\(\int e^x \text {sech}(2 x) \tanh (2 x) \, dx\) [87]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 12, antiderivative size = 88 \[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=-\frac {e^{3 x}}{1+e^{4 x}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} e^x}{1+e^{2 x}}\right )}{2 \sqrt {2}} \] Output:

-exp(3*x)/(1+exp(4*x))+1/4*arctan(-1+2^(1/2)*exp(x))*2^(1/2)+1/4*arctan(1+ 
2^(1/2)*exp(x))*2^(1/2)-1/4*arctanh(2^(1/2)*exp(x)/(1+exp(2*x)))*2^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.48 \[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=\frac {2}{3} e^{3 x} \left (\operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-e^{4 x}\right )-2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},2,\frac {7}{4},-e^{4 x}\right )\right ) \] Input:

Integrate[E^x*Sech[2*x]*Tanh[2*x],x]
 

Output:

(2*E^(3*x)*(Hypergeometric2F1[3/4, 1, 7/4, -E^(4*x)] - 2*Hypergeometric2F1 
[3/4, 2, 7/4, -E^(4*x)]))/3
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.44, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {2720, 27, 957, 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.

\(\displaystyle \int e^x \tanh (2 x) \text {sech}(2 x) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \int -\frac {2 e^{2 x} \left (1-e^{4 x}\right )}{\left (e^{4 x}+1\right )^2}de^x\)

\(\Big \downarrow \) 27

\(\displaystyle -2 \int \frac {e^{2 x} \left (1-e^{4 x}\right )}{\left (1+e^{4 x}\right )^2}de^x\)

\(\Big \downarrow \) 957

\(\displaystyle -2 \left (\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}-\frac {1}{2} \int \frac {e^{2 x}}{1+e^{4 x}}de^x\right )\)

\(\Big \downarrow \) 826

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x-\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

\(\Big \downarrow \) 1476

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^x+e^{2 x}}de^x-\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

\(\Big \downarrow \) 1082

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 x}}d\left (1+\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 x}}d\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

\(\Big \downarrow \) 1479

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {1}{2} \int \frac {1+\sqrt {2} e^x}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

\(\Big \downarrow \) 1103

\(\displaystyle -2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}\right )\right )+\frac {e^{3 x}}{2 \left (e^{4 x}+1\right )}\right )\)

Input:

Int[E^x*Sech[2*x]*Tanh[2*x],x]
 

Output:

-2*(E^(3*x)/(2*(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)/2)
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217
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])
 

rule 826
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]]))
 

rule 957
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)]))
 

rule 1082
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]
 

rule 1103
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]
 

rule 1476
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]
 

rule 1479
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]
 

rule 2720
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]]]
 
Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45

method result size
risch \(-\frac {{\mathrm e}^{3 x}}{{\mathrm e}^{4 x}+1}+2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4096 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (512 \textit {\_R}^{3}+{\mathrm e}^{x}\right )\right )\) \(40\)
default \(\frac {\tanh \left (\frac {x}{2}\right )^{3}-3 \tanh \left (\frac {x}{2}\right )^{2}-\tanh \left (\frac {x}{2}\right )-1}{\tanh \left (\frac {x}{2}\right )^{4}+6 \tanh \left (\frac {x}{2}\right )^{2}+1}+\frac {\sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3-2 \sqrt {2}\right )}{8}-\frac {\left (\sqrt {2}-2\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2 \sqrt {2}-2}\right )}{2 \left (2 \sqrt {2}-2\right )}-\frac {\sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3+2 \sqrt {2}\right )}{8}+\frac {\left (2+\sqrt {2}\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{4+4 \sqrt {2}}\) \(148\)

Input:

int(exp(x)*sech(2*x)*tanh(2*x),x,method=_RETURNVERBOSE)
 

Output:

-exp(x)^3/(exp(x)^4+1)+2*sum(_R*ln(512*_R^3+exp(x)),_R=RootOf(4096*_Z^4+1) 
)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 365 vs. \(2 (63) = 126\).

Time = 0.09 (sec) , antiderivative size = 365, normalized size of antiderivative = 4.15 \[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=-\frac {8 \, \cosh \left (x\right )^{3} + 24 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 24 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 8 \, \sinh \left (x\right )^{3} - 2 \, {\left (\sqrt {2} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} \sinh \left (x\right )^{4} + \sqrt {2}\right )} \arctan \left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right ) + 1\right ) - 2 \, {\left (\sqrt {2} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} \sinh \left (x\right )^{4} + \sqrt {2}\right )} \arctan \left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right ) - 1\right ) + {\left (\sqrt {2} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} \sinh \left (x\right )^{4} + \sqrt {2}\right )} \log \left (\frac {\sqrt {2} + 2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (\sqrt {2} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \sqrt {2} \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \sqrt {2} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} \sinh \left (x\right )^{4} + \sqrt {2}\right )} \log \left (-\frac {\sqrt {2} - 2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\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 )}} \] Input:

integrate(exp(x)*sech(2*x)*tanh(2*x),x, algorithm="fricas")
 

Output:

