3.215 \(\int e^x \tanh (2 x) \, dx\)

Optimal. Leaf size=95 \[ e^x+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{\sqrt {2}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {2282, 388, 211, 1165, 628, 1162, 617, 204} \[ e^x+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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
]]))

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 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]

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 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 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]]]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 95, normalized size = 1.00 \[ e^x+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 113, normalized size = 1.19 \[ \sqrt {2} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + \sqrt {2} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - \frac {1}{4} \, \sqrt {2} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*arctan(-sqrt(2)*e^x + sqrt(2)*sqrt(sqrt(2)*e^x + e^(2*x) + 1) - 1) + sqrt(2)*arctan(-sqrt(2)*e^x + 1/2
*sqrt(2)*sqrt(-4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 1) - 1/4*sqrt(2)*log(4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 1/4*sqrt
(2)*log(-4*sqrt(2)*e^x + 4*e^(2*x) + 4) + e^x

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giac [A]  time = 0.11, size = 78, normalized size = 0.82 \[ -\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/4*
sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) + e^x

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maple [C]  time = 0.18, size = 24, normalized size = 0.25 \[ {\mathrm e}^{x}+\left (\munderset {\textit {\_R} =\RootOf \left (16 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-2 \textit {\_R} \right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)+sum(_R*ln(exp(x)-2*_R),_R=RootOf(16*_Z^4+1))

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maxima [A]  time = 0.46, size = 78, normalized size = 0.82 \[ -\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/4*
sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/4*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) + e^x

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mupad [B]  time = 1.26, size = 81, normalized size = 0.85 \[ {\mathrm {e}}^x-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}{2}\right )}{2}+\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}^2+2\right )}{4}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}{2}\right )}{2}-\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}^2+2\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x) - (2^(1/2)*atan((2^(1/2)*(2*exp(x) - 2^(1/2)))/2))/2 + (2^(1/2)*log((2*exp(x) - 2^(1/2))^2 + 2))/4 - (2
^(1/2)*atan((2^(1/2)*(2*exp(x) + 2^(1/2)))/2))/2 - (2^(1/2)*log((2*exp(x) + 2^(1/2))^2 + 2))/4

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \tanh {\left (2 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*tanh(2*x), x)

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