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

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

[Out]

exp(x)+exp(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 = 0.09, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {2282, 390, 288, 211, 1165, 628, 1162, 617, 204} \[ e^x+\frac {e^x}{e^{4 x}+1}+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{4 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^x + E^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 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 288

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] - Dist[(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]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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(2 x) \, dx &=\operatorname {Subst}\left (\int \frac {\left (1-x^4\right )^2}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (1-\frac {4 x^4}{\left (1+x^4\right )^2}\right ) \, dx,x,e^x\right )\\ &=e^x-4 \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{1+e^{4 x}}-\operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,e^x\right )\\ &=e^x+\frac {e^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 )\\ &=e^x+\frac {e^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}}\\ &=e^x+\frac {e^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}}\\ &=e^x+\frac {e^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]  time = 0.05, size = 48, normalized size = 0.42 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^4+1\& ,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}^3}\& \right ]+e^x+\frac {e^x}{e^{4 x}+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^x + E^x/(1 + E^(4*x)) + RootSum[1 + #1^4 & , (x - Log[E^x - #1])/#1^3 & ]/4

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fricas [B]  time = 1.14, size = 168, normalized size = 1.49 \[ \frac {4 \, {\left (\sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 4 \, {\left (\sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - {\left (\sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + {\left (\sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 8 \, e^{\left (5 \, x\right )} + 16 \, e^{x}}{8 \, {\left (e^{\left (4 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(2*x)^2,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*(sqr
t(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^(5*x) + 16*e^x)/(e^(4*x) + 1)

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giac [A]  time = 0.13, size = 89, normalized size = 0.79 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {e^{x}}{e^{\left (4 \, x\right )} + 1} + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(2*x)^2,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*
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^x/(e^(4*x) + 1) + e^x

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maple [C]  time = 0.21, size = 35, normalized size = 0.31 \[ {\mathrm e}^{x}+\frac {{\mathrm e}^{x}}{1+{\mathrm e}^{4 x}}+\left (\munderset {\textit {\_R} =\RootOf \left (256 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-4 \textit {\_R} \right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)+exp(x)/(1+exp(4*x))+sum(_R*ln(exp(x)-4*_R),_R=RootOf(256*_Z^4+1))

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maxima [A]  time = 0.41, size = 89, normalized size = 0.79 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {e^{x}}{e^{\left (4 \, x\right )} + 1} + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(2*x)^2,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*
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^x/(e^(4*x) + 1) + e^x

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mupad [B]  time = 0.25, size = 86, normalized size = 0.76 \[ {\mathrm {e}}^x+\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{4\,x}+1}-\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\left ({\mathrm {e}}^x-\frac {\sqrt {2}}{2}\right )\right )}{4}-\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\left ({\mathrm {e}}^x+\frac {\sqrt {2}}{2}\right )\right )}{4}+\frac {\sqrt {2}\,\ln \left ({\left ({\mathrm {e}}^x-\frac {\sqrt {2}}{2}\right )}^2+\frac {1}{2}\right )}{8}-\frac {\sqrt {2}\,\ln \left ({\left ({\mathrm {e}}^x+\frac {\sqrt {2}}{2}\right )}^2+\frac {1}{2}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*tanh(2*x)**2, x)

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