3.222 \(\int e^x \tanh ^2(4 x) \, dx\)

Optimal. Leaf size=382 \[ e^x+\frac {e^x}{2 \left (e^{8 x}+1\right )}+\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (-\sqrt {2-\sqrt {2}} e^x+e^{2 x}+1\right )-\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (\sqrt {2-\sqrt {2}} e^x+e^{2 x}+1\right )+\frac {1}{32} \sqrt {2+\sqrt {2}} \log \left (-\sqrt {2+\sqrt {2}} e^x+e^{2 x}+1\right )-\frac {1}{32} \sqrt {2+\sqrt {2}} \log \left (\sqrt {2+\sqrt {2}} e^x+e^{2 x}+1\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 e^x}{\sqrt {2+\sqrt {2}}}\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 e^x}{\sqrt {2-\sqrt {2}}}\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 e^x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 e^x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}} \]

[Out]

exp(x)+1/2*exp(x)/(1+exp(8*x))+1/32*ln(1+exp(2*x)-exp(x)*(2-2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)-1/32*ln(1+exp(2*
x)+exp(x)*(2-2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)+1/8*arctan((-2*exp(x)+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))/(4-
2*2^(1/2))^(1/2)-1/8*arctan((2*exp(x)+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)+1/32*ln(1+exp(
2*x)-exp(x)*(2+2^(1/2))^(1/2))*(2+2^(1/2))^(1/2)-1/32*ln(1+exp(2*x)+exp(x)*(2+2^(1/2))^(1/2))*(2+2^(1/2))^(1/2
)+1/8*arctan((-2*exp(x)+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)-1/8*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 = 0.41, antiderivative size = 382, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {2282, 390, 288, 213, 1169, 634, 618, 204, 628} \[ e^x+\frac {e^x}{2 \left (e^{8 x}+1\right )}+\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (-\sqrt {2-\sqrt {2}} e^x+e^{2 x}+1\right )-\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (\sqrt {2-\sqrt {2}} e^x+e^{2 x}+1\right )+\frac {1}{32} \sqrt {2+\sqrt {2}} \log \left (-\sqrt {2+\sqrt {2}} e^x+e^{2 x}+1\right )-\frac {1}{32} \sqrt {2+\sqrt {2}} \log \left (\sqrt {2+\sqrt {2}} e^x+e^{2 x}+1\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 e^x}{\sqrt {2+\sqrt {2}}}\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 e^x}{\sqrt {2-\sqrt {2}}}\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 e^x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{8 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 e^x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Tanh[4*x]^2,x]

[Out]

E^x + E^x/(2*(1 + E^(8*x))) + ArcTan[(Sqrt[2 - Sqrt[2]] - 2*E^x)/Sqrt[2 + Sqrt[2]]]/(8*Sqrt[2*(2 - Sqrt[2])])
+ ArcTan[(Sqrt[2 + Sqrt[2]] - 2*E^x)/Sqrt[2 - Sqrt[2]]]/(8*Sqrt[2*(2 + Sqrt[2])]) - ArcTan[(Sqrt[2 - Sqrt[2]]
+ 2*E^x)/Sqrt[2 + Sqrt[2]]]/(8*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(Sqrt[2 + Sqrt[2]] + 2*E^x)/Sqrt[2 - Sqrt[2]]]/
(8*Sqrt[2*(2 + Sqrt[2])]) + (Sqrt[2 - Sqrt[2]]*Log[1 - Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)])/32 - (Sqrt[2 - Sqrt[2
]]*Log[1 + Sqrt[2 - Sqrt[2]]*E^x + E^(2*x)])/32 + (Sqrt[2 + Sqrt[2]]*Log[1 - Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)])
/32 - (Sqrt[2 + Sqrt[2]]*Log[1 + Sqrt[2 + Sqrt[2]]*E^x + E^(2*x)])/32

