3.10.37 \(\int e^x \text {sech}(2 x) \tanh (2 x) \, dx\) [937]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {2320, 12, 468, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\text {ArcTan}\left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}+\frac {\text {ArcTan}\left (\sqrt {2} e^x+1\right )}{2 \sqrt {2}}-\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}} \end {gather*}

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 12

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

Rule 210

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 303

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 468

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 631

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 642

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]

Rule 1176

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 1179

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 2320

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]]]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 42, normalized size = 0.37 \begin {gather*} \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 ) \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.39, size = 40, normalized size = 0.35

method	result	size
risch	\(-\frac {{\mathrm e}^{3 x}}{1+{\mathrm e}^{4 x}}+2 \left (\munderset {\textit {\_R} =\RootOf \left (4096 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (512 \textit {\_R}^{3}+{\mathrm e}^{x}\right )\right )\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.49, size = 90, normalized size = 0.80 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (80) = 160\).
time = 0.34, size = 164, normalized size = 1.45 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int e^{x} \tanh {\left (2 x \right )} \operatorname {sech}{\left (2 x \right )}\, dx \end {gather*}

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 = 0.42, size = 90, normalized size = 0.80 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 0.28, size = 91, normalized size = 0.81 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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