3.217 \(\int e^x \coth ^2(2 x) \, dx\)

Optimal. Leaf size=35 \[ e^x+\frac {e^x}{1-e^{4 x}}-\frac {1}{2} \tan ^{-1}\left (e^x\right )-\frac {1}{2} \tanh ^{-1}\left (e^x\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2282, 390, 288, 212, 206, 203} \[ e^x+\frac {e^x}{1-e^{4 x}}-\frac {1}{2} \tan ^{-1}\left (e^x\right )-\frac {1}{2} \tanh ^{-1}\left (e^x\right ) \]

Antiderivative was successfully verified.

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

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Mathematica [C]  time = 1.62, size = 113, normalized size = 3.23 \[ \frac {16}{585} e^{5 x} \left (e^{4 x}+1\right )^2 \, _4F_3\left (\frac {5}{4},2,2,2;1,1,\frac {17}{4};e^{4 x}\right )+\frac {1}{640} e^{-7 x} \left (5 \left (1208 e^{4 x}+102 e^{8 x}-248 e^{12 x}+e^{16 x}+729\right ) \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{4 x}\right )-6769 e^{4 x}-1483 e^{8 x}+681 e^{12 x}-3645\right ) \]

Warning: Unable to verify antiderivative.

[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

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fricas [B]  time = 0.53, size = 230, normalized size = 6.57 \[ \frac {4 \, \cosh \relax (x)^{5} + 40 \, \cosh \relax (x)^{3} \sinh \relax (x)^{2} + 40 \, \cosh \relax (x)^{2} \sinh \relax (x)^{3} + 20 \, \cosh \relax (x) \sinh \relax (x)^{4} + 4 \, \sinh \relax (x)^{5} - 2 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} \sinh \relax (x) + 6 \, \cosh \relax (x)^{2} \sinh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} - 1\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) - {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} \sinh \relax (x) + 6 \, \cosh \relax (x)^{2} \sinh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} - 1\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} \sinh \relax (x) + 6 \, \cosh \relax (x)^{2} \sinh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} - 1\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right ) + 4 \, {\left (5 \, \cosh \relax (x)^{4} - 2\right )} \sinh \relax (x) - 8 \, \cosh \relax (x)}{4 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x)^{3} \sinh \relax (x) + 6 \, \cosh \relax (x)^{2} \sinh \relax (x)^{2} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.12, size = 35, normalized size = 1.00 \[ -\frac {e^{x}}{e^{\left (4 \, x\right )} - 1} - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.22, size = 48, normalized size = 1.37 \[ {\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{{\mathrm e}^{4 x}-1}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{4}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{4}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{4}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 34, normalized size = 0.97 \[ -\frac {e^{x}}{e^{\left (4 \, x\right )} - 1} - \frac {1}{2} \, \arctan \left (e^{x}\right ) + e^{x} - \frac {1}{4} \, \log \left (e^{x} + 1\right ) + \frac {1}{4} \, \log \left (e^{x} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.18, size = 38, normalized size = 1.09 \[ \frac {\ln \left (1-{\mathrm {e}}^x\right )}{4}-\frac {\ln \left (-{\mathrm {e}}^x-1\right )}{4}-\frac {\mathrm {atan}\left ({\mathrm {e}}^x\right )}{2}+{\mathrm {e}}^x-\frac {{\mathrm {e}}^x}{{\mathrm {e}}^{4\,x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*coth(2*x)**2, x)

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