3.1.8 \(\int \frac {1}{(b \coth (c+d x))^{7/2}} \, dx\) [8]
Optimal. Leaf size=100 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {2}{5 b d (b \coth (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \coth (c+d x)}} \]
[Out]
-arctan((b*coth(d*x+c))^(1/2)/b^(1/2))/b^(7/2)/d+arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))/b^(7/2)/d-2/5/b/d/(b*c
oth(d*x+c))^(5/2)-2/b^3/d/(b*coth(d*x+c))^(1/2)
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Rubi [A]
time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps
used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3555, 3557,
335, 304, 209, 212} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {2}{b^3 d \sqrt {b \coth (c+d x)}}-\frac {2}{5 b d (b \coth (c+d x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(b*Coth[c + d*x])^(-7/2),x]
[Out]
-(ArcTan[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(7/2)*d)) + ArcTanh[Sqrt[b*Coth[c + d*x]]/Sqrt[b]]/(b^(7/2)*d) - 2/
(5*b*d*(b*Coth[c + d*x])^(5/2)) - 2/(b^3*d*Sqrt[b*Coth[c + d*x]])
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 212
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] && (GtQ[a, 0] || LtQ[b, 0])
Rule 304
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 335
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 3555
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x])^(n + 1)/(b*d*(n + 1)), x] - Dist[
1/b^2, Int[(b*Tan[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
Rule 3557
Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Rubi steps
\begin {align*} \int \frac {1}{(b \coth (c+d x))^{7/2}} \, dx &=-\frac {2}{5 b d (b \coth (c+d x))^{5/2}}+\frac {\int \frac {1}{(b \coth (c+d x))^{3/2}} \, dx}{b^2}\\ &=-\frac {2}{5 b d (b \coth (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \coth (c+d x)}}+\frac {\int \sqrt {b \coth (c+d x)} \, dx}{b^4}\\ &=-\frac {2}{5 b d (b \coth (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \coth (c+d x)}}-\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{-b^2+x^2} \, dx,x,b \coth (c+d x)\right )}{b^3 d}\\ &=-\frac {2}{5 b d (b \coth (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \coth (c+d x)}}-\frac {2 \text {Subst}\left (\int \frac {x^2}{-b^2+x^4} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b^3 d}\\ &=-\frac {2}{5 b d (b \coth (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \coth (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b^3 d}-\frac {\text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \coth (c+d x)}\right )}{b^3 d}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {b \coth (c+d x)}}{\sqrt {b}}\right )}{b^{7/2} d}-\frac {2}{5 b d (b \coth (c+d x))^{5/2}}-\frac {2}{b^3 d \sqrt {b \coth (c+d x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in
optimal.
time = 0.07, size = 38, normalized size = 0.38 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};\coth ^2(c+d x)\right )}{5 b d (b \coth (c+d x))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(b*Coth[c + d*x])^(-7/2),x]
[Out]
(-2*Hypergeometric2F1[-5/4, 1, -1/4, Coth[c + d*x]^2])/(5*b*d*(b*Coth[c + d*x])^(5/2))
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Maple [A]
time = 2.04, size = 77, normalized size = 0.77
| | |
| method |
result |
size |
| | |
| derivativedivides |
\(-\frac {2 b \left (-\frac {\arctanh \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2 b^{\frac {9}{2}}}+\frac {\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2 b^{\frac {9}{2}}}+\frac {1}{b^{4} \sqrt {b \coth \left (d x +c \right )}}+\frac {1}{5 b^{2} \left (b \coth \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}\) |
\(77\) |
| default |
\(-\frac {2 b \left (-\frac {\arctanh \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2 b^{\frac {9}{2}}}+\frac {\arctan \left (\frac {\sqrt {b \coth \left (d x +c \right )}}{\sqrt {b}}\right )}{2 b^{\frac {9}{2}}}+\frac {1}{b^{4} \sqrt {b \coth \left (d x +c \right )}}+\frac {1}{5 b^{2} \left (b \coth \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}\) |
\(77\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(b*coth(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
[Out]
-2/d*b*(-1/2/b^(9/2)*arctanh((b*coth(d*x+c))^(1/2)/b^(1/2))+1/2/b^(9/2)*arctan((b*coth(d*x+c))^(1/2)/b^(1/2))+
1/b^4/(b*coth(d*x+c))^(1/2)+1/5/b^2/(b*coth(d*x+c))^(5/2))
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c))^(7/2),x, algorithm="maxima")
[Out]
integrate((b*coth(d*x + c))^(-7/2), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1041 vs.
