3.1.96 \(\int (a+b \sinh (c+d x))^5 \, dx\) [96]

Optimal. Leaf size=183 \[ \frac {1}{8} a \left (8 a^4-40 a^2 b^2+15 b^4\right ) x+\frac {b \left (107 a^4-192 a^2 b^2+16 b^4\right ) \cosh (c+d x)}{30 d}+\frac {7 a b^2 \left (22 a^2-23 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{120 d}+\frac {b \left (47 a^2-16 b^2\right ) \cosh (c+d x) (a+b \sinh (c+d x))^2}{60 d}+\frac {9 a b \cosh (c+d x) (a+b \sinh (c+d x))^3}{20 d}+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^4}{5 d} \]

[Out]

1/8*a*(8*a^4-40*a^2*b^2+15*b^4)*x+1/30*b*(107*a^4-192*a^2*b^2+16*b^4)*cosh(d*x+c)/d+7/120*a*b^2*(22*a^2-23*b^2
)*cosh(d*x+c)*sinh(d*x+c)/d+1/60*b*(47*a^2-16*b^2)*cosh(d*x+c)*(a+b*sinh(d*x+c))^2/d+9/20*a*b*cosh(d*x+c)*(a+b
*sinh(d*x+c))^3/d+1/5*b*cosh(d*x+c)*(a+b*sinh(d*x+c))^4/d

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Rubi [A]
time = 0.19, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2735, 2832, 2813} \begin {gather*} \frac {b \left (47 a^2-16 b^2\right ) \cosh (c+d x) (a+b \sinh (c+d x))^2}{60 d}+\frac {7 a b^2 \left (22 a^2-23 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{120 d}+\frac {b \left (107 a^4-192 a^2 b^2+16 b^4\right ) \cosh (c+d x)}{30 d}+\frac {1}{8} a x \left (8 a^4-40 a^2 b^2+15 b^4\right )+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^4}{5 d}+\frac {9 a b \cosh (c+d x) (a+b \sinh (c+d x))^3}{20 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[c + d*x])^5,x]

[Out]

(a*(8*a^4 - 40*a^2*b^2 + 15*b^4)*x)/8 + (b*(107*a^4 - 192*a^2*b^2 + 16*b^4)*Cosh[c + d*x])/(30*d) + (7*a*b^2*(
22*a^2 - 23*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(120*d) + (b*(47*a^2 - 16*b^2)*Cosh[c + d*x]*(a + b*Sinh[c + d*x
])^2)/(60*d) + (9*a*b*Cosh[c + d*x]*(a + b*Sinh[c + d*x])^3)/(20*d) + (b*Cosh[c + d*x]*(a + b*Sinh[c + d*x])^4
)/(5*d)

Rule 2735

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
- 1)/(d*n)), x] + Dist[1/n, Int[(a + b*Sin[c + d*x])^(n - 2)*Simp[a^2*n + b^2*(n - 1) + a*b*(2*n - 1)*Sin[c +
d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]

Rule 2813

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(2*a*c +
b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Cos[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2832

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d
)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(f*(m + 1))), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Sim
p[b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int (a+b \sinh (c+d x))^5 \, dx &=\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^4}{5 d}+\frac {1}{5} \int (a+b \sinh (c+d x))^3 \left (5 a^2-4 b^2+9 a b \sinh (c+d x)\right ) \, dx\\ &=\frac {9 a b \cosh (c+d x) (a+b \sinh (c+d x))^3}{20 d}+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^4}{5 d}+\frac {1}{20} \int (a+b \sinh (c+d x))^2 \left (a \left (20 a^2-43 b^2\right )+b \left (47 a^2-16 b^2\right ) \sinh (c+d x)\right ) \, dx\\ &=\frac {b \left (47 a^2-16 b^2\right ) \cosh (c+d x) (a+b \sinh (c+d x))^2}{60 d}+\frac {9 a b \cosh (c+d x) (a+b \sinh (c+d x))^3}{20 d}+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^4}{5 d}+\frac {1}{60} \int (a+b \sinh (c+d x)) \left (60 a^4-223 a^2 b^2+32 b^4+7 a b \left (22 a^2-23 b^2\right ) \sinh (c+d x)\right ) \, dx\\ &=\frac {1}{8} a \left (8 a^4-40 a^2 b^2+15 b^4\right ) x+\frac {b \left (107 a^4-192 a^2 b^2+16 b^4\right ) \cosh (c+d x)}{30 d}+\frac {7 a b^2 \left (22 a^2-23 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{120 d}+\frac {b \left (47 a^2-16 b^2\right ) \cosh (c+d x) (a+b \sinh (c+d x))^2}{60 d}+\frac {9 a b \cosh (c+d x) (a+b \sinh (c+d x))^3}{20 d}+\frac {b \cosh (c+d x) (a+b \sinh (c+d x))^4}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.47, size = 138, normalized size = 0.75 \begin {gather*} \frac {300 b \left (8 a^4-12 a^2 b^2+b^4\right ) \cosh (c+d x)+50 \left (8 a^2 b^3-b^5\right ) \cosh (3 (c+d x))+6 b^5 \cosh (5 (c+d x))+15 a \left (4 \left (8 a^4-40 a^2 b^2+15 b^4\right ) (c+d x)+40 \left (2 a^2 b^2-b^4\right ) \sinh (2 (c+d x))+5 b^4 \sinh (4 (c+d x))\right )}{480 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[c + d*x])^5,x]

