### 3.18 $$\int \sinh ^3(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=105 $\frac{b^2 (3 a+4 b) \text{sech}^3(c+d x)}{3 d}+\frac{(a+b)^3 \cosh ^3(c+d x)}{3 d}-\frac{(a+b)^2 (a+4 b) \cosh (c+d x)}{d}-\frac{3 b (a+b) (a+2 b) \text{sech}(c+d x)}{d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d}$

[Out]

-(((a + b)^2*(a + 4*b)*Cosh[c + d*x])/d) + ((a + b)^3*Cosh[c + d*x]^3)/(3*d) - (3*b*(a + b)*(a + 2*b)*Sech[c +
d*x])/d + (b^2*(3*a + 4*b)*Sech[c + d*x]^3)/(3*d) - (b^3*Sech[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.12258, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.087, Rules used = {3664, 448} $\frac{b^2 (3 a+4 b) \text{sech}^3(c+d x)}{3 d}+\frac{(a+b)^3 \cosh ^3(c+d x)}{3 d}-\frac{(a+b)^2 (a+4 b) \cosh (c+d x)}{d}-\frac{3 b (a+b) (a+2 b) \text{sech}(c+d x)}{d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

-(((a + b)^2*(a + 4*b)*Cosh[c + d*x])/d) + ((a + b)^3*Cosh[c + d*x]^3)/(3*d) - (3*b*(a + b)*(a + 2*b)*Sech[c +
d*x])/d + (b^2*(3*a + 4*b)*Sech[c + d*x]^3)/(3*d) - (b^3*Sech[c + d*x]^5)/(5*d)

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 448

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

Rubi steps

\begin{align*} \int \sinh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right ) \left (a+b-b x^2\right )^3}{x^4} \, dx,x,\text{sech}(c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (3 (-a-2 b) b (a+b)-\frac{(a+b)^3}{x^4}+\frac{(a+b)^2 (a+4 b)}{x^2}+b^2 (3 a+4 b) x^2-b^3 x^4\right ) \, dx,x,\text{sech}(c+d x)\right )}{d}\\ &=-\frac{(a+b)^2 (a+4 b) \cosh (c+d x)}{d}+\frac{(a+b)^3 \cosh ^3(c+d x)}{3 d}-\frac{3 b (a+b) (a+2 b) \text{sech}(c+d x)}{d}+\frac{b^2 (3 a+4 b) \text{sech}^3(c+d x)}{3 d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.290715, size = 91, normalized size = 0.87 $\frac{20 b^2 (3 a+4 b) \text{sech}^3(c+d x)-45 (a+b)^2 (a+5 b) \cosh (c+d x)+5 (a+b)^3 \cosh (3 (c+d x))-180 b (a+b) (a+2 b) \text{sech}(c+d x)-12 b^3 \text{sech}^5(c+d x)}{60 d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(-45*(a + b)^2*(a + 5*b)*Cosh[c + d*x] + 5*(a + b)^3*Cosh[3*(c + d*x)] - 180*b*(a + b)*(a + 2*b)*Sech[c + d*x]
+ 20*b^2*(3*a + 4*b)*Sech[c + d*x]^3 - 12*b^3*Sech[c + d*x]^5)/(60*d)

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Maple [B]  time = 0.052, size = 287, normalized size = 2.7 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( dx+c \right ) +3\,{a}^{2}b \left ( 1/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{\cosh \left ( dx+c \right ) }}+4/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{\cosh \left ( dx+c \right ) }}-8/3\,\cosh \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( 1/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-8/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+16/3\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{\cosh \left ( dx+c \right ) }}-16/3\,\cosh \left ( dx+c \right ) \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{8}}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{8\, \left ( \sinh \left ( dx+c \right ) \right ) ^{6}}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-16\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{64\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{5\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{128\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{128\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15\,\cosh \left ( dx+c \right ) }}-{\frac{128\,\cosh \left ( dx+c \right ) }{15}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(-2/3+1/3*sinh(d*x+c)^2)*cosh(d*x+c)+3*a^2*b*(1/3*sinh(d*x+c)^4/cosh(d*x+c)+4/3*sinh(d*x+c)^2/cosh(d*
x+c)-8/3*cosh(d*x+c))+3*a*b^2*(1/3*sinh(d*x+c)^6/cosh(d*x+c)^3-2*sinh(d*x+c)^4/cosh(d*x+c)^3-8/3*sinh(d*x+c)^2
/cosh(d*x+c)^3+16/3*sinh(d*x+c)^2/cosh(d*x+c)-16/3*cosh(d*x+c))+b^3*(1/3*sinh(d*x+c)^8/cosh(d*x+c)^5-8/3*sinh(
d*x+c)^6/cosh(d*x+c)^5-16*sinh(d*x+c)^4/cosh(d*x+c)^5-64/5*sinh(d*x+c)^2/cosh(d*x+c)^5+128/15*sinh(d*x+c)^2/co
sh(d*x+c)^3+128/15*sinh(d*x+c)^2/cosh(d*x+c)-128/15*cosh(d*x+c)))

