3.8 \(\int (c+d x)^4 \cosh ^2(a+b x) \, dx\)

Optimal. Leaf size=162 \[ -\frac{3 d^3 (c+d x) \cosh ^2(a+b x)}{2 b^4}+\frac{3 d^2 (c+d x)^2 \sinh (a+b x) \cosh (a+b x)}{2 b^3}-\frac{d (c+d x)^3 \cosh ^2(a+b x)}{b^2}+\frac{3 d^4 \sinh (a+b x) \cosh (a+b x)}{4 b^5}+\frac{(c+d x)^4 \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac{d (c+d x)^3}{2 b^2}+\frac{3 d^4 x}{4 b^4}+\frac{(c+d x)^5}{10 d} \]

[Out]

(3*d^4*x)/(4*b^4) + (d*(c + d*x)^3)/(2*b^2) + (c + d*x)^5/(10*d) - (3*d^3*(c + d*x)*Cosh[a + b*x]^2)/(2*b^4) -
(d*(c + d*x)^3*Cosh[a + b*x]^2)/b^2 + (3*d^4*Cosh[a + b*x]*Sinh[a + b*x])/(4*b^5) + (3*d^2*(c + d*x)^2*Cosh[a
+ b*x]*Sinh[a + b*x])/(2*b^3) + ((c + d*x)^4*Cosh[a + b*x]*Sinh[a + b*x])/(2*b)

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Rubi [A]  time = 0.10147, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3311, 32, 2635, 8} \[ -\frac{3 d^3 (c+d x) \cosh ^2(a+b x)}{2 b^4}+\frac{3 d^2 (c+d x)^2 \sinh (a+b x) \cosh (a+b x)}{2 b^3}-\frac{d (c+d x)^3 \cosh ^2(a+b x)}{b^2}+\frac{3 d^4 \sinh (a+b x) \cosh (a+b x)}{4 b^5}+\frac{(c+d x)^4 \sinh (a+b x) \cosh (a+b x)}{2 b}+\frac{d (c+d x)^3}{2 b^2}+\frac{3 d^4 x}{4 b^4}+\frac{(c+d x)^5}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4*Cosh[a + b*x]^2,x]

[Out]

(3*d^4*x)/(4*b^4) + (d*(c + d*x)^3)/(2*b^2) + (c + d*x)^5/(10*d) - (3*d^3*(c + d*x)*Cosh[a + b*x]^2)/(2*b^4) -
(d*(c + d*x)^3*Cosh[a + b*x]^2)/b^2 + (3*d^4*Cosh[a + b*x]*Sinh[a + b*x])/(4*b^5) + (3*d^2*(c + d*x)^2*Cosh[a
+ b*x]*Sinh[a + b*x])/(2*b^3) + ((c + d*x)^4*Cosh[a + b*x]*Sinh[a + b*x])/(2*b)

