Integrand size = 7, antiderivative size = 73 \[ \int \cosh (x) \text {sech}(5 x) \, dx=-\sqrt {\frac {2}{5 \left (5+\sqrt {5}\right )}} \arctan \left (\sqrt {5-2 \sqrt {5}} \tanh (x)\right )+\frac {1}{5} \sqrt {\frac {1}{2} \left (5+\sqrt {5}\right )} \arctan \left (\sqrt {5+2 \sqrt {5}} \tanh (x)\right ) \] Output:
-2^(1/2)/(25+5*5^(1/2))^(1/2)*arctan((5-2*5^(1/2))^(1/2)*tanh(x))+1/10*(10 +2*5^(1/2))^(1/2)*arctan((5+2*5^(1/2))^(1/2)*tanh(x))
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.15 \[ \int \cosh (x) \text {sech}(5 x) \, dx=\frac {\sqrt {5+\sqrt {5}} \arctan \left (\frac {\left (5+\sqrt {5}\right ) \tanh (x)}{\sqrt {10-2 \sqrt {5}}}\right )+\sqrt {5-\sqrt {5}} \arctan \left (\frac {\left (-5+\sqrt {5}\right ) \tanh (x)}{\sqrt {2 \left (5+\sqrt {5}\right )}}\right )}{5 \sqrt {2}} \] Input:
Integrate[Cosh[x]*Sech[5*x],x]
Output:
(Sqrt[5 + Sqrt[5]]*ArcTan[((5 + Sqrt[5])*Tanh[x])/Sqrt[10 - 2*Sqrt[5]]] + Sqrt[5 - Sqrt[5]]*ArcTan[((-5 + Sqrt[5])*Tanh[x])/Sqrt[2*(5 + Sqrt[5])]])/ (5*Sqrt[2])
Time = 0.32 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.22, 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.
\(\displaystyle \int \cosh (x) \text {sech}(5 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (i x)}{\cos (5 i x)}dx\) |
\(\Big \downarrow \) 4889 |
\(\displaystyle \int \frac {1-\tanh ^2(x)}{5 \tanh ^4(x)+10 \tanh ^2(x)+1}d\tanh (x)\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle -\frac {1}{2} \left (1-\sqrt {5}\right ) \int \frac {1}{5 \tanh ^2(x)-2 \sqrt {5}+5}d\tanh (x)-\frac {1}{2} \left (1+\sqrt {5}\right ) \int \frac {1}{5 \tanh ^2(x)+2 \sqrt {5}+5}d\tanh (x)\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\left (1+\sqrt {5}\right ) \arctan \left (\sqrt {5-2 \sqrt {5}} \tanh (x)\right )}{2 \sqrt {5 \left (5+2 \sqrt {5}\right )}}-\frac {\left (1-\sqrt {5}\right ) \arctan \left (\sqrt {5+2 \sqrt {5}} \tanh (x)\right )}{2 \sqrt {5 \left (5-2 \sqrt {5}\right )}}\) |
Input:
Int[Cosh[x]*Sech[5*x],x]
Output:
-1/2*((1 + Sqrt[5])*ArcTan[Sqrt[5 - 2*Sqrt[5]]*Tanh[x]])/Sqrt[5*(5 + 2*Sqr t[5])] - ((1 - Sqrt[5])*ArcTan[Sqrt[5 + 2*Sqrt[5]]*Tanh[x]])/(2*Sqrt[5*(5 - 2*Sqrt[5])])
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 = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.56
method | result | size |
risch | \(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (32000 \textit {\_Z}^{4}+400 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-4000 \textit {\_R}^{3}+200 \textit {\_R}^{2}+{\mathrm e}^{2 x}-30 \textit {\_R} +1\right )\right )\) | \(41\) |
Input:
int(cosh(x)*sech(5*x),x,method=_RETURNVERBOSE)
Output:
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))
Leaf count of result is larger than twice the leaf count of optimal. 115 vs. \(2 (52) = 104\).
Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.58 \[ \int \cosh (x) \text {sech}(5 x) \, dx=\frac {1}{5} \, \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} \arctan \left (\frac {1}{10} \, {\left (4 \, \sqrt {5} \cosh \left (x\right )^{2} + 8 \, \sqrt {5} \cosh \left (x\right ) \sinh \left (x\right ) + 4 \, \sqrt {5} \sinh \left (x\right )^{2} - \sqrt {5} - 5\right )} \sqrt {\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}\right ) - \frac {1}{5} \, \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}} \arctan \left (\frac {1}{10} \, {\left (4 \, \sqrt {5} \cosh \left (x\right )^{2} + 8 \, \sqrt {5} \cosh \left (x\right ) \sinh \left (x\right ) + 4 \, \sqrt {5} \sinh \left (x\right )^{2} - \sqrt {5} + 5\right )} \sqrt {-\frac {1}{2} \, \sqrt {5} + \frac {5}{2}}\right ) \] Input:
integrate(cosh(x)*sech(5*x),x, algorithm="fricas")
Output:
1/5*sqrt(1/2*sqrt(5) + 5/2)*arctan(1/10*(4*sqrt(5)*cosh(x)^2 + 8*sqrt(5)*c osh(x)*sinh(x) + 4*sqrt(5)*sinh(x)^2 - sqrt(5) - 5)*sqrt(1/2*sqrt(5) + 5/2 )) - 1/5*sqrt(-1/2*sqrt(5) + 5/2)*arctan(1/10*(4*sqrt(5)*cosh(x)^2 + 8*sqr t(5)*cosh(x)*sinh(x) + 4*sqrt(5)*sinh(x)^2 - sqrt(5) + 5)*sqrt(-1/2*sqrt(5 ) + 5/2))
\[ \int \cosh (x) \text {sech}(5 x) \, dx=\int \cosh {\left (x \right )} \operatorname {sech}{\left (5 x \right )}\, dx \] Input:
integrate(cosh(x)*sech(5*x),x)
Output:
Integral(cosh(x)*sech(5*x), x)
\[ \int \cosh (x) \text {sech}(5 x) \, dx=\int { \cosh \left (x\right ) \operatorname {sech}\left (5 \, x\right ) \,d x } \] Input:
integrate(cosh(x)*sech(5*x),x, algorithm="maxima")
Output:
1/5*sqrt(5)*arctan((sqrt(5) + 4*e^(-2*x) - 1)/sqrt(2*sqrt(5) + 10))/sqrt(2 *sqrt(5) + 10) - 1/5*sqrt(5)*arctan(-(sqrt(5) - 4*e^(-2*x) + 1)/sqrt(-2*sq rt(5) + 10))/sqrt(-2*sqrt(5) + 10) - 1/10*log(-(sqrt(5) + 1)*e^(-2*x) + 2* e^(-4*x) + 2)/(sqrt(5) + 1) + 1/10*log((sqrt(5) - 1)*e^(-2*x) + 2*e^(-4*x) + 2)/(sqrt(5) - 1) - 1/5*integrate((e^(7*x) - 2*e^(5*x) - 2*e^(3*x) + e^x )*e^x/(e^(8*x) - e^(6*x) + e^(4*x) - e^(2*x) + 1), x) + 1/10*log(e^(2*x) + 1) - 1/10*log(e^(-2*x) + 1)
Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.93 \[ \int \cosh (x) \text {sech}(5 x) \, dx=-\frac {1}{10} \, \sqrt {-2 \, \sqrt {5} + 10} \arctan \left (\frac {\sqrt {5} + 4 \, e^{\left (2 \, x\right )} - 1}{\sqrt {2 \, \sqrt {5} + 10}}\right ) + \frac {1}{10} \, \sqrt {2 \, \sqrt {5} + 10} \arctan \left (-\frac {\sqrt {5} - 4 \, e^{\left (2 \, x\right )} + 1}{\sqrt {-2 \, \sqrt {5} + 10}}\right ) \] Input:
integrate(cosh(x)*sech(5*x),x, algorithm="giac")
Output:
-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)/sqr t(-2*sqrt(5) + 10))
Time = 3.55 (sec) , antiderivative size = 297, normalized size of antiderivative = 4.07 \[ \int \cosh (x) \text {sech}(5 x) \, dx=\ln \left (1-\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (48\,{\mathrm {e}}^{2\,x}+\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-72\right )-8\right )-{\mathrm {e}}^{2\,x}\right )\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}-\ln \left (\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-48\,{\mathrm {e}}^{2\,x}+72\right )-8\right )-{\mathrm {e}}^{2\,x}+1\right )\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}-\ln \left (\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-48\,{\mathrm {e}}^{2\,x}+72\right )-8\right )-{\mathrm {e}}^{2\,x}+1\right )\,\sqrt {-\frac {\sqrt {5}}{200}-\frac {1}{40}}+\ln \left (1-\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (4\,{\mathrm {e}}^{2\,x}+\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (48\,{\mathrm {e}}^{2\,x}+\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}}\,\left (360\,{\mathrm {e}}^{2\,x}-360\right )-72\right )-8\right )-{\mathrm {e}}^{2\,x}\right )\,\sqrt {\frac {\sqrt {5}}{200}-\frac {1}{40}} \] Input:
int(cosh(x)/cosh(5*x),x)
Output:
log(1 - (- 5^(1/2)/200 - 1/40)^(1/2)*(4*exp(2*x) + (- 5^(1/2)/200 - 1/40)^ (1/2)*(48*exp(2*x) + (- 5^(1/2)/200 - 1/40)^(1/2)*(360*exp(2*x) - 360) - 7 2) - 8) - exp(2*x))*(- 5^(1/2)/200 - 1/40)^(1/2) - log((5^(1/2)/200 - 1/40 )^(1/2)*(4*exp(2*x) + (5^(1/2)/200 - 1/40)^(1/2)*((5^(1/2)/200 - 1/40)^(1/ 2)*(360*exp(2*x) - 360) - 48*exp(2*x) + 72) - 8) - exp(2*x) + 1)*(5^(1/2)/ 200 - 1/40)^(1/2) - log((- 5^(1/2)/200 - 1/40)^(1/2)*(4*exp(2*x) + (- 5^(1 /2)/200 - 1/40)^(1/2)*((- 5^(1/2)/200 - 1/40)^(1/2)*(360*exp(2*x) - 360) - 48*exp(2*x) + 72) - 8) - exp(2*x) + 1)*(- 5^(1/2)/200 - 1/40)^(1/2) + log (1 - (5^(1/2)/200 - 1/40)^(1/2)*(4*exp(2*x) + (5^(1/2)/200 - 1/40)^(1/2)*( 48*exp(2*x) + (5^(1/2)/200 - 1/40)^(1/2)*(360*exp(2*x) - 360) - 72) - 8) - exp(2*x))*(5^(1/2)/200 - 1/40)^(1/2)
\[ \int \cosh (x) \text {sech}(5 x) \, dx=\int \cosh \left (x \right ) \mathrm {sech}\left (5 x \right )d x \] Input:
int(cosh(x)*sech(5*x),x)
Output:
int(cosh(x)*sech(5*x),x)