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 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 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 ^2(4 x) \, dx &=\operatorname {Subst}\left (\int \frac {\left (1-x^8\right )^2}{\left (1+x^8\right )^2} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (1-\frac {4 x^8}{\left (1+x^8\right )^2}\right ) \, dx,x,e^x\right )\\ &=e^x-4 \operatorname {Subst}\left (\int \frac {x^8}{\left (1+x^8\right )^2} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1+e^{8 x}\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^8} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1+e^{8 x}\right )}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx,x,e^x\right )}{4 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+x^2}{1+\sqrt {2} x^2+x^4} \, dx,x,e^x\right )}{4 \sqrt {2}}\\ &=e^x+\frac {e^x}{2 \left (1+e^{8 x}\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 )}{8 \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 )}{8 \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 )}{8 \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 )}{8 \sqrt {2 \left (2+\sqrt {2}\right )}}\\ &=e^x+\frac {e^x}{2 \left (1+e^{8 x}\right )}-\frac {1}{16} \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}{16} \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}{32} \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}{32} \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}{32} \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}{32} \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}{16} \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}{16} \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 {e^x}{2 \left (1+e^{8 x}\right )}+\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (1-\sqrt {2-\sqrt {2}} e^x+e^{2 x}\right )-\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (1+\sqrt {2-\sqrt {2}} e^x+e^{2 x}\right )+\frac {1}{32} \sqrt {2+\sqrt {2}} \log \left (1-\sqrt {2+\sqrt {2}} e^x+e^{2 x}\right )-\frac {1}{32} \sqrt {2+\sqrt {2}} \log \left (1+\sqrt {2+\sqrt {2}} e^x+e^{2 x}\right )+\frac {1}{8} \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}{8} \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}{8} \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}{8} \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 {e^x}{2 \left (1+e^{8 x}\right )}+\frac {1}{16} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 e^x}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{16} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 e^x}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}+2 e^x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{16} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}+2 e^x}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (1-\sqrt {2-\sqrt {2}} e^x+e^{2 x}\right )-\frac {1}{32} \sqrt {2-\sqrt {2}} \log \left (1+\sqrt {2-\sqrt {2}} e^x+e^{2 x}\right )+\frac {1}{32} \sqrt {2+\sqrt {2}} \log \left (1-\sqrt {2+\sqrt {2}} e^x+e^{2 x}\right )-\frac {1}{32} \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 = 0.05, size = 51, normalized size = 0.13 \[ \frac {1}{16} \text {RootSum}\left [\text {$\#$1}^8+1\& ,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}^7}\& \right ]+e^x+\frac {e^x}{2 \left (e^{8 x}+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Tanh[4*x]^2,x]

[Out]

E^x + E^x/(2*(1 + E^(8*x))) + RootSum[1 + #1^8 & , (x - Log[E^x - #1])/#1^7 & ]/16

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fricas [B]  time = 0.57, size = 1373, normalized size = 3.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x)^2,x, algorithm="fricas")

[Out]

1/128*(8*sqrt(-sqrt(2) + 2)*(e^(8*x) + 1)*arctan((2*sqrt(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) - sqrt(sqrt(2) +
2) - 2*e^x)/sqrt(-sqrt(2) + 2)) + 8*sqrt(-sqrt(2) + 2)*(e^(8*x) + 1)*arctan((2*sqrt(-sqrt(sqrt(2) + 2)*e^x +
e^(2*x) + 1) + sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) - 2*sqrt(-sqrt(2) + 2)*(e^(8*x) + 1)*log(sqrt(-s
qrt(2) + 2)*e^x + e^(2*x) + 1) + 2*sqrt(-sqrt(2) + 2)*(e^(8*x) + 1)*log(-sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1)
+ 8*(sqrt(sqrt(2) + 2)*e^(8*x) + sqrt(sqrt(2) + 2))*arctan((2*sqrt(sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) - sq
rt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) + 8*(sqrt(sqrt(2) + 2)*e^(8*x) + sqrt(sqrt(2) + 2))*arctan((2*sqr
t(-sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) + 4*(sqrt(2)*sqrt(sq
rt(2) + 2)*e^(8*x) + (sqrt(2)*e^(8*x) + 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))) + 4*(sqrt(2)*sqrt(sqrt(2) +
2)*e^(8*x) + (sqrt(2)*e^(8*x) + 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))) - 4*(sqrt(2)*sqrt(sqrt(2) + 2)*e^
(8*x) - (sqrt(2)*e^(8*x) + sqrt(2))*sqrt(-sqrt(2) + 2) + sqrt(2)*sqrt(sqrt(2) + 2))*arctan((2*sqrt(2)*e^x - sq
rt(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))) - 4*(sqrt(2)*sqrt(sqrt(2) + 2)*e^(8*x) -
(sqrt(2)*e^(8*x) + sqrt(2))*sqrt(-sqrt(2) + 2) + sqrt(2)*sqrt(sqrt(2) + 2))*arctan((2*sqrt(2)*e^x - sqrt(2)*sq
rt(-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))) - (sqrt(2)*sqrt(sqrt(2) + 2)*e^(8*x) + (sqrt(2)*
e^(8*x) + sqrt(2))*sqrt(-sqrt(2) + 2) + sqrt(2)*sqrt(sqrt(2) + 2))*log(2*sqrt(2)*sqrt(sqrt(2) + 2)*e^x + 2*sqr
t(2)*sqrt(-sqrt(2) + 2)*e^x + 4*e^(2*x) + 4) - (sqrt(2)*sqrt(sqrt(2) + 2)*e^(8*x) - (sqrt(2)*e^(8*x) + 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) + (sqrt(2)*sqrt(sqrt(2) + 2)*e^(8*x) - (sqrt(2)*e^(8*x) + 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) + (sqrt(2)*sqrt(sqrt(2) + 2)*e^(8*x) + (sqrt(2)*e^(8*x) + sqrt(2))*sqrt(-sqrt(2) + 2) + sqrt(2)*sq
rt(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) - 2*
(sqrt(sqrt(2) + 2)*e^(8*x) + sqrt(sqrt(2) + 2))*log(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) + 2*(sqrt(sqrt(2) + 2
)*e^(8*x) + sqrt(sqrt(2) + 2))*log(-sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) + 128*e^(9*x) + 192*e^x)/(e^(8*x) + 1
)

