\(\int x (a+b x)^2 \cosh (c+d x) \, dx\) [11]

Optimal result

Integrand size = 15, antiderivative size = 134 \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=-\frac {6 b^2 \cosh (c+d x)}{d^4}-\frac {a^2 \cosh (c+d x)}{d^2}-\frac {4 a b x \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {4 a b \sinh (c+d x)}{d^3}+\frac {6 b^2 x \sinh (c+d x)}{d^3}+\frac {a^2 x \sinh (c+d x)}{d}+\frac {2 a b x^2 \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d} \]

[Out]

-6*b^2*cosh(d*x+c)/d^4-a^2*cosh(d*x+c)/d^2-4*a*b*x*cosh(d*x+c)/d^2-3*b^2*x^2*cosh(d*x+c)/d^2+4*a*b*sinh(d*x+c)
/d^3+6*b^2*x*sinh(d*x+c)/d^3+a^2*x*sinh(d*x+c)/d+2*a*b*x^2*sinh(d*x+c)/d+b^2*x^3*sinh(d*x+c)/d

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6874, 3377, 2718, 2717} \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=-\frac {a^2 \cosh (c+d x)}{d^2}+\frac {a^2 x \sinh (c+d x)}{d}+\frac {4 a b \sinh (c+d x)}{d^3}-\frac {4 a b x \cosh (c+d x)}{d^2}+\frac {2 a b x^2 \sinh (c+d x)}{d}-\frac {6 b^2 \cosh (c+d x)}{d^4}+\frac {6 b^2 x \sinh (c+d x)}{d^3}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {b^2 x^3 \sinh (c+d x)}{d} \]

[In]

