Integrand size = 8, antiderivative size = 91 \[ \int e^x \coth (4 x) \, dx=e^x-\frac {\arctan \left (e^x\right )}{2}+\frac {\arctan \left (1-\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\arctan \left (1+\sqrt {2} e^x\right )}{2 \sqrt {2}}-\frac {\text {arctanh}\left (e^x\right )}{2}-\frac {\text {arctanh}\left (\frac {\sqrt {2} e^x}{1+e^{2 x}}\right )}{2 \sqrt {2}} \] Output:
exp(x)-1/2*arctan(exp(x))-1/4*arctan(-1+2^(1/2)*exp(x))*2^(1/2)-1/4*arctan (1+2^(1/2)*exp(x))*2^(1/2)-1/2*arctanh(exp(x))-1/4*arctanh(2^(1/2)*exp(x)/ (1+exp(2*x)))*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.24 \[ \int e^x \coth (4 x) \, dx=e^x-2 e^x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},1,\frac {9}{8},e^{8 x}\right ) \] Input:
Integrate[E^x*Coth[4*x],x]
Output:
E^x - 2*E^x*Hypergeometric2F1[1/8, 1, 9/8, E^(8*x)]
Time = 0.38 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.47, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.875, Rules used = {2720, 25, 913, 758, 755, 756, 216, 219, 1476, 1082, 217, 1479, 25, 27, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^x \coth (4 x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \int -\frac {e^{8 x}+1}{1-e^{8 x}}de^x\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1+e^{8 x}}{1-e^{8 x}}de^x\) |
\(\Big \downarrow \) 913 |
\(\displaystyle e^x-2 \int \frac {1}{1-e^{8 x}}de^x\) |
\(\Big \downarrow \) 758 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \int \frac {1}{1-e^{4 x}}de^x+\frac {1}{2} \int \frac {1}{1+e^{4 x}}de^x\right )\) |
\(\Big \downarrow \) 755 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \int \frac {1}{1-e^{4 x}}de^x+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-e^{2 x}}de^x+\frac {1}{2} \int \frac {1}{1+e^{2 x}}de^x\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-e^{2 x}}de^x+\frac {\arctan \left (e^x\right )}{2}\right )+\frac {1}{2} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \int \frac {1+e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^x+e^{2 x}}de^x+\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 x}}d\left (1-\sqrt {2} e^x\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 x}}d\left (1+\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1-e^{2 x}}{1+e^{4 x}}de^x+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {\int -\frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^x\right )}{1+\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {1}{2} \left (\frac {\int \frac {\sqrt {2}-2 e^x}{1-\sqrt {2} e^x+e^{2 x}}de^x}{2 \sqrt {2}}+\frac {1}{2} \int \frac {1+\sqrt {2} e^x}{1+\sqrt {2} e^x+e^{2 x}}de^x\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle e^x-2 \left (\frac {1}{2} \left (\frac {\arctan \left (e^x\right )}{2}+\frac {\text {arctanh}\left (e^x\right )}{2}\right )+\frac {1}{2} \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^x+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^x\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{2 \sqrt {2}}\right )\right )\right )\) |
Input:
Int[E^x*Coth[4*x],x]
Output:
E^x - 2*((ArcTan[E^x]/2 + ArcTanh[E^x]/2)/2 + ((-(ArcTan[1 - Sqrt[2]*E^x]/ Sqrt[2]) + ArcTan[1 + Sqrt[2]*E^x]/Sqrt[2])/2 + (-1/2*Log[1 - Sqrt[2]*E^x + E^(2*x)]/Sqrt[2] + Log[1 + Sqrt[2]*E^x + E^(2*x)]/(2*Sqrt[2]))/2)/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] ], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b , 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^(n/2)), x], x] + Simp[r/(2*a) Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a, b}, x] && IGtQ[n/4, 1] && !GtQ[a/b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( p + 1) + 1))/(b*(n*(p + 1) + 1)) Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b , c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.56 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62
method | result | size |
risch | \({\mathrm e}^{x}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (256 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-4 \textit {\_R} \right )\right )+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{4}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{4}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{4}\) | \(56\) |
Input:
int(exp(x)*coth(4*x),x,method=_RETURNVERBOSE)