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giac [A]  time = 0.14, size = 263, normalized size = 0.69 \[ -\frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {\sqrt {\sqrt {2} + 2} + 2 \, e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \arctan \left (-\frac {\sqrt {\sqrt {2} + 2} - 2 \, e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {\sqrt {-\sqrt {2} + 2} + 2 \, e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \arctan \left (-\frac {\sqrt {-\sqrt {2} + 2} - 2 \, e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{32} \, \sqrt {\sqrt {2} + 2} \log \left (\sqrt {\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {\sqrt {2} + 2} \log \left (-\sqrt {\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{32} \, \sqrt {-\sqrt {2} + 2} \log \left (\sqrt {-\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {-\sqrt {2} + 2} \log \left (-\sqrt {-\sqrt {2} + 2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} + 1\right )}} + e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x)^2,x, algorithm="giac")

[Out]

-1/16*sqrt(-sqrt(2) + 2)*arctan((sqrt(sqrt(2) + 2) + 2*e^x)/sqrt(-sqrt(2) + 2)) - 1/16*sqrt(-sqrt(2) + 2)*arct
an(-(sqrt(sqrt(2) + 2) - 2*e^x)/sqrt(-sqrt(2) + 2)) - 1/16*sqrt(sqrt(2) + 2)*arctan((sqrt(-sqrt(2) + 2) + 2*e^
x)/sqrt(sqrt(2) + 2)) - 1/16*sqrt(sqrt(2) + 2)*arctan(-(sqrt(-sqrt(2) + 2) - 2*e^x)/sqrt(sqrt(2) + 2)) - 1/32*
sqrt(sqrt(2) + 2)*log(sqrt(sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/32*sqrt(sqrt(2) + 2)*log(-sqrt(sqrt(2) + 2)*e^x
+ e^(2*x) + 1) - 1/32*sqrt(-sqrt(2) + 2)*log(sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/32*sqrt(-sqrt(2) + 2)*
log(-sqrt(-sqrt(2) + 2)*e^x + e^(2*x) + 1) + 1/2*e^x/(e^(8*x) + 1) + e^x

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maple [C]  time = 0.25, size = 36, normalized size = 0.09 \[ {\mathrm e}^{x}+\frac {{\mathrm e}^{x}}{2+2 \,{\mathrm e}^{8 x}}+\left (\munderset {\textit {\_R} =\RootOf \left (4294967296 \textit {\_Z}^{8}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-16 \textit {\_R} \right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*tanh(4*x)^2,x)

