### 3.19 $$\int \sinh ^2(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx$$

Optimal. Leaf size=122 $\frac{b^2 (3 a+2 b) \tanh ^3(c+d x)}{3 d}+\frac{3 b (a+b)^2 \tanh (c+d x)}{d}+\frac{(a+b)^3}{4 d (1-\tanh (c+d x))}-\frac{(a+b)^3}{4 d (\tanh (c+d x)+1)}-\frac{1}{2} x (a+b)^2 (a+7 b)+\frac{b^3 \tanh ^5(c+d x)}{5 d}$

[Out]

-((a + b)^2*(a + 7*b)*x)/2 + (a + b)^3/(4*d*(1 - Tanh[c + d*x])) + (3*b*(a + b)^2*Tanh[c + d*x])/d + (b^2*(3*a
+ 2*b)*Tanh[c + d*x]^3)/(3*d) + (b^3*Tanh[c + d*x]^5)/(5*d) - (a + b)^3/(4*d*(1 + Tanh[c + d*x]))

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Rubi [A]  time = 0.185146, antiderivative size = 139, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {3663, 467, 528, 388, 206} $\frac{b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{\sinh (c+d x) \cosh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{1}{2} x (a+b)^2 (a+7 b)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a + b)^2*(a + 7*b)*x)/2 + (b*(81*a^2 + 190*a*b + 105*b^2)*Tanh[c + d*x])/(30*d) + (b*(33*a + 35*b)*Tanh[c +
d*x]*(a + b*Tanh[c + d*x]^2))/(30*d) + (7*b*Tanh[c + d*x]*(a + b*Tanh[c + d*x]^2)^2)/(10*d) + (Cosh[c + d*x]*
Sinh[c + d*x]*(a + b*Tanh[c + d*x]^2)^3)/(2*d)

Rule 3663

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

Rule 467

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

Rule 528

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

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^3}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^2 \left (a+7 b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right ) \left (-a (5 a+7 b)-b (33 a+35 b) x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{10 d}\\ &=\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{\operatorname{Subst}\left (\int \frac{a \left (15 a^2+54 a b+35 b^2\right )+b \left (81 a^2+190 a b+105 b^2\right ) x^2}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{30 d}\\ &=\frac{b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}-\frac{\left ((a+b)^2 (a+7 b)\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=-\frac{1}{2} (a+b)^2 (a+7 b) x+\frac{b \left (81 a^2+190 a b+105 b^2\right ) \tanh (c+d x)}{30 d}+\frac{b (33 a+35 b) \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )}{30 d}+\frac{7 b \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^2}{10 d}+\frac{\cosh (c+d x) \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3}{2 d}\\ \end{align*}

Mathematica [A]  time = 2.15865, size = 95, normalized size = 0.78 $\frac{4 b \tanh (c+d x) \left (45 a^2-b (15 a+16 b) \text{sech}^2(c+d x)+105 a b+3 b^2 \text{sech}^4(c+d x)+58 b^2\right )-30 (a+7 b) (a+b)^2 (c+d x)+15 (a+b)^3 \sinh (2 (c+d x))}{60 d}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-30*(a + b)^2*(a + 7*b)*(c + d*x) + 15*(a + b)^3*Sinh[2*(c + d*x)] + 4*b*(45*a^2 + 105*a*b + 58*b^2 - b*(15*a
+ 16*b)*Sech[c + d*x]^2 + 3*b^2*Sech[c + d*x]^4)*Tanh[c + d*x])/(60*d)

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Maple [A]  time = 0.053, size = 180, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) +3\,{a}^{2}b \left ( 1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{\cosh \left ( dx+c \right ) }}-3/2\,dx-3/2\,c+3/2\,\tanh \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( 1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-5/2\,dx-5/2\,c+5/2\,\tanh \left ( dx+c \right ) +5/6\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3} \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{7\,dx}{2}}-{\frac{7\,c}{2}}+{\frac{7\,\tanh \left ( dx+c \right ) }{2}}+{\frac{7\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{6}}+{\frac{7\, \left ( \tanh \left ( dx+c \right ) \right ) ^{5}}{10}} \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+3*a^2*b*(1/2*sinh(d*x+c)^3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*
tanh(d*x+c))+3*a*b^2*(1/2*sinh(d*x+c)^5/cosh(d*x+c)^3-5/2*d*x-5/2*c+5/2*tanh(d*x+c)+5/6*tanh(d*x+c)^3)+b^3*(1/
2*sinh(d*x+c)^7/cosh(d*x+c)^5-7/2*d*x-7/2*c+7/2*tanh(d*x+c)+7/6*tanh(d*x+c)^3+7/10*tanh(d*x+c)^5))

