Integrand size = 16, antiderivative size = 149 \[ \int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx=\frac {4 e^{5 x}}{3 \left (1+e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1+e^{4 x}\right )^2}-\frac {3 e^x}{8 \left (1+e^{4 x}\right )}-\frac {3 \arctan \left (1-\sqrt {2} e^x\right )}{16 \sqrt {2}}+\frac {3 \arctan \left (1+\sqrt {2} e^x\right )}{16 \sqrt {2}}-\frac {3 \log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{32 \sqrt {2}}+\frac {3 \log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{32 \sqrt {2}} \]
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)+exp (x)*2^(1/2))*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.43 \[ \int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx=\frac {1}{96} \left (-\frac {4 e^x \left (9+6 e^{4 x}+29 e^{8 x}\right )}{\left (1+e^{4 x}\right )^3}-9 \text {RootSum}\left [1+\text {$\#$1}^4\&,\frac {x-\log \left (e^x-\text {$\#$1}\right )}{\text {$\#$1}^3}\&\right ]\right ) \]
((-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
Time = 0.40 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.15, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {2720, 27, 963, 27, 957, 817, 755, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int e^x \tanh ^2(2 x) \text {sech}^2(2 x) \, dx\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \int \frac {4 e^{4 x} \left (1-e^{4 x}\right )^2}{\left (e^{4 x}+1\right )^4}de^x\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \int \frac {e^{4 x} \left (1-e^{4 x}\right )^2}{\left (1+e^{4 x}\right )^4}de^x\) |
| \(\Big \downarrow \) 963 |
| \(\displaystyle 4 \left (\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}-\frac {1}{12} \int \frac {4 e^{4 x} \left (2-3 e^{4 x}\right )}{\left (1+e^{4 x}\right )^3}de^x\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}-\frac {1}{3} \int \frac {e^{4 x} \left (2-3 e^{4 x}\right )}{\left (1+e^{4 x}\right )^3}de^x\right )\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \int \frac {e^{4 x}}{\left (1+e^{4 x}\right )^2}de^x-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \int \frac {1}{1+e^{4 x}}de^x-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^x+e^{2 x}}de^x+\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 x}}d\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 x}}d\left (1+\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {1}{2} \int \frac {1+\sqrt {2} e^x}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle 4 \left (\frac {1}{3} \left (\frac {9}{8} \left (\frac {1}{4} \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}\right )\right )-\frac {e^x}{4 \left (e^{4 x}+1\right )}\right )-\frac {5 e^{5 x}}{8 \left (e^{4 x}+1\right )^2}\right )+\frac {e^{5 x}}{3 \left (e^{4 x}+1\right )^3}\right )\) |
4*(E^(5*x)/(3*(1 + E^(4*x))^3) + ((-5*E^(5*x))/(8*(1 + E^(4*x))^2) + (9*(- 1/4*E^x/(1 + E^(4*x)) + ((-(ArcTan[1 - Sqrt[2]*E^x]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*E^x]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*E^x + E^(2*x)]/Sqrt[2] + L og[1 + Sqrt[2]*E^x + E^(2*x)]/(2*Sqrt[2]))/2)/4))/8)/3)
3.10.40.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[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]]))
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] - Simp[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]
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] - Simp[(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] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
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] + Simp[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]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[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])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 13.56 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.34
| method | result | size |
| risch | \(-\frac {{\mathrm e}^{x} \left (29 \,{\mathrm e}^{8 x}+6 \,{\mathrm e}^{4 x}+9\right )}{24 \left (1+{\mathrm e}^{4 x}\right )^{3}}+4 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (268435456 \textit {\_Z}^{4}+81\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+\frac {128 \textit {\_R}}{3}\right )\right )\) | \(50\) |
| default | \(\frac {\frac {3 \tanh \left (\frac {x}{2}\right )^{11}}{8}-\frac {3 \tanh \left (\frac {x}{2}\right )^{10}}{8}-\frac {109 \tanh \left (\frac {x}{2}\right )^{9}}{24}-\frac {173 \tanh \left (\frac {x}{2}\right )^{8}}{8}-\frac {49 \tanh \left (\frac {x}{2}\right )^{7}}{4}-\frac {231 \tanh \left (\frac {x}{2}\right )^{6}}{4}+\frac {49 \tanh \left (\frac {x}{2}\right )^{5}}{4}-\frac {117 \tanh \left (\frac {x}{2}\right )^{4}}{4}+\frac {109 \tanh \left (\frac {x}{2}\right )^{3}}{24}-\frac {63 \tanh \left (\frac {x}{2}\right )^{2}}{8}-\frac {3 \tanh \left (\frac {x}{2}\right )}{8}-\frac {11}{24}}{\left (\tanh \left (\frac {x}{2}\right )^{4}+6 \tanh \left (\frac {x}{2}\right )^{2}+1\right )^{3}}+\frac {3 \sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3+2 \sqrt {2}\right )}{64}+\frac {3 \left (2+\sqrt {2}\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2+2 \sqrt {2}}\right )}{16 \left (2+2 \sqrt {2}\right )}-\frac {3 \sqrt {2}\, \ln \left (\tanh \left (\frac {x}{2}\right )^{2}+3-2 \sqrt {2}\right )}{64}-\frac {3 \left (-2+\sqrt {2}\right ) \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right )}{2 \sqrt {2}-2}\right )}{16 \left (2 \sqrt {2}-2\right )}\) | \(212\) |
