Integrand size = 14, antiderivative size = 129 \[ \int e^x \text {sech}^2(2 x) \tanh (2 x) \, dx=-\frac {e^{5 x}}{\left (1+e^{4 x}\right )^2}-\frac {e^x}{4 \left (1+e^{4 x}\right )}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}} \]
-exp(5*x)/(1+exp(4*x))^2-1/4*exp(x)/(1+exp(4*x))+1/16*arctan(-1+exp(x)*2^( 1/2))*2^(1/2)+1/16*arctan(1+exp(x)*2^(1/2))*2^(1/2)-1/32*ln(1+exp(2*x)-exp (x)*2^(1/2))*2^(1/2)+1/32*ln(1+exp(2*x)+exp(x)*2^(1/2))*2^(1/2)
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.93 \[ \int e^x \text {sech}^2(2 x) \tanh (2 x) \, dx=\frac {1}{32} \left (\frac {32 e^x}{\left (1+e^{4 x}\right )^2}-\frac {40 e^x}{1+e^{4 x}}-2 \sqrt {2} \arctan \left (1-\sqrt {2} e^x\right )+2 \sqrt {2} \arctan \left (1+\sqrt {2} e^x\right )-\sqrt {2} \log \left (1-\sqrt {2} e^x+e^{2 x}\right )+\sqrt {2} \log \left (1+\sqrt {2} e^x+e^{2 x}\right )\right ) \]
((32*E^x)/(1 + E^(4*x))^2 - (40*E^x)/(1 + E^(4*x)) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*E^x] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*E^x] - Sqrt[2]*Log[1 - Sqrt[2] *E^x + E^(2*x)] + Sqrt[2]*Log[1 + Sqrt[2]*E^x + E^(2*x)])/32
Time = 0.36 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.15, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {2720, 27, 957, 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.
| \(\displaystyle \int e^x \tanh (2 x) \text {sech}^2(2 x) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \int -\frac {4 e^{4 x} \left (1-e^{4 x}\right )}{\left (e^{4 x}+1\right )^3}de^x\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -4 \int \frac {e^{4 x} \left (1-e^{4 x}\right )}{\left (1+e^{4 x}\right )^3}de^x\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle -4 \left (\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}-\frac {1}{4} \int \frac {e^{4 x}}{\left (1+e^{4 x}\right )^2}de^x\right )\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {e^x}{4 \left (e^{4 x}+1\right )}-\frac {1}{4} \int \frac {1}{1+e^{4 x}}de^x\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x-\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^x+e^{2 x}}de^x-\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 x}}d\left (1+\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 x}}d\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {1}{2} \int \frac {1+\sqrt {2} e^x}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle -4 \left (\frac {1}{4} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}\right )\right )+\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )+\frac {e^{5 x}}{4 \left (e^{4 x}+1\right )^2}\right )\) |
-4*(E^(5*x)/(4*(1 + E^(4*x))^2) + (E^x/(4*(1 + E^(4*x))) + ((ArcTan[1 - Sq rt[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)/4)/4)
3.10.38.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
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 = 5.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.34
| method | result | size |
| risch | \(-\frac {{\mathrm e}^{x} \left (5 \,{\mathrm e}^{4 x}+1\right )}{4 \left (1+{\mathrm e}^{4 x}\right )^{2}}+4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (16777216 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+64 \textit {\_R} \right )\right )\) | \(44\) |
| default | \(\frac {\frac {\tanh \left (\frac {x}{2}\right )^{7}}{4}-\frac {17 \tanh \left (\frac {x}{2}\right )^{6}}{4}-\frac {11 \tanh \left (\frac {x}{2}\right )^{5}}{4}-\frac {57 \tanh \left (\frac {x}{2}\right )^{4}}{4}+\frac {11 \tanh \left (\frac {x}{2}\right )^{3}}{4}-\frac {19 \tanh \left (\frac {x}{2}\right )^{2}}{4}-\frac {\tanh \left (\frac {x}{2}\right )}{4}-\frac {3}{4}}{\left (\tanh \left (\frac {x}{2}\right )^{4}+6 \tanh \left (\frac {x}{2}\right )^{2}+1\right )^{2}}+\frac {\sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3+2 \sqrt {2}\right )}{32}+\frac {\left (2+\sqrt {2}\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{16+16 \sqrt {2}}-\frac {\sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3-2 \sqrt {2}\right )}{32}-\frac {\left (-2+\sqrt {2}\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2 \sqrt {2}-2}\right )}{8 \left (2 \sqrt {2}-2\right )}\) | \(180\) |