Int[x*(a + b*x)^2*Cosh[c + d*x],x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^2 x \cosh (c+d x)+2 a b x^2 \cosh (c+d x)+b^2 x^3 \cosh (c+d x)\right ) \, dx \\ & = a^2 \int x \cosh (c+d x) \, dx+(2 a b) \int x^2 \cosh (c+d x) \, dx+b^2 \int x^3 \cosh (c+d x) \, dx \\ & = \frac {a^2 x \sinh (c+d x)}{d}+\frac {2 a b x^2 \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}-\frac {a^2 \int \sinh (c+d x) \, dx}{d}-\frac {(4 a b) \int x \sinh (c+d x) \, dx}{d}-\frac {\left (3 b^2\right ) \int x^2 \sinh (c+d x) \, dx}{d} \\ & = -\frac {a^2 \cosh (c+d x)}{d^2}-\frac {4 a b x \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {a^2 x \sinh (c+d x)}{d}+\frac {2 a b x^2 \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}+\frac {(4 a b) \int \cosh (c+d x) \, dx}{d^2}+\frac {\left (6 b^2\right ) \int x \cosh (c+d x) \, dx}{d^2} \\ & = -\frac {a^2 \cosh (c+d x)}{d^2}-\frac {4 a b x \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {4 a b \sinh (c+d x)}{d^3}+\frac {6 b^2 x \sinh (c+d x)}{d^3}+\frac {a^2 x \sinh (c+d x)}{d}+\frac {2 a b x^2 \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d}-\frac {\left (6 b^2\right ) \int \sinh (c+d x) \, dx}{d^3} \\ & = -\frac {6 b^2 \cosh (c+d x)}{d^4}-\frac {a^2 \cosh (c+d x)}{d^2}-\frac {4 a b x \cosh (c+d x)}{d^2}-\frac {3 b^2 x^2 \cosh (c+d x)}{d^2}+\frac {4 a b \sinh (c+d x)}{d^3}+\frac {6 b^2 x \sinh (c+d x)}{d^3}+\frac {a^2 x \sinh (c+d x)}{d}+\frac {2 a b x^2 \sinh (c+d x)}{d}+\frac {b^2 x^3 \sinh (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.65 \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=\frac {-\left (\left (a^2 d^2+4 a b d^2 x+3 b^2 \left (2+d^2 x^2\right )\right ) \cosh (c+d x)\right )+d \left (a^2 d^2 x+2 a b \left (2+d^2 x^2\right )+b^2 x \left (6+d^2 x^2\right )\right ) \sinh (c+d x)}{d^4} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(\frac {4 \left (\frac {3 b x}{4}+a \right ) d^{2} x b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 d \left (x \left (b x +a \right )^{2} d^{2}+6 b^{2} x +4 a b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (3 x^{2} b^{2}+4 a b x +2 a^{2}\right ) d^{2}+12 b^{2}}{d^{4} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}\)	\(108\)
risch	\(\frac {\left (b^{2} d^{3} x^{3}+2 a b \,d^{3} x^{2}+a^{2} d^{3} x -3 x^{2} d^{2} b^{2}-4 a b \,d^{2} x -a^{2} d^{2}+6 b^{2} d x +4 d a b -6 b^{2}\right ) {\mathrm e}^{d x +c}}{2 d^{4}}-\frac {\left (b^{2} d^{3} x^{3}+2 a b \,d^{3} x^{2}+a^{2} d^{3} x +3 x^{2} d^{2} b^{2}+4 a b \,d^{2} x +a^{2} d^{2}+6 b^{2} d x +4 d a b +6 b^{2}\right ) {\mathrm e}^{-d x -c}}{2 d^{4}}\)	\(172\)
parts	\(\frac {b^{2} x^{3} \sinh \left (d x +c \right )}{d}+\frac {2 a b \,x^{2} \sinh \left (d x +c \right )}{d}+\frac {a^{2} x \sinh \left (d x +c \right )}{d}-\frac {\frac {3 b^{2} \left (\left (d x +c \right )^{2} \cosh \left (d x +c \right )-2 \left (d x +c \right ) \sinh \left (d x +c \right )+2 \cosh \left (d x +c \right )\right )}{d^{2}}-\frac {6 b^{2} c \left (\left (d x +c \right ) \cosh \left (d x +c \right )-\sinh \left (d x +c \right )\right )}{d^{2}}+\frac {4 b a \left (\left (d x +c \right ) \cosh \left (d x +c \right )-\sinh \left (d x +c \right )\right )}{d}+\frac {3 b^{2} c^{2} \cosh \left (d x +c \right )}{d^{2}}-\frac {4 b c a \cosh \left (d x +c \right )}{d}+\cosh \left (d x +c \right ) a^{2}}{d^{2}}\)	\(197\)
meijerg	\(\frac {8 b^{2} \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {3}{4 \sqrt {\pi }}-\frac {\left (\frac {3 x^{2} d^{2}}{2}+3\right ) \cosh \left (d x \right )}{4 \sqrt {\pi }}+\frac {d x \left (\frac {x^{2} d^{2}}{2}+3\right ) \sinh \left (d x \right )}{4 \sqrt {\pi }}\right )}{d^{4}}-\frac {8 i b^{2} \sinh \left (c \right ) \sqrt {\pi }\, \left (\frac {i x d \left (\frac {5 x^{2} d^{2}}{2}+15\right ) \cosh \left (d x \right )}{20 \sqrt {\pi }}-\frac {i \left (\frac {15 x^{2} d^{2}}{2}+15\right ) \sinh \left (d x \right )}{20 \sqrt {\pi }}\right )}{d^{4}}+\frac {8 i a b \cosh \left (c \right ) \sqrt {\pi }\, \left (\frac {i x d \cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {i \left (\frac {3 x^{2} d^{2}}{2}+3\right ) \sinh \left (d x \right )}{6 \sqrt {\pi }}\right )}{d^{3}}+\frac {8 b a \sinh \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (\frac {x^{2} d^{2}}{2}+1\right ) \cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {d x \sinh \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{3}}-\frac {2 a^{2} \cosh \left (c \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (d x \right )}{2 \sqrt {\pi }}-\frac {d x \sinh \left (d x \right )}{2 \sqrt {\pi }}\right )}{d^{2}}+\frac {a^{2} \sinh \left (c \right ) \left (\cosh \left (d x \right ) x d -\sinh \left (d x \right )\right )}{d^{2}}\)	\(274\)
derivativedivides	\(\frac {\frac {b^{2} \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{2}}-\frac {3 b^{2} c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{2}}+\frac {2 b a \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d}+\frac {3 b^{2} c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{2}}-\frac {4 b c a \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d}+a^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )-\frac {b^{2} c^{3} \sinh \left (d x +c \right )}{d^{2}}+\frac {2 b \,c^{2} a \sinh \left (d x +c \right )}{d}-c \,a^{2} \sinh \left (d x +c \right )}{d^{2}}\)	\(283\)
default	\(\frac {\frac {b^{2} \left (\left (d x +c \right )^{3} \sinh \left (d x +c \right )-3 \left (d x +c \right )^{2} \cosh \left (d x +c \right )+6 \left (d x +c \right ) \sinh \left (d x +c \right )-6 \cosh \left (d x +c \right )\right )}{d^{2}}-\frac {3 b^{2} c \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d^{2}}+\frac {2 b a \left (\left (d x +c \right )^{2} \sinh \left (d x +c \right )-2 \left (d x +c \right ) \cosh \left (d x +c \right )+2 \sinh \left (d x +c \right )\right )}{d}+\frac {3 b^{2} c^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d^{2}}-\frac {4 b c a \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )}{d}+a^{2} \left (\left (d x +c \right ) \sinh \left (d x +c \right )-\cosh \left (d x +c \right )\right )-\frac {b^{2} c^{3} \sinh \left (d x +c \right )}{d^{2}}+\frac {2 b \,c^{2} a \sinh \left (d x +c \right )}{d}-c \,a^{2} \sinh \left (d x +c \right )}{d^{2}}\)	\(283\)

