### 3.25 $$\int \frac{\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx$$

Optimal. Leaf size=118 $\frac{x \left (3 a^2-6 a b-b^2\right )}{8 (a+b)^3}+\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{d (a+b)^3}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 d (a+b)}-\frac{(5 a+b) \sinh (c+d x) \cosh (c+d x)}{8 d (a+b)^2}$

[Out]

((3*a^2 - 6*a*b - b^2)*x)/(8*(a + b)^3) + (a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/((a + b)^3
*d) - ((5*a + b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*(a + b)^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*(a + b)*d)

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Rubi [A]  time = 0.171706, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.261, Rules used = {3663, 470, 527, 522, 206, 205} $\frac{x \left (3 a^2-6 a b-b^2\right )}{8 (a+b)^3}+\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{d (a+b)^3}+\frac{\sinh (c+d x) \cosh ^3(c+d x)}{4 d (a+b)}-\frac{(5 a+b) \sinh (c+d x) \cosh (c+d x)}{8 d (a+b)^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[c + d*x]^4/(a + b*Tanh[c + d*x]^2),x]

[Out]

((3*a^2 - 6*a*b - b^2)*x)/(8*(a + b)^3) + (a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/((a + b)^3
*d) - ((5*a + b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*(a + b)^2*d) + (Cosh[c + d*x]^3*Sinh[c + d*x])/(4*(a + b)*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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 522

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

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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\sinh ^4(c+d x)}{a+b \tanh ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^3 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}-\frac{\operatorname{Subst}\left (\int \frac{a+(4 a+b) x^2}{\left (1-x^2\right )^2 \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{4 (a+b) d}\\ &=-\frac{(5 a+b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}-\frac{\operatorname{Subst}\left (\int \frac{-a (3 a-b)+b (5 a+b) x^2}{\left (1-x^2\right ) \left (a+b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 (a+b)^2 d}\\ &=-\frac{(5 a+b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}+\frac{\left (a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tanh (c+d x)\right )}{(a+b)^3 d}+\frac{\left (3 a^2-6 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 (a+b)^3 d}\\ &=\frac{\left (3 a^2-6 a b-b^2\right ) x}{8 (a+b)^3}+\frac{a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )}{(a+b)^3 d}-\frac{(5 a+b) \cosh (c+d x) \sinh (c+d x)}{8 (a+b)^2 d}+\frac{\cosh ^3(c+d x) \sinh (c+d x)}{4 (a+b) d}\\ \end{align*}

Mathematica [A]  time = 0.253554, size = 93, normalized size = 0.79 $\frac{4 \left (3 a^2-6 a b-b^2\right ) (c+d x)+32 a^{3/2} \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a}}\right )+(a+b)^2 \sinh (4 (c+d x))-8 a (a+b) \sinh (2 (c+d x))}{32 d (a+b)^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[c + d*x]^4/(a + b*Tanh[c + d*x]^2),x]

[Out]

(4*(3*a^2 - 6*a*b - b^2)*(c + d*x) + 32*a^(3/2)*Sqrt[b]*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]] - 8*a*(a + b)*
Sinh[2*(c + d*x)] + (a + b)^2*Sinh[4*(c + d*x)])/(32*(a + b)^3*d)

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Maple [B]  time = 0.108, size = 865, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^2),x)

[Out]

