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]

exp(x)+1/8*ln(1+exp(2*x)-exp(x)*(2-2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)-1/8*ln(1+exp(2*x)+exp(x)*(2-2^(1/2))^(1/2
))*(2-2^(1/2))^(1/2)+1/2*arctan((-2*exp(x)+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)-1/2*arcta
n((2*exp(x)+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)+1/8*ln(1+exp(2*x)-exp(x)*(2+2^(1/2))^(1/
2))*(2+2^(1/2))^(1/2)-1/8*ln(1+exp(2*x)+exp(x)*(2+2^(1/2))^(1/2))*(2+2^(1/2))^(1/2)+1/2*arctan((-2*exp(x)+(2+2
^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)-1/2*arctan((2*exp(x)+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2
))/(4+2*2^(1/2))^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 366, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 8, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, 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 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 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 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 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 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 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 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 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 (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*}

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Mathematica [C]  time = 0.01, 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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fricas [B]  time = 0.65, size = 1089, normalized size = 2.98 \[ \text {result too large to display} \]

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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giac [A]  time = 0.15, size = 251, normalized size = 0.69 \[ -\frac {1}{4} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {\sqrt {\sqrt {2} + 2} + 2 \, e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {-\sqrt {2} + 2} \arctan \left (-\frac {\sqrt {\sqrt {2} + 2} - 2 \, e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {\sqrt {-\sqrt {2} + 2} + 2 \, e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {\sqrt {2} + 2} \arctan \left (-\frac {\sqrt {-\sqrt {2} + 2} - 2 \, e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \log \left (\sqrt {\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \log \left (-\sqrt {\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \log \left (\sqrt {-\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \log \left (-\sqrt {-\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + e^{x} \]

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

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maple [C]  time = 0.20, size = 24, normalized size = 0.07 \[ {\mathrm e}^{x}+\left (\munderset {\textit {\_R} =\RootOf \left (65536 \textit {\_Z}^{8}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-4 \textit {\_R} \right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*tanh(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 = 0.00, size = 0, normalized size = 0.00 \[ e^{x} - 2 \, \int \frac {e^{x}}{e^{\left (8 \, x\right )} + 1}\,{d x} \]

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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mupad [B]  time = 3.82, size = 457, normalized size = 1.25 \[ {\mathrm {e}}^x-\ln \left (2\,{\mathrm {e}}^x+\sqrt {\sqrt {2}+2}+\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )+\ln \left (2\,{\mathrm {e}}^x+\sqrt {2-\sqrt {2}}-\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}\right )\,\left (-\frac {\sqrt {2-\sqrt {2}}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )+\ln \left (2\,{\mathrm {e}}^x-\sqrt {\sqrt {2}+2}-\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )-\ln \left (2\,{\mathrm {e}}^x-\sqrt {2-\sqrt {2}}+\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}\right )\,\left (-\frac {\sqrt {2-\sqrt {2}}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )+\sqrt {2}\,\ln \left (2\,{\mathrm {e}}^x+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (2\,{\mathrm {e}}^x+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (2\,{\mathrm {e}}^x+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (4-4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (2\,{\mathrm {e}}^x+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (4+4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{8}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )\,\left (-\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \tanh {\left (4 x \right )}\, dx \]

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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