3.944 \(\int e^x \coth ^2(2 x) \text {csch}^2(2 x) \, dx\)
Optimal. Leaf size=75 \[ \frac {3 e^x}{8 \left (1-e^{4 x}\right )}-\frac {5 e^{5 x}}{6 \left (1-e^{4 x}\right )^2}+\frac {4 e^{5 x}}{3 \left (1-e^{4 x}\right )^3}-\frac {3}{16} \tan ^{-1}\left (e^x\right )-\frac {3}{16} \tanh ^{-1}\left (e^x\right ) \]
[Out]
4/3*exp(5*x)/(1-exp(4*x))^3-5/6*exp(5*x)/(1-exp(4*x))^2+3/8*exp(x)/(1-exp(4*x))-3/16*arctan(exp(x))-3/16*arcta
nh(exp(x))
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Rubi [A] time = 0.07, antiderivative size = 75, normalized size of antiderivative = 1.00,
number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used =
{2282, 12, 463, 457, 288, 212, 206, 203} \[ \frac {3 e^x}{8 \left (1-e^{4 x}\right )}-\frac {5 e^{5 x}}{6 \left (1-e^{4 x}\right )^2}+\frac {4 e^{5 x}}{3 \left (1-e^{4 x}\right )^3}-\frac {3}{16} \tan ^{-1}\left (e^x\right )-\frac {3}{16} \tanh ^{-1}\left (e^x\right ) \]
Antiderivative was successfully verified.
[In]
Int[E^x*Coth[2*x]^2*Csch[2*x]^2,x]
[Out]
(4*E^(5*x))/(3*(1 - E^(4*x))^3) - (5*E^(5*x))/(6*(1 - E^(4*x))^2) + (3*E^x)/(8*(1 - E^(4*x))) - (3*ArcTan[E^x]
)/16 - (3*ArcTanh[E^x])/16
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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 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 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) \text {csch}^2(2 x) \, dx &=\operatorname {Subst}\left (\int \frac {4 x^4 \left (1+x^4\right )^2}{\left (1-x^4\right )^4} \, dx,x,e^x\right )\\ &=4 \operatorname {Subst}\left (\int \frac {x^4 \left (1+x^4\right )^2}{\left (1-x^4\right )^4} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1-e^{4 x}\right )^3}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^4 \left (8+12 x^4\right )}{\left (1-x^4\right )^3} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1-e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1-e^{4 x}\right )^2}+\frac {3}{2} \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^4\right )^2} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1-e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1-e^{4 x}\right )^2}+\frac {3 e^x}{8 \left (1-e^{4 x}\right )}-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1-e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1-e^{4 x}\right )^2}+\frac {3 e^x}{8 \left (1-e^{4 x}\right )}-\frac {3}{16} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^x\right )-\frac {3}{16} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^x\right )\\ &=\frac {4 e^{5 x}}{3 \left (1-e^{4 x}\right )^3}-\frac {5 e^{5 x}}{6 \left (1-e^{4 x}\right )^2}+\frac {3 e^x}{8 \left (1-e^{4 x}\right )}-\frac {3}{16} \tan ^{-1}\left (e^x\right )-\frac {3}{16} \tanh ^{-1}\left (e^x\right )\\ \end {align*}
