3.4.16 \(\int e^x \text {csch}(3 x) \, dx\) [316]

Optimal. Leaf size=54 \[ \frac {\text {ArcTan}\left (\frac {1+2 e^{2 x}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (1-e^{2 x}\right )-\frac {1}{6} \log \left (1+e^{2 x}+e^{4 x}\right ) \]

[Out]

1/3*ln(1-exp(2*x))-1/6*ln(1+exp(2*x)+exp(4*x))+1/3*arctan(1/3*(1+2*exp(2*x))*3^(1/2))*3^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {2320, 12, 281, 298, 31, 648, 632, 210, 642} \begin {gather*} \frac {\text {ArcTan}\left (\frac {2 e^{2 x}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (1-e^{2 x}\right )-\frac {1}{6} \log \left (e^{2 x}+e^{4 x}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

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*} \int e^x \text {csch}(3 x) \, dx &=\text {Subst}\left (\int \frac {2 x^3}{-1+x^6} \, dx,x,e^x\right )\\ &=2 \text {Subst}\left (\int \frac {x^3}{-1+x^6} \, dx,x,e^x\right )\\ &=\text {Subst}\left (\int \frac {x}{-1+x^3} \, dx,x,e^{2 x}\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{-1+x} \, dx,x,e^{2 x}\right )-\frac {1}{3} \text {Subst}\left (\int \frac {-1+x}{1+x+x^2} \, dx,x,e^{2 x}\right )\\ &=\frac {1}{3} \log \left (1-e^{2 x}\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,e^{2 x}\right )+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,e^{2 x}\right )\\ &=\frac {1}{3} \log \left (1-e^{2 x}\right )-\frac {1}{6} \log \left (1+e^{2 x}+e^{4 x}\right )-\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 e^{2 x}\right )\\ &=\frac {\tan ^{-1}\left (\frac {1+2 e^{2 x}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (1-e^{2 x}\right )-\frac {1}{6} \log \left (1+e^{2 x}+e^{4 x}\right )\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 22, normalized size = 0.41 \begin {gather*} -\frac {1}{2} e^{4 x} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};e^{6 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x*Csch[3*x],x]

[Out]

-1/2*(E^(4*x)*Hypergeometric2F1[2/3, 1, 5/3, E^(6*x)])

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Maple [C] Result contains complex when optimal does not.
time = 0.57, size = 79, normalized size = 1.46

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{2 x}-1\right )}{3}-\frac {\ln \left ({\mathrm e}^{2 x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right )}{6}+\frac {i \ln \left ({\mathrm e}^{2 x}+\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6}-\frac {\ln \left ({\mathrm e}^{2 x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{6}-\frac {i \ln \left ({\mathrm e}^{2 x}+\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {3}}{6}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*csch(3*x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.47, size = 73, normalized size = 1.35 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} + 1\right )}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{x} - 1\right )}\right ) - \frac {1}{6} \, \log \left (e^{\left (2 \, x\right )} + e^{x} + 1\right ) - \frac {1}{6} \, \log \left (e^{\left (2 \, x\right )} - e^{x} + 1\right ) + \frac {1}{3} \, \log \left (e^{x} + 1\right ) + \frac {1}{3} \, \log \left (e^{x} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.49, size = 83, normalized size = 1.54 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {3 \, \sqrt {3} \cosh \left (x\right ) + \sqrt {3} \sinh \left (x\right )}{3 \, {\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) - \frac {1}{6} \, \log \left (\frac {2 \, \cosh \left (x\right )^{2} + 2 \, \sinh \left (x\right )^{2} + 1}{\cosh \left (x\right )^{2} - 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}}\right ) + \frac {1}{3} \, \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*arctan(-1/3*(3*sqrt(3)*cosh(x) + sqrt(3)*sinh(x))/(cosh(x) - sinh(x))) - 1/6*log((2*cosh(x)^2 + 2
*sinh(x)^2 + 1)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1/3*log(2*sinh(x)/(cosh(x) - sinh(x)))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int e^{x} \operatorname {csch}{\left (3 x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*csch(3*x), x)

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Giac [A]
time = 0.41, size = 43, normalized size = 0.80 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (2 \, x\right )} + 1\right )}\right ) - \frac {1}{6} \, \log \left (e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{3} \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.18, size = 65, normalized size = 1.20 \begin {gather*} \frac {\ln \left (8\,{\mathrm {e}}^{2\,x}-8\right )}{3}+\ln \left (24\,{\mathrm {e}}^{2\,x}\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-8\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (-24\,{\mathrm {e}}^{2\,x}\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-8\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(8*exp(2*x) - 8)/3 + log(24*exp(2*x)*((3^(1/2)*1i)/6 - 1/6) - 8)*((3^(1/2)*1i)/6 - 1/6) - log(- 24*exp(2*x)
*((3^(1/2)*1i)/6 + 1/6) - 8)*((3^(1/2)*1i)/6 + 1/6)

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