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Maxima [B]  time = 1.1, size = 509, normalized size = 4.17 \begin{align*} -\frac{1}{8} \, a^{3}{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{120} \, b^{3}{\left (\frac{420 \,{\left (d x + c\right )}}{d} + \frac{15 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{1003 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3350 \, e^{\left (-4 \, d x - 4 \, c\right )} + 5590 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3915 \, e^{\left (-8 \, d x - 8 \, c\right )} + 1455 \, e^{\left (-10 \, d x - 10 \, c\right )} + 15}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 5 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 10 \, e^{\left (-8 \, d x - 8 \, c\right )} + 5 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )}\right )}}\right )} - \frac{1}{8} \, a b^{2}{\left (\frac{60 \,{\left (d x + c\right )}}{d} + \frac{3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} - \frac{3}{8} \, a^{2} b{\left (\frac{12 \,{\left (d x + c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^3*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/120*b^3*(420*(d*x + c)/d + 15*e^(-2*d*x - 2*c)/d -
(1003*e^(-2*d*x - 2*c) + 3350*e^(-4*d*x - 4*c) + 5590*e^(-6*d*x - 6*c) + 3915*e^(-8*d*x - 8*c) + 1455*e^(-10*
d*x - 10*c) + 15)/(d*(e^(-2*d*x - 2*c) + 5*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 10*e^(-8*d*x - 8*c) + 5*e^
(-10*d*x - 10*c) + e^(-12*d*x - 12*c)))) - 1/8*a*b^2*(60*(d*x + c)/d + 3*e^(-2*d*x - 2*c)/d - (121*e^(-2*d*x -
2*c) + 201*e^(-4*d*x - 4*c) + 147*e^(-6*d*x - 6*c) + 3)/(d*(e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 3*e^(-6*d
*x - 6*c) + e^(-8*d*x - 8*c)))) - 3/8*a^2*b*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (17*e^(-2*d*x - 2*c) + 1)/(
d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c))))

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Fricas [B]  time = 2.10247, size = 1854, normalized size = 15.2 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^7 - 4*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^
2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c)^5 - 20*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b
^2 + 7*b^3)*d*x)*cosh(d*x + c)*sinh(d*x + c)^4 + (75*a^3 + 585*a^2*b + 1065*a*b^2 + 539*b^3 + 315*(a^3 + 3*a^2
*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^5 - 20*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b
+ 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c)^3 + 5*(105*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 27*a^3 + 2
97*a^2*b + 489*a*b^2 + 203*b^3 + 2*(75*a^3 + 585*a^2*b + 1065*a*b^2 + 539*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^
3 - 20*(2*(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c)^3 + 3*(90
*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^2 - 40*
(90*a^2*b + 210*a*b^2 + 116*b^3 + 15*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x)*cosh(d*x + c) + 5*(21*(a^3 + 3*a^
2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^6 + (75*a^3 + 585*a^2*b + 1065*a*b^2 + 539*b^3)*cosh(d*x + c)^4 + 15*a^3 +
189*a^2*b + 285*a*b^2 + 175*b^3 + 3*(27*a^3 + 297*a^2*b + 489*a*b^2 + 203*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))
/(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*c
osh(d*x + c))*sinh(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)**2*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 2.1757, size = 533, normalized size = 4.37 \begin{align*} -\frac{60 \,{\left (a^{3} + 9 \, a^{2} b + 15 \, a b^{2} + 7 \, b^{3}\right )} d x - 15 \,{\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 18 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 14 \, 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 (-2 \, d x - 2 \, c\right )} - 15 \,{\left (a^{3} e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, d x + 16 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 16 \, c\right )} + b^{3} e^{\left (2 \, d x + 16 \, c\right )}\right )} e^{\left (-14 \, c\right )} + \frac{16 \,{\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 135 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 450 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 240 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 600 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 340 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 390 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 200 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 105 \, a b^{2} + 58 \, b^{3}\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)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

-1/120*(60*(a^3 + 9*a^2*b + 15*a*b^2 + 7*b^3)*d*x - 15*(2*a^3*e^(2*d*x + 2*c) + 18*a^2*b*e^(2*d*x + 2*c) + 30*
a*b^2*e^(2*d*x + 2*c) + 14*b^3*e^(2*d*x + 2*c) - a^3 - 3*a^2*b - 3*a*b^2 - b^3)*e^(-2*d*x - 2*c) - 15*(a^3*e^(
2*d*x + 16*c) + 3*a^2*b*e^(2*d*x + 16*c) + 3*a*b^2*e^(2*d*x + 16*c) + b^3*e^(2*d*x + 16*c))*e^(-14*c) + 16*(45
*a^2*b*e^(8*d*x + 8*c) + 135*a*b^2*e^(8*d*x + 8*c) + 90*b^3*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x + 6*c) + 450*
a*b^2*e^(6*d*x + 6*c) + 240*b^3*e^(6*d*x + 6*c) + 270*a^2*b*e^(4*d*x + 4*c) + 600*a*b^2*e^(4*d*x + 4*c) + 340*
b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) + 390*a*b^2*e^(2*d*x + 2*c) + 200*b^3*e^(2*d*x + 2*c) + 45*a^2
*b + 105*a*b^2 + 58*b^3)/(e^(2*d*x + 2*c) + 1)^5)/d