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Mathematica [C] time = 5.54, size = 310, normalized size = 4.13 \[ \frac {e^{-7 x} \left (1280 e^{16 x} \left (1346 e^{4 x}+557 e^{8 x}+821\right ) \, _4F_3\left (2,2,2,\frac {9}{4};1,1,\frac {21}{4};e^{4 x}\right )+10240 e^{16 x} \left (42 e^{4 x}+19 e^{8 x}+23\right ) \, _5F_4\left (2,2,2,2,\frac {9}{4};1,1,1,\frac {21}{4};e^{4 x}\right )+20480 e^{16 x} \, _6F_5\left (2,2,2,2,2,\frac {9}{4};1,1,1,1,\frac {21}{4};e^{4 x}\right )+40960 e^{20 x} \, _6F_5\left (2,2,2,2,2,\frac {9}{4};1,1,1,1,\frac {21}{4};e^{4 x}\right )+20480 e^{24 x} \, _6F_5\left (2,2,2,2,2,\frac {9}{4};1,1,1,1,\frac {21}{4};e^{4 x}\right )+732349800 e^{4 x} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{4 x}\right )-635067810 e^{8 x} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{4 x}\right )-384831720 e^{12 x} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{4 x}\right )+60913125 e^{16 x} \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{4 x}\right )+1070609085 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};e^{4 x}\right )-946471617 e^{4 x}+369641285 e^{8 x}+351173641 e^{12 x}-23818496 e^{16 x}-1070609085\right )}{3818880} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[E^x*Coth[2*x]^2*Csch[2*x]^2,x]
[Out]
(-1070609085 - 946471617*E^(4*x) + 369641285*E^(8*x) + 351173641*E^(12*x) - 23818496*E^(16*x) + 1070609085*Hyp
ergeometric2F1[1/4, 1, 5/4, E^(4*x)] + 732349800*E^(4*x)*Hypergeometric2F1[1/4, 1, 5/4, E^(4*x)] - 635067810*E
^(8*x)*Hypergeometric2F1[1/4, 1, 5/4, E^(4*x)] - 384831720*E^(12*x)*Hypergeometric2F1[1/4, 1, 5/4, E^(4*x)] +
60913125*E^(16*x)*Hypergeometric2F1[1/4, 1, 5/4, E^(4*x)] + 1280*E^(16*x)*(821 + 1346*E^(4*x) + 557*E^(8*x))*H
ypergeometricPFQ[{2, 2, 2, 9/4}, {1, 1, 21/4}, E^(4*x)] + 10240*E^(16*x)*(23 + 42*E^(4*x) + 19*E^(8*x))*Hyperg
eometricPFQ[{2, 2, 2, 2, 9/4}, {1, 1, 1, 21/4}, E^(4*x)] + 20480*E^(16*x)*HypergeometricPFQ[{2, 2, 2, 2, 2, 9/
4}, {1, 1, 1, 1, 21/4}, E^(4*x)] + 40960*E^(20*x)*HypergeometricPFQ[{2, 2, 2, 2, 2, 9/4}, {1, 1, 1, 1, 21/4},
E^(4*x)] + 20480*E^(24*x)*HypergeometricPFQ[{2, 2, 2, 2, 2, 9/4}, {1, 1, 1, 1, 21/4}, E^(4*x)])/(3818880*E^(7*
x))
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fricas [B] time = 0.53, size = 992, normalized size = 13.23 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(x)*coth(2*x)^2*csch(2*x)^2,x, algorithm="fricas")
[Out]
-1/96*(116*cosh(x)^9 + 9744*cosh(x)^3*sinh(x)^6 + 4176*cosh(x)^2*sinh(x)^7 + 1044*cosh(x)*sinh(x)^8 + 116*sinh
(x)^9 + 24*(609*cosh(x)^4 - 1)*sinh(x)^5 - 24*cosh(x)^5 + 24*(609*cosh(x)^5 - 5*cosh(x))*sinh(x)^4 + 48*(203*c
osh(x)^6 - 5*cosh(x)^2)*sinh(x)^3 + 48*(87*cosh(x)^7 - 5*cosh(x)^3)*sinh(x)^2 + 18*(cosh(x)^12 + 220*cosh(x)^3
*sinh(x)^9 + 66*cosh(x)^2*sinh(x)^10 + 12*cosh(x)*sinh(x)^11 + sinh(x)^12 + 3*(165*cosh(x)^4 - 1)*sinh(x)^8 -