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (c+d x)^4 \cosh ^2(a+b x) \, dx &=-\frac{d (c+d x)^3 \cosh ^2(a+b x)}{b^2}+\frac{(c+d x)^4 \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac{1}{2} \int (c+d x)^4 \, dx+\frac{\left (3 d^2\right ) \int (c+d x)^2 \cosh ^2(a+b x) \, dx}{b^2}\\ &=\frac{(c+d x)^5}{10 d}-\frac{3 d^3 (c+d x) \cosh ^2(a+b x)}{2 b^4}-\frac{d (c+d x)^3 \cosh ^2(a+b x)}{b^2}+\frac{3 d^2 (c+d x)^2 \cosh (a+b x) \sinh (a+b x)}{2 b^3}+\frac{(c+d x)^4 \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac{\left (3 d^2\right ) \int (c+d x)^2 \, dx}{2 b^2}+\frac{\left (3 d^4\right ) \int \cosh ^2(a+b x) \, dx}{2 b^4}\\ &=\frac{d (c+d x)^3}{2 b^2}+\frac{(c+d x)^5}{10 d}-\frac{3 d^3 (c+d x) \cosh ^2(a+b x)}{2 b^4}-\frac{d (c+d x)^3 \cosh ^2(a+b x)}{b^2}+\frac{3 d^4 \cosh (a+b x) \sinh (a+b x)}{4 b^5}+\frac{3 d^2 (c+d x)^2 \cosh (a+b x) \sinh (a+b x)}{2 b^3}+\frac{(c+d x)^4 \cosh (a+b x) \sinh (a+b x)}{2 b}+\frac{\left (3 d^4\right ) \int 1 \, dx}{4 b^4}\\ &=\frac{3 d^4 x}{4 b^4}+\frac{d (c+d x)^3}{2 b^2}+\frac{(c+d x)^5}{10 d}-\frac{3 d^3 (c+d x) \cosh ^2(a+b x)}{2 b^4}-\frac{d (c+d x)^3 \cosh ^2(a+b x)}{b^2}+\frac{3 d^4 \cosh (a+b x) \sinh (a+b x)}{4 b^5}+\frac{3 d^2 (c+d x)^2 \cosh (a+b x) \sinh (a+b x)}{2 b^3}+\frac{(c+d x)^4 \cosh (a+b x) \sinh (a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.615663, size = 132, normalized size = 0.81 \[ \frac{10 \sinh (2 (a+b x)) \left (6 b^2 d^2 (c+d x)^2+2 b^4 (c+d x)^4+3 d^4\right )-20 b d (c+d x) \cosh (2 (a+b x)) \left (2 b^2 (c+d x)^2+3 d^2\right )+8 b^5 x \left (10 c^2 d^2 x^2+10 c^3 d x+5 c^4+5 c d^3 x^3+d^4 x^4\right )}{80 b^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b^5*x*(5*c^4 + 10*c^3*d*x + 10*c^2*d^2*x^2 + 5*c*d^3*x^3 + d^4*x^4) - 20*b*d*(c + d*x)*(3*d^2 + 2*b^2*(c +
d*x)^2)*Cosh[2*(a + b*x)] + 10*(3*d^4 + 6*b^2*d^2*(c + d*x)^2 + 2*b^4*(c + d*x)^4)*Sinh[2*(a + b*x)])/(80*b^5)

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Maple [B]  time = 0.01, size = 910, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^4*cosh(b*x+a)^2,x)

[Out]

1/b*(1/b^4*d^4*(1/2*(b*x+a)^4*cosh(b*x+a)*sinh(b*x+a)+1/10*(b*x+a)^5-(b*x+a)^3*cosh(b*x+a)^2+3/2*(b*x+a)^2*cos
h(b*x+a)*sinh(b*x+a)+1/2*(b*x+a)^3-3/2*(b*x+a)*cosh(b*x+a)^2+3/4*cosh(b*x+a)*sinh(b*x+a)+3/4*b*x+3/4*a)+1/b^4*
d^4*a^4*(1/2*cosh(b*x+a)*sinh(b*x+a)+1/2*b*x+1/2*a)+12/b^3*d^3*c*a^2*(1/2*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+1/4*
(b*x+a)^2-1/4*cosh(b*x+a)^2)-12/b^2*d^2*c^2*a*(1/2*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+1/4*(b*x+a)^2-1/4*cosh(b*x+
a)^2)-12/b^3*d^3*c*a*(1/2*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)+1/6*(b*x+a)^3-1/2*(b*x+a)*cosh(b*x+a)^2+1/4*cosh(b
*x+a)*sinh(b*x+a)+1/4*b*x+1/4*a)+4/b*d*c^3*(1/2*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+1/4*(b*x+a)^2-1/4*cosh(b*x+a)^
2)-4/b^3*d^3*a^3*c*(1/2*cosh(b*x+a)*sinh(b*x+a)+1/2*b*x+1/2*a)+6/b^2*d^2*a^2*c^2*(1/2*cosh(b*x+a)*sinh(b*x+a)+
1/2*b*x+1/2*a)-4/b*d*a*c^3*(1/2*cosh(b*x+a)*sinh(b*x+a)+1/2*b*x+1/2*a)-4/b^4*d^4*a^3*(1/2*(b*x+a)*cosh(b*x+a)*
sinh(b*x+a)+1/4*(b*x+a)^2-1/4*cosh(b*x+a)^2)+4/b^3*d^3*c*(1/2*(b*x+a)^3*cosh(b*x+a)*sinh(b*x+a)+1/8*(b*x+a)^4-
3/4*(b*x+a)^2*cosh(b*x+a)^2+3/4*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+3/8*(b*x+a)^2-3/8*cosh(b*x+a)^2)+6/b^2*d^2*c^2
*(1/2*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)+1/6*(b*x+a)^3-1/2*(b*x+a)*cosh(b*x+a)^2+1/4*cosh(b*x+a)*sinh(b*x+a)+1/
4*b*x+1/4*a)+6/b^4*d^4*a^2*(1/2*(b*x+a)^2*cosh(b*x+a)*sinh(b*x+a)+1/6*(b*x+a)^3-1/2*(b*x+a)*cosh(b*x+a)^2+1/4*
cosh(b*x+a)*sinh(b*x+a)+1/4*b*x+1/4*a)-4/b^4*d^4*a*(1/2*(b*x+a)^3*cosh(b*x+a)*sinh(b*x+a)+1/8*(b*x+a)^4-3/4*(b
*x+a)^2*cosh(b*x+a)^2+3/4*(b*x+a)*cosh(b*x+a)*sinh(b*x+a)+3/8*(b*x+a)^2-3/8*cosh(b*x+a)^2)+c^4*(1/2*cosh(b*x+a
)*sinh(b*x+a)+1/2*b*x+1/2*a))

