\(\int e^x \coth ^2(2 x) \, dx\) [217]

Optimal result

Integrand size = 10, antiderivative size = 35 \[ \int e^x \coth ^2(2 x) \, dx=e^x+\frac {e^x}{1-e^{4 x}}-\frac {\arctan \left (e^x\right )}{2}-\frac {\text {arctanh}\left (e^x\right )}{2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2320, 398, 294, 218, 212, 209} \[ \int e^x \coth ^2(2 x) \, dx=-\frac {1}{2} \arctan \left (e^x\right )-\frac {\text {arctanh}\left (e^x\right )}{2}+e^x+\frac {e^x}{1-e^{4 x}} \]

[In]

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

[Out]

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

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 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

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 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^4\right )^2}{\left (1-x^4\right )^2} \, dx,x,e^x\right ) \\ & = \text {Subst}\left (\int \left (1+\frac {4 x^4}{\left (1-x^4\right )^2}\right ) \, dx,x,e^x\right ) \\ & = e^x+4 \text {Subst}\left (\int \frac {x^4}{\left (1-x^4\right )^2} \, dx,x,e^x\right ) \\ & = e^x+\frac {e^x}{1-e^{4 x}}-\text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,e^x\right ) \\ & = e^x+\frac {e^x}{1-e^{4 x}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right ) \\ & = e^x+\frac {e^x}{1-e^{4 x}}-\frac {\arctan \left (e^x\right )}{2}-\frac {\text {arctanh}\left (e^x\right )}{2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.72 (sec) , antiderivative size = 113, normalized size of antiderivative = 3.23 \[ \int e^x \coth ^2(2 x) \, dx=\frac {1}{640} e^{-7 x} \left (-3645-6769 e^{4 x}-1483 e^{8 x}+681 e^{12 x}+5 \left (729+1208 e^{4 x}+102 e^{8 x}-248 e^{12 x}+e^{16 x}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},e^{4 x}\right )\right )+\frac {16}{585} e^{5 x} \left (1+e^{4 x}\right )^2 \, _4F_3\left (\frac {5}{4},2,2,2;1,1,\frac {17}{4};e^{4 x}\right ) \]

[In]

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

[Out]

(-3645 - 6769*E^(4*x) - 1483*E^(8*x) + 681*E^(12*x) + 5*(729 + 1208*E^(4*x) + 102*E^(8*x) - 248*E^(12*x) + E^(
16*x))*Hypergeometric2F1[1/4, 1, 5/4, E^(4*x)])/(640*E^(7*x)) + (16*E^(5*x)*(1 + E^(4*x))^2*HypergeometricPFQ[
{5/4, 2, 2, 2}, {1, 1, 17/4}, E^(4*x)])/585

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.37

method	result	size
risch	\({\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{{\mathrm e}^{4 x}-1}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{4}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{4}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{4}\)	\(48\)

[In]

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

[Out]

exp(x)-exp(x)/(exp(4*x)-1)-1/4*ln(exp(x)+1)+1/4*I*ln(exp(x)-I)-1/4*I*ln(exp(x)+I)+1/4*ln(exp(x)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 230, normalized size of antiderivative = 6.57 \[ \int e^x \coth ^2(2 x) \, dx=\frac {4 \, \cosh \left (x\right )^{5} + 40 \, \cosh \left (x\right )^{3} \sinh \left (x\right )^{2} + 40 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{3} + 20 \, \cosh \left (x\right ) \sinh \left (x\right )^{4} + 4 \, \sinh \left (x\right )^{5} - 2 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 4 \, {\left (5 \, \cosh \left (x\right )^{4} - 2\right )} \sinh \left (x\right ) - 8 \, \cosh \left (x\right )}{4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 1\right )}} \]

[In]

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

[Out]

1/4*(4*cosh(x)^5 + 40*cosh(x)^3*sinh(x)^2 + 40*cosh(x)^2*sinh(x)^3 + 20*cosh(x)*sinh(x)^4 + 4*sinh(x)^5 - 2*(c
osh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)*arctan(cosh(x) +
sinh(x)) - (cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)*lo
g(cosh(x) + sinh(x) + 1) + (cosh(x)^4 + 4*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + si
nh(x)^4 - 1)*log(cosh(x) + sinh(x) - 1) + 4*(5*cosh(x)^4 - 2)*sinh(x) - 8*cosh(x))/(cosh(x)^4 + 4*cosh(x)^3*si
nh(x) + 6*cosh(x)^2*sinh(x)^2 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 - 1)

Sympy [F]

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

[In]

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

[Out]

Integral(exp(x)*coth(2*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.97 \[ \int e^x \coth ^2(2 x) \, dx=-\frac {e^{x}}{e^{\left (4 \, x\right )} - 1} - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left (e^{x} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int e^x \coth ^2(2 x) \, dx=-\frac {e^{x}}{e^{\left (4 \, x\right )} - 1} - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.09 \[ \int e^x \coth ^2(2 x) \, dx=\frac {\ln \left (1-{\mathrm {e}}^x\right )}{4}-\frac {\ln \left (-{\mathrm {e}}^x-1\right )}{4}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{2}+{\mathrm {e}}^x-\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{4\,x}-1} \]

[In]

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

[Out]

log(1 - exp(x))/4 - log(- exp(x) - 1)/4 - atan(exp(x))/2 + exp(x) - exp(x)/(exp(4*x) - 1)