3*cosh(x)^8 + 24*(33*cosh(x)^5 - cosh(x))*sinh(x)^7 + 84*(11*cosh(x)^6 - cosh(x)^2)*sinh(x)^6 + 24*(33*cosh(x)
^7 - 7*cosh(x)^3)*sinh(x)^5 + 3*(165*cosh(x)^8 - 70*cosh(x)^4 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(55*cosh(x)^9 -
42*cosh(x)^5 + 3*cosh(x))*sinh(x)^3 + 6*(11*cosh(x)^10 - 14*cosh(x)^6 + 3*cosh(x)^2)*sinh(x)^2 + 12*(cosh(x)^
11 - 2*cosh(x)^7 + cosh(x)^3)*sinh(x) - 1)*arctan(cosh(x) + sinh(x)) + 9*(cosh(x)^12 + 220*cosh(x)^3*sinh(x)^9
+ 66*cosh(x)^2*sinh(x)^10 + 12*cosh(x)*sinh(x)^11 + sinh(x)^12 + 3*(165*cosh(x)^4 - 1)*sinh(x)^8 - 3*cosh(x)^
8 + 24*(33*cosh(x)^5 - cosh(x))*sinh(x)^7 + 84*(11*cosh(x)^6 - cosh(x)^2)*sinh(x)^6 + 24*(33*cosh(x)^7 - 7*cos
h(x)^3)*sinh(x)^5 + 3*(165*cosh(x)^8 - 70*cosh(x)^4 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(55*cosh(x)^9 - 42*cosh(x
)^5 + 3*cosh(x))*sinh(x)^3 + 6*(11*cosh(x)^10 - 14*cosh(x)^6 + 3*cosh(x)^2)*sinh(x)^2 + 12*(cosh(x)^11 - 2*cos
h(x)^7 + cosh(x)^3)*sinh(x) - 1)*log(cosh(x) + sinh(x) + 1) - 9*(cosh(x)^12 + 220*cosh(x)^3*sinh(x)^9 + 66*cos
h(x)^2*sinh(x)^10 + 12*cosh(x)*sinh(x)^11 + sinh(x)^12 + 3*(165*cosh(x)^4 - 1)*sinh(x)^8 - 3*cosh(x)^8 + 24*(3
3*cosh(x)^5 - cosh(x))*sinh(x)^7 + 84*(11*cosh(x)^6 - cosh(x)^2)*sinh(x)^6 + 24*(33*cosh(x)^7 - 7*cosh(x)^3)*s
inh(x)^5 + 3*(165*cosh(x)^8 - 70*cosh(x)^4 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(55*cosh(x)^9 - 42*cosh(x)^5 + 3*c
osh(x))*sinh(x)^3 + 6*(11*cosh(x)^10 - 14*cosh(x)^6 + 3*cosh(x)^2)*sinh(x)^2 + 12*(cosh(x)^11 - 2*cosh(x)^7 +
cosh(x)^3)*sinh(x) - 1)*log(cosh(x) + sinh(x) - 1) + 12*(87*cosh(x)^8 - 10*cosh(x)^4 + 3)*sinh(x) + 36*cosh(x)
)/(cosh(x)^12 + 220*cosh(x)^3*sinh(x)^9 + 66*cosh(x)^2*sinh(x)^10 + 12*cosh(x)*sinh(x)^11 + sinh(x)^12 + 3*(16
5*cosh(x)^4 - 1)*sinh(x)^8 - 3*cosh(x)^8 + 24*(33*cosh(x)^5 - cosh(x))*sinh(x)^7 + 84*(11*cosh(x)^6 - cosh(x)^
2)*sinh(x)^6 + 24*(33*cosh(x)^7 - 7*cosh(x)^3)*sinh(x)^5 + 3*(165*cosh(x)^8 - 70*cosh(x)^4 + 1)*sinh(x)^4 + 3*
cosh(x)^4 + 4*(55*cosh(x)^9 - 42*cosh(x)^5 + 3*cosh(x))*sinh(x)^3 + 6*(11*cosh(x)^10 - 14*cosh(x)^6 + 3*cosh(x
)^2)*sinh(x)^2 + 12*(cosh(x)^11 - 2*cosh(x)^7 + cosh(x)^3)*sinh(x) - 1)
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giac [A] time = 0.12, size = 48, normalized size = 0.64 \[ -\frac {29 \, e^{\left (9 \, x\right )} - 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \, {\left (e^{\left (4 \, x\right )} - 1\right )}^{3}} - \frac {3}{16} \, \arctan \left (e^{x}\right ) - \frac {3}{32} \, \log \left (e^{x} + 1\right ) + \frac {3}{32} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(x)*coth(2*x)^2*csch(2*x)^2,x, algorithm="giac")
[Out]
-1/24*(29*e^(9*x) - 6*e^(5*x) + 9*e^x)/(e^(4*x) - 1)^3 - 3/16*arctan(e^x) - 3/32*log(e^x + 1) + 3/32*log(abs(e
^x - 1))