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Maxima [B]  time = 1.12478, size = 516, normalized size = 3.19 \begin{align*} \frac{1}{4} \,{\left (4 \, x^{2} + \frac{{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{2}} - \frac{{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{2}}\right )} c^{3} d + \frac{1}{8} \,{\left (8 \, x^{3} + \frac{3 \,{\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{3}} - \frac{3 \,{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{3}}\right )} c^{2} d^{2} + \frac{1}{8} \,{\left (4 \, x^{4} + \frac{{\left (4 \, b^{3} x^{3} e^{\left (2 \, a\right )} - 6 \, b^{2} x^{2} e^{\left (2 \, a\right )} + 6 \, b x e^{\left (2 \, a\right )} - 3 \, e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{4}} - \frac{{\left (4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{4}}\right )} c d^{3} + \frac{1}{80} \,{\left (8 \, x^{5} + \frac{5 \,{\left (2 \, b^{4} x^{4} e^{\left (2 \, a\right )} - 4 \, b^{3} x^{3} e^{\left (2 \, a\right )} + 6 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 6 \, b x e^{\left (2 \, a\right )} + 3 \, e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{b^{5}} - \frac{5 \,{\left (2 \, b^{4} x^{4} + 4 \, b^{3} x^{3} + 6 \, b^{2} x^{2} + 6 \, b x + 3\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{b^{5}}\right )} d^{4} + \frac{1}{8} \, c^{4}{\left (4 \, x + \frac{e^{\left (2 \, b x + 2 \, a\right )}}{b} - \frac{e^{\left (-2 \, b x - 2 \, a\right )}}{b}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*x^2 + (2*b*x*e^(2*a) - e^(2*a))*e^(2*b*x)/b^2 - (2*b*x + 1)*e^(-2*b*x - 2*a)/b^2)*c^3*d + 1/8*(8*x^3 +
3*(2*b^2*x^2*e^(2*a) - 2*b*x*e^(2*a) + e^(2*a))*e^(2*b*x)/b^3 - 3*(2*b^2*x^2 + 2*b*x + 1)*e^(-2*b*x - 2*a)/b^3
)*c^2*d^2 + 1/8*(4*x^4 + (4*b^3*x^3*e^(2*a) - 6*b^2*x^2*e^(2*a) + 6*b*x*e^(2*a) - 3*e^(2*a))*e^(2*b*x)/b^4 - (
4*b^3*x^3 + 6*b^2*x^2 + 6*b*x + 3)*e^(-2*b*x - 2*a)/b^4)*c*d^3 + 1/80*(8*x^5 + 5*(2*b^4*x^4*e^(2*a) - 4*b^3*x^
3*e^(2*a) + 6*b^2*x^2*e^(2*a) - 6*b*x*e^(2*a) + 3*e^(2*a))*e^(2*b*x)/b^5 - 5*(2*b^4*x^4 + 4*b^3*x^3 + 6*b^2*x^
2 + 6*b*x + 3)*e^(-2*b*x - 2*a)/b^5)*d^4 + 1/8*c^4*(4*x + e^(2*b*x + 2*a)/b - e^(-2*b*x - 2*a)/b)