-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+81))
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 1616, normalized size of antiderivative = 10.85 \[ \int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx=\text {Too large to display} \]
-1/192*(232*cosh(x)^9 + 19488*cosh(x)^3*sinh(x)^6 + 8352*cosh(x)^2*sinh(x) ^7 + 2088*cosh(x)*sinh(x)^8 + 232*sinh(x)^9 + 48*(609*cosh(x)^4 + 1)*sinh( x)^5 + 48*cosh(x)^5 + 48*(609*cosh(x)^5 + 5*cosh(x))*sinh(x)^4 + 96*(203*c osh(x)^6 + 5*cosh(x)^2)*sinh(x)^3 + 96*(87*cosh(x)^7 + 5*cosh(x)^3)*sinh(x )^2 + 9*(-(I + 1)*sqrt(2)*cosh(x)^12 - (220*I + 220)*sqrt(2)*cosh(x)^3*sin h(x)^9 - (66*I + 66)*sqrt(2)*cosh(x)^2*sinh(x)^10 - (12*I + 12)*sqrt(2)*co sh(x)*sinh(x)^11 - (I + 1)*sqrt(2)*sinh(x)^12 + 3*(-(165*I + 165)*sqrt(2)* cosh(x)^4 - (I + 1)*sqrt(2))*sinh(x)^8 - (3*I + 3)*sqrt(2)*cosh(x)^8 + 24* (-(33*I + 33)*sqrt(2)*cosh(x)^5 - (I + 1)*sqrt(2)*cosh(x))*sinh(x)^7 + 84* (-(11*I + 11)*sqrt(2)*cosh(x)^6 - (I + 1)*sqrt(2)*cosh(x)^2)*sinh(x)^6 + 2 4*(-(33*I + 33)*sqrt(2)*cosh(x)^7 - (7*I + 7)*sqrt(2)*cosh(x)^3)*sinh(x)^5 + 3*(-(165*I + 165)*sqrt(2)*cosh(x)^8 - (70*I + 70)*sqrt(2)*cosh(x)^4 - ( I + 1)*sqrt(2))*sinh(x)^4 - (3*I + 3)*sqrt(2)*cosh(x)^4 + 4*(-(55*I + 55)* sqrt(2)*cosh(x)^9 - (42*I + 42)*sqrt(2)*cosh(x)^5 - (3*I + 3)*sqrt(2)*cosh (x))*sinh(x)^3 + 6*(-(11*I + 11)*sqrt(2)*cosh(x)^10 - (14*I + 14)*sqrt(2)* cosh(x)^6 - (3*I + 3)*sqrt(2)*cosh(x)^2)*sinh(x)^2 + 12*(-(I + 1)*sqrt(2)* cosh(x)^11 - (2*I + 2)*sqrt(2)*cosh(x)^7 - (I + 1)*sqrt(2)*cosh(x)^3)*sinh (x) - (I + 1)*sqrt(2))*log((I + 1)*sqrt(2) + 2*cosh(x) + 2*sinh(x)) + 9*(( I - 1)*sqrt(2)*cosh(x)^12 + (220*I - 220)*sqrt(2)*cosh(x)^3*sinh(x)^9 + (6 6*I - 66)*sqrt(2)*cosh(x)^2*sinh(x)^10 + (12*I - 12)*sqrt(2)*cosh(x)*si...
\[ \int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx=\int e^{x} \tanh ^{2}{\left (2 x \right )} \operatorname {sech}^{2}{\left (2 x \right )}\, dx \]
Time = 0.28 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.77 \[ \int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx=\frac {3}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {3}{32} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {3}{64} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {3}{64} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {29 \, e^{\left (9 \, x\right )} + 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \, {\left (e^{\left (12 \, x\right )} + 3 \, e^{\left (8 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 1\right )}} \]
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)) + 3/64*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)
Time = 0.27 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.69 \[ \int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx=\frac {3}{32} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) + \frac {3}{32} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) + \frac {3}{64} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {3}{64} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {29 \, e^{\left (9 \, x\right )} + 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \, {\left (e^{\left (4 \, x\right )} + 1\right )}^{3}} \]
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)) + 3/64*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
Time = 2.83 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.03 \[ \int e^x \text {sech}^2(2 x) \tanh ^2(2 x) \, dx=\frac {5\,{\mathrm {e}}^x}{6\,\left (2\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{8\,x}+1\right )}-\frac {\frac {{\mathrm {e}}^{9\,x}}{3}-\frac {2\,{\mathrm {e}}^{5\,x}}{3}+\frac {{\mathrm {e}}^x}{3}}{3\,{\mathrm {e}}^{4\,x}+3\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{12\,x}+1}-\frac {7\,{\mathrm {e}}^x}{8\,\left ({\mathrm {e}}^{4\,x}+1\right )}+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (-\frac {3}{16}-\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (\frac {3}{64}+\frac {3}{64}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (-\frac {3}{16}+\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (\frac {3}{64}-\frac {3}{64}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (\frac {3}{16}-\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{64}+\frac {3}{64}{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\sqrt {2}\,\left (\frac {3}{16}+\frac {3}{16}{}\mathrm {i}\right )\right )\,\left (-\frac {3}{64}-\frac {3}{64}{}\mathrm {i}\right ) \]
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))