-1/8*(8*cosh(x)^3 + 24*cosh(x)^2*sinh(x) + 24*cosh(x)*sinh(x)^2 + 8*sinh(x 
)^3 - 2*(sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)^3*sinh(x) + 6*sqrt(2)*cosh( 
x)^2*sinh(x)^2 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + sqrt(2) 
)*arctan(sqrt(2)*cosh(x) + sqrt(2)*sinh(x) + 1) - 2*(sqrt(2)*cosh(x)^4 + 4 
*sqrt(2)*cosh(x)^3*sinh(x) + 6*sqrt(2)*cosh(x)^2*sinh(x)^2 + 4*sqrt(2)*cos 
h(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + sqrt(2))*arctan(sqrt(2)*cosh(x) + sqr 
t(2)*sinh(x) - 1) + (sqrt(2)*cosh(x)^4 + 4*sqrt(2)*cosh(x)^3*sinh(x) + 6*s 
qrt(2)*cosh(x)^2*sinh(x)^2 + 4*sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x) 
^4 + sqrt(2))*log((sqrt(2) + 2*cosh(x))/(cosh(x) - sinh(x))) - (sqrt(2)*co 
sh(x)^4 + 4*sqrt(2)*cosh(x)^3*sinh(x) + 6*sqrt(2)*cosh(x)^2*sinh(x)^2 + 4* 
sqrt(2)*cosh(x)*sinh(x)^3 + sqrt(2)*sinh(x)^4 + sqrt(2))*log(-(sqrt(2) - 2 
*cosh(x))/(cosh(x) - 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)
 

Sympy [F]

\[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=\int e^{x} \tanh {\left (2 x \right )} \operatorname {sech}{\left (2 x \right )}\, dx \] Input:

integrate(exp(x)*sech(2*x)*tanh(2*x),x)
 

Output:

Integral(exp(x)*tanh(2*x)*sech(2*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{\left (3 \, x\right )}}{e^{\left (4 \, x\right )} + 1} \] Input:

integrate(exp(x)*sech(2*x)*tanh(2*x),x, algorithm="maxima")
 

Output:

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^(3*x)/(e^(4*x) + 1)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.02 \[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{\left (3 \, x\right )}}{e^{\left (4 \, x\right )} + 1} \] Input:

integrate(exp(x)*sech(2*x)*tanh(2*x),x, algorithm="giac")
 

Output:

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^(3*x)/(e^(4*x) + 1)
 

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.03 \[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=-\frac {{\mathrm {e}}^{3\,x}}{{\mathrm {e}}^{4\,x}+1}+\sqrt {2}\,\ln \left (1+\sqrt {2}\,{\mathrm {e}}^x\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (1+\sqrt {2}\,{\mathrm {e}}^x\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (1+\sqrt {2}\,{\mathrm {e}}^x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (1+\sqrt {2}\,{\mathrm {e}}^x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \] Input:

int((tanh(2*x)*exp(x))/cosh(2*x),x)
 

Output:

2^(1/2)*log(1 - 2^(1/2)*exp(x)*(1/2 + 1i/2))*(1/8 + 1i/8) + 2^(1/2)*log(1 
- 2^(1/2)*exp(x)*(1/2 - 1i/2))*(1/8 - 1i/8) - 2^(1/2)*log(2^(1/2)*exp(x)*( 
1/2 - 1i/2) + 1)*(1/8 - 1i/8) - 2^(1/2)*log(2^(1/2)*exp(x)*(1/2 + 1i/2) + 
1)*(1/8 + 1i/8) - exp(3*x)/(exp(4*x) + 1)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.14 \[ \int e^x \text {sech}(2 x) \tanh (2 x) \, dx=\frac {2 e^{4 x} \sqrt {2}\, \mathit {atan} \left (\frac {2 e^{x}-\sqrt {2}}{\sqrt {2}}\right )+2 \sqrt {2}\, \mathit {atan} \left (\frac {2 e^{x}-\sqrt {2}}{\sqrt {2}}\right )+2 e^{4 x} \sqrt {2}\, \mathit {atan} \left (\frac {2 e^{x}+\sqrt {2}}{\sqrt {2}}\right )+2 \sqrt {2}\, \mathit {atan} \left (\frac {2 e^{x}+\sqrt {2}}{\sqrt {2}}\right )+e^{4 x} \sqrt {2}\, \mathrm {log}\left (e^{2 x}-e^{x} \sqrt {2}+1\right )-e^{4 x} \sqrt {2}\, \mathrm {log}\left (e^{2 x}+e^{x} \sqrt {2}+1\right )-8 e^{3 x}+\sqrt {2}\, \mathrm {log}\left (e^{2 x}-e^{x} \sqrt {2}+1\right )-\sqrt {2}\, \mathrm {log}\left (e^{2 x}+e^{x} \sqrt {2}+1\right )}{8 e^{4 x}+8} \] Input:

int(exp(x)*sech(2*x)*tanh(2*x),x)
 

Output:

(2*e**(4*x)*sqrt(2)*atan((2*e**x - sqrt(2))/sqrt(2)) + 2*sqrt(2)*atan((2*e 
**x - sqrt(2))/sqrt(2)) + 2*e**(4*x)*sqrt(2)*atan((2*e**x + sqrt(2))/sqrt( 
2)) + 2*sqrt(2)*atan((2*e**x + sqrt(2))/sqrt(2)) + e**(4*x)*sqrt(2)*log(e* 
*(2*x) - e**x*sqrt(2) + 1) - e**(4*x)*sqrt(2)*log(e**(2*x) + e**x*sqrt(2) 
+ 1) - 8*e**(3*x) + sqrt(2)*log(e**(2*x) - e**x*sqrt(2) + 1) - sqrt(2)*log 
(e**(2*x) + e**x*sqrt(2) + 1))/(8*(e**(4*x) + 1))