\(\int e^x \tanh ^2(3 x) \, dx\) [218]

Optimal result

Integrand size = 10, antiderivative size = 113 \[ \int e^x \tanh ^2(3 x) \, dx=e^x+\frac {2 e^x}{3 \left (1+e^{6 x}\right )}-\frac {2 \arctan \left (e^x\right )}{9}+\frac {1}{9} \arctan \left (\sqrt {3}-2 e^x\right )-\frac {1}{9} \arctan \left (\sqrt {3}+2 e^x\right )+\frac {\log \left (1-\sqrt {3} e^x+e^{2 x}\right )}{6 \sqrt {3}}-\frac {\log \left (1+\sqrt {3} e^x+e^{2 x}\right )}{6 \sqrt {3}} \]

[Out]

exp(x)+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)

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {2320, 398, 294, 215, 648, 632, 210, 642, 209} \[ \int e^x \tanh ^2(3 x) \, dx=-\frac {2}{9} \arctan \left (e^x\right )+\frac {1}{9} \arctan \left (\sqrt {3}-2 e^x\right )-\frac {1}{9} \arctan \left (2 e^x+\sqrt {3}\right )+e^x+\frac {2 e^x}{3 \left (e^{6 x}+1\right )}+\frac {\log \left (-\sqrt {3} e^x+e^{2 x}+1\right )}{6 \sqrt {3}}-\frac {\log \left (\sqrt {3} e^x+e^{2 x}+1\right )}{6 \sqrt {3}} \]

[In]

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

[Out]

E^x + (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 209

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

Rule 210

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])

Rule 215

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/(a*n))*Int[1/
(r^2 + s^2*x^2), x] + 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 294

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 398

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 632

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 642

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 648

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 2320

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*} \text {integral}& = \text {Subst}\left (\int \frac {\left (1-x^6\right )^2}{\left (1+x^6\right )^2} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (1-\frac {4 x^6}{\left (1+x^6\right )^2}\right ) \, dx,x,e^x\right ) \\ & = e^x-4 \text {Subst}\left (\int \frac {x^6}{\left (1+x^6\right )^2} \, dx,x,e^x\right ) \\ & = e^x+\frac {2 e^x}{3 \left (1+e^{6 x}\right )}-\frac {2}{3} \text {Subst}\left (\int \frac {1}{1+x^6} \, dx,x,e^x\right ) \\ & = e^x+\frac {2 e^x}{3 \left (1+e^{6 x}\right )}-\frac {2}{9} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )-\frac {2}{9} \text {Subst}\left (\int \frac {1-\frac {\sqrt {3} x}{2}}{1-\sqrt {3} x+x^2} \, dx,x,e^x\right )-\frac {2}{9} \text {Subst}\left (\int \frac {1+\frac {\sqrt {3} x}{2}}{1+\sqrt {3} x+x^2} \, dx,x,e^x\right ) \\ & = e^x+\frac {2 e^x}{3 \left (1+e^{6 x}\right )}-\frac {2 \arctan \left (e^x\right )}{9}-\frac {1}{18} \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,e^x\right )-\frac {1}{18} \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,e^x\right )+\frac {\text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,e^x\right )}{6 \sqrt {3}}-\frac {\text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,e^x\right )}{6 \sqrt {3}} \\ & = e^x+\frac {2 e^x}{3 \left (1+e^{6 x}\right )}-\frac {2 \arctan \left (e^x\right )}{9}+\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} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 e^x\right )+\frac {1}{9} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 e^x\right ) \\ & = e^x+\frac {2 e^x}{3 \left (1+e^{6 x}\right )}-\frac {2 \arctan \left (e^x\right )}{9}+\frac {1}{9} \arctan \left (\sqrt {3}-2 e^x\right )-\frac {1}{9} \arctan \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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.10 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.86 \[ \int e^x \tanh ^2(3 x) \, dx=e^x+\frac {2 e^x}{3 \left (1+e^{6 x}\right )}-\frac {2 \arctan \left (e^x\right )}{9}-\frac {1}{9} \text {RootSum}\left [1-\text {$\#$1}^2+\text {$\#$1}^4\&,\frac {-2 x+2 \log \left (e^x-\text {$\#$1}\right )+x \text {$\#$1}^2-\log \left (e^x-\text {$\#$1}\right ) \text {$\#$1}^2}{-\text {$\#$1}+2 \text {$\#$1}^3}\&\right ] \]

[In]

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

[Out]

