Integrand size = 12, antiderivative size = 73 \[ \int \frac {1}{(3+5 \cosh (c+d x))^3} \, dx=\frac {43 \arctan \left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{1024 d}+\frac {5 \sinh (c+d x)}{32 d (3+5 \cosh (c+d x))^2}-\frac {45 \sinh (c+d x)}{512 d (3+5 \cosh (c+d x))} \]
43/1024*arctan(1/2*tanh(1/2*d*x+1/2*c))/d+5/32*sinh(d*x+c)/d/(3+5*cosh(d*x +c))^2-45/512*sinh(d*x+c)/d/(3+5*cosh(d*x+c))
Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(3+5 \cosh (c+d x))^3} \, dx=-\frac {43 \arctan \left (2 \coth \left (\frac {1}{2} (c+d x)\right )\right )+\frac {10 (11+45 \cosh (c+d x)) \sinh (c+d x)}{(3+5 \cosh (c+d x))^2}}{1024 d} \]
-1/1024*(43*ArcTan[2*Coth[(c + d*x)/2]] + (10*(11 + 45*Cosh[c + d*x])*Sinh [c + d*x])/(3 + 5*Cosh[c + d*x])^2)/d
Time = 0.39 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3143, 25, 3042, 3233, 27, 3042, 3138, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(5 \cosh (c+d x)+3)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (3+5 \sin \left (i c+i d x+\frac {\pi }{2}\right )\right )^3}dx\) |
\(\Big \downarrow \) 3143 |
\(\displaystyle \frac {1}{32} \int -\frac {6-5 \cosh (c+d x)}{(5 \cosh (c+d x)+3)^2}dx+\frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}-\frac {1}{32} \int \frac {6-5 \cosh (c+d x)}{(5 \cosh (c+d x)+3)^2}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}-\frac {1}{32} \int \frac {6-5 \sin \left (i c+i d x+\frac {\pi }{2}\right )}{\left (5 \sin \left (i c+i d x+\frac {\pi }{2}\right )+3\right )^2}dx\) |
\(\Big \downarrow \) 3233 |
\(\displaystyle \frac {1}{32} \left (-\frac {1}{16} \int -\frac {43}{5 \cosh (c+d x)+3}dx-\frac {45 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}\right )+\frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{32} \left (\frac {43}{16} \int \frac {1}{5 \cosh (c+d x)+3}dx-\frac {45 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}\right )+\frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}+\frac {1}{32} \left (-\frac {45 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}+\frac {43}{16} \int \frac {1}{5 \sin \left (i c+i d x+\frac {\pi }{2}\right )+3}dx\right )\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}+\frac {1}{32} \left (-\frac {45 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}-\frac {43 i \int \frac {1}{2 \tanh ^2\left (\frac {1}{2} (c+d x)\right )+8}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{32} \left (\frac {43 \arctan \left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{32 d}-\frac {45 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}\right )+\frac {5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2}\) |
(5*Sinh[c + d*x])/(32*d*(3 + 5*Cosh[c + d*x])^2) + ((43*ArcTan[Tanh[(c + d *x)/2]/2])/(32*d) - (45*Sinh[c + d*x])/(16*d*(3 + 5*Cosh[c + d*x])))/32
3.1.73.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 - b^2))), x] + Simp [1/((n + 1)*(a^2 - b^2)) Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*( m + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]
Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.85
method | result | size |