[In]

int(x*(b*x+a)^2*cosh(d*x+c),x,method=_RETURNVERBOSE)

[Out]

(4*(3/4*b*x+a)*d^2*x*b*tanh(1/2*d*x+1/2*c)^2-2*d*(x*(b*x+a)^2*d^2+6*b^2*x+4*a*b)*tanh(1/2*d*x+1/2*c)+(3*b^2*x^
2+4*a*b*x+2*a^2)*d^2+12*b^2)/d^4/(tanh(1/2*d*x+1/2*c)^2-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.71 \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=-\frac {{\left (3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} + 6 \, b^{2}\right )} \cosh \left (d x + c\right ) - {\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + 4 \, a b d + {\left (a^{2} d^{3} + 6 \, b^{2} d\right )} x\right )} \sinh \left (d x + c\right )}{d^{4}} \]

[In]

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

[Out]

-((3*b^2*d^2*x^2 + 4*a*b*d^2*x + a^2*d^2 + 6*b^2)*cosh(d*x + c) - (b^2*d^3*x^3 + 2*a*b*d^3*x^2 + 4*a*b*d + (a^
2*d^3 + 6*b^2*d)*x)*sinh(d*x + c))/d^4

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.28 \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=\begin {cases} \frac {a^{2} x \sinh {\left (c + d x \right )}}{d} - \frac {a^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {2 a b x^{2} \sinh {\left (c + d x \right )}}{d} - \frac {4 a b x \cosh {\left (c + d x \right )}}{d^{2}} + \frac {4 a b \sinh {\left (c + d x \right )}}{d^{3}} + \frac {b^{2} x^{3} \sinh {\left (c + d x \right )}}{d} - \frac {3 b^{2} x^{2} \cosh {\left (c + d x \right )}}{d^{2}} + \frac {6 b^{2} x \sinh {\left (c + d x \right )}}{d^{3}} - \frac {6 b^{2} \cosh {\left (c + d x \right )}}{d^{4}} & \text {for}\: d \neq 0 \\\left (\frac {a^{2} x^{2}}{2} + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{4}}{4}\right ) \cosh {\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**2*x*sinh(c + d*x)/d - a**2*cosh(c + d*x)/d**2 + 2*a*b*x**2*sinh(c + d*x)/d - 4*a*b*x*cosh(c + d*
x)/d**2 + 4*a*b*sinh(c + d*x)/d**3 + b**2*x**3*sinh(c + d*x)/d - 3*b**2*x**2*cosh(c + d*x)/d**2 + 6*b**2*x*sin
h(c + d*x)/d**3 - 6*b**2*cosh(c + d*x)/d**4, Ne(d, 0)), ((a**2*x**2/2 + 2*a*b*x**3/3 + b**2*x**4/4)*cosh(c), T
rue))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (134) = 268\).

