Integrand size = 7, antiderivative size = 75 \[ \int \text {csch}(5 x) \sinh (x) \, dx=\frac {1}{5} \sqrt {\frac {1}{2} \left (5-\sqrt {5}\right )} \arctan \left (\frac {\tanh (x)}{\sqrt {5-2 \sqrt {5}}}\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\frac {\tanh (x)}{\sqrt {5+2 \sqrt {5}}}\right ) \]
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)
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.12 \[ \int \text {csch}(5 x) \sinh (x) \, dx=\frac {\sqrt {5+\sqrt {5}} \arctan \left (\frac {\left (-3+\sqrt {5}\right ) \tanh (x)}{\sqrt {10-2 \sqrt {5}}}\right )+\sqrt {5-\sqrt {5}} \arctan \left (\frac {\left (3+\sqrt {5}\right ) \tanh (x)}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )}{5 \sqrt {2}} \]
(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])
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4889, 1480, 216}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sinh (x) \text {csch}(5 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (i x)}{\sin (5 i x)}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {1-\tanh ^2(x)}{\tanh ^4(x)+10 \tanh ^2(x)+5}d\tanh (x)\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle -\frac {1}{10} \left (5-3 \sqrt {5}\right ) \int \frac {1}{\tanh ^2(x)-2 \sqrt {5}+5}d\tanh (x)-\frac {1}{10} \left (5+3 \sqrt {5}\right ) \int \frac {1}{\tanh ^2(x)+2 \sqrt {5}+5}d\tanh (x)\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\left (5-3 \sqrt {5}\right ) \arctan \left (\frac {\tanh (x)}{\sqrt {5-2 \sqrt {5}}}\right )}{10 \sqrt {5-2 \sqrt {5}}}-\frac {\left (5+3 \sqrt {5}\right ) \arctan \left (\frac {\tanh (x)}{\sqrt {5+2 \sqrt {5}}}\right )}{10 \sqrt {5+2 \sqrt {5}}}\) |
-1/10*((5 - 3*Sqrt[5])*ArcTan[Tanh[x]/Sqrt[5 - 2*Sqrt[5]]])/Sqrt[5 - 2*Sqr t[5]] - ((5 + 3*Sqrt[5])*ArcTan[Tanh[x]/Sqrt[5 + 2*Sqrt[5]]])/(10*Sqrt[5 + 2*Sqrt[5]])
3.3.19.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(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]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55
method | result | size |
risch | \(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (32000 \textit {\_Z}^{4}+400 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (4000 \textit {\_R}^{3}-200 \textit {\_R}^{2}+{\mathrm e}^{2 x}+30 \textit {\_R} -1\right )\right )\) | \(41\) |
Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (49) = 98\).
Time = 0.26 (sec) , antiderivative size = 243, normalized size of antiderivative = 3.24 \[ \int \text {csch}(5 x) \sinh (x) \, dx=\frac {1}{20} \, \sqrt {2} \sqrt {\sqrt {5} - 5} \log \left (8 \, \cosh \left (x\right )^{2} + 16 \, \cosh \left (x\right ) \sinh \left (x\right ) + 8 \, \sinh \left (x\right )^{2} + {\left (\sqrt {5} \sqrt {2} + \sqrt {2}\right )} \sqrt {\sqrt {5} - 5} - 2 \, \sqrt {5} + 2\right ) - \frac {1}{20} \, \sqrt {2} \sqrt {\sqrt {5} - 5} \log \left (8 \, \cosh \left (x\right )^{2} + 16 \, \cosh \left (x\right ) \sinh \left (x\right ) + 8 \, \sinh \left (x\right )^{2} - {\left (\sqrt {5} \sqrt {2} + \sqrt {2}\right )} \sqrt {\sqrt {5} - 5} - 2 \, \sqrt {5} + 2\right ) - \frac {1}{20} \, \sqrt {2} \sqrt {-\sqrt {5} - 5} \log \left (8 \, \cosh \left (x\right )^{2} + 16 \, \cosh \left (x\right ) \sinh \left (x\right ) + 8 \, \sinh \left (x\right )^{2} + {\left (\sqrt {5} \sqrt {2} - \sqrt {2}\right )} \sqrt {-\sqrt {5} - 5} + 2 \, \sqrt {5} + 2\right ) + \frac {1}{20} \, \sqrt {2} \sqrt {-\sqrt {5} - 5} \log \left (8 \, \cosh \left (x\right )^{2} + 16 \, \cosh \left (x\right ) \sinh \left (x\right ) + 8 \, \sinh \left (x\right )^{2} - {\left (\sqrt {5} \sqrt {2} - \sqrt {2}\right )} \sqrt {-\sqrt {5} - 5} + 2 \, \sqrt {5} + 2\right ) \]