\(2 (82) = 164\).
time = 0.51, size = 2132, normalized size = 21.32 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c))^(7/2),x, algorithm="fricas")
[Out]
[-1/20*(10*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + 3*(5*cosh(d*x + c)^2 + 1)*si
nh(d*x + c)^4 + 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 + 3*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*cosh(d*x +
c)^4 + 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c)^2 + 6*(cosh(d*x + c)^5 + 2*cosh(d*x + c)^3 + c
osh(d*x + c))*sinh(d*x + c) + 1)*sqrt(-b)*arctan((cosh(d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x +
c)^2)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*s
inh(d*x + c)^2 + b)) + 5*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + 3*(5*cosh(d*x
+ c)^2 + 1)*sinh(d*x + c)^4 + 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 + 3*cosh(d*x + c))*sinh(d*x + c)^3 + 3*
(5*cosh(d*x + c)^4 + 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c)^2 + 6*(cosh(d*x + c)^5 + 2*cosh(
d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(-b)*log(-(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)^3*sinh(d*x
+ c) + 6*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 - 2*(cosh(
d*x + c)^2 + 2*cosh(d*x + c)*sinh(d*x + c) + sinh(d*x + c)^2 - 1)*sqrt(-b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))
- 2*b)/(cosh(d*x + c)^4 + 4*cosh(d*x + c)^3*sinh(d*x + c) + 6*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*cosh(d*x +
c)*sinh(d*x + c)^3 + sinh(d*x + c)^4)) + 16*(3*cosh(d*x + c)^6 + 18*cosh(d*x + c)*sinh(d*x + c)^5 + 3*sinh(d*x
+ c)^6 + (45*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^4 + cosh(d*x + c)^4 + 4*(15*cosh(d*x + c)^3 + cosh(d*x + c))*
sinh(d*x + c)^3 + (45*cosh(d*x + c)^4 + 6*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(9*cosh(d
*x + c)^5 + 2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) - 3)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/(b^4*d*
cosh(d*x + c)^6 + 6*b^4*d*cosh(d*x + c)*sinh(d*x + c)^5 + b^4*d*sinh(d*x + c)^6 + 3*b^4*d*cosh(d*x + c)^4 + 3*
b^4*d*cosh(d*x + c)^2 + b^4*d + 3*(5*b^4*d*cosh(d*x + c)^2 + b^4*d)*sinh(d*x + c)^4 + 4*(5*b^4*d*cosh(d*x + c)
^3 + 3*b^4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^4*d*cosh(d*x + c)^4 + 6*b^4*d*cosh(d*x + c)^2 + b^4*d)*si
nh(d*x + c)^2 + 6*(b^4*d*cosh(d*x + c)^5 + 2*b^4*d*cosh(d*x + c)^3 + b^4*d*cosh(d*x + c))*sinh(d*x + c)), -1/2
0*(10*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5 + sinh(d*x + c)^6 + 3*(5*cosh(d*x + c)^2 + 1)*sinh(d*
x + c)^4 + 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 + 3*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*cosh(d*x + c)^4
+ 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x + c)^2 + 6*(cosh(d*x + c)^5 + 2*cosh(d*x + c)^3 + cosh(d
*x + c))*sinh(d*x + c) + 1)*sqrt(b)*arctan(sqrt(b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c))/(b*cosh(d*x + c)^2 + 2*
b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + b)) - 5*(cosh(d*x + c)^6 + 6*cosh(d*x + c)*sinh(d*x + c)^5
+ sinh(d*x + c)^6 + 3*(5*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^4 + 3*cosh(d*x + c)^4 + 4*(5*cosh(d*x + c)^3 + 3*
cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*cosh(d*x + c)^4 + 6*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^2 + 3*cosh(d*x +
c)^2 + 6*(cosh(d*x + c)^5 + 2*cosh(d*x + c)^3 + cosh(d*x + c))*sinh(d*x + c) + 1)*sqrt(b)*log(2*b*cosh(d*x + c
)^4 + 8*b*cosh(d*x + c)^3*sinh(d*x + c) + 12*b*cosh(d*x + c)^2*sinh(d*x + c)^2 + 8*b*cosh(d*x + c)*sinh(d*x +
c)^3 + 2*b*sinh(d*x + c)^4 + 2*(cosh(d*x + c)^4 + 4*cosh(d*x + c)*sinh(d*x + c)^3 + sinh(d*x + c)^4 + (6*cosh(
d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c))*sqrt(