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maple [C] time = 0.46, size = 60, normalized size = 0.80 \[ -\frac {{\mathrm e}^{x} \left (29 \,{\mathrm e}^{8 x}-6 \,{\mathrm e}^{4 x}+9\right )}{24 \left ({\mathrm e}^{4 x}-1\right )^{3}}+\frac {3 i \ln \left ({\mathrm e}^{x}-i\right )}{32}-\frac {3 i \ln \left ({\mathrm e}^{x}+i\right )}{32}+\frac {3 \ln \left ({\mathrm e}^{x}-1\right )}{32}-\frac {3 \ln \left ({\mathrm e}^{x}+1\right )}{32} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(exp(x)*coth(2*x)^2*csch(2*x)^2,x)
[Out]
-1/24*exp(x)*(29*exp(8*x)-6*exp(4*x)+9)/(exp(4*x)-1)^3+3/32*I*ln(exp(x)-I)-3/32*I*ln(exp(x)+I)+3/32*ln(exp(x)-
1)-3/32*ln(exp(x)+1)
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maxima [A] time = 0.44, size = 59, normalized size = 0.79 \[ -\frac {29 \, e^{\left (9 \, x\right )} - 6 \, e^{\left (5 \, x\right )} + 9 \, e^{x}}{24 \, {\left (e^{\left (12 \, x\right )} - 3 \, e^{\left (8 \, x\right )} + 3 \, e^{\left (4 \, x\right )} - 1\right )}} - \frac {3}{16} \, \arctan \left (e^{x}\right ) - \frac {3}{32} \, \log \left (e^{x} + 1\right ) + \frac {3}{32} \, \log \left (e^{x} - 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(x)*coth(2*x)^2*csch(2*x)^2,x, algorithm="maxima")
[Out]
-1/24*(29*e^(9*x) - 6*e^(5*x) + 9*e^x)/(e^(12*x) - 3*e^(8*x) + 3*e^(4*x) - 1) - 3/16*arctan(e^x) - 3/32*log(e^
x + 1) + 3/32*log(e^x - 1)
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mupad [B] time = 2.18, size = 114, normalized size = 1.52 \[ \frac {3\,\ln \left (\frac {3}{8}-\frac {3\,{\mathrm {e}}^x}{8}\right )}{32}-\frac {3\,\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}-\frac {3}{8}\right )}{32}-\frac {7\,{\mathrm {e}}^x}{8\,\left ({\mathrm {e}}^{4\,x}-1\right )}-\frac {\frac {2\,{\mathrm {e}}^{5\,x}}{3}+\frac {{\mathrm {e}}^{9\,x}}{3}+\frac {{\mathrm {e}}^x}{3}}{3\,{\mathrm {e}}^{4\,x}-3\,{\mathrm {e}}^{8\,x}+{\mathrm {e}}^{12\,x}-1}-\frac {5\,{\mathrm {e}}^x}{6\,\left ({\mathrm {e}}^{8\,x}-2\,{\mathrm {e}}^{4\,x}+1\right )}-\frac {\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}-\frac {3}{8}{}\mathrm {i}\right )\,3{}\mathrm {i}}{32}+\frac {\ln \left (-\frac {3\,{\mathrm {e}}^x}{8}+\frac {3}{8}{}\mathrm {i}\right )\,3{}\mathrm {i}}{32} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((coth(2*x)^2*exp(x))/sinh(2*x)^2,x)
[Out]
(3*log(3/8 - (3*exp(x))/8))/32 - (3*log(- (3*exp(x))/8 - 3/8))/32 - (log(- (3*exp(x))/8 - 3i/8)*3i)/32 + (log(
3i/8 - (3*exp(x))/8)*3i)/32 - (7*exp(x))/(8*(exp(4*x) - 1)) - ((2*exp(5*x))/3 + exp(9*x)/3 + exp(x)/3)/(3*exp(
4*x) - 3*exp(8*x) + exp(12*x) - 1) - (5*exp(x))/(6*(exp(8*x) - 2*exp(4*x) + 1))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{x} \coth ^{2}{\left (2 x \right )} \operatorname {csch}^{2}{\left (2 x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(x)*coth(2*x)**2*csch(2*x)**2,x)
[Out]
Integral(exp(x)*coth(2*x)**2*csch(2*x)**2, x)
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