Output:
exp(x)+sum(_R*ln(exp(x)-4*_R),_R=RootOf(256*_Z^4+1))+1/4*ln(exp(x)-1)-1/4* ln(exp(x)+1)+1/4*I*ln(exp(x)-I)-1/4*I*ln(exp(x)+I)
Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.32 \[ \int e^x \coth (4 x) \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right ) + 1\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right ) - 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\frac {\sqrt {2} + 2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - 2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - \frac {1}{2} \, \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + \cosh \left (x\right ) - \frac {1}{4} \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \frac {1}{4} \, \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + \sinh \left (x\right ) \] Input:
integrate(exp(x)*coth(4*x),x, algorithm="fricas")
Output:
-1/4*sqrt(2)*arctan(sqrt(2)*cosh(x) + sqrt(2)*sinh(x) + 1) - 1/4*sqrt(2)*a rctan(sqrt(2)*cosh(x) + sqrt(2)*sinh(x) - 1) - 1/8*sqrt(2)*log((sqrt(2) + 2*cosh(x))/(cosh(x) - sinh(x))) + 1/8*sqrt(2)*log(-(sqrt(2) - 2*cosh(x))/( cosh(x) - sinh(x))) - 1/2*arctan(cosh(x) + sinh(x)) + cosh(x) - 1/4*log(co sh(x) + sinh(x) + 1) + 1/4*log(cosh(x) + sinh(x) - 1) + sinh(x)
\[ \int e^x \coth (4 x) \, dx=\int e^{x} \coth {\left (4 x \right )}\, dx \] Input:
integrate(exp(x)*coth(4*x),x)
Output:
Integral(exp(x)*coth(4*x), x)
Time = 0.12 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.07 \[ \int e^x \coth (4 x) \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left (e^{x} - 1\right ) \] Input:
integrate(exp(x)*coth(4*x),x, algorithm="maxima")
Output:
-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/4*sqrt(2)*arctan(-1 /2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/8*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/8*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/2*arctan(e^x) + e^x - 1 /4*log(e^x + 1) + 1/4*log(e^x - 1)
Time = 0.11 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.08 \[ \int e^x \coth (4 x) \, dx=-\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \] Input:
integrate(exp(x)*coth(4*x),x, algorithm="giac")
Output:
-1/4*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/4*sqrt(2)*arctan(-1 /2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/8*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/8*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/2*arctan(e^x) + e^x - 1 /4*log(e^x + 1) + 1/4*log(abs(e^x - 1))
Time = 2.46 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int e^x \coth (4 x) \, dx=\frac {\ln \left (2-2\,{\mathrm {e}}^x\right )}{4}-\frac {\ln \left (-2\,{\mathrm {e}}^x-2\right )}{4}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{2}+{\mathrm {e}}^x-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}{2}\right )}{4}+\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x-\sqrt {2}\right )}^2+2\right )}{8}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}{2}\right )}{4}-\frac {\sqrt {2}\,\ln \left ({\left (2\,{\mathrm {e}}^x+\sqrt {2}\right )}^2+2\right )}{8} \] Input:
int(coth(4*x)*exp(x),x)
Output:
log(2 - 2*exp(x))/4 - log(- 2*exp(x) - 2)/4 - atan(exp(x))/2 + exp(x) - (2 ^(1/2)*atan((2^(1/2)*(2*exp(x) - 2^(1/2)))/2))/4 + (2^(1/2)*log((2*exp(x) - 2^(1/2))^2 + 2))/8 - (2^(1/2)*atan((2^(1/2)*(2*exp(x) + 2^(1/2)))/2))/4 - (2^(1/2)*log((2*exp(x) + 2^(1/2))^2 + 2))/8
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.11 \[ \int e^x \coth (4 x) \, dx=-\frac {\mathit {atan} \left (e^{x}\right )}{2}-\frac {\sqrt {2}\, \mathit {atan} \left (\frac {2 e^{x}-\sqrt {2}}{\sqrt {2}}\right )}{4}-\frac {\sqrt {2}\, \mathit {atan} \left (\frac {2 e^{x}+\sqrt {2}}{\sqrt {2}}\right )}{4}+e^{x}+\frac {\sqrt {2}\, \mathrm {log}\left (e^{2 x}-e^{x} \sqrt {2}+1\right )}{8}-\frac {\sqrt {2}\, \mathrm {log}\left (e^{2 x}+e^{x} \sqrt {2}+1\right )}{8}+\frac {\mathrm {log}\left (e^{x}-1\right )}{4}-\frac {\mathrm {log}\left (e^{x}+1\right )}{4} \] Input:
int(exp(x)*coth(4*x),x)
Output:
( - 4*atan(e**x) - 2*sqrt(2)*atan((2*e**x - sqrt(2))/sqrt(2)) - 2*sqrt(2)* atan((2*e**x + sqrt(2))/sqrt(2)) + 8*e**x + sqrt(2)*log(e**(2*x) - e**x*sq rt(2) + 1) - sqrt(2)*log(e**(2*x) + e**x*sqrt(2) + 1) + 2*log(e**x - 1) - 2*log(e**x + 1))/8