### 3.943 $$\int e^x \coth ^2(2 x) \text{csch}(2 x) \, dx$$

Optimal. Leaf size=55 $\frac{3 e^{3 x}}{4 \left (1-e^{4 x}\right )}-\frac{e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac{5}{8} \tan ^{-1}\left (e^x\right )-\frac{5}{8} \tanh ^{-1}\left (e^x\right )$

[Out]

-(E^(3*x)/(1 - E^(4*x))^2) + (3*E^(3*x))/(4*(1 - E^(4*x))) + (5*ArcTan[E^x])/8 - (5*ArcTanh[E^x])/8

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Rubi [A]  time = 0.0480443, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 14, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {2282, 12, 463, 457, 298, 203, 206} $\frac{3 e^{3 x}}{4 \left (1-e^{4 x}\right )}-\frac{e^{3 x}}{\left (1-e^{4 x}\right )^2}+\frac{5}{8} \tan ^{-1}\left (e^x\right )-\frac{5}{8} \tanh ^{-1}\left (e^x\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(E^(3*x)/(1 - E^(4*x))^2) + (3*E^(3*x))/(4*(1 - E^(4*x))) + (5*ArcTan[E^x])/8 - (5*ArcTanh[E^x])/8

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]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

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 298

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

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])

Rubi steps

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

Mathematica [C]  time = 3.15484, size = 161, normalized size = 2.93 $-\frac{16 e^{7 x} \left (e^{4 x}+1\right )^2 \, _5F_4\left (\frac{7}{4},2,2,2,2;1,1,1,\frac{19}{4};e^{4 x}\right )}{1155}-\frac{8 e^{7 x} \left (26 e^{4 x}+11 e^{8 x}+15\right ) \, _4F_3\left (\frac{7}{4},2,2,2;1,1,\frac{19}{4};e^{4 x}\right )}{1155}+\frac{e^{-5 x} \left (-7 \left (24152 e^{4 x}-10058 e^{8 x}-9048 e^{12 x}+513 e^{16 x}+25289\right ) \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};e^{4 x}\right )+244931 e^{4 x}+43161 e^{8 x}-26091 e^{12 x}+177023\right )}{10752}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(177023 + 244931*E^(4*x) + 43161*E^(8*x) - 26091*E^(12*x) - 7*(25289 + 24152*E^(4*x) - 10058*E^(8*x) - 9048*E^
(12*x) + 513*E^(16*x))*Hypergeometric2F1[3/4, 1, 7/4, E^(4*x)])/(10752*E^(5*x)) - (8*E^(7*x)*(15 + 26*E^(4*x)
+ 11*E^(8*x))*HypergeometricPFQ[{7/4, 2, 2, 2}, {1, 1, 19/4}, E^(4*x)])/1155 - (16*E^(7*x)*(1 + E^(4*x))^2*Hyp
ergeometricPFQ[{7/4, 2, 2, 2, 2}, {1, 1, 1, 19/4}, E^(4*x)])/1155

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Maple [C]  time = 0.097, size = 56, normalized size = 1. \begin{align*} -{\frac{{{\rm e}^{3\,x}} \left ( 3\,{{\rm e}^{4\,x}}+1 \right ) }{4\, \left ({{\rm e}^{4\,x}}-1 \right ) ^{2}}}-{\frac{5\,\ln \left ({{\rm e}^{x}}+1 \right ) }{16}}+{\frac{5\,i}{16}}\ln \left ({{\rm e}^{x}}+i \right ) -{\frac{5\,i}{16}}\ln \left ({{\rm e}^{x}}-i \right ) +{\frac{5\,\ln \left ({{\rm e}^{x}}-1 \right ) }{16}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*exp(3*x)*(3*exp(4*x)+1)/(exp(4*x)-1)^2-5/16*ln(exp(x)+1)+5/16*I*ln(exp(x)+I)-5/16*I*ln(exp(x)-I)+5/16*ln(
exp(x)-1)

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Maxima [A]  time = 1.53406, size = 63, normalized size = 1.15 \begin{align*} -\frac{3 \, e^{\left (7 \, x\right )} + e^{\left (3 \, x\right )}}{4 \,{\left (e^{\left (8 \, x\right )} - 2 \, e^{\left (4 \, x\right )} + 1\right )}} + \frac{5}{8} \, \arctan \left (e^{x}\right ) - \frac{5}{16} \, \log \left (e^{x} + 1\right ) + \frac{5}{16} \, \log \left (e^{x} - 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*e^(7*x) + e^(3*x))/(e^(8*x) - 2*e^(4*x) + 1) + 5/8*arctan(e^x) - 5/16*log(e^x + 1) + 5/16*log(e^x - 1)

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Fricas [B]  time = 1.89445, size = 1868, normalized size = 33.96 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(12*cosh(x)^7 + 420*cosh(x)^3*sinh(x)^4 + 252*cosh(x)^2*sinh(x)^5 + 84*cosh(x)*sinh(x)^6 + 12*sinh(x)^7
+ 4*(105*cosh(x)^4 + 1)*sinh(x)^3 + 4*cosh(x)^3 + 12*(21*cosh(x)^5 + cosh(x))*sinh(x)^2 - 10*(cosh(x)^8 + 56*c
osh(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)*arctan(cosh(x) + sinh(x)) + 5*(cosh(x)^8 + 56*cosh(x)^3*sinh(x)^5 + 28*cosh(x)^2*sin
h(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 - cos
h(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) - 5*(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) + 12*(7*cosh(x)^6 +
cosh(x)^2)*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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.15557, size = 57, normalized size = 1.04 \begin{align*} -\frac{3 \, e^{\left (7 \, x\right )} + e^{\left (3 \, x\right )}}{4 \,{\left (e^{\left (4 \, x\right )} - 1\right )}^{2}} + \frac{5}{8} \, \arctan \left (e^{x}\right ) - \frac{5}{16} \, \log \left (e^{x} + 1\right ) + \frac{5}{16} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*e^(7*x) + e^(3*x))/(e^(4*x) - 1)^2 + 5/8*arctan(e^x) - 5/16*log(e^x + 1) + 5/16*log(abs(e^x - 1))