-8/d/(32*a+32*b)/(tanh(1/2*d*x+1/2*c)+1)^4+32/d/(64*a+64*b)/(tanh(1/2*d*x+1/2*c)+1)^3+1/8/d/(a+b)^2/(tanh(1/2*
d*x+1/2*c)+1)^2*a-3/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)+1)^2*b-3/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)+1)*a+1/8/d/(a+b
)^2/(tanh(1/2*d*x+1/2*c)+1)*b+3/8/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)+1)*a^2-3/4/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)
+1)*a*b-1/8/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)+1)*b^2-1/d*a^3*b/(a+b)^3/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*
b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))+1/d*a^2*b/(a+b)^3/((2*(b*(a+b))
^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2))-1/d*a^2*b^2/(a+b)^3/
(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)-a-2*b)*a)^(1/2)*arctanh(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)-a-2*b)*a
)^(1/2))-1/d*a^3*b/(a+b)^3/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((
2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/d*a^2*b/(a+b)^3/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2)*arctan(a*tanh(1/2*d*x
+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))-1/d*a^2*b^2/(a+b)^3/(b*(a+b))^(1/2)/((2*(b*(a+b))^(1/2)+a+2*b)*a)
^(1/2)*arctan(a*tanh(1/2*d*x+1/2*c)/((2*(b*(a+b))^(1/2)+a+2*b)*a)^(1/2))+8/d/(32*a+32*b)/(tanh(1/2*d*x+1/2*c)-
1)^4+32/d/(64*a+64*b)/(tanh(1/2*d*x+1/2*c)-1)^3-1/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)^2*a+3/8/d/(a+b)^2/(tanh(
1/2*d*x+1/2*c)-1)^2*b-3/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)*a+1/8/d/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)*b-3/8/d/(a
+b)^3*ln(tanh(1/2*d*x+1/2*c)-1)*a^2+3/4/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)-1)*a*b+1/8/d/(a+b)^3*ln(tanh(1/2*d*x+
1/2*c)-1)*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.31437, size = 5158, normalized size = 43.71 \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)^4/(a+b*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/64*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^7 + (a^2 + 2*a*
b + b^2)*sinh(d*x + c)^8 + 8*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c)^4 - 8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7*
(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^
3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(3*a^2 - 6*a*
b - b^2)*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 4*
(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c) - 20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*b)*cosh
(d*x + c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 12*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c)^2 - 30*(a^
2 + a*b)*cosh(d*x + c)^4 + 2*a^2 + 2*a*b)*sinh(d*x + c)^2 + 32*(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)^3*sinh(d
*x + c) + 6*a*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4)*sqrt(-a
*b)*log(((a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2*
a*b + b^2)*sinh(d*x + c)^4 + 2*(a^2 - b^2)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 -
b^2)*sinh(d*x + c)^2 + a^2 - 6*a*b + b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 - b^2)*cosh(d*x + c))
*sinh(d*x + c) + 4*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2
+ a - b)*sqrt(-a*b))/((a + b)*cosh(d*x + c)^4 + 4*(a + b)*cosh(d*x + c)*sinh(d*x + c)^3 + (a + b)*sinh(d*x + c
)^4 + 2*(a - b)*cosh(d*x + c)^2 + 2*(3*(a + b)*cosh(d*x + c)^2 + a - b)*sinh(d*x + c)^2 + 4*((a + b)*cosh(d*x
+ c)^3 + (a - b)*cosh(d*x + c))*sinh(d*x + c) + a + b)) - a^2 - 2*a*b - b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(d*x
+ c)^7 + 4*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c)^5 + 2*(a^2 + a*b)*cosh(d*x
+ c))*sinh(d*x + c))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*
cosh(d*x + c)^3*sinh(d*x + c) + 6*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a^3 +
3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d*x + c)^4)
, 1/64*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^8 + 8*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^7 + (a^2 + 2*a
*b + b^2)*sinh(d*x + c)^8 + 8*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c)^4 - 8*(a^2 + a*b)*cosh(d*x + c)^6 + 4*(7
*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 - 2*a^2 - 2*a*b)*sinh(d*x + c)^6 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)
^3 - 6*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 2*(35*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 4*(3*a^2 - 6*a
*b - b^2)*d*x - 60*(a^2 + a*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 8*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 4
*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c) - 20*(a^2 + a*b)*cosh(d*x + c)^3)*sinh(d*x + c)^3 + 8*(a^2 + a*b)*cos
h(d*x + c)^2 + 4*(7*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 12*(3*a^2 - 6*a*b - b^2)*d*x*cosh(d*x + c)^2 - 30*(a
^2 + a*b)*cosh(d*x + c)^4 + 2*a^2 + 2*a*b)*sinh(d*x + c)^2 + 64*(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)^3*sinh(
d*x + c) + 6*a*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4)*sqrt(a
*b)*arctan(1/2*((a + b)*cosh(d*x + c)^2 + 2*(a + b)*cosh(d*x + c)*sinh(d*x + c) + (a + b)*sinh(d*x + c)^2 + a
- b)*sqrt(a*b)/(a*b)) - a^2 - 2*a*b - b^2 + 8*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^7 + 4*(3*a^2 - 6*a*b - b^2)*d
*x*cosh(d*x + c)^3 - 6*(a^2 + a*b)*cosh(d*x + c)^5 + 2*(a^2 + a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 3*a^2
*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^4 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*
(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d*x + c)^2*sinh(d*x + c)^2 + 4*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*cosh(d
*x + c)*sinh(d*x + c)^3 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*sinh(d*x + c)^4)]

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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)**4/(a+b*tanh(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 2.39497, size = 406, normalized size = 3.44 \begin{align*} \frac{\frac{64 \, a^{2} b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt{a b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt{a b}} + \frac{8 \,{\left (3 \, a^{2} - 6 \, a b - b^{2}\right )} d x}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac{{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 36 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + 2 \, a b + b^{2}\right )} e^{\left (-4 \, d x\right )}}{a^{3} e^{\left (4 \, c\right )} + 3 \, a^{2} b e^{\left (4 \, c\right )} + 3 \, a b^{2} e^{\left (4 \, c\right )} + b^{3} e^{\left (4 \, c\right )}} + \frac{a e^{\left (4 \, d x + 20 \, c\right )} + b e^{\left (4 \, d x + 20 \, c\right )} - 8 \, a e^{\left (2 \, d x + 18 \, c\right )}}{a^{2} e^{\left (16 \, c\right )} + 2 \, a b e^{\left (16 \, c\right )} + b^{2} e^{\left (16 \, c\right )}}}{64 \, d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^2),x, algorithm="giac")

[Out]

1/64*(64*a^2*b*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^3 + 3*a^2*b + 3*a*b^2
+ b^3)*sqrt(a*b)) + 8*(3*a^2 - 6*a*b - b^2)*d*x/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) - (18*a^2*e^(4*d*x + 4*c) - 3
6*a*b*e^(4*d*x + 4*c) - 6*b^2*e^(4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) - 8*a*b*e^(2*d*x + 2*c) + a^2 + 2*a*b +
b^2)*e^(-4*d*x)/(a^3*e^(4*c) + 3*a^2*b*e^(4*c) + 3*a*b^2*e^(4*c) + b^3*e^(4*c)) + (a*e^(4*d*x + 20*c) + b*e^(4
*d*x + 20*c) - 8*a*e^(2*d*x + 18*c))/(a^2*e^(16*c) + 2*a*b*e^(16*c) + b^2*e^(16*c)))/d