3.940 \(\int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx\)

Optimal. Leaf size=149 \[ -\frac {3 e^x}{8 \left (e^{4 x}+1\right )}-\frac {5 e^{5 x}}{6 \left (e^{4 x}+1\right )^2}+\frac {4 e^{5 x}}{3 \left (e^{4 x}+1\right )^3}-\frac {3 \log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{32 \sqrt {2}}+\frac {3 \log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{32 \sqrt {2}}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} e^x\right )}{16 \sqrt {2}}+\frac {3 \tan ^{-1}\left (\sqrt {2} e^x+1\right )}{16 \sqrt {2}} \]

[Out]

4/3*exp(5*x)/(1+exp(4*x))^3-5/6*exp(5*x)/(1+exp(4*x))^2-3/8*exp(x)/(1+exp(4*x))+3/32*arctan(-1+exp(x)*2^(1/2))
*2^(1/2)+3/32*arctan(1+exp(x)*2^(1/2))*2^(1/2)-3/64*ln(1+exp(2*x)-exp(x)*2^(1/2))*2^(1/2)+3/64*ln(1+exp(2*x)+e
xp(x)*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {2282, 12, 463, 457, 288, 211, 1165, 628, 1162, 617, 204} \[ -\frac {3 e^x}{8 \left (e^{4 x}+1\right )}-\frac {5 e^{5 x}}{6 \left (e^{4 x}+1\right )^2}+\frac {4 e^{5 x}}{3 \left (e^{4 x}+1\right )^3}-\frac {3 \log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{32 \sqrt {2}}+\frac {3 \log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{32 \sqrt {2}}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} e^x\right )}{16 \sqrt {2}}+\frac {3 \tan ^{-1}\left (\sqrt {2} e^x+1\right )}{16 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*E^(5*x))/(3*(1 + E^(4*x))^3) - (5*E^(5*x))/(6*(1 + E^(4*x))^2) - (3*E^x)/(8*(1 + E^(4*x))) - (3*ArcTan[1 -
Sqrt[2]*E^x])/(16*Sqrt[2]) + (3*ArcTan[1 + Sqrt[2]*E^x])/(16*Sqrt[2]) - (3*Log[1 - Sqrt[2]*E^x + E^(2*x)])/(32
*Sqrt[2]) + (3*Log[1 + Sqrt[2]*E^x + E^(2*x)])/(32*Sqrt[2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

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 \text {sech}^2(2 x) \tanh ^2(2 x) \, dx &=\operatorname {Subst}\left (\int \frac {4 x^4 \left (1-x^4\right )^2}{\left (1+x^4\right )^4} \, dx,x,e^x\right )\\ &=4 \operatorname {Subst}\left (\int \frac {x^4 \left (1-x^4\right )^2}{\left (1+x^4\right )^4} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^4 \left (8-12 x^4\right )}{\left (1+x^4\right )^3} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^4\right )^2} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac {3 e^x}{8 \left (1+e^{4 x}\right )}+\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac {3 e^x}{8 \left (1+e^{4 x}\right )}+\frac {3}{16} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,e^x\right )+\frac {3}{16} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac {3 e^x}{8 \left (1+e^{4 x}\right )}+\frac {3}{32} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,e^x\right )+\frac {3}{32} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,e^x\right )-\frac {3 \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,e^x\right )}{32 \sqrt {2}}-\frac {3 \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,e^x\right )}{32 \sqrt {2}}\\ &=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac {3 e^x}{8 \left (1+e^{4 x}\right )}-\frac {3 \log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{32 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{32 \sqrt {2}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^x\right )}{16 \sqrt {2}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^x\right )}{16 \sqrt {2}}\\ &=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac {3 e^x}{8 \left (1+e^{4 x}\right )}-\frac {3 \tan ^{-1}\left (1-\sqrt {2} e^x\right )}{16 \sqrt {2}}+\frac {3 \tan ^{-1}\left (1+\sqrt {2} e^x\right )}{16 \sqrt {2}}-\frac {3 \log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{32 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{32 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.07, size = 64, normalized size = 0.43 \[ \frac {1}{96} \left (-9 \text {RootSum}\left [\text {$\#$1}^4+1\& ,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}^3}\& \right ]-\frac {4 e^x \left (6 e^{4 x}+29 e^{8 x}+9\right )}{\left (e^{4 x}+1\right )^3}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*E^x*(9 + 6*E^(4*x) + 29*E^(8*x)))/(1 + E^(4*x))^3 - 9*RootSum[1 + #1^4 & , (x - Log[E^x - #1])/#1^3 & ])/
96