1/20*sqrt(2)*sqrt(sqrt(5) - 5)*log(8*cosh(x)^2 + 16*cosh(x)*sinh(x) + 8*si nh(x)^2 + (sqrt(5)*sqrt(2) + sqrt(2))*sqrt(sqrt(5) - 5) - 2*sqrt(5) + 2) - 1/20*sqrt(2)*sqrt(sqrt(5) - 5)*log(8*cosh(x)^2 + 16*cosh(x)*sinh(x) + 8*s inh(x)^2 - (sqrt(5)*sqrt(2) + sqrt(2))*sqrt(sqrt(5) - 5) - 2*sqrt(5) + 2) - 1/20*sqrt(2)*sqrt(-sqrt(5) - 5)*log(8*cosh(x)^2 + 16*cosh(x)*sinh(x) + 8 *sinh(x)^2 + (sqrt(5)*sqrt(2) - sqrt(2))*sqrt(-sqrt(5) - 5) + 2*sqrt(5) + 2) + 1/20*sqrt(2)*sqrt(-sqrt(5) - 5)*log(8*cosh(x)^2 + 16*cosh(x)*sinh(x) + 8*sinh(x)^2 - (sqrt(5)*sqrt(2) - sqrt(2))*sqrt(-sqrt(5) - 5) + 2*sqrt(5) + 2)
\[ \int \text {csch}(5 x) \sinh (x) \, dx=\int \sinh {\left (x \right )} \operatorname {csch}{\left (5 x \right )}\, dx \]
\[ \int \text {csch}(5 x) \sinh (x) \, dx=\int { \operatorname {csch}\left (5 \, x\right ) \sinh \left (x\right ) \,d x } \]
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*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*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)
Time = 0.28 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.91 \[ \int \text {csch}(5 x) \sinh (x) \, dx=\frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (-\frac {\sqrt {5} - 4 \, e^{\left (2 \, x\right )} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) - \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (\frac {\sqrt {5} + 4 \, e^{\left (2 \, x\right )} + 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) \]
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))
Time = 4.99 (sec) , antiderivative size = 282, normalized size of antiderivative = 3.76 \[ \int \text {csch}(5 x) \sinh (x) \, dx=2\,\mathrm {atan}\left (\frac {\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {9\,\sqrt {5}}{25}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}+\frac {4}{5}}{5\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\frac {9\,\sqrt {5}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}}{5}+\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\frac {9\,\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}}{5}}\right )\,\sqrt {\frac {\sqrt {5}}{200}+\frac {1}{40}}+\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,\left (\ln \left (\frac {9\,\sqrt {5}}{25}-\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}-\frac {4}{5}-{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,5{}\mathrm {i}+\frac {\sqrt {5}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}-\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,1{}\mathrm {i}+\frac {\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}\right )\,1{}\mathrm {i}-\ln \left (\frac {9\,\sqrt {5}}{25}-\frac {{\mathrm {e}}^{2\,x}}{5}+\frac {6\,\sqrt {5}\,{\mathrm {e}}^{2\,x}}{25}-\frac {4}{5}+{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,5{}\mathrm {i}-\frac {\sqrt {5}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}+\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,1{}\mathrm {i}-\frac {\sqrt {5}\,{\mathrm {e}}^{2\,x}\,\sqrt {\frac {1}{40}-\frac {\sqrt {5}}{200}}\,9{}\mathrm {i}}{5}\right )\,1{}\mathrm {i}\right ) \]
2*atan((exp(2*x)/5 + (9*5^(1/2))/25 + (6*5^(1/2)*exp(2*x))/25 + 4/5)/(5*ex p(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 + 1/40) ^(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)