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Maxima [B]  time = 1.10065, size = 593, normalized size = 5.65 \begin{align*} -\frac{1}{120} \, b^{3}{\left (\frac{5 \,{\left (45 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{d} + \frac{200 \, e^{\left (-2 \, d x - 2 \, c\right )} + 2515 \, e^{\left (-4 \, d x - 4 \, c\right )} + 6680 \, e^{\left (-6 \, d x - 6 \, c\right )} + 9073 \, e^{\left (-8 \, d x - 8 \, c\right )} + 5600 \, e^{\left (-10 \, d x - 10 \, c\right )} + 1665 \, e^{\left (-12 \, d x - 12 \, c\right )} - 5}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + 5 \, e^{\left (-5 \, d x - 5 \, c\right )} + 10 \, e^{\left (-7 \, d x - 7 \, c\right )} + 10 \, e^{\left (-9 \, d x - 9 \, c\right )} + 5 \, e^{\left (-11 \, d x - 11 \, c\right )} + e^{\left (-13 \, d x - 13 \, c\right )}\right )}}\right )} - \frac{1}{8} \, a b^{2}{\left (\frac{33 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{30 \, e^{\left (-2 \, d x - 2 \, c\right )} + 240 \, e^{\left (-4 \, d x - 4 \, c\right )} + 322 \, e^{\left (-6 \, d x - 6 \, c\right )} + 177 \, e^{\left (-8 \, d x - 8 \, c\right )} - 1}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} + 3 \, e^{\left (-7 \, d x - 7 \, c\right )} + e^{\left (-9 \, d x - 9 \, c\right )}\right )}}\right )} - \frac{1}{8} \, a^{2} b{\left (\frac{21 \, e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{20 \, e^{\left (-2 \, d x - 2 \, c\right )} + 69 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1}{d{\left (e^{\left (-3 \, d x - 3 \, c\right )} + e^{\left (-5 \, d x - 5 \, c\right )}\right )}}\right )} + \frac{1}{24} \, a^{3}{\left (\frac{e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac{9 \, e^{\left (d x + c\right )}}{d} - \frac{9 \, e^{\left (-d x - c\right )}}{d} + \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

-1/120*b^3*(5*(45*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + (200*e^(-2*d*x - 2*c) + 2515*e^(-4*d*x - 4*c) + 6680*e^
(-6*d*x - 6*c) + 9073*e^(-8*d*x - 8*c) + 5600*e^(-10*d*x - 10*c) + 1665*e^(-12*d*x - 12*c) - 5)/(d*(e^(-3*d*x
- 3*c) + 5*e^(-5*d*x - 5*c) + 10*e^(-7*d*x - 7*c) + 10*e^(-9*d*x - 9*c) + 5*e^(-11*d*x - 11*c) + e^(-13*d*x -
13*c)))) - 1/8*a*b^2*((33*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + (30*e^(-2*d*x - 2*c) + 240*e^(-4*d*x - 4*c) + 3
22*e^(-6*d*x - 6*c) + 177*e^(-8*d*x - 8*c) - 1)/(d*(e^(-3*d*x - 3*c) + 3*e^(-5*d*x - 5*c) + 3*e^(-7*d*x - 7*c)
+ e^(-9*d*x - 9*c)))) - 1/8*a^2*b*((21*e^(-d*x - c) - e^(-3*d*x - 3*c))/d + (20*e^(-2*d*x - 2*c) + 69*e^(-4*d
*x - 4*c) - 1)/(d*(e^(-3*d*x - 3*c) + e^(-5*d*x - 5*c)))) + 1/24*a^3*(e^(3*d*x + 3*c)/d - 9*e^(d*x + c)/d - 9*
e^(-d*x - c)/d + e^(-3*d*x - 3*c)/d)

