {1}{5}} + 4 \, e^{\left (-2 \, x\right )}}{\sqrt {5} \left (-1\right )^{\frac {1}{5}} + \left (-1\right )^{\frac {1}{5}} \sqrt {-2 \, \sqrt {5} - 10} - \left (-1\right )^{\frac {1}{5}} + 4 \, e^{\left (-2 \, x\right )}}\right )}{10 \, \sqrt {-2 \, \sqrt {5} - 10}} - \frac {\log \left (-{\left (\sqrt {5} \left (-1\right )^{\frac {1}{5}} + \left (-1\right )^{\frac {1}{5}}\right )} e^{\left (-2 \, x\right )} + 2 \, \left (-1\right )^{\frac {2}{5}} + 2 \, e^{\left (-4 \, x\right )}\right )}{10 \, {\left (\sqrt {5} \left (-1\right )^{\frac {2}{5}} + \left (-1\right )^{\frac {2}{5}}\right )}} + \frac {\log \left ({\left (\sqrt {5} \left (-1\right )^{\frac {1}{5}} - \left (-1\right )^{\frac {1}{5}}\right )} e^{\left (-2 \, x\right )} + 2 \, \left (-1\right )^{\frac {2}{5}} + 2 \, e^{\left (-4 \, x\right )}\right )}{10 \, {\left (\sqrt {5} \left (-1\right )^{\frac {2}{5}} - \left (-1\right )^{\frac {2}{5}}\right )}} - \frac {1}{10} \, \int \frac {{\left (e^{\left (3 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 3 \, e^{x} + 4\right )} e^{x}}{e^{\left (4 \, x\right )} + e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} + e^{x} + 1}\,{d x} - \frac {1}{10} \, \int \frac {{\left (e^{\left (3 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 3 \, e^{x} - 4\right )} e^{x}}{e^{\left (4 \, x\right )} - e^{\left (3 \, x\right )} + e^{\left (2 \, x\right )} - e^{x} + 1}\,{d x} + \frac {1}{10} \, \log \left (e^{x} + 1\right ) + \frac {1}{10} \, \log \left (e^{x} - 1\right ) \]

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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mupad [B]  time = 4.26, size = 282, normalized size = 3.76 \[ 2\,\mathrm {atan}\left (\frac {\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {9\,\sqrt {5}}{25}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}+\frac {4}{5}}{5\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\frac {9\,\sqrt {5}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}}{5}+\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\frac {9\,\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}}{5}}\right )\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,\left (\ln \left (\frac {9\,\sqrt {5}}{25}-\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}-\frac {4}{5}-{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,5{}\mathrm {i}+\frac {\sqrt {5}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}-\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,1{}\mathrm {i}+\frac {\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}\right )\,1{}\mathrm {i}-\ln \left (\frac {9\,\sqrt {5}}{25}-\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}-\frac {4}{5}+{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,5{}\mathrm {i}-\frac {\sqrt {5}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}+\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,1{}\mathrm {i}-\frac {\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}\right )\,1{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*atan((exp(2*x)/5 + (9*5^(1/2))/25 + (6*5^(1/2)*exp(2*x))/25 + 4/5)/(5*exp(2*x)*(5^(1/2)/200 + 1/40)^(1/2) +
(9*5^(1/2)*(5^(1/2)/200 + 1/40)^(1/2))/5 + (5^(1/2)/200 + 1/40)^(1/2) + (9*5^(1/2)*exp(2*x)*(5^(1/2)/200 + 1/4
0)^(1/2))/5))*(5^(1/2)/200 + 1/40)^(1/2) + (1/40 - 5^(1/2)/200)^(1/2)*(log((5^(1/2)*(1/40 - 5^(1/2)/200)^(1/2)
*9i)/5 - exp(2*x)*(1/40 - 5^(1/2)/200)^(1/2)*5i - exp(2*x)/5 + (9*5^(1/2))/25 - (1/40 - 5^(1/2)/200)^(1/2)*1i
+ (6*5^(1/2)*exp(2*x))/25 + (5^(1/2)*exp(2*x)*(1/40 - 5^(1/2)/200)^(1/2)*9i)/5 - 4/5)*1i - log(exp(2*x)*(1/40
- 5^(1/2)/200)^(1/2)*5i - exp(2*x)/5 - (5^(1/2)*(1/40 - 5^(1/2)/200)^(1/2)*9i)/5 + (9*5^(1/2))/25 + (1/40 - 5^
(1/2)/200)^(1/2)*1i + (6*5^(1/2)*exp(2*x))/25 - (5^(1/2)*exp(2*x)*(1/40 - 5^(1/2)/200)^(1/2)*9i)/5 - 4/5)*1i)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\relax (x )} \operatorname {csch}{\left (5 x \right )}\, dx \]

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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