3.225 \(\int e^x \coth ^2(4 x) \, dx\)

Optimal. Leaf size=134 \[ e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {1}{8} \tan ^{-1}\left (e^x\right )+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{8 \sqrt {2}}-\frac {1}{8} \tanh ^{-1}\left (e^x\right ) \]

[Out]

exp(x)+1/2*exp(x)/(1-exp(8*x))-1/8*arctan(exp(x))-1/8*arctanh(exp(x))-1/16*arctan(-1+exp(x)*2^(1/2))*2^(1/2)-1
/16*arctan(1+exp(x)*2^(1/2))*2^(1/2)+1/32*ln(1+exp(2*x)-exp(x)*2^(1/2))*2^(1/2)-1/32*ln(1+exp(2*x)+exp(x)*2^(1
/2))*2^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.300, Rules used = {2282, 390, 288, 214, 212, 206, 203, 211, 1165, 628, 1162, 617, 204} \[ e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}+\frac {\log \left (-\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^x+e^{2 x}+1\right )}{16 \sqrt {2}}-\frac {1}{8} \tan ^{-1}\left (e^x\right )+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} e^x+1\right )}{8 \sqrt {2}}-\frac {1}{8} \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x*Coth[4*x]^2,x]

[Out]

E^x + E^x/(2*(1 - E^(8*x))) - ArcTan[E^x]/8 + ArcTan[1 - Sqrt[2]*E^x]/(8*Sqrt[2]) - ArcTan[1 + Sqrt[2]*E^x]/(8
*Sqrt[2]) - ArcTanh[E^x]/8 + Log[1 - Sqrt[2]*E^x + E^(2*x)]/(16*Sqrt[2]) - Log[1 + Sqrt[2]*E^x + E^(2*x)]/(16*
Sqrt[2])

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 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 211

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

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 214

Int[((a_) + (b_.)*(x_)^(n_))^(-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^(n/2)), x], x] + Dist[r/(2*a), Int[1/(r + s*x^(n/2)), x], x]] /; FreeQ[{a,
b}, x] && IGtQ[n/4, 1] &&  !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 390

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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(4 x) \, dx &=\operatorname {Subst}\left (\int \frac {\left (1+x^8\right )^2}{\left (1-x^8\right )^2} \, dx,x,e^x\right )\\ &=\operatorname {Subst}\left (\int \left (1+\frac {4 x^8}{\left (1-x^8\right )^2}\right ) \, dx,x,e^x\right )\\ &=e^x+4 \operatorname {Subst}\left (\int \frac {x^8}{\left (1-x^8\right )^2} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^8} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,e^x\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 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 {1}{8} \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,e^x\right )-\frac {1}{8} \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,e^x\right )\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{8} \tan ^{-1}\left (e^x\right )-\frac {1}{8} \tanh ^{-1}\left (e^x\right )-\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,e^x\right )-\frac {1}{16} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,e^x\right )+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,e^x\right )}{16 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,e^x\right )}{16 \sqrt {2}}\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{8} \tan ^{-1}\left (e^x\right )-\frac {1}{8} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} e^x\right )}{8 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} e^x\right )}{8 \sqrt {2}}\\ &=e^x+\frac {e^x}{2 \left (1-e^{8 x}\right )}-\frac {1}{8} \tan ^{-1}\left (e^x\right )+\frac {\tan ^{-1}\left (1-\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} e^x\right )}{8 \sqrt {2}}-\frac {1}{8} \tanh ^{-1}\left (e^x\right )+\frac {\log \left (1-\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} e^x+e^{2 x}\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 1.88, size = 113, normalized size = 0.84 \[ \frac {64 e^{9 x} \left (e^{8 x}+1\right )^2 \, _4F_3\left (\frac {9}{8},2,2,2;1,1,\frac {33}{8};e^{8 x}\right )}{3825}+\frac {e^{-15 x} \left (9 \left (8368 e^{8 x}+1486 e^{16 x}-1456 e^{24 x}+e^{32 x}+4913\right ) \, _2F_1\left (\frac {1}{8},1;\frac {9}{8};e^{8 x}\right )-80225 e^{8 x}-15127 e^{16 x}+9361 e^{24 x}-44217\right )}{9216} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^x*Coth[4*x]^2,x]

[Out]

(-44217 - 80225*E^(8*x) - 15127*E^(16*x) + 9361*E^(24*x) + 9*(4913 + 8368*E^(8*x) + 1486*E^(16*x) - 1456*E^(24
*x) + E^(32*x))*Hypergeometric2F1[1/8, 1, 9/8, E^(8*x)])/(9216*E^(15*x)) + (64*E^(9*x)*(1 + E^(8*x))^2*Hyperge
ometricPFQ[{9/8, 2, 2, 2}, {1, 1, 33/8}, E^(8*x)])/3825

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fricas [B]  time = 0.53, size = 213, normalized size = 1.59 \[ \frac {4 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \sqrt {2} \sqrt {\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1} - 1\right ) + 4 \, {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \arctan \left (-\sqrt {2} e^{x} + \frac {1}{2} \, \sqrt {2} \sqrt {-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4} + 1\right ) - 4 \, {\left (e^{\left (8 \, x\right )} - 1\right )} \arctan \left (e^{x}\right ) - {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \log \left (4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) + {\left (\sqrt {2} e^{\left (8 \, x\right )} - \sqrt {2}\right )} \log \left (-4 \, \sqrt {2} e^{x} + 4 \, e^{\left (2 \, x\right )} + 4\right ) - 2 \, {\left (e^{\left (8 \, x\right )} - 1\right )} \log \left (e^{x} + 1\right ) + 2 \, {\left (e^{\left (8 \, x\right )} - 1\right )} \log \left (e^{x} - 1\right ) + 32 \, e^{\left (9 \, x\right )} - 48 \, e^{x}}{32 \, {\left (e^{\left (8 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(4*x)^2,x, algorithm="fricas")

