Integrand size = 10, antiderivative size = 111 \[ \int e^x \text {sech}^2(2 x) \, dx=-\frac {e^x}{1+e^{4 x}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{4 \sqrt {2}} \]
-exp(x)/(1+exp(4*x))+1/4*arctan(-1+exp(x)*2^(1/2))*2^(1/2)+1/4*arctan(1+ex p(x)*2^(1/2))*2^(1/2)-1/8*ln(1+exp(2*x)-exp(x)*2^(1/2))*2^(1/2)+1/8*ln(1+e xp(2*x)+exp(x)*2^(1/2))*2^(1/2)
Time = 0.07 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95 \[ \int e^x \text {sech}^2(2 x) \, dx=\frac {1}{8} \left (-\frac {8 e^x}{1+e^{4 x}}-2 \sqrt {2} \arctan \left (1-\sqrt {2} e^x\right )+2 \sqrt {2} \arctan \left (1+\sqrt {2} e^x\right )-\sqrt {2} \log \left (1-\sqrt {2} e^x+e^{2 x}\right )+\sqrt {2} \log \left (1+\sqrt {2} e^x+e^{2 x}\right )\right ) \]
((-8*E^x)/(1 + E^(4*x)) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*E^x] + 2*Sqrt[2]*Ar cTan[1 + Sqrt[2]*E^x] - Sqrt[2]*Log[1 - Sqrt[2]*E^x + E^(2*x)] + Sqrt[2]*L og[1 + Sqrt[2]*E^x + E^(2*x)])/8
Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, Rules used = {2720, 27, 817, 755, 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.
| \(\displaystyle \int e^x \text {sech}^2(2 x) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \int \frac {4 e^{4 x}}{\left (e^{4 x}+1\right )^2}de^x\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \int \frac {e^{4 x}}{\left (1+e^{4 x}\right )^2}de^x\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle 4 \left (\frac {1}{4} \int \frac {1}{1+e^{4 x}}de^x-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^x+e^{2 x}}de^x+\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 x}}d\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 x}}d\left (1+\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {1}{2} \int \frac {1+\sqrt {2} e^x}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle 4 \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )\) |
4*(-1/4*E^x/(1 + E^(4*x)) + ((-(ArcTan[1 - Sqrt[2]*E^x]/Sqrt[2]) + ArcTan[ 1 + Sqrt[2]*E^x]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*E^x + E^(2*x)]/Sqrt[2] + Log[1 + Sqrt[2]*E^x + E^(2*x)]/(2*Sqrt[2]))/2)/4)
3.3.76.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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 = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.32
| method | result | size |
| risch | \(-\frac {{\mathrm e}^{x}}{1+{\mathrm e}^{4 x}}+4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (65536 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+16 \textit {\_R} \right )\right )\) | \(36\) |
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 349, normalized size of antiderivative = 3.14 \[ \int e^x \text {sech}^2(2 x) \, dx=\frac {{\left (\left (i + 1\right ) \, \sqrt {2} \cosh \left (x\right )^{4} + \left (4 i + 4\right ) \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) + \left (6 i + 6\right ) \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + \left (4 i + 4\right ) \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \left (i + 1\right ) \, \sqrt {2} \sinh \left (x\right )^{4} + \left (i + 1\right ) \, \sqrt {2}\right )} \log \left (\left (i + 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + {\left (-\left (i - 1\right ) \, \sqrt {2} \cosh \left (x\right )^{4} - \left (4 i - 4\right ) \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) - \left (6 i - 6\right ) \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} - \left (4 i - 4\right ) \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} - \left (i - 1\right ) \, \sqrt {2} \sinh \left (x\right )^{4} - \left (i - 1\right ) \, \sqrt {2}\right )} \log \left (-\left (i - 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + {\left (\left (i - 1\right ) \, \sqrt {2} \cosh \left (x\right )^{4} + \left (4 i - 