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Fricas [B]  time = 1.87598, size = 662, normalized size = 4.09 \begin{align*} \frac{2 \, b^{5} d^{4} x^{5} + 10 \, b^{5} c d^{3} x^{4} + 20 \, b^{5} c^{2} d^{2} x^{3} + 20 \, b^{5} c^{3} d x^{2} + 10 \, b^{5} c^{4} x - 5 \,{\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 2 \, b^{3} c^{3} d + 3 \, b c d^{3} + 3 \,{\left (2 \, b^{3} c^{2} d^{2} + b d^{4}\right )} x\right )} \cosh \left (b x + a\right )^{2} + 5 \,{\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 2 \, b^{4} c^{4} + 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4} + 6 \,{\left (2 \, b^{4} c^{2} d^{2} + b^{2} d^{4}\right )} x^{2} + 4 \,{\left (2 \, b^{4} c^{3} d + 3 \, b^{2} c d^{3}\right )} x\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - 5 \,{\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 2 \, b^{3} c^{3} d + 3 \, b c d^{3} + 3 \,{\left (2 \, b^{3} c^{2} d^{2} + b d^{4}\right )} x\right )} \sinh \left (b x + a\right )^{2}}{20 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*cosh(b*x+a)^2,x, algorithm="fricas")

[Out]

1/20*(2*b^5*d^4*x^5 + 10*b^5*c*d^3*x^4 + 20*b^5*c^2*d^2*x^3 + 20*b^5*c^3*d*x^2 + 10*b^5*c^4*x - 5*(2*b^3*d^4*x
^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d + 3*b*c*d^3 + 3*(2*b^3*c^2*d^2 + b*d^4)*x)*cosh(b*x + a)^2 + 5*(2*b^4*d^4*x
^4 + 8*b^4*c*d^3*x^3 + 2*b^4*c^4 + 6*b^2*c^2*d^2 + 3*d^4 + 6*(2*b^4*c^2*d^2 + b^2*d^4)*x^2 + 4*(2*b^4*c^3*d +
3*b^2*c*d^3)*x)*cosh(b*x + a)*sinh(b*x + a) - 5*(2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d + 3*b*c*d^3 + 3
*(2*b^3*c^2*d^2 + b*d^4)*x)*sinh(b*x + a)^2)/b^5

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Sympy [A]  time = 6.52616, size = 660, normalized size = 4.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4*cosh(b*x+a)**2,x)

[Out]