Time = 0.19 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.05 \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=-\frac {1}{24} \, d {\left (\frac {6 \, {\left (d^{2} x^{2} e^{c} - 2 \, d x e^{c} + 2 \, e^{c}\right )} a^{2} e^{\left (d x\right )}}{d^{3}} + \frac {6 \, {\left (d^{2} x^{2} + 2 \, d x + 2\right )} a^{2} e^{\left (-d x - c\right )}}{d^{3}} + \frac {8 \, {\left (d^{3} x^{3} e^{c} - 3 \, d^{2} x^{2} e^{c} + 6 \, d x e^{c} - 6 \, e^{c}\right )} a b e^{\left (d x\right )}}{d^{4}} + \frac {8 \, {\left (d^{3} x^{3} + 3 \, d^{2} x^{2} + 6 \, d x + 6\right )} a b e^{\left (-d x - c\right )}}{d^{4}} + \frac {3 \, {\left (d^{4} x^{4} e^{c} - 4 \, d^{3} x^{3} e^{c} + 12 \, d^{2} x^{2} e^{c} - 24 \, d x e^{c} + 24 \, e^{c}\right )} b^{2} e^{\left (d x\right )}}{d^{5}} + \frac {3 \, {\left (d^{4} x^{4} + 4 \, d^{3} x^{3} + 12 \, d^{2} x^{2} + 24 \, d x + 24\right )} b^{2} e^{\left (-d x - c\right )}}{d^{5}}\right )} + \frac {1}{12} \, {\left (3 \, b^{2} x^{4} + 8 \, a b x^{3} + 6 \, a^{2} x^{2}\right )} \cosh \left (d x + c\right ) \]

[In]

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

[Out]

-1/24*d*(6*(d^2*x^2*e^c - 2*d*x*e^c + 2*e^c)*a^2*e^(d*x)/d^3 + 6*(d^2*x^2 + 2*d*x + 2)*a^2*e^(-d*x - c)/d^3 +
8*(d^3*x^3*e^c - 3*d^2*x^2*e^c + 6*d*x*e^c - 6*e^c)*a*b*e^(d*x)/d^4 + 8*(d^3*x^3 + 3*d^2*x^2 + 6*d*x + 6)*a*b*
e^(-d*x - c)/d^4 + 3*(d^4*x^4*e^c - 4*d^3*x^3*e^c + 12*d^2*x^2*e^c - 24*d*x*e^c + 24*e^c)*b^2*e^(d*x)/d^5 + 3*
(d^4*x^4 + 4*d^3*x^3 + 12*d^2*x^2 + 24*d*x + 24)*b^2*e^(-d*x - c)/d^5) + 1/12*(3*b^2*x^4 + 8*a*b*x^3 + 6*a^2*x
^2)*cosh(d*x + c)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.28 \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=\frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + a^{2} d^{3} x - 3 \, b^{2} d^{2} x^{2} - 4 \, a b d^{2} x - a^{2} d^{2} + 6 \, b^{2} d x + 4 \, a b d - 6 \, b^{2}\right )} e^{\left (d x + c\right )}}{2 \, d^{4}} - \frac {{\left (b^{2} d^{3} x^{3} + 2 \, a b d^{3} x^{2} + a^{2} d^{3} x + 3 \, b^{2} d^{2} x^{2} + 4 \, a b d^{2} x + a^{2} d^{2} + 6 \, b^{2} d x + 4 \, a b d + 6 \, b^{2}\right )} e^{\left (-d x - c\right )}}{2 \, d^{4}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.83 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.93 \[ \int x (a+b x)^2 \cosh (c+d x) \, dx=\frac {b^2\,x^3\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {3\,b^2\,x^2\,\mathrm {cosh}\left (c+d\,x\right )}{d^2}-\frac {\mathrm {cosh}\left (c+d\,x\right )\,\left (a^2\,d^2+6\,b^2\right )}{d^4}+\frac {4\,a\,b\,\mathrm {sinh}\left (c+d\,x\right )}{d^3}+\frac {x\,\mathrm {sinh}\left (c+d\,x\right )\,\left (a^2\,d^2+6\,b^2\right )}{d^3}+\frac {2\,a\,b\,x^2\,\mathrm {sinh}\left (c+d\,x\right )}{d}-\frac {4\,a\,b\,x\,\mathrm {cosh}\left (c+d\,x\right )}{d^2} \]

[In]

int(x*cosh(c + d*x)*(a + b*x)^2,x)

[Out]

(b^2*x^3*sinh(c + d*x))/d - (3*b^2*x^2*cosh(c + d*x))/d^2 - (cosh(c + d*x)*(6*b^2 + a^2*d^2))/d^4 + (4*a*b*sin
h(c + d*x))/d^3 + (x*sinh(c + d*x)*(6*b^2 + a^2*d^2))/d^3 + (2*a*b*x^2*sinh(c + d*x))/d - (4*a*b*x*cosh(c + d*
x))/d^2