[Out]

exp(x)+1/2*exp(x)/(1+exp(8*x))+sum(_R*ln(exp(x)-16*_R),_R=RootOf(4294967296*_Z^8+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, e^{\left (9 \, x\right )} + 3 \, e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} + 1\right )}} - \int \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x)^2,x, algorithm="maxima")

[Out]

1/2*(2*e^(9*x) + 3*e^x)/(e^(8*x) + 1) - integrate(1/2*e^x/(e^(8*x) + 1), x)

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mupad [B]  time = 3.98, size = 474, normalized size = 1.24 \[ {\mathrm {e}}^x+\frac {{\mathrm {e}}^x}{2\,\left ({\mathrm {e}}^{8\,x}+1\right )}+\ln \left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {\sqrt {2}+2}}{4}-\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )-\ln \left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {\sqrt {2}+2}}{4}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )+\ln \left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {2-\sqrt {2}}}{4}-\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {\sqrt {2-\sqrt {2}}}{32}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{32}\right )-\ln \left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {2-\sqrt {2}}}{4}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {\sqrt {2-\sqrt {2}}}{32}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{32}\right )+\sqrt {2}\,\ln \left (\frac {{\mathrm {e}}^x}{2}+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\frac {{\mathrm {e}}^x}{2}+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\frac {{\mathrm {e}}^x}{2}+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )\,\left (4-4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )\,\left (-\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\frac {{\mathrm {e}}^x}{2}+\sqrt {2}\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\right )\,\left (4+4{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}+2}}{32}+\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{32}\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)^2*exp(x),x)

[Out]

exp(x) + exp(x)/(2*(exp(8*x) + 1)) + log(exp(x)/2 - (2^(1/2) + 2)^(1/2)/4 - ((2 - 2^(1/2))^(1/2)*1i)/4)*((2^(1
/2) + 2)^(1/2)/32 + ((2 - 2^(1/2))^(1/2)*1i)/32) - log(exp(x)/2 + (2^(1/2) + 2)^(1/2)/4 + ((2 - 2^(1/2))^(1/2)
*1i)/4)*((2^(1/2) + 2)^(1/2)/32 + ((2 - 2^(1/2))^(1/2)*1i)/32) + log(exp(x)/2 - ((2^(1/2) + 2)^(1/2)*1i)/4 + (
2 - 2^(1/2))^(1/2)/4)*(((2^(1/2) + 2)^(1/2)*1i)/32 - (2 - 2^(1/2))^(1/2)/32) - log(exp(x)/2 + ((2^(1/2) + 2)^(
1/2)*1i)/4 - (2 - 2^(1/2))^(1/2)/4)*(((2^(1/2) + 2)^(1/2)*1i)/32 - (2 - 2^(1/2))^(1/2)/32) + 2^(1/2)*log(exp(x
)/2 - 2^(1/2)*((2^(1/2) + 2)^(1/2)/32 + ((2 - 2^(1/2))^(1/2)*1i)/32)*(4 + 4i))*((2^(1/2) + 2)^(1/2)/32 + ((2 -
2^(1/2))^(1/2)*1i)/32)*(1/2 + 1i/2) + 2^(1/2)*log(exp(x)/2 - 2^(1/2)*((2^(1/2) + 2)^(1/2)/32 + ((2 - 2^(1/2))
^(1/2)*1i)/32)*(4 - 4i))*((2^(1/2) + 2)^(1/2)/32 + ((2 - 2^(1/2))^(1/2)*1i)/32)*(1/2 - 1i/2) - 2^(1/2)*log(exp
(x)/2 + 2^(1/2)*((2^(1/2) + 2)^(1/2)/32 + ((2 - 2^(1/2))^(1/2)*1i)/32)*(4 - 4i))*((2^(1/2) + 2)^(1/2)/32 + ((2
- 2^(1/2))^(1/2)*1i)/32)*(1/2 - 1i/2) - 2^(1/2)*log(exp(x)/2 + 2^(1/2)*((2^(1/2) + 2)^(1/2)/32 + ((2 - 2^(1/2
))^(1/2)*1i)/32)*(4 + 4i))*((2^(1/2) + 2)^(1/2)/32 + ((2 - 2^(1/2))^(1/2)*1i)/32)*(1/2 + 1i/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \tanh ^{2}{\left (4 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*tanh(4*x)**2,x)

[Out]

Integral(exp(x)*tanh(4*x)**2, x)

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