4\right ) \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) + \left (6 i - 6\right ) \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + \left (4 i - 4\right ) \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \left (i - 1\right ) \, \sqrt {2} \sinh \left (x\right )^{4} + \left (i - 1\right ) \, \sqrt {2}\right )} \log \left (\left (i - 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) + {\left (-\left (i + 1\right ) \, \sqrt {2} \cosh \left (x\right )^{4} - \left (4 i + 4\right ) \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) - \left (6 i + 6\right ) \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} - \left (4 i + 4\right ) \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} - \left (i + 1\right ) \, \sqrt {2} \sinh \left (x\right )^{4} - \left (i + 1\right ) \, \sqrt {2}\right )} \log \left (-\left (i + 1\right ) \, \sqrt {2} + 2 \, \cosh \left (x\right ) + 2 \, \sinh \left (x\right )\right ) - 8 \, \cosh \left (x\right ) - 8 \, \sinh \left (x\right )}{8 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 1\right )}} \]
1/8*(((I + 1)*sqrt(2)*cosh(x)^4 + (4*I + 4)*sqrt(2)*cosh(x)^3*sinh(x) + (6 *I + 6)*sqrt(2)*cosh(x)^2*sinh(x)^2 + (4*I + 4)*sqrt(2)*cosh(x)*sinh(x)^3 + (I + 1)*sqrt(2)*sinh(x)^4 + (I + 1)*sqrt(2))*log((I + 1)*sqrt(2) + 2*cos h(x) + 2*sinh(x)) + (-(I - 1)*sqrt(2)*cosh(x)^4 - (4*I - 4)*sqrt(2)*cosh(x )^3*sinh(x) - (6*I - 6)*sqrt(2)*cosh(x)^2*sinh(x)^2 - (4*I - 4)*sqrt(2)*co sh(x)*sinh(x)^3 - (I - 1)*sqrt(2)*sinh(x)^4 - (I - 1)*sqrt(2))*log(-(I - 1 )*sqrt(2) + 2*cosh(x) + 2*sinh(x)) + ((I - 1)*sqrt(2)*cosh(x)^4 + (4*I - 4 )*sqrt(2)*cosh(x)^3*sinh(x) + (6*I - 6)*sqrt(2)*cosh(x)^2*sinh(x)^2 + (4*I - 4)*sqrt(2)*cosh(x)*sinh(x)^3 + (I - 1)*sqrt(2)*sinh(x)^4 + (I - 1)*sqrt (2))*log((I - 1)*sqrt(2) + 2*cosh(x) + 2*sinh(x)) + (-(I + 1)*sqrt(2)*cosh (x)^4 - (4*I + 4)*sqrt(2)*cosh(x)^3*sinh(x) - (6*I + 6)*sqrt(2)*cosh(x)^2* sinh(x)^2 - (4*I + 4)*sqrt(2)*cosh(x)*sinh(x)^3 - (I + 1)*sqrt(2)*sinh(x)^ 4 - (I + 1)*sqrt(2))*log(-(I + 1)*sqrt(2) + 2*cosh(x) + 2*sinh(x)) - 8*cos h(x) - 8*sinh(x))/(cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 1)
\[ \int e^x \text {sech}^2(2 x) \, dx=\int e^{x} \operatorname {sech}^{2}{\left (2 x \right )}\, dx \]
Time = 0.27 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.79 \[ \int e^x \text {sech}^2(2 x) \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{e^{\left (4 \, x\right )} + 1} \]
1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/4*sqrt(2)*arctan(-1/ 2*sqrt(2)*(sqrt(2) - 2*e^x)) + 1/8*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) - 1/8*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - e^x/(e^(4*x) + 1)
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.79 \[ \int e^x \text {sech}^2(2 x) \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{e^{\left (4 \, x\right )} + 1} \]
1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/4*sqrt(2)*arctan(-1/ 2*sqrt(2)*(sqrt(2) - 2*e^x)) + 1/8*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) - 1/8*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - e^x/(e^(4*x) + 1)
Time = 1.80 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.77 \[ \int e^x \text {sech}^2(2 x) \, dx=\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\left ({\mathrm {e}}^x-\frac {\sqrt {2}}{2}\right )\right )}{4}-\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{4\,x}+1}+\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\left ({\mathrm {e}}^x+\frac {\sqrt {2}}{2}\right )\right )}{4}-\frac {\sqrt {2}\,\ln \left ({\left ({\mathrm {e}}^x-\frac {\sqrt {2}}{2}\right )}^2+\frac {1}{2}\right )}{8}+\frac {\sqrt {2}\,\ln \left ({\left ({\mathrm {e}}^x+\frac {\sqrt {2}}{2}\right )}^2+\frac {1}{2}\right )}{8} \]