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Fricas [B]  time = 1.96694, size = 1388, normalized size = 13.22 \begin{align*} \frac{5 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{8} + 5 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{8} - 20 \,{\left (a^{3} + 12 \, a^{2} b + 21 \, a b^{2} + 10 \, b^{3}\right )} \cosh \left (d x + c\right )^{6} - 20 \,{\left (a^{3} + 12 \, a^{2} b + 21 \, a b^{2} + 10 \, b^{3} - 7 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{6} - 20 \,{\left (11 \, a^{3} + 123 \, a^{2} b + 249 \, a b^{2} + 137 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 10 \,{\left (35 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} - 22 \, a^{3} - 246 \, a^{2} b - 498 \, a b^{2} - 274 \, b^{3} - 30 \,{\left (a^{3} + 12 \, a^{2} b + 21 \, a b^{2} + 10 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{4} - 425 \, a^{3} - 5235 \, a^{2} b - 10395 \, a b^{2} - 5649 \, b^{3} - 20 \,{\left (31 \, a^{3} + 372 \, a^{2} b + 747 \, a b^{2} + 390 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 20 \,{\left (7 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{6} - 15 \,{\left (a^{3} + 12 \, a^{2} b + 21 \, a b^{2} + 10 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} - 31 \, a^{3} - 372 \, a^{2} b - 747 \, a b^{2} - 390 \, b^{3} - 6 \,{\left (11 \, a^{3} + 123 \, a^{2} b + 249 \, a b^{2} + 137 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2}}{120 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/120*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^8 + 5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^8 -
20*(a^3 + 12*a^2*b + 21*a*b^2 + 10*b^3)*cosh(d*x + c)^6 - 20*(a^3 + 12*a^2*b + 21*a*b^2 + 10*b^3 - 7*(a^3 + 3
*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 - 20*(11*a^3 + 123*a^2*b + 249*a*b^2 + 137*b^3)*cosh(
d*x + c)^4 + 10*(35*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 - 22*a^3 - 246*a^2*b - 498*a*b^2 - 274*b^3
- 30*(a^3 + 12*a^2*b + 21*a*b^2 + 10*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 - 425*a^3 - 5235*a^2*b - 10395*a*b
^2 - 5649*b^3 - 20*(31*a^3 + 372*a^2*b + 747*a*b^2 + 390*b^3)*cosh(d*x + c)^2 + 20*(7*(a^3 + 3*a^2*b + 3*a*b^2
+ b^3)*cosh(d*x + c)^6 - 15*(a^3 + 12*a^2*b + 21*a*b^2 + 10*b^3)*cosh(d*x + c)^4 - 31*a^3 - 372*a^2*b - 747*a
*b^2 - 390*b^3 - 6*(11*a^3 + 123*a^2*b + 249*a*b^2 + 137*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2)/(d*cosh(d*x +
c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*s
inh(d*x + c)^2 + 10*d*cosh(d*x + c))

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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(sinh(d*x+c)**3*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 2.43728, size = 597, normalized size = 5.69 \begin{align*} -\frac{5 \,{\left (9 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 63 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 99 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} e^{\left (-3 \, d x - 3 \, c\right )} - 5 \,{\left (a^{3} e^{\left (3 \, d x + 48 \, c\right )} + 3 \, a^{2} b e^{\left (3 \, d x + 48 \, c\right )} + 3 \, a b^{2} e^{\left (3 \, d x + 48 \, c\right )} + b^{3} e^{\left (3 \, d x + 48 \, c\right )} - 9 \, a^{3} e^{\left (d x + 46 \, c\right )} - 63 \, a^{2} b e^{\left (d x + 46 \, c\right )} - 99 \, a b^{2} e^{\left (d x + 46 \, c\right )} - 45 \, b^{3} e^{\left (d x + 46 \, c\right )}\right )} e^{\left (-45 \, c\right )} + \frac{16 \,{\left (45 \, a^{2} b e^{\left (9 \, d x + 9 \, c\right )} + 135 \, a b^{2} e^{\left (9 \, d x + 9 \, c\right )} + 90 \, b^{3} e^{\left (9 \, d x + 9 \, c\right )} + 180 \, a^{2} b e^{\left (7 \, d x + 7 \, c\right )} + 480 \, a b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 280 \, b^{3} e^{\left (7 \, d x + 7 \, c\right )} + 270 \, a^{2} b e^{\left (5 \, d x + 5 \, c\right )} + 690 \, a b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 428 \, b^{3} e^{\left (5 \, d x + 5 \, c\right )} + 180 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} + 480 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 280 \, b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 45 \, a^{2} b e^{\left (d x + c\right )} + 135 \, a b^{2} e^{\left (d x + c\right )} + 90 \, b^{3} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{120 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

-1/120*(5*(9*a^3*e^(2*d*x + 2*c) + 63*a^2*b*e^(2*d*x + 2*c) + 99*a*b^2*e^(2*d*x + 2*c) + 45*b^3*e^(2*d*x + 2*c
) - a^3 - 3*a^2*b - 3*a*b^2 - b^3)*e^(-3*d*x - 3*c) - 5*(a^3*e^(3*d*x + 48*c) + 3*a^2*b*e^(3*d*x + 48*c) + 3*a
*b^2*e^(3*d*x + 48*c) + b^3*e^(3*d*x + 48*c) - 9*a^3*e^(d*x + 46*c) - 63*a^2*b*e^(d*x + 46*c) - 99*a*b^2*e^(d*
x + 46*c) - 45*b^3*e^(d*x + 46*c))*e^(-45*c) + 16*(45*a^2*b*e^(9*d*x + 9*c) + 135*a*b^2*e^(9*d*x + 9*c) + 90*b
^3*e^(9*d*x + 9*c) + 180*a^2*b*e^(7*d*x + 7*c) + 480*a*b^2*e^(7*d*x + 7*c) + 280*b^3*e^(7*d*x + 7*c) + 270*a^2
*b*e^(5*d*x + 5*c) + 690*a*b^2*e^(5*d*x + 5*c) + 428*b^3*e^(5*d*x + 5*c) + 180*a^2*b*e^(3*d*x + 3*c) + 480*a*b
^2*e^(3*d*x + 3*c) + 280*b^3*e^(3*d*x + 3*c) + 45*a^2*b*e^(d*x + c) + 135*a*b^2*e^(d*x + c) + 90*b^3*e^(d*x +
c))/(e^(2*d*x + 2*c) + 1)^5)/d