3.317 \(\int e^x \text {csch}^2(3 x) \, dx\)

Optimal. Leaf size=105 \[ \frac {2 e^x}{3 \left (1-e^{6 x}\right )}+\frac {1}{18} \log \left (-e^x+e^{2 x}+1\right )-\frac {1}{18} \log \left (e^x+e^{2 x}+1\right )+\frac {\tan ^{-1}\left (\frac {1-2 e^x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2 e^x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2}{9} \tanh ^{-1}\left (e^x\right ) \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 105, 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, 210, 634, 618, 204, 628, 206} \[ \frac {2 e^x}{3 \left (1-e^{6 x}\right )}+\frac {1}{18} \log \left (-e^x+e^{2 x}+1\right )-\frac {1}{18} \log \left (e^x+e^{2 x}+1\right )+\frac {\tan ^{-1}\left (\frac {1-2 e^x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {2 e^x+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2}{9} \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^x)/(3*(1 - E^(6*x))) + ArcTan[(1 - 2*E^x)/Sqrt[3]]/(3*Sqrt[3]) - ArcTan[(1 + 2*E^x)/Sqrt[3]]/(3*Sqrt[3])
- (2*ArcTanh[E^x])/9 + Log[1 - E^x + E^(2*x)]/18 - Log[1 + E^x + E^(2*x)]/18

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 206

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

Rule 210

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

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Mathematica [C]  time = 0.03, size = 34, normalized size = 0.32 \[ \frac {2}{3} e^x \left (\frac {1}{1-e^{6 x}}-\, _2F_1\left (\frac {1}{6},1;\frac {7}{6};e^{6 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Csch[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 [B]  time = 0.85, size = 560, normalized size = 5.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*(sqrt(3)*cosh(x)^6 + 6*sqrt(3)*cosh(x)^5*sinh(x) + 15*sqrt(3)*cosh(x)^4*sinh(x)^2 + 20*sqrt(3)*cosh(x
)^3*sinh(x)^3 + 15*sqrt(3)*cosh(x)^2*sinh(x)^4 + 6*sqrt(3)*cosh(x)*sinh(x)^5 + sqrt(3)*sinh(x)^6 - sqrt(3))*ar
ctan(2/3*sqrt(3)*cosh(x) + 2/3*sqrt(3)*sinh(x) + 1/3*sqrt(3)) + 2*(sqrt(3)*cosh(x)^6 + 6*sqrt(3)*cosh(x)^5*sin
h(x) + 15*sqrt(3)*cosh(x)^4*sinh(x)^2 + 20*sqrt(3)*cosh(x)^3*sinh(x)^3 + 15*sqrt(3)*cosh(x)^2*sinh(x)^4 + 6*sq
rt(3)*cosh(x)*sinh(x)^5 + sqrt(3)*sinh(x)^6 - sqrt(3))*arctan(2/3*sqrt(3)*cosh(x) + 2/3*sqrt(3)*sinh(x) - 1/3*
sqrt(3)) + (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*s
inh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 - 1)*log((2*cosh(x) + 1)/(cosh(x) - 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)*log((2*cosh(x) - 1)/(cosh(x) - sinh(x))) + 2*(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)*log(cosh(x
) + sinh(x) + 1) - 2*(cosh(x)^6 + 6*cosh(x)^5*sinh(x) + 15*cosh(x)^4*sinh(x)^2 + 20*cosh(x)^3*sinh(x)^3 + 15*c
osh(x)^2*sinh(x)^4 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 - 1)*log(cosh(x) + sinh(x) - 1) + 12*cosh(x) + 12*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)

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giac [A]  time = 0.13, size = 86, normalized size = 0.82 \[ -\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} + 1\right )}\right ) - \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} - 1\right )}\right ) - \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} - 1\right )}} - \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac {1}{9} \, \log \left (e^{x} + 1\right ) + \frac {1}{9} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.10, size = 148, normalized size = 1.41 \[ -\frac {2 \,{\mathrm e}^{x}}{3 \left ({\mathrm e}^{6 x}-1\right )}+\frac {\ln \left ({\mathrm e}^{x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{18}+\frac {i \ln \left ({\mathrm e}^{x}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{18}+\frac {\ln \left ({\mathrm e}^{x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{18}-\frac {i \ln \left ({\mathrm e}^{x}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{18}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{9}-\frac {\ln \left ({\mathrm e}^{x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{18}+\frac {i \ln \left ({\mathrm e}^{x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{18}-\frac {\ln \left ({\mathrm e}^{x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{18}-\frac {i \ln \left ({\mathrm e}^{x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{18}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 85, normalized size = 0.81 \[ -\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} + 1\right )}\right ) - \frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} - 1\right )}\right ) - \frac {2 \, e^{x}}{3 \, {\left (e^{\left (6 \, x\right )} - 1\right )}} - \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) + \frac {1}{18} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) - \frac {1}{9} \, \log \left (e^{x} + 1\right ) + \frac {1}{9} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.40, size = 91, normalized size = 0.87 \[ \frac {\ln \left (\frac {2}{3}-\frac {2\,{\mathrm {e}}^x}{3}\right )}{9}-\frac {\ln \left (-\frac {2\,{\mathrm {e}}^x}{3}-\frac {2}{3}\right )}{9}+\frac {\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}-\frac {1}{3}\right )}^2+\frac {1}{3}\right )}{18}-\frac {\ln \left ({\left (\frac {2\,{\mathrm {e}}^x}{3}+\frac {1}{3}\right )}^2+\frac {1}{3}\right )}{18}-\frac {2\,{\mathrm {e}}^x}{3\,\left ({\mathrm {e}}^{6\,x}-1\right )}-\frac {\sqrt {3}\,\mathrm {atan}\left (\sqrt {3}\,\left (\frac {2\,{\mathrm {e}}^x}{3}-\frac {1}{3}\right )\right )}{9}-\frac {\sqrt {3}\,\mathrm {atan}\left (\sqrt {3}\,\left (\frac {2\,{\mathrm {e}}^x}{3}+\frac {1}{3}\right )\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*csch(3*x)**2, x)

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