3.280 \(\int e^x \text {sech}^2(3 x) \, dx\)

Optimal. Leaf size=110 \[ -\frac {2 e^x}{3 \left (e^{6 x}+1\right )}-\frac {\log \left (-\sqrt {3} e^x+e^{2 x}+1\right )}{6 \sqrt {3}}+\frac {\log \left (\sqrt {3} e^x+e^{2 x}+1\right )}{6 \sqrt {3}}+\frac {2}{9} \tan ^{-1}\left (e^x\right )-\frac {1}{9} \tan ^{-1}\left (\sqrt {3}-2 e^x\right )+\frac {1}{9} \tan ^{-1}\left (2 e^x+\sqrt {3}\right ) \]

[Out]

-2/3*exp(x)/(1+exp(6*x))+2/9*arctan(exp(x))+1/9*arctan(2*exp(x)-3^(1/2))+1/9*arctan(2*exp(x)+3^(1/2))-1/18*ln(
1+exp(2*x)-exp(x)*3^(1/2))*3^(1/2)+1/18*ln(1+exp(2*x)+exp(x)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {2282, 12, 288, 209, 634, 618, 204, 628, 203} \[ -\frac {2 e^x}{3 \left (e^{6 x}+1\right )}-\frac {\log \left (-\sqrt {3} e^x+e^{2 x}+1\right )}{6 \sqrt {3}}+\frac {\log \left (\sqrt {3} e^x+e^{2 x}+1\right )}{6 \sqrt {3}}+\frac {2}{9} \tan ^{-1}\left (e^x\right )-\frac {1}{9} \tan ^{-1}\left (\sqrt {3}-2 e^x\right )+\frac {1}{9} \tan ^{-1}\left (2 e^x+\sqrt {3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sech[3*x]^2,x]

[Out]

(-2*E^x)/(3*(1 + E^(6*x))) + (2*ArcTan[E^x])/9 - ArcTan[Sqrt[3] - 2*E^x]/9 + ArcTan[Sqrt[3] + 2*E^x]/9 - Log[1
- Sqrt[3]*E^x + E^(2*x)]/(6*Sqrt[3]) + Log[1 + Sqrt[3]*E^x + E^(2*x)]/(6*Sqrt[3])

Rule 12

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 209

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u, v}, Simp[u = Int[(r - s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x] +
Int[(r + s*Cos[((2*k - 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 +
s^2*x^2), x])/(a*n) + Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)
/4, 0] && PosQ[a/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 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 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(3 x) \, dx &=\operatorname {Subst}\left (\int \frac {4 x^6}{\left (1+x^6\right )^2} \, dx,x,e^x\right )\\ &=4 \operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^6\right )^2} \, dx,x,e^x\right )\\ &=-\frac {2 e^x}{3 \left (1+e^{6 x}\right )}+\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,e^x\right )\\ &=-\frac {2 e^x}{3 \left (1+e^{6 x}\right )}+\frac {2}{9} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )+\frac {2}{9} \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,e^x\right )+\frac {2}{9} \operatorname {Subst}\left (\int \frac {1+\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,e^x\right )\\ &=-\frac {2 e^x}{3 \left (1+e^{6 x}\right )}+\frac {2}{9} \tan ^{-1}\left (e^x\right )+\frac {1}{18} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,e^x\right )+\frac {1}{18} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,e^x\right )-\frac {\operatorname {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,e^x\right )}{6 \sqrt {3}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,e^x\right )}{6 \sqrt {3}}\\ &=-\frac {2 e^x}{3 \left (1+e^{6 x}\right )}+\frac {2}{9} \tan ^{-1}\left (e^x\right )-\frac {\log \left (1-\sqrt {3} e^x+e^{2 x}\right )}{6 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} e^x+e^{2 x}\right )}{6 \sqrt {3}}-\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 e^x\right )-\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 e^x\right )\\ &=-\frac {2 e^x}{3 \left (1+e^{6 x}\right )}+\frac {2}{9} \tan ^{-1}\left (e^x\right )-\frac {1}{9} \tan ^{-1}\left (\sqrt {3}-2 e^x\right )+\frac {1}{9} \tan ^{-1}\left (\sqrt {3}+2 e^x\right )-\frac {\log \left (1-\sqrt {3} e^x+e^{2 x}\right )}{6 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} e^x+e^{2 x}\right )}{6 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 34, normalized size = 0.31 \[ \frac {2}{3} e^x \left (\, _2F_1\left (\frac {1}{6},1;\frac {7}{6};-e^{6 x}\right )-\frac {1}{e^{6 x}+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sech[3*x]^2,x]

[Out]

(2*E^x*(-(1 + E^(6*x))^(-1) + Hypergeometric2F1[1/6, 1, 7/6, -E^(6*x)]))/3

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fricas [A]  time = 0.77, size = 154, normalized size = 1.40 \[ -\frac {4 \, {\left (e^{\left (6 \, x\right )} + 1\right )} \arctan \left (\sqrt {3} + \sqrt {-4 \, \sqrt {3} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} - 2 \, e^{x}\right ) + 4 \, {\left (e^{\left (6 \, x\right )} + 1\right )} \arctan \left (-\sqrt {3} + 2 \, \sqrt {\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1} - 2 \, e^{x}\right ) - 4 \, {\left (e^{\left (6 \, x\right )} + 1\right )} \arctan \left (e^{x}\right ) - {\left (\sqrt {3} e^{\left (6 \, x\right )} + \sqrt {3}\right )} \log \left (4 \, \sqrt {3} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + {\left (\sqrt {3} e^{\left (6 \, x\right )} + \sqrt {3}\right )} \log \left (-4 \, \sqrt {3} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + 12 \, e^{x}}{18 \, {\left (e^{\left (6 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x)^2,x, algorithm="fricas")

