3.942 \(\int e^x \coth (2 x) \text {csch}^2(2 x) \, dx\)

Optimal. Leaf size=53 \[ \frac {e^x}{4 \left (1-e^{4 x}\right )}-\frac {e^{5 x}}{\left (1-e^{4 x}\right )^2}-\frac {1}{8} \tan ^{-1}\left (e^x\right )-\frac {1}{8} \tanh ^{-1}\left (e^x\right ) \]

[Out]

-exp(5*x)/(1-exp(4*x))^2+1/4*exp(x)/(1-exp(4*x))-1/8*arctan(exp(x))-1/8*arctanh(exp(x))

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Rubi [A]  time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2282, 12, 457, 288, 212, 206, 203} \[ \frac {e^x}{4 \left (1-e^{4 x}\right )}-\frac {e^{5 x}}{\left (1-e^{4 x}\right )^2}-\frac {1}{8} \tan ^{-1}\left (e^x\right )-\frac {1}{8} \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(5*x)/(1 - E^(4*x))^2) + E^x/(4*(1 - E^(4*x))) - ArcTan[E^x]/8 - ArcTanh[E^x]/8

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 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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

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 457

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] - Dist[(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] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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 \coth (2 x) \text {csch}^2(2 x) \, dx &=\operatorname {Subst}\left (\int \frac {4 x^4 \left (-1-x^4\right )}{\left (1-x^4\right )^3} \, dx,x,e^x\right )\\ &=4 \operatorname {Subst}\left (\int \frac {x^4 \left (-1-x^4\right )}{\left (1-x^4\right )^3} \, dx,x,e^x\right )\\ &=-\frac {e^{5 x}}{\left (1-e^{4 x}\right )^2}+\operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^4\right )^2} \, dx,x,e^x\right )\\ &=-\frac {e^{5 x}}{\left (1-e^{4 x}\right )^2}+\frac {e^x}{4 \left (1-e^{4 x}\right )}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,e^x\right )\\ &=-\frac {e^{5 x}}{\left (1-e^{4 x}\right )^2}+\frac {e^x}{4 \left (1-e^{4 x}\right )}-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )\\ &=-\frac {e^{5 x}}{\left (1-e^{4 x}\right )^2}+\frac {e^x}{4 \left (1-e^{4 x}\right )}-\frac {1}{8} \tan ^{-1}\left (e^x\right )-\frac {1}{8} \tanh ^{-1}\left (e^x\right )\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 54, normalized size = 1.02 \[ -\frac {-2 e^x+10 e^{5 x}+\left (e^{4 x}-1\right )^2 \tan ^{-1}\left (e^x\right )+\left (e^{4 x}-1\right )^2 \tanh ^{-1}\left (e^x\right )}{8 \left (e^{4 x}-1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Coth[2*x]*Csch[2*x]^2,x]

[Out]

-1/8*(-2*E^x + 10*E^(5*x) + (-1 + E^(4*x))^2*ArcTan[E^x] + (-1 + E^(4*x))^2*ArcTanh[E^x])/(-1 + E^(4*x))^2