E^x + (2*E^x)/(3*(1 + E^(6*x))) - (2*ArcTan[E^x])/9 - RootSum[1 - #1^2 + #1^4 & , (-2*x + 2*Log[E^x - #1] + x*
#1^2 - Log[E^x - #1]*#1^2)/(-#1 + 2*#1^3) & ]/9

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.52

method	result	size
risch	\({\mathrm e}^{x}+\frac {2 \,{\mathrm e}^{x}}{3 \left (1+{\mathrm e}^{6 x}\right )}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{9}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{9}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (6561 \textit {\_Z}^{4}-81 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-9 \textit {\_R} \right )\right )\)	\(59\)

[In]

int(exp(x)*tanh(3*x)^2,x,method=_RETURNVERBOSE)

[Out]

exp(x)+2/3*exp(x)/(1+exp(6*x))+1/9*I*ln(exp(x)-I)-1/9*I*ln(exp(x)+I)+sum(_R*ln(exp(x)-9*_R),_R=RootOf(6561*_Z^
4-81*_Z^2+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 547, normalized size of antiderivative = 4.84 \[ \int e^x \tanh ^2(3 x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/18*(18*cosh(x)^7 + 378*cosh(x)^5*sinh(x)^2 + 630*cosh(x)^4*sinh(x)^3 + 630*cosh(x)^3*sinh(x)^4 + 378*cosh(x)
^2*sinh(x)^5 + 126*cosh(x)*sinh(x)^6 + 18*sinh(x)^7 - (cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^
2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 1)*sqrt(2*I*sqrt(3) +
2)*log(sqrt(2*I*sqrt(3) + 2) + 2*cosh(x) + 2*sinh(x)) + (cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x
)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 1)*sqrt(2*I*sqrt(3)
+ 2)*log(-sqrt(2*I*sqrt(3) + 2) + 2*cosh(x) + 2*sinh(x)) - (cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sin
h(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 1)*sqrt(-2*I*sqrt
(3) + 2)*log(sqrt(-2*I*sqrt(3) + 2) + 2*cosh(x) + 2*sinh(x)) + (cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4
*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 1)*sqrt(-2*I*
sqrt(3) + 2)*log(-sqrt(-2*I*sqrt(3) + 2) + 2*cosh(x) + 2*sinh(x)) - 4*(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*co
sh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 1)*arc
tan(cosh(x) + sinh(x)) + 6*(21*cosh(x)^6 + 5)*sinh(x) + 30*cosh(x))/(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh
(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*cosh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 1)

Sympy [F]

\[ \int e^x \tanh ^2(3 x) \, dx=\int e^{x} \tanh ^{2}{\left (3 x \right )}\, dx \]

[In]

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

[Out]

Integral(exp(x)*tanh(3*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int e^x \tanh ^2(3 x) \, dx=-\frac {1}{18} \, \sqrt {3} \log \left (\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{18} \, \sqrt {3} \log \left (-\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} + 1\right )}} - \frac {1}{9} \, \arctan \left (\sqrt {3} + 2 \, e^{x}\right ) - \frac {1}{9} \, \arctan \left (-\sqrt {3} + 2 \, e^{x}\right ) - \frac {2}{9} \, \arctan \left (e^{x}\right ) + e^{x} \]

[In]

integrate(exp(x)*tanh(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) + e^x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.72 \[ \int e^x \tanh ^2(3 x) \, dx=-\frac {1}{18} \, \sqrt {3} \log \left (\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{18} \, \sqrt {3} \log \left (-\sqrt {3} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} + 1\right )}} - \frac {1}{9} \, \arctan \left (\sqrt {3} + 2 \, e^{x}\right ) - \frac {1}{9} \, \arctan \left (-\sqrt {3} + 2 \, e^{x}\right ) - \frac {2}{9} \, \arctan \left (e^{x}\right ) + e^{x} \]

[In]

integrate(exp(x)*tanh(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) + e^x

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.76 \[ \int e^x \tanh ^2(3 x) \, dx={\mathrm {e}}^x-\frac {\mathrm {atan}\left (2\,{\mathrm {e}}^x+\sqrt {3}\right )}{9}-\frac {\mathrm {atan}\left (2\,{\mathrm {e}}^x-\sqrt {3}\right )}{9}-\frac {2\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{9}+\frac {2\,{\mathrm {e}}^x}{3\,\left ({\mathrm {e}}^{6\,x}+1\right )}+\frac {\sqrt {3}\,\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}-\frac {\sqrt {3}}{3}\right )}^2+\frac {1}{9}\right )}{18}-\frac {\sqrt {3}\,\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}+\frac {\sqrt {3}}{3}\right )}^2+\frac {1}{9}\right )}{18} \]

[In]

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

[Out]

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