-1/4*exp(x)*(5*exp(4*x)+1)/(1+exp(4*x))^2+4*sum(_R*ln(exp(x)+64*_R),_R=Roo tOf(16777216*_Z^4+1))
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 882, normalized size of antiderivative = 6.84 \[ \int e^x \text {sech}^2(2 x) \tanh (2 x) \, dx=\text {Too large to display} \]
-1/32*(40*cosh(x)^5 + 400*cosh(x)^3*sinh(x)^2 + 400*cosh(x)^2*sinh(x)^3 + 200*cosh(x)*sinh(x)^4 + 40*sinh(x)^5 - ((I + 1)*sqrt(2)*cosh(x)^8 + (56*I + 56)*sqrt(2)*cosh(x)^3*sinh(x)^5 + (28*I + 28)*sqrt(2)*cosh(x)^2*sinh(x)^ 6 + (8*I + 8)*sqrt(2)*cosh(x)*sinh(x)^7 + (I + 1)*sqrt(2)*sinh(x)^8 - 2*(- (35*I + 35)*sqrt(2)*cosh(x)^4 - (I + 1)*sqrt(2))*sinh(x)^4 + (2*I + 2)*sqr t(2)*cosh(x)^4 - 8*(-(7*I + 7)*sqrt(2)*cosh(x)^5 - (I + 1)*sqrt(2)*cosh(x) )*sinh(x)^3 - 4*(-(7*I + 7)*sqrt(2)*cosh(x)^6 - (3*I + 3)*sqrt(2)*cosh(x)^ 2)*sinh(x)^2 - 8*(-(I + 1)*sqrt(2)*cosh(x)^7 - (I + 1)*sqrt(2)*cosh(x)^3)* sinh(x) + (I + 1)*sqrt(2))*log((I + 1)*sqrt(2) + 2*cosh(x) + 2*sinh(x)) - (-(I - 1)*sqrt(2)*cosh(x)^8 - (56*I - 56)*sqrt(2)*cosh(x)^3*sinh(x)^5 - (2 8*I - 28)*sqrt(2)*cosh(x)^2*sinh(x)^6 - (8*I - 8)*sqrt(2)*cosh(x)*sinh(x)^ 7 - (I - 1)*sqrt(2)*sinh(x)^8 - 2*((35*I - 35)*sqrt(2)*cosh(x)^4 + (I - 1) *sqrt(2))*sinh(x)^4 - (2*I - 2)*sqrt(2)*cosh(x)^4 - 8*((7*I - 7)*sqrt(2)*c osh(x)^5 + (I - 1)*sqrt(2)*cosh(x))*sinh(x)^3 - 4*((7*I - 7)*sqrt(2)*cosh( x)^6 + (3*I - 3)*sqrt(2)*cosh(x)^2)*sinh(x)^2 - 8*((I - 1)*sqrt(2)*cosh(x) ^7 + (I - 1)*sqrt(2)*cosh(x)^3)*sinh(x) - (I - 1)*sqrt(2))*log(-(I - 1)*sq rt(2) + 2*cosh(x) + 2*sinh(x)) - ((I - 1)*sqrt(2)*cosh(x)^8 + (56*I - 56)* sqrt(2)*cosh(x)^3*sinh(x)^5 + (28*I - 28)*sqrt(2)*cosh(x)^2*sinh(x)^6 + (8 *I - 8)*sqrt(2)*cosh(x)*sinh(x)^7 + (I - 1)*sqrt(2)*sinh(x)^8 - 2*(-(35*I - 35)*sqrt(2)*cosh(x)^4 - (I - 1)*sqrt(2))*sinh(x)^4 + (2*I - 2)*sqrt(2...
\[ \int e^x \text {sech}^2(2 x) \tanh (2 x) \, dx=\int e^{x} \tanh {\left (2 x \right )} \operatorname {sech}^{2}{\left (2 x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.78 \[ \int e^x \text {sech}^2(2 x) \tanh (2 x) \, dx=\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {5 \, e^{\left (5 \, x\right )} + e^{x}}{4 \, {\left (e^{\left (8 \, x\right )} + 2 \, e^{\left (4 \, x\right )} + 1\right )}} \]
1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/16*sqrt(2)*arctan(- 1/2*sqrt(2)*(sqrt(2) - 2*e^x)) + 1/32*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) - 1/32*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/4*(5*e^(5*x) + e^x)/ (e^(8*x) + 2*e^(4*x) + 1)
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74 \[ \int e^x \text {sech}^2(2 x) \tanh (2 x) \, dx=\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {5 \, e^{\left (5 \, x\right )} + e^{x}}{4 \, {\left (e^{\left (4 \, x\right )} + 1\right )}^{2}} \]
1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 1/16*sqrt(2)*arctan(- 1/2*sqrt(2)*(sqrt(2) - 2*e^x)) + 1/32*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) - 1/32*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/4*(5*e^(5*x) + e^x)/ (e^(4*x) + 1)^2
Time = 2.65 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95 \[ \int e^x \text {sech}^2(2 x) \tanh (2 x) \, dx=-\frac {\frac {{\mathrm {e}}^{5\,x}}{2}-\frac {{\mathrm {e}}^x}{2}}{2\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{8\,x}+1}-\frac {3\,{\mathrm {e}}^x}{4\,\left ({\mathrm {e}}^{4\,x}+1\right )}+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (\frac {1}{32}+\frac {1}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (\frac {1}{32}-\frac {1}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{32}+\frac {1}{32}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {{\mathrm {e}}^x}{4}+\sqrt {2}\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{32}-\frac {1}{32}{}\mathrm {i}\right ) \]
2^(1/2)*log(- exp(x)/4 - 2^(1/2)*(1/8 + 1i/8))*(1/32 + 1i/32) - (3*exp(x)) /(4*(exp(4*x) + 1)) - (exp(5*x)/2 - exp(x)/2)/(2*exp(4*x) + exp(8*x) + 1) + 2^(1/2)*log(- exp(x)/4 - 2^(1/2)*(1/8 - 1i/8))*(1/32 - 1i/32) - 2^(1/2)* log(2^(1/2)*(1/8 - 1i/8) - exp(x)/4)*(1/32 - 1i/32) - 2^(1/2)*log(2^(1/2)* (1/8 + 1i/8) - exp(x)/4)*(1/32 + 1i/32)