b)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)) - b) + 16*(3*cosh(d*x + c)^6 + 18*cosh(d*x + c)*sinh(d*x + c)^5 + 3*sin
h(d*x + c)^6 + (45*cosh(d*x + c)^2 + 1)*sinh(d*x + c)^4 + cosh(d*x + c)^4 + 4*(15*cosh(d*x + c)^3 + cosh(d*x +
c))*sinh(d*x + c)^3 + (45*cosh(d*x + c)^4 + 6*cosh(d*x + c)^2 - 1)*sinh(d*x + c)^2 - cosh(d*x + c)^2 + 2*(9*c
osh(d*x + c)^5 + 2*cosh(d*x + c)^3 - cosh(d*x + c))*sinh(d*x + c) - 3)*sqrt(b*cosh(d*x + c)/sinh(d*x + c)))/(b
^4*d*cosh(d*x + c)^6 + 6*b^4*d*cosh(d*x + c)*sinh(d*x + c)^5 + b^4*d*sinh(d*x + c)^6 + 3*b^4*d*cosh(d*x + c)^4
+ 3*b^4*d*cosh(d*x + c)^2 + b^4*d + 3*(5*b^4*d*cosh(d*x + c)^2 + b^4*d)*sinh(d*x + c)^4 + 4*(5*b^4*d*cosh(d*x
+ c)^3 + 3*b^4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(5*b^4*d*cosh(d*x + c)^4 + 6*b^4*d*cosh(d*x + c)^2 + b^4*
d)*sinh(d*x + c)^2 + 6*(b^4*d*cosh(d*x + c)^5 + 2*b^4*d*cosh(d*x + c)^3 + b^4*d*cosh(d*x + c))*sinh(d*x + c))]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (b \coth {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c))**(7/2),x)
[Out]
Integral((b*coth(c + d*x))**(-7/2), x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 333 vs.
\(2 (82) = 164\).
time = 0.55, size = 333, normalized size = 3.33 \begin {gather*} \frac {\frac {10 \, \arctan \left (-\frac {\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}}{\sqrt {b}}\right )}{b^{\frac {7}{2}} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} - \frac {5 \, \log \left ({\left | -\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} + \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} \right |}\right )}{b^{\frac {7}{2}} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} - \frac {16 \, {\left (5 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{4} + 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{3} \sqrt {b} + 20 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )}^{2} b + 10 \, {\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b}\right )} b^{\frac {3}{2}} + 3 \, b^{2}\right )}}{{\left (\sqrt {b} e^{\left (2 \, d x + 2 \, c\right )} - \sqrt {b e^{\left (4 \, d x + 4 \, c\right )} - b} + \sqrt {b}\right )}^{5} b^{3} \mathrm {sgn}\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}}{10 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(b*coth(d*x+c))^(7/2),x, algorithm="giac")
[Out]
1/10*(10*arctan(-(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))/sqrt(b))/(b^(7/2)*sgn(e^(2*d*x + 2*c)
- 1)) - 5*log(abs(-sqrt(b)*e^(2*d*x + 2*c) + sqrt(b*e^(4*d*x + 4*c) - b)))/(b^(7/2)*sgn(e^(2*d*x + 2*c) - 1))
- 16*(5*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^4 + 10*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4
*d*x + 4*c) - b))^3*sqrt(b) + 20*(sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))^2*b + 10*(sqrt(b)*e^(
2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c) - b))*b^(3/2) + 3*b^2)/((sqrt(b)*e^(2*d*x + 2*c) - sqrt(b*e^(4*d*x + 4*c
) - b) + sqrt(b))^5*b^3*sgn(e^(2*d*x + 2*c) - 1)))/d
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Mupad [B]
time = 1.60, size = 80, normalized size = 0.80 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{7/2}\,d}-\frac {\mathrm {atan}\left (\frac {\sqrt {b\,\mathrm {coth}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{7/2}\,d}-\frac {\frac {2}{5\,b}+\frac {2\,{\mathrm {coth}\left (c+d\,x\right )}^2}{b}}{d\,{\left (b\,\mathrm {coth}\left (c+d\,x\right )\right )}^{5/2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(b*coth(c + d*x))^(7/2),x)
[Out]
atanh((b*coth(c + d*x))^(1/2)/b^(1/2))/(b^(7/2)*d) - atan((b*coth(c + d*x))^(1/2)/b^(1/2))/(b^(7/2)*d) - (2/(5
*b) + (2*coth(c + d*x)^2)/b)/(d*(b*coth(c + d*x))^(5/2))
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