3.223 \(\int e^x \tanh (4 x) \, dx\)

Optimal. Leaf size=366 \[ e^x+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}} \]

[Out]

E^x + ArcTan[(Sqrt[2 - Sqrt[2]] - 2*E^x)/Sqrt[2 + Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) + ArcTan[(Sqrt[2 + Sqrt[
2]] - 2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 + Sqrt[2])]) - ArcTan[(Sqrt[2 - Sqrt[2]] + 2*E^x)/Sqrt[2 + Sqrt[2
]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(Sqrt[2 + Sqrt[2]] + 2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 + Sqrt[2])]
) + (Sqrt[2 - Sqrt[2]]*Log[1 - Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)])/8 - (Sqrt[2 - Sqrt[2]]*Log[1 + Sqrt[2 - Sqrt[
2]]*E^x + E^(2*x)])/8 + (Sqrt[2 + Sqrt[2]]*Log[1 - Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)])/8 - (Sqrt[2 + Sqrt[2]]*Lo
g[1 + Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)])/8

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Rubi [A]  time = 0.244167, antiderivative size = 366, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 8, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {2282, 388, 213, 1169, 634, 618, 204, 628} \[ e^x+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (-\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (\sqrt{2-\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (-\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (\sqrt{2+\sqrt{2}} e^x+e^{2 x}+1\right )+\frac{\tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\tan ^{-1}\left (\frac{2 e^x+\sqrt{2+\sqrt{2}}}{\sqrt{2-\sqrt{2}}}\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Tanh[4*x],x]

[Out]

E^x + ArcTan[(Sqrt[2 - Sqrt[2]] - 2*E^x)/Sqrt[2 + Sqrt[2]]]/(2*Sqrt[2*(2 - Sqrt[2])]) + ArcTan[(Sqrt[2 + Sqrt[
2]] - 2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 + Sqrt[2])]) - ArcTan[(Sqrt[2 - Sqrt[2]] + 2*E^x)/Sqrt[2 + Sqrt[2
]]]/(2*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(Sqrt[2 + Sqrt[2]] + 2*E^x)/Sqrt[2 - Sqrt[2]]]/(2*Sqrt[2*(2 + Sqrt[2])]
) + (Sqrt[2 - Sqrt[2]]*Log[1 - Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)])/8 - (Sqrt[2 - Sqrt[2]]*Log[1 + Sqrt[2 - Sqrt[
2]]*E^x + E^(2*x)])/8 + (Sqrt[2 + Sqrt[2]]*Log[1 - Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)])/8 - (Sqrt[2 + Sqrt[2]]*Lo
g[1 + Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)])/8

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 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 213

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 4]], s = Denominator[Rt[a/b, 4]]},
 Dist[r/(2*Sqrt[2]*a), Int[(Sqrt[2]*r - s*x^(n/4))/(r^2 - Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x] + Dist[r/
(2*Sqrt[2]*a), Int[(Sqrt[2]*r + s*x^(n/4))/(r^2 + Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x]] /; FreeQ[{a, b},
 x] && IGtQ[n/4, 1] && GtQ[a/b, 0]