[Out]

1/32*(4*(sqrt(2)*e^(8*x) - sqrt(2))*arctan(-sqrt(2)*e^x + sqrt(2)*sqrt(sqrt(2)*e^x + e^(2*x) + 1) - 1) + 4*(sq
rt(2)*e^(8*x) - sqrt(2))*arctan(-sqrt(2)*e^x + 1/2*sqrt(2)*sqrt(-4*sqrt(2)*e^x + 4*e^(2*x) + 4) + 1) - 4*(e^(8
*x) - 1)*arctan(e^x) - (sqrt(2)*e^(8*x) - sqrt(2))*log(4*sqrt(2)*e^x + 4*e^(2*x) + 4) + (sqrt(2)*e^(8*x) - sqr
t(2))*log(-4*sqrt(2)*e^x + 4*e^(2*x) + 4) - 2*(e^(8*x) - 1)*log(e^x + 1) + 2*(e^(8*x) - 1)*log(e^x - 1) + 32*e
^(9*x) - 48*e^x)/(e^(8*x) - 1)

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giac [A]  time = 0.14, size = 110, normalized size = 0.82 \[ -\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(4*x)^2,x, algorithm="giac")

[Out]

-1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/16*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/
32*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/32*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/2*e^x/(e^(8*x) -
1) - 1/8*arctan(e^x) + e^x - 1/16*log(e^x + 1) + 1/16*log(abs(e^x - 1))

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maple [C]  time = 0.27, size = 68, normalized size = 0.51 \[ {\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{2 \left ({\mathrm e}^{8 x}-1\right )}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{16}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{16}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{16}+\left (\munderset {\textit {\_R} =\RootOf \left (65536 \textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{x}-16 \textit {\_R} \right )\right )-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*coth(4*x)^2,x)

[Out]

exp(x)-1/2*exp(x)/(exp(8*x)-1)+1/16*ln(exp(x)-1)+1/16*I*ln(exp(x)-I)-1/16*I*ln(exp(x)+I)+sum(_R*ln(exp(x)-16*_
R),_R=RootOf(65536*_Z^4+1))-1/16*ln(exp(x)+1)

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maxima [A]  time = 0.50, size = 109, normalized size = 0.81 \[ -\frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{x}\right )}\right ) - \frac {1}{16} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{x}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (-\sqrt {2} e^{x} + e^{\left (2 \, x\right )} + 1\right ) - \frac {e^{x}}{2 \, {\left (e^{\left (8 \, x\right )} - 1\right )}} - \frac {1}{8} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{16} \, \log \left (e^{x} + 1\right ) + \frac {1}{16} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(4*x)^2,x, algorithm="maxima")

[Out]

-1/16*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/16*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*e^x)) - 1/
32*sqrt(2)*log(sqrt(2)*e^x + e^(2*x) + 1) + 1/32*sqrt(2)*log(-sqrt(2)*e^x + e^(2*x) + 1) - 1/2*e^x/(e^(8*x) -
1) - 1/8*arctan(e^x) + e^x - 1/16*log(e^x + 1) + 1/16*log(e^x - 1)

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mupad [B]  time = 1.39, size = 122, normalized size = 0.91 \[ \frac {\ln \left (\frac {1}{2}-\frac {{\mathrm {e}}^x}{2}\right )}{16}-\frac {\ln \left (-\frac {{\mathrm {e}}^x}{2}-\frac {1}{2}\right )}{16}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{8}+{\mathrm {e}}^x-\frac {{\mathrm {e}}^x}{2\,\left ({\mathrm {e}}^{8\,x}-1\right )}-\frac {\sqrt {2}\,\mathrm {atan}\left (2\,\sqrt {2}\,\left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {2}}{4}\right )\right )}{16}-\frac {\sqrt {2}\,\mathrm {atan}\left (2\,\sqrt {2}\,\left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {2}}{4}\right )\right )}{16}+\frac {\sqrt {2}\,\ln \left ({\left (\frac {{\mathrm {e}}^x}{2}-\frac {\sqrt {2}}{4}\right )}^2+\frac {1}{8}\right )}{32}-\frac {\sqrt {2}\,\ln \left ({\left (\frac {{\mathrm {e}}^x}{2}+\frac {\sqrt {2}}{4}\right )}^2+\frac {1}{8}\right )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(4*x)^2*exp(x),x)

[Out]

log(1/2 - exp(x)/2)/16 - log(- exp(x)/2 - 1/2)/16 - atan(exp(x))/8 + exp(x) - exp(x)/(2*(exp(8*x) - 1)) - (2^(
1/2)*atan(2*2^(1/2)*(exp(x)/2 - 2^(1/2)/4)))/16 - (2^(1/2)*atan(2*2^(1/2)*(exp(x)/2 + 2^(1/2)/4)))/16 + (2^(1/
2)*log((exp(x)/2 - 2^(1/2)/4)^2 + 1/8))/32 - (2^(1/2)*log((exp(x)/2 + 2^(1/2)/4)^2 + 1/8))/32

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \coth ^{2}{\left (4 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*coth(4*x)**2,x)

[Out]

Integral(exp(x)*coth(4*x)**2, x)

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