[Out]

-1/18*(4*(e^(6*x) + 1)*arctan(sqrt(3) + sqrt(-4*sqrt(3)*e^x + 4*e^(2*x) + 4) - 2*e^x) + 4*(e^(6*x) + 1)*arctan
(-sqrt(3) + 2*sqrt(sqrt(3)*e^x + e^(2*x) + 1) - 2*e^x) - 4*(e^(6*x) + 1)*arctan(e^x) - (sqrt(3)*e^(6*x) + sqrt
(3))*log(4*sqrt(3)*e^x + 4*e^(2*x) + 4) + (sqrt(3)*e^(6*x) + sqrt(3))*log(-4*sqrt(3)*e^x + 4*e^(2*x) + 4) + 12
*e^x)/(e^(6*x) + 1)

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giac [A]  time = 0.12, size = 79, normalized size = 0.72 \[ \frac {1}{18} \, \sqrt {3} \log \left (\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{18} \, \sqrt {3} \log \left (-\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} + 1\right )}} + \frac {1}{9} \, \arctan \left (\sqrt {3} + 2 \, e^{x}\right ) + \frac {1}{9} \, \arctan \left (-\sqrt {3} + 2 \, e^{x}\right ) + \frac {2}{9} \, \arctan \left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x)^2,x, algorithm="giac")

[Out]

1/18*sqrt(3)*log(sqrt(3)*e^x + e^(2*x) + 1) - 1/18*sqrt(3)*log(-sqrt(3)*e^x + e^(2*x) + 1) - 2/3*e^x/(e^(6*x)
+ 1) + 1/9*arctan(sqrt(3) + 2*e^x) + 1/9*arctan(-sqrt(3) + 2*e^x) + 2/9*arctan(e^x)

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maple [C]  time = 0.24, size = 59, normalized size = 0.54 \[ -\frac {2 \,{\mathrm e}^{x}}{3 \left (1+{\mathrm e}^{6 x}\right )}+\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{9}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{9}+4 \left (\munderset {\textit {\_R} =\RootOf \left (1679616 \textit {\_Z}^{4}-1296 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}+36 \textit {\_R} \right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sech(3*x)^2,x)

[Out]

-2/3*exp(x)/(1+exp(6*x))+1/9*I*ln(exp(x)+I)-1/9*I*ln(exp(x)-I)+4*sum(_R*ln(exp(x)+36*_R),_R=RootOf(1679616*_Z^
4-1296*_Z^2+1))

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maxima [A]  time = 0.42, size = 79, normalized size = 0.72 \[ \frac {1}{18} \, \sqrt {3} \log \left (\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{18} \, \sqrt {3} \log \left (-\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} + 1\right )}} + \frac {1}{9} \, \arctan \left (\sqrt {3} + 2 \, e^{x}\right ) + \frac {1}{9} \, \arctan \left (-\sqrt {3} + 2 \, e^{x}\right ) + \frac {2}{9} \, \arctan \left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x)^2,x, algorithm="maxima")

[Out]

1/18*sqrt(3)*log(sqrt(3)*e^x + e^(2*x) + 1) - 1/18*sqrt(3)*log(-sqrt(3)*e^x + e^(2*x) + 1) - 2/3*e^x/(e^(6*x)
+ 1) + 1/9*arctan(sqrt(3) + 2*e^x) + 1/9*arctan(-sqrt(3) + 2*e^x) + 2/9*arctan(e^x)

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mupad [B]  time = 0.31, size = 84, normalized size = 0.76 \[ \frac {2\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{9}+\frac {\mathrm {atan}\left (2\,{\mathrm {e}}^x+\sqrt {3}\right )}{9}+\frac {\mathrm {atan}\left (2\,{\mathrm {e}}^x-\sqrt {3}\right )}{9}-\frac {2\,{\mathrm {e}}^x}{3\,\left ({\mathrm {e}}^{6\,x}+1\right )}-\frac {\sqrt {3}\,\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}-\frac {\sqrt {3}}{3}\right )}^2+\frac {1}{9}\right )}{18}+\frac {\sqrt {3}\,\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}+\frac {\sqrt {3}}{3}\right )}^2+\frac {1}{9}\right )}{18} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/cosh(3*x)^2,x)

[Out]

(2*atan(exp(x)))/9 + atan(2*exp(x) + 3^(1/2))/9 + atan(2*exp(x) - 3^(1/2))/9 - (2*exp(x))/(3*(exp(6*x) + 1)) -
(3^(1/2)*log(((2*exp(x))/3 - 3^(1/2)/3)^2 + 1/9))/18 + (3^(1/2)*log(((2*exp(x))/3 + 3^(1/2)/3)^2 + 1/9))/18

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \operatorname {sech}^{2}{\left (3 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sech(3*x)**2,x)

[Out]

Integral(exp(x)*sech(3*x)**2, x)

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