Rule 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
 b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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 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 \tanh (4 x) \, dx &=\operatorname{Subst}\left (\int \frac{-1+x^8}{1+x^8} \, dx,x,e^x\right )\\ &=e^x-2 \operatorname{Subst}\left (\int \frac{1}{1+x^8} \, dx,x,e^x\right )\\ &=e^x-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-x^2}{1-\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{\sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+x^2}{1+\sqrt{2} x^2+x^4} \, dx,x,e^x\right )}{\sqrt{2}}\\ &=e^x-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2-\sqrt{2}\right )}-\left (-1+\sqrt{2}\right ) x}{1-\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2-\sqrt{2}\right )}+\left (-1+\sqrt{2}\right ) x}{1+\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2-\sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2+\sqrt{2}\right )}-\left (1+\sqrt{2}\right ) x}{1-\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2 \left (2+\sqrt{2}\right )}+\left (1+\sqrt{2}\right ) x}{1+\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )}{2 \sqrt{2 \left (2+\sqrt{2}\right )}}\\ &=e^x-\frac{1}{4} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{4} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )+\frac{1}{8} \sqrt{2-\sqrt{2}} \operatorname{Subst}\left (\int \frac{-\sqrt{2-\sqrt{2}}+2 x}{1-\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \operatorname{Subst}\left (\int \frac{\sqrt{2-\sqrt{2}}+2 x}{1+\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \operatorname{Subst}\left (\int \frac{-\sqrt{2+\sqrt{2}}+2 x}{1-\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \operatorname{Subst}\left (\int \frac{\sqrt{2+\sqrt{2}}+2 x}{1+\sqrt{2+\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{4} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )-\frac{1}{4} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2-\sqrt{2}} x+x^2} \, dx,x,e^x\right )\\ &=e^x+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2+\sqrt{2}-x^2} \, dx,x,-\sqrt{2+\sqrt{2}}+2 e^x\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3-2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2+\sqrt{2}-x^2} \, dx,x,\sqrt{2+\sqrt{2}}+2 e^x\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2-\sqrt{2}-x^2} \, dx,x,-\sqrt{2-\sqrt{2}}+2 e^x\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (3+2 \sqrt{2}\right )} \operatorname{Subst}\left (\int \frac{1}{-2-\sqrt{2}-x^2} \, dx,x,\sqrt{2-\sqrt{2}}+2 e^x\right )\\ &=e^x+\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}-2 e^x}{\sqrt{2+\sqrt{2}}}\right )+\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}-2 e^x}{\sqrt{2-\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2-\sqrt{2}}+2 e^x}{\sqrt{2+\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{\sqrt{2+\sqrt{2}}+2 e^x}{\sqrt{2-\sqrt{2}}}\right )+\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1-\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2-\sqrt{2}} \log \left (1+\sqrt{2-\sqrt{2}} e^x+e^{2 x}\right )+\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1-\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )-\frac{1}{8} \sqrt{2+\sqrt{2}} \log \left (1+\sqrt{2+\sqrt{2}} e^x+e^{2 x}\right )\\ \end{align*}

Mathematica [C]  time = 0.0117819, size = 24, normalized size = 0.07 \[ e^x-2 e^x \, _2F_1\left (\frac{1}{8},1;\frac{9}{8};-e^{8 x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Tanh[4*x],x]

[Out]

E^x - 2*E^x*Hypergeometric2F1[1/8, 1, 9/8, -E^(8*x)]

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Maple [C]  time = 0.059, size = 24, normalized size = 0.1 \begin{align*}{{\rm e}^{x}}+\sum _{{\it \_R}={\it RootOf} \left ( 65536\,{{\it \_Z}}^{8}+1 \right ) }{\it \_R}\,\ln \left ({{\rm e}^{x}}-4\,{\it \_R} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*tanh(4*x),x)

[Out]

exp(x)+sum(_R*ln(exp(x)-4*_R),_R=RootOf(65536*_Z^8+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{x} - 2 \, \int \frac{e^{x}}{e^{\left (8 \, x\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x),x, algorithm="maxima")

[Out]

e^x - 2*integrate(e^x/(e^(8*x) + 1), x)

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Fricas [B]  time = 2.39636, size = 3330, normalized size = 9.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x),x, algorithm="fricas")

[Out]