Piecewise((-c**4*x*sinh(a + b*x)**2/2 + c**4*x*cosh(a + b*x)**2/2 - c**3*d*x**2*sinh(a + b*x)**2 + c**3*d*x**2
*cosh(a + b*x)**2 - c**2*d**2*x**3*sinh(a + b*x)**2 + c**2*d**2*x**3*cosh(a + b*x)**2 - c*d**3*x**4*sinh(a + b
*x)**2/2 + c*d**3*x**4*cosh(a + b*x)**2/2 - d**4*x**5*sinh(a + b*x)**2/10 + d**4*x**5*cosh(a + b*x)**2/10 + c*
*4*sinh(a + b*x)*cosh(a + b*x)/(2*b) + 2*c**3*d*x*sinh(a + b*x)*cosh(a + b*x)/b + 3*c**2*d**2*x**2*sinh(a + b*
x)*cosh(a + b*x)/b + 2*c*d**3*x**3*sinh(a + b*x)*cosh(a + b*x)/b + d**4*x**4*sinh(a + b*x)*cosh(a + b*x)/(2*b)
- c**3*d*sinh(a + b*x)**2/b**2 - 3*c**2*d**2*x*sinh(a + b*x)**2/(2*b**2) - 3*c**2*d**2*x*cosh(a + b*x)**2/(2*
b**2) - 3*c*d**3*x**2*sinh(a + b*x)**2/(2*b**2) - 3*c*d**3*x**2*cosh(a + b*x)**2/(2*b**2) - d**4*x**3*sinh(a +
b*x)**2/(2*b**2) - d**4*x**3*cosh(a + b*x)**2/(2*b**2) + 3*c**2*d**2*sinh(a + b*x)*cosh(a + b*x)/(2*b**3) + 3
*c*d**3*x*sinh(a + b*x)*cosh(a + b*x)/b**3 + 3*d**4*x**2*sinh(a + b*x)*cosh(a + b*x)/(2*b**3) - 3*c*d**3*sinh(
a + b*x)**2/(2*b**4) - 3*d**4*x*sinh(a + b*x)**2/(4*b**4) - 3*d**4*x*cosh(a + b*x)**2/(4*b**4) + 3*d**4*sinh(a
+ b*x)*cosh(a + b*x)/(4*b**5), Ne(b, 0)), ((c**4*x + 2*c**3*d*x**2 + 2*c**2*d**2*x**3 + c*d**3*x**4 + d**4*x*
*5/5)*cosh(a)**2, True))

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Giac [B]  time = 1.17811, size = 502, normalized size = 3.1 \begin{align*} \frac{1}{10} \, d^{4} x^{5} + \frac{1}{2} \, c d^{3} x^{4} + c^{2} d^{2} x^{3} + c^{3} d x^{2} + \frac{1}{2} \, c^{4} x + \frac{{\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 12 \, b^{4} c^{2} d^{2} x^{2} - 4 \, b^{3} d^{4} x^{3} + 8 \, b^{4} c^{3} d x - 12 \, b^{3} c d^{3} x^{2} + 2 \, b^{4} c^{4} - 12 \, b^{3} c^{2} d^{2} x + 6 \, b^{2} d^{4} x^{2} - 4 \, b^{3} c^{3} d + 12 \, b^{2} c d^{3} x + 6 \, b^{2} c^{2} d^{2} - 6 \, b d^{4} x - 6 \, b c d^{3} + 3 \, d^{4}\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{5}} - \frac{{\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 12 \, b^{4} c^{2} d^{2} x^{2} + 4 \, b^{3} d^{4} x^{3} + 8 \, b^{4} c^{3} d x + 12 \, b^{3} c d^{3} x^{2} + 2 \, b^{4} c^{4} + 12 \, b^{3} c^{2} d^{2} x + 6 \, b^{2} d^{4} x^{2} + 4 \, b^{3} c^{3} d + 12 \, b^{2} c d^{3} x + 6 \, b^{2} c^{2} d^{2} + 6 \, b d^{4} x + 6 \, b c d^{3} + 3 \, d^{4}\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*d^4*x^5 + 1/2*c*d^3*x^4 + c^2*d^2*x^3 + c^3*d*x^2 + 1/2*c^4*x + 1/16*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 1
2*b^4*c^2*d^2*x^2 - 4*b^3*d^4*x^3 + 8*b^4*c^3*d*x - 12*b^3*c*d^3*x^2 + 2*b^4*c^4 - 12*b^3*c^2*d^2*x + 6*b^2*d^
4*x^2 - 4*b^3*c^3*d + 12*b^2*c*d^3*x + 6*b^2*c^2*d^2 - 6*b*d^4*x - 6*b*c*d^3 + 3*d^4)*e^(2*b*x + 2*a)/b^5 - 1/
16*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 12*b^4*c^2*d^2*x^2 + 4*b^3*d^4*x^3 + 8*b^4*c^3*d*x + 12*b^3*c*d^3*x^2 +
2*b^4*c^4 + 12*b^3*c^2*d^2*x + 6*b^2*d^4*x^2 + 4*b^3*c^3*d + 12*b^2*c*d^3*x + 6*b^2*c^2*d^2 + 6*b*d^4*x + 6*b*
c*d^3 + 3*d^4)*e^(-2*b*x - 2*a)/b^5