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fricas [B]  time = 0.54, size = 522, normalized size = 9.85 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(20*cosh(x)^5 + 200*cosh(x)^3*sinh(x)^2 + 200*cosh(x)^2*sinh(x)^3 + 100*cosh(x)*sinh(x)^4 + 20*sinh(x)^5
+ 2*(cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sinh(x)^6 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 2*(35*co
sh(x)^4 - 1)*sinh(x)^4 - 2*cosh(x)^4 + 8*(7*cosh(x)^5 - cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 3*cosh(x)^2)*sin
h(x)^2 + 8*(cosh(x)^7 - cosh(x)^3)*sinh(x) + 1)*arctan(cosh(x) + sinh(x)) + (cosh(x)^8 + 56*cosh(x)^3*sinh(x)^
5 + 28*cosh(x)^2*sinh(x)^6 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + 2*(35*cosh(x)^4 - 1)*sinh(x)^4 - 2*cosh(x)^4 +
8*(7*cosh(x)^5 - cosh(x))*sinh(x)^3 + 4*(7*cosh(x)^6 - 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 - cosh(x)^3)*sinh
(x) + 1)*log(cosh(x) + sinh(x) + 1) - (cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sinh(x)^6 + 8*cosh(x)
*sinh(x)^7 + sinh(x)^8 + 2*(35*cosh(x)^4 - 1)*sinh(x)^4 - 2*cosh(x)^4 + 8*(7*cosh(x)^5 - cosh(x))*sinh(x)^3 +
4*(7*cosh(x)^6 - 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 - cosh(x)^3)*sinh(x) + 1)*log(cosh(x) + sinh(x) - 1) +
4*(25*cosh(x)^4 - 1)*sinh(x) - 4*cosh(x))/(cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sinh(x)^6 + 8*cos
h(x)*sinh(x)^7 + sinh(x)^8 + 2*(35*cosh(x)^4 - 1)*sinh(x)^4 - 2*cosh(x)^4 + 8*(7*cosh(x)^5 - cosh(x))*sinh(x)^
3 + 4*(7*cosh(x)^6 - 3*cosh(x)^2)*sinh(x)^2 + 8*(cosh(x)^7 - cosh(x)^3)*sinh(x) + 1)

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giac [A]  time = 0.12, size = 42, normalized size = 0.79 \[ -\frac {5 \, e^{\left (5 \, x\right )} - e^{x}}{4 \, {\left (e^{\left (4 \, x\right )} - 1\right )}^{2}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(5*e^(5*x) - e^x)/(e^(4*x) - 1)^2 - 1/8*arctan(e^x) - 1/16*log(e^x + 1) + 1/16*log(abs(e^x - 1))

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maple [C]  time = 0.38, size = 54, normalized size = 1.02 \[ -\frac {{\mathrm e}^{x} \left (5 \,{\mathrm e}^{4 x}-1\right )}{4 \left ({\mathrm e}^{4 x}-1\right )^{2}}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{16}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{16}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{16}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*exp(x)*(5*exp(4*x)-1)/(exp(4*x)-1)^2+1/16*I*ln(exp(x)-I)-1/16*I*ln(exp(x)+I)-1/16*ln(exp(x)+1)+1/16*ln(ex
p(x)-1)

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maxima [A]  time = 0.58, size = 47, normalized size = 0.89 \[ -\frac {5 \, e^{\left (5 \, x\right )} - e^{x}}{4 \, {\left (e^{\left (8 \, x\right )} - 2 \, e^{\left (4 \, x\right )} + 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(5*e^(5*x) - e^x)/(e^(8*x) - 2*e^(4*x) + 1) - 1/8*arctan(e^x) - 1/16*log(e^x + 1) + 1/16*log(e^x - 1)

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mupad [B]  time = 2.00, size = 80, normalized size = 1.51 \[ \frac {\ln \left (\frac {1}{4}-\frac {{\mathrm {e}}^x}{4}\right )}{16}-\frac {\ln \left (\frac {{\mathrm {e}}^x}{4}+\frac {1}{4}\right )}{16}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{8}-\frac {{\mathrm {e}}^{5\,x}}{2\,\left ({\mathrm {e}}^{8\,x}-2\,{\mathrm {e}}^{4\,x}+1\right )}-\frac {3\,{\mathrm {e}}^x}{4\,\left ({\mathrm {e}}^{4\,x}-1\right )}-\frac {{\mathrm {e}}^x}{2\,\left ({\mathrm {e}}^{8\,x}-2\,{\mathrm {e}}^{4\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1/4 - exp(x)/4)/16 - log(exp(x)/4 + 1/4)/16 - atan(exp(x))/8 - exp(5*x)/(2*(exp(8*x) - 2*exp(4*x) + 1)) -
(3*exp(x))/(4*(exp(4*x) - 1)) - exp(x)/(2*(exp(8*x) - 2*exp(4*x) + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*coth(2*x)*csch(2*x)**2, x)

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