1/8*(sqrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*arctan(-(2*sqrt(2)*e^x - sqrt(2)*sqrt(2*sqrt(2)*s
qrt(sqrt(2) + 2)*e^x - 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) +
 2))/(sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))) + 1/8*(sqrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*
arctan(-(2*sqrt(2)*e^x - sqrt(2)*sqrt(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*
e^(2*x) + 4) - sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))/(sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))) - 1/8*(sqrt(2
)*sqrt(sqrt(2) + 2) - sqrt(2)*sqrt(-sqrt(2) + 2))*arctan((2*sqrt(2)*e^x - sqrt(2)*sqrt(2*sqrt(2)*sqrt(sqrt(2)
+ 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))/(sqrt(s
qrt(2) + 2) - sqrt(-sqrt(2) + 2))) - 1/8*(sqrt(2)*sqrt(sqrt(2) + 2) - sqrt(2)*sqrt(-sqrt(2) + 2))*arctan((2*sq
rt(2)*e^x - sqrt(2)*sqrt(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x - 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4)
- sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))/(sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))) - 1/32*(sqrt(2)*sqrt(sqrt(
2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*log(2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x +
 4*e^(2*x) + 4) - 1/32*(sqrt(2)*sqrt(sqrt(2) + 2) - sqrt(2)*sqrt(-sqrt(2) + 2))*log(2*sqrt(2)*sqrt(sqrt(2) + 2
)*e^x - 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + 1/32*(sqrt(2)*sqrt(sqrt(2) + 2) - sqrt(2)*sqrt(-sq
rt(2) + 2))*log(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x + 2*sqrt(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + 1/32*(s
qrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*log(-2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x - 2*sqrt(2)*sqrt(-
sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) + 1/4*sqrt(-sqrt(2) + 2)*arctan((2*sqrt(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1)
 - sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) + 1/4*sqrt(-sqrt(2) + 2)*arctan((2*sqrt(-sqrt(sqrt(2) + 2)*e
^x + e^(2*x) + 1) + sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) + 1/4*sqrt(sqrt(2) + 2)*arctan((2*sqrt(sqrt
(-sqrt(2) + 2)*e^x + e^(2*x) + 1) - sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) + 1/4*sqrt(sqrt(2) + 2)*arc
tan((2*sqrt(-sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) - 1/16*sqr
t(sqrt(2) + 2)*log(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/16*sqrt(sqrt(2) + 2)*log(-sqrt(sqrt(2) + 2)*e^x +
e^(2*x) + 1) - 1/16*sqrt(-sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/16*sqrt(-sqrt(2) + 2)*log
(-sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + e^x

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \tanh{\left (4 x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x),x)

[Out]

Integral(exp(x)*tanh(4*x), x)

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Giac [A]  time = 1.44803, size = 339, normalized size = 0.93 \begin{align*} -\frac{1}{4} \, \sqrt{-\sqrt{2} + 2} \arctan \left (\frac{\sqrt{\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{-\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{-\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{\sqrt{2} + 2} \arctan \left (\frac{\sqrt{-\sqrt{2} + 2} + 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{4} \, \sqrt{\sqrt{2} + 2} \arctan \left (-\frac{\sqrt{-\sqrt{2} + 2} - 2 \, e^{x}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{8} \, \sqrt{\sqrt{2} + 2} \log \left (-\sqrt{\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac{1}{8} \, \sqrt{-\sqrt{2} + 2} \log \left (-\sqrt{-\sqrt{2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x),x, algorithm="giac")

[Out]

-1/4*sqrt(-sqrt(2) + 2)*arctan((sqrt(sqrt(2) + 2) + 2*e^x)/sqrt(-sqrt(2) + 2)) - 1/4*sqrt(-sqrt(2) + 2)*arctan
(-(sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) - 1/4*sqrt(sqrt(2) + 2)*arctan((sqrt(-sqrt(2) + 2) + 2*e^x)/
sqrt(sqrt(2) + 2)) - 1/4*sqrt(sqrt(2) + 2)*arctan(-(sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) - 1/8*sqrt(
sqrt(2) + 2)*log(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/8*sqrt(sqrt(2) + 2)*log(-sqrt(sqrt(2) + 2)*e^x + e^(
2*x) + 1) - 1/8*sqrt(-sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/8*sqrt(-sqrt(2) + 2)*log(-sqr
t(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + e^x