derivativedivides | \(\frac {\frac {-\frac {85 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128}-\frac {35 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}}{4 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{2}}+\frac {43 \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1024}}{d}\) | \(62\) |
default | \(\frac {\frac {-\frac {85 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{128}-\frac {35 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{32}}{4 \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+4\right )^{2}}+\frac {43 \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1024}}{d}\) | \(62\) |
risch | \(\frac {215 \,{\mathrm e}^{3 d x +3 c}+387 \,{\mathrm e}^{2 d x +2 c}+325 \,{\mathrm e}^{d x +c}+225}{256 d \left (5 \,{\mathrm e}^{2 d x +2 c}+6 \,{\mathrm e}^{d x +c}+5\right )^{2}}+\frac {43 i \ln \left ({\mathrm e}^{d x +c}+\frac {3}{5}+\frac {4 i}{5}\right )}{2048 d}-\frac {43 i \ln \left ({\mathrm e}^{d x +c}+\frac {3}{5}-\frac {4 i}{5}\right )}{2048 d}\) | \(96\) |
parallelrisch | \(\frac {43 i \left (-43-25 \cosh \left (2 d x +2 c \right )-60 \cosh \left (d x +c \right )\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 i\right )+43 i \left (43+25 \cosh \left (2 d x +2 c \right )+60 \cosh \left (d x +c \right )\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2 i\right )-440 \sinh \left (d x +c \right )-900 \sinh \left (2 d x +2 c \right )}{2048 d \left (43+25 \cosh \left (2 d x +2 c \right )+60 \cosh \left (d x +c \right )\right )}\) | \(123\) |
1/d*(1/4*(-85/128*tanh(1/2*d*x+1/2*c)^3-35/32*tanh(1/2*d*x+1/2*c))/(tanh(1 /2*d*x+1/2*c)^2+4)^2+43/1024*arctan(1/2*tanh(1/2*d*x+1/2*c)))
Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (64) = 128\).
Time = 0.25 (sec) , antiderivative size = 408, normalized size of antiderivative = 5.59 \[ \int \frac {1}{(3+5 \cosh (c+d x))^3} \, dx=\frac {860 \, \cosh \left (d x + c\right )^{3} + 516 \, {\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right )^{2} + 860 \, \sinh \left (d x + c\right )^{3} + 43 \, {\left (25 \, \cosh \left (d x + c\right )^{4} + 20 \, {\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right )^{3} + 25 \, \sinh \left (d x + c\right )^{4} + 60 \, \cosh \left (d x + c\right )^{3} + 2 \, {\left (75 \, \cosh \left (d x + c\right )^{2} + 90 \, \cosh \left (d x + c\right ) + 43\right )} \sinh \left (d x + c\right )^{2} + 86 \, \cosh \left (d x + c\right )^{2} + 4 \, {\left (25 \, \cosh \left (d x + c\right )^{3} + 45 \, \cosh \left (d x + c\right )^{2} + 43 \, \cosh \left (d x + c\right ) + 15\right )} \sinh \left (d x + c\right ) + 60 \, \cosh \left (d x + c\right ) + 25\right )} \arctan \left (\frac {5}{4} \, \cosh \left (d x + c\right ) + \frac {5}{4} \, \sinh \left (d x + c\right ) + \frac {3}{4}\right ) + 1548 \, \cosh \left (d x + c\right )^{2} + 4 \, {\left (645 \, \cosh \left (d x + c\right )^{2} + 774 \, \cosh \left (d x + c\right ) + 325\right )} \sinh \left (d x + c\right ) + 1300 \, \cosh \left (d x + c\right ) + 900}{1024 \, {\left (25 \, d \cosh \left (d x + c\right )^{4} + 25 \, d \sinh \left (d x + c\right )^{4} + 60 \, d \cosh \left (d x + c\right )^{3} + 20 \, {\left (5 \, d \cosh \left (d x + c\right ) + 3 \, d\right )} \sinh \left (d x + c\right )^{3} + 86 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (75 \, d \cosh \left (d x + c\right )^{2} + 90 \, d \cosh \left (d x + c\right ) + 43 \, d\right )} \sinh \left (d x + c\right )^{2} + 60 \, d \cosh \left (d x + c\right ) + 4 \, {\left (25 \, d \cosh \left (d x + c\right )^{3} + 45 \, d \cosh \left (d x + c\right )^{2} + 43 \, d \cosh \left (d x + c\right ) + 15 \, d\right )} \sinh \left (d x + c\right ) + 25 \, d\right )}} \]
1/1024*(860*cosh(d*x + c)^3 + 516*(5*cosh(d*x + c) + 3)*sinh(d*x + c)^2 + 860*sinh(d*x + c)^3 + 43*(25*cosh(d*x + c)^4 + 20*(5*cosh(d*x + c) + 3)*si nh(d*x + c)^3 + 25*sinh(d*x + c)^4 + 60*cosh(d*x + c)^3 + 2*(75*cosh(d*x + c)^2 + 90*cosh(d*x + c) + 43)*sinh(d*x + c)^2 + 86*cosh(d*x + c)^2 + 4*(2 5*cosh(d*x + c)^3 + 45*cosh(d*x + c)^2 + 43*cosh(d*x + c) + 15)*sinh(d*x + c) + 60*cosh(d*x + c) + 25)*arctan(5/4*cosh(d*x + c) + 5/4*sinh(d*x + c) + 3/4) + 1548*cosh(d*x + c)^2 + 4*(645*cosh(d*x + c)^2 + 774*cosh(d*x + c) + 325)*sinh(d*x + c) + 1300*cosh(d*x + c) + 900)/(25*d*cosh(d*x + c)^4 + 25*d*sinh(d*x + c)^4 + 60*d*cosh(d*x + c)^3 + 20*(5*d*cosh(d*x + c) + 3*d) *sinh(d*x + c)^3 + 86*d*cosh(d*x + c)^2 + 2*(75*d*cosh(d*x + c)^2 + 90*d*c osh(d*x + c) + 43*d)*sinh(d*x + c)^2 + 60*d*cosh(d*x + c) + 4*(25*d*cosh(d *x + c)^3 + 45*d*cosh(d*x + c)^2 + 43*d*cosh(d*x + c) + 15*d)*sinh(d*x + c ) + 25*d)
Result contains complex when optimal does not.
Time = 2.04 (sec) , antiderivative size = 507, normalized size of antiderivative = 6.95 \[ \int \frac {1}{(3+5 \cosh (c+d x))^3} \, dx=\begin {cases} - \frac {\log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )}}{125 d \cosh ^{3}{\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 225 d \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 135 d \cosh {\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 27 d} & \text {for}\: c = \log {\left (\left (-3 - 4 i\right ) e^{- d x} \right )} - \log {\left (5 \right )} \\\frac {x}{125 \cosh ^{3}{\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 225 \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 135 \cosh {\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\left (5 \right )} \right )} + 27} & \text {for}\: c = \log {\left (\left (-3 + 4 i\right ) e^{- d x} \right )} - \log {\left (5 \right )} \\\frac {x}{\left (5 \cosh {\left (c \right )} + 3\right )^{3}} & \text {for}\: d = 0 \\\frac {43 \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} - \frac {170 \tanh ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} + \frac {344 \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} - \frac {280 \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} + \frac {688 \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} & \text {otherwise} \end {cases} \]
