### 3.937 $$\int e^x \text{sech}(2 x) \tanh (2 x) \, dx$$

Optimal. Leaf size=113 $-\frac{e^{3 x}}{e^{4 x}+1}+\frac{\log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{4 \sqrt{2}}-\frac{\log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} e^x+1\right )}{2 \sqrt{2}}$

[Out]

-(E^(3*x)/(1 + E^(4*x))) - ArcTan[1 - Sqrt[2]*E^x]/(2*Sqrt[2]) + ArcTan[1 + Sqrt[2]*E^x]/(2*Sqrt[2]) + Log[1 -
Sqrt[2]*E^x + E^(2*x)]/(4*Sqrt[2]) - Log[1 + Sqrt[2]*E^x + E^(2*x)]/(4*Sqrt[2])

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Rubi [A]  time = 0.0820164, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.75, Rules used = {2282, 12, 457, 297, 1162, 617, 204, 1165, 628} $-\frac{e^{3 x}}{e^{4 x}+1}+\frac{\log \left (-\sqrt{2} e^x+e^{2 x}+1\right )}{4 \sqrt{2}}-\frac{\log \left (\sqrt{2} e^x+e^{2 x}+1\right )}{4 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} e^x+1\right )}{2 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(E^(3*x)/(1 + E^(4*x))) - ArcTan[1 - Sqrt[2]*E^x]/(2*Sqrt[2]) + ArcTan[1 + Sqrt[2]*E^x]/(2*Sqrt[2]) + Log[1 -
Sqrt[2]*E^x + E^(2*x)]/(4*Sqrt[2]) - Log[1 + Sqrt[2]*E^x + E^(2*x)]/(4*Sqrt[2])

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 457

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] - Dist[(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] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]

Rubi steps

\begin{align*} \int e^x \text{sech}(2 x) \tanh (2 x) \, dx &=\operatorname{Subst}\left (\int \frac{2 x^2 \left (-1+x^4\right )}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x^2 \left (-1+x^4\right )}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=-\frac{e^{3 x}}{1+e^{4 x}}+\operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,e^x\right )\\ &=-\frac{e^{3 x}}{1+e^{4 x}}-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,e^x\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=-\frac{e^{3 x}}{1+e^{4 x}}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,e^x\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,e^x\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,e^x\right )}{4 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,e^x\right )}{4 \sqrt{2}}\\ &=-\frac{e^{3 x}}{1+e^{4 x}}+\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}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} e^x\right )}{2 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} e^x\right )}{2 \sqrt{2}}\\ &=-\frac{e^{3 x}}{1+e^{4 x}}-\frac{\tan ^{-1}\left (1-\sqrt{2} e^x\right )}{2 \sqrt{2}}+\frac{\tan ^{-1}\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}}\\ \end{align*}

Mathematica [C]  time = 0.0275269, size = 42, normalized size = 0.37 $\frac{2}{3} e^{3 x} \left (\, _2F_1\left (\frac{3}{4},1;\frac{7}{4};-e^{4 x}\right )-2 \, _2F_1\left (\frac{3}{4},2;\frac{7}{4};-e^{4 x}\right )\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [C]  time = 0.052, size = 40, normalized size = 0.4 \begin{align*} -{\frac{{{\rm e}^{3\,x}}}{1+{{\rm e}^{4\,x}}}}+2\,\sum _{{\it \_R}={\it RootOf} \left ( 4096\,{{\it \_Z}}^{4}+1 \right ) }{\it \_R}\,\ln \left ( 512\,{{\it \_R}}^{3}+{{\rm e}^{x}} \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.49453, size = 122, normalized size = 1.08 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*s
qrt(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)

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Fricas [B]  time = 1.99851, size = 500, normalized size = 4.42 \begin{align*} -\frac{4 \,{\left (\sqrt{2} e^{\left (4 \, x\right )} + \sqrt{2}\right )} \arctan \left (-\sqrt{2} e^{x} + \sqrt{2} \sqrt{\sqrt{2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 4 \,{\left (\sqrt{2} e^{\left (4 \, x\right )} + \sqrt{2}\right )} \arctan \left (-\sqrt{2} e^{x} + \frac{1}{2} \, \sqrt{2} \sqrt{-4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) +{\left (\sqrt{2} e^{\left (4 \, x\right )} + \sqrt{2}\right )} \log \left (4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) -{\left (\sqrt{2} e^{\left (4 \, x\right )} + \sqrt{2}\right )} \log \left (-4 \, \sqrt{2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 8 \, e^{\left (3 \, x\right )}}{8 \,{\left (e^{\left (4 \, x\right )} + 1\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*(sqrt(2)*e^(4*x) + sqrt(2))*arctan(-sqrt(2)*e^x + sqrt(2)*sqrt(sqrt(2)*e^x + e^(2*x) + 1) - 1) + 4*(sq
rt(2)*e^(4*x) + sqrt(2))*arctan(-sqrt(2)*e^x + 1/2*sqrt(2)*sqrt(-4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 1) + (sqrt(2
)*e^(4*x) + sqrt(2))*log(4*sqrt(2)*e^x + 4*e^(2*x) + 4) - (sqrt(2)*e^(4*x) + sqrt(2))*log(-4*sqrt(2)*e^x + 4*e
^(2*x) + 4) + 8*e^(3*x))/(e^(4*x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \tanh{\left (2 x \right )} \operatorname{sech}{\left (2 x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16349, size = 122, normalized size = 1.08 \begin{align*} \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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*s
qrt(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)