Integrand size = 8, antiderivative size = 65 \[ \int e^x \text {sech}(2 x) \, dx=-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} e^x}{1+e^{2 x}}\right )}{\sqrt {2}} \] Output:
1/2*arctan(-1+2^(1/2)*exp(x))*2^(1/2)+1/2*arctan(1+2^(1/2)*exp(x))*2^(1/2) -1/2*arctanh(2^(1/2)*exp(x)/(1+exp(2*x)))*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.37 \[ \int e^x \text {sech}(2 x) \, dx=\frac {2}{3} e^{3 x} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-e^{4 x}\right ) \] Input:
Integrate[E^x*Sech[2*x],x]
Output:
(2*E^(3*x)*Hypergeometric2F1[3/4, 1, 7/4, -E^(4*x)])/3
Time = 0.32 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.60, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {2720, 27, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
\(\displaystyle \int e^x \text {sech}(2 x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int \frac {2 e^{2 x}}{e^{4 x}+1}de^x\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{2 x}}{1+e^{4 x}}de^x\) |
\(\Big \downarrow \) 826 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^x+e^{2 x}}de^x+\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 x}}d\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 x}}d\left (1+\sqrt {2} e^x\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {1}{2} \int \frac {1+\sqrt {2} e^x}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}\right )\right )\) |
Input:
Int[E^x*Sech[2*x],x]
Output:
2*((-(ArcTan[1 - Sqrt[2]*E^x]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*E^x]/Sqrt[2])/ 2 + (Log[1 - Sqrt[2]*E^x + E^(2*x)]/(2*Sqrt[2]) - Log[1 + Sqrt[2]*E^x + E^ (2*x)]/(2*Sqrt[2]))/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.64 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.38
method | result | size |
risch | \(2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (64 \textit {\_R}^{3}+{\mathrm e}^{x}\right )\right )\) | \(25\) |
Input:
int(exp(x)*sech(2*x),x,method=_RETURNVERBOSE)
Output:
2*sum(_R*ln(64*_R^3+exp(x)),_R=RootOf(256*_Z^4+1))
Time = 0.08 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38 \[ \int e^x \text {sech}(2 x) \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right ) + 1\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right ) - 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\frac {\sqrt {2} + 2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - 2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) \] Input:
integrate(exp(x)*sech(2*x),x, algorithm="fricas")
Output:
1/2*sqrt(2)*arctan(sqrt(2)*cosh(x) + sqrt(2)*sinh(x) + 1) + 1/2*sqrt(2)*ar ctan(sqrt(2)*cosh(x) + sqrt(2)*sinh(x) - 1) - 1/4*sqrt(2)*log((sqrt(2) + 2 *cosh(x))/(cosh(x) - sinh(x))) + 1/4*sqrt(2)*log(-(sqrt(2) - 2*cosh(x))/(c osh(x) - sinh(x)))
\[ \int e^x \text {sech}(2 x) \, dx=\int e^{x} \operatorname {sech}{\left (2 x \right )}\, dx \] Input:
integrate(exp(x)*sech(2*x),x)
Output:
Integral(exp(x)*sech(2*x), x)
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int e^x \text {sech}(2 x) \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) \] Input:
integrate(exp(x)*sech(2*x),x, algorithm="maxima")
Output:
1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/2*sqrt(2)*arctan(-1/ 2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/4*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1)
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int e^x \text {sech}(2 x) \, dx=\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) \] Input:
integrate(exp(x)*sech(2*x),x, algorithm="giac")
Output:
1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/2*sqrt(2)*arctan(-1/ 2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/4*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1)
Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.18 \[ \int e^x \text {sech}(2 x) \, dx=\sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (-2-2{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (-2+2{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (2-2{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (4+\sqrt {2}\,{\mathrm {e}}^x\,\left (2+2{}\mathrm {i}\right )\right )\,\left (-\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \] Input:
int(exp(x)/cosh(2*x),x)
Output:
2^(1/2)*log(4 - 2^(1/2)*exp(x)*(2 + 2i))*(1/4 + 1i/4) + 2^(1/2)*log(4 - 2^ (1/2)*exp(x)*(2 - 2i))*(1/4 - 1i/4) - 2^(1/2)*log(2^(1/2)*exp(x)*(2 - 2i) + 4)*(1/4 - 1i/4) - 2^(1/2)*log(2^(1/2)*exp(x)*(2 + 2i) + 4)*(1/4 + 1i/4)
Time = 0.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int e^x \text {sech}(2 x) \, dx=\frac {\sqrt {2}\, \left (2 \mathit {atan} \left (\frac {2 e^{x}-\sqrt {2}}{\sqrt {2}}\right )+2 \mathit {atan} \left (\frac {2 e^{x}+\sqrt {2}}{\sqrt {2}}\right )+\mathrm {log}\left (e^{2 x}-e^{x} \sqrt {2}+1\right )-\mathrm {log}\left (e^{2 x}+e^{x} \sqrt {2}+1\right )\right )}{4} \] Input:
int(exp(x)*sech(2*x),x)
Output:
(sqrt(2)*(2*atan((2*e**x - sqrt(2))/sqrt(2)) + 2*atan((2*e**x + sqrt(2))/s qrt(2)) + log(e**(2*x) - e**x*sqrt(2) + 1) - log(e**(2*x) + e**x*sqrt(2) + 1)))/4