Piecewise((-log(-3*exp(-d*x) - 4*I*exp(-d*x))/(125*d*cosh(d*x + log(-3*exp (-d*x) - 4*I*exp(-d*x)) - log(5))**3 + 225*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp(-d*x)) - log(5))**2 + 135*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp (-d*x)) - log(5)) + 27*d), Eq(c, log((-3 - 4*I)*exp(-d*x)) - log(5))), (x/ (125*cosh(d*x + log(-3*exp(-d*x) + 4*I*exp(-d*x)) - log(5))**3 + 225*cosh( d*x + log(-3*exp(-d*x) + 4*I*exp(-d*x)) - log(5))**2 + 135*cosh(d*x + log( -3*exp(-d*x) + 4*I*exp(-d*x)) - log(5)) + 27), Eq(c, log((-3 + 4*I)*exp(-d *x)) - log(5))), (x/(5*cosh(c) + 3)**3, Eq(d, 0)), (43*tanh(c/2 + d*x/2)** 4*atan(tanh(c/2 + d*x/2)/2)/(1024*d*tanh(c/2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) - 170*tanh(c/2 + d*x/2)**3/(1024*d*tanh(c/2 + d*x/ 2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) + 344*tanh(c/2 + d*x/2)**2* atan(tanh(c/2 + d*x/2)/2)/(1024*d*tanh(c/2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) - 280*tanh(c/2 + d*x/2)/(1024*d*tanh(c/2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) + 688*atan(tanh(c/2 + d*x/2)/2)/ (1024*d*tanh(c/2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d), Tru e))
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \frac {1}{(3+5 \cosh (c+d x))^3} \, dx=-\frac {43 \, \arctan \left (\frac {5}{4} \, e^{\left (-d x - c\right )} + \frac {3}{4}\right )}{1024 \, d} - \frac {325 \, e^{\left (-d x - c\right )} + 387 \, e^{\left (-2 \, d x - 2 \, c\right )} + 215 \, e^{\left (-3 \, d x - 3 \, c\right )} + 225}{256 \, d {\left (60 \, e^{\left (-d x - c\right )} + 86 \, e^{\left (-2 \, d x - 2 \, c\right )} + 60 \, e^{\left (-3 \, d x - 3 \, c\right )} + 25 \, e^{\left (-4 \, d x - 4 \, c\right )} + 25\right )}} \]
-43/1024*arctan(5/4*e^(-d*x - c) + 3/4)/d - 1/256*(325*e^(-d*x - c) + 387* e^(-2*d*x - 2*c) + 215*e^(-3*d*x - 3*c) + 225)/(d*(60*e^(-d*x - c) + 86*e^ (-2*d*x - 2*c) + 60*e^(-3*d*x - 3*c) + 25*e^(-4*d*x - 4*c) + 25))
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(3+5 \cosh (c+d x))^3} \, dx=\frac {\frac {4 \, {\left (215 \, e^{\left (3 \, d x + 3 \, c\right )} + 387 \, e^{\left (2 \, d x + 2 \, c\right )} + 325 \, e^{\left (d x + c\right )} + 225\right )}}{{\left (5 \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} + 5\right )}^{2}} + 43 \, \arctan \left (\frac {5}{4} \, e^{\left (d x + c\right )} + \frac {3}{4}\right )}{1024 \, d} \]
1/1024*(4*(215*e^(3*d*x + 3*c) + 387*e^(2*d*x + 2*c) + 325*e^(d*x + c) + 2 25)/(5*e^(2*d*x + 2*c) + 6*e^(d*x + c) + 5)^2 + 43*arctan(5/4*e^(d*x + c) + 3/4))/d
Time = 1.73 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(3+5 \cosh (c+d x))^3} \, dx=\frac {\frac {43\,{\mathrm {e}}^{c+d\,x}}{256\,d}+\frac {129}{1280\,d}}{6\,{\mathrm {e}}^{c+d\,x}+5\,{\mathrm {e}}^{2\,c+2\,d\,x}+5}-\frac {\frac {7\,{\mathrm {e}}^{c+d\,x}}{40\,d}-\frac {3}{8\,d}}{60\,{\mathrm {e}}^{c+d\,x}+86\,{\mathrm {e}}^{2\,c+2\,d\,x}+60\,{\mathrm {e}}^{3\,c+3\,d\,x}+25\,{\mathrm {e}}^{4\,c+4\,d\,x}+25}+\frac {43\,\mathrm {atan}\left (\left (\frac {3}{4\,d}+\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4\,d}\right )\,\sqrt {d^2}\right )}{1024\,\sqrt {d^2}} \]
((43*exp(c + d*x))/(256*d) + 129/(1280*d))/(6*exp(c + d*x) + 5*exp(2*c + 2 *d*x) + 5) - ((7*exp(c + d*x))/(40*d) - 3/(8*d))/(60*exp(c + d*x) + 86*exp (2*c + 2*d*x) + 60*exp(3*c + 3*d*x) + 25*exp(4*c + 4*d*x) + 25) + (43*atan ((3/(4*d) + (5*exp(d*x)*exp(c))/(4*d))*(d^2)^(1/2)))/(1024*(d^2)^(1/2))