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fricas [B]  time = 0.51, size = 259, normalized size = 1.74 \[ -\frac {36 \, {\left (\sqrt {2} e^{\left (12 \, x\right )} + 3 \, \sqrt {2} e^{\left (8 \, x\right )} + 3 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 36 \, {\left (\sqrt {2} e^{\left (12 \, x\right )} + 3 \, \sqrt {2} e^{\left (8 \, x\right )} + 3 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - 9 \, {\left (\sqrt {2} e^{\left (12 \, x\right )} + 3 \, \sqrt {2} e^{\left (8 \, x\right )} + 3 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 9 \, {\left (\sqrt {2} e^{\left (12 \, x\right )} + 3 \, \sqrt {2} e^{\left (8 \, x\right )} + 3 \, \sqrt {2} e^{\left (4 \, x\right )} + \sqrt {2}\right )} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 232 \, e^{\left (9 \, x\right )} + 48 \, e^{\left (5 \, x\right )} + 72 \, e^{x}}{192 \, {\left (e^{\left (12 \, x\right )} + 3 \, e^{\left (8 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(36*(sqrt(2)*e^(12*x) + 3*sqrt(2)*e^(8*x) + 3*sqrt(2)*e^(4*x) + sqrt(2))*arctan(-sqrt(2)*e^x + sqrt(2)*
sqrt(sqrt(2)*e^x + e^(2*x) + 1) - 1) + 36*(sqrt(2)*e^(12*x) + 3*sqrt(2)*e^(8*x) + 3*sqrt(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) - 9*(sqrt(2)*e^(12*x) + 3*sqrt(2)
*e^(8*x) + 3*sqrt(2)*e^(4*x) + sqrt(2))*log(4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 9*(sqrt(2)*e^(12*x) + 3*sqrt(2)*e
^(8*x) + 3*sqrt(2)*e^(4*x) + sqrt(2))*log(-4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 232*e^(9*x) + 48*e^(5*x) + 72*e^x)
/(e^(12*x) + 3*e^(8*x) + 3*e^(4*x) + 1)

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giac [A]  time = 0.12, size = 103, normalized size = 0.69 \[ \frac {3}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {3}{32} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {3}{64} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {3}{64} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {29 \, e^{\left (9 \, x\right )} + 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \, {\left (e^{\left (4 \, x\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 3/32*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) + 3/6
4*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) - 3/64*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/24*(29*e^(9*x) + 6
*e^(5*x) + 9*e^x)/(e^(4*x) + 1)^3

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maple [C]  time = 0.37, size = 50, normalized size = 0.34 \[ -\frac {{\mathrm e}^{x} \left (29 \,{\mathrm e}^{8 x}+6 \,{\mathrm e}^{4 x}+9\right )}{24 \left (1+{\mathrm e}^{4 x}\right )^{3}}+4 \left (\munderset {\textit {\_R} =\RootOf \left (268435456 \textit {\_Z}^{4}+81\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\frac {128 \textit {\_R}}{3}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*exp(x)*(29*exp(8*x)+6*exp(4*x)+9)/(1+exp(4*x))^3+4*sum(_R*ln(exp(x)+128/3*_R),_R=RootOf(268435456*_Z^4+8
1))

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maxima [A]  time = 0.41, size = 115, normalized size = 0.77 \[ \frac {3}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {3}{32} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {3}{64} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {3}{64} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {29 \, e^{\left (9 \, x\right )} + 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \, {\left (e^{\left (12 \, x\right )} + 3 \, e^{\left (8 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) + 3/32*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) + 3/6
4*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) - 3/64*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/24*(29*e^(9*x) + 6
*e^(5*x) + 9*e^x)/(e^(12*x) + 3*e^(8*x) + 3*e^(4*x) + 1)

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mupad [B]  time = 2.17, size = 154, normalized size = 1.03 \[ \frac {5\,{\mathrm {e}}^x}{6\,\left (2\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{8\,x}+1\right )}-\frac {\frac {{\mathrm {e}}^{9\,x}}{3}-\frac {2\,{\mathrm {e}}^{5\,x}}{3}+\frac {{\mathrm {e}}^x}{3}}{3\,{\mathrm {e}}^{4\,x}+3\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{12\,x}+1}-\frac {7\,{\mathrm {e}}^x}{8\,\left ({\mathrm {e}}^{4\,x}+1\right )}+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (-\frac {3}{16}-\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (\frac {3}{64}+\frac {3}{64}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (-\frac {3}{16}+\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (\frac {3}{64}-\frac {3}{64}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (\frac {3}{16}-\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{64}+\frac {3}{64}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (\frac {3}{16}+\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{64}-\frac {3}{64}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2^(1/2)*log(- (3*exp(x))/8 - 2^(1/2)*(3/16 + 3i/16))*(3/64 + 3i/64) - (exp(9*x)/3 - (2*exp(5*x))/3 + exp(x)/3)
/(3*exp(4*x) + 3*exp(8*x) + exp(12*x) + 1) - (7*exp(x))/(8*(exp(4*x) + 1)) + 2^(1/2)*log(- (3*exp(x))/8 - 2^(1
/2)*(3/16 - 3i/16))*(3/64 - 3i/64) - 2^(1/2)*log(2^(1/2)*(3/16 - 3i/16) - (3*exp(x))/8)*(3/64 - 3i/64) - 2^(1/
2)*log(2^(1/2)*(3/16 + 3i/16) - (3*exp(x))/8)*(3/64 + 3i/64) + (5*exp(x))/(6*(2*exp(4*x) + exp(8*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*tanh(2*x)**2*sech(2*x)**2, x)

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