[Out]

(300*b*(8*a^4 - 12*a^2*b^2 + b^4)*Cosh[c + d*x] + 50*(8*a^2*b^3 - b^5)*Cosh[3*(c + d*x)] + 6*b^5*Cosh[5*(c + d
*x)] + 15*a*(4*(8*a^4 - 40*a^2*b^2 + 15*b^4)*(c + d*x) + 40*(2*a^2*b^2 - b^4)*Sinh[2*(c + d*x)] + 5*b^4*Sinh[4
*(c + d*x)]))/(480*d)

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Maple [A]
time = 1.04, size = 145, normalized size = 0.79

method	result	size
default	\(a^{5} x +\frac {\left (-\frac {5}{16} b^{5}+\frac {5}{2} a^{2} b^{3}\right ) \cosh \left (3 d x +3 c \right )}{3 d}+\frac {\left (-\frac {5}{2} a \,b^{4}+5 a^{3} b^{2}\right ) \sinh \left (2 d x +2 c \right )}{2 d}+\frac {\left (\frac {5}{8} b^{5}-\frac {15}{2} a^{2} b^{3}+5 a^{4} b \right ) \cosh \left (d x +c \right )}{d}+\frac {15 x a \,b^{4}}{8}-5 x \,a^{3} b^{2}+\frac {b^{5} \cosh \left (5 d x +5 c \right )}{80 d}+\frac {5 a \,b^{4} \sinh \left (4 d x +4 c \right )}{32 d}\)	\(145\)
risch	\(a^{5} x -5 x \,a^{3} b^{2}+\frac {15 x a \,b^{4}}{8}+\frac {b^{5} {\mathrm e}^{5 d x +5 c}}{160 d}+\frac {5 a \,b^{4} {\mathrm e}^{4 d x +4 c}}{64 d}+\frac {5 b^{3} {\mathrm e}^{3 d x +3 c} a^{2}}{12 d}-\frac {5 b^{5} {\mathrm e}^{3 d x +3 c}}{96 d}+\frac {5 b^{2} a^{3} {\mathrm e}^{2 d x +2 c}}{4 d}-\frac {5 b^{4} a \,{\mathrm e}^{2 d x +2 c}}{8 d}+\frac {5 b \,{\mathrm e}^{d x +c} a^{4}}{2 d}-\frac {15 b^{3} {\mathrm e}^{d x +c} a^{2}}{4 d}+\frac {5 b^{5} {\mathrm e}^{d x +c}}{16 d}+\frac {5 b \,{\mathrm e}^{-d x -c} a^{4}}{2 d}-\frac {15 b^{3} {\mathrm e}^{-d x -c} a^{2}}{4 d}+\frac {5 b^{5} {\mathrm e}^{-d x -c}}{16 d}-\frac {5 b^{2} a^{3} {\mathrm e}^{-2 d x -2 c}}{4 d}+\frac {5 b^{4} a \,{\mathrm e}^{-2 d x -2 c}}{8 d}+\frac {5 b^{3} {\mathrm e}^{-3 d x -3 c} a^{2}}{12 d}-\frac {5 b^{5} {\mathrm e}^{-3 d x -3 c}}{96 d}-\frac {5 a \,b^{4} {\mathrm e}^{-4 d x -4 c}}{64 d}+\frac {b^{5} {\mathrm e}^{-5 d x -5 c}}{160 d}\)	\(344\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sinh(d*x+c))^5,x,method=_RETURNVERBOSE)

[Out]

a^5*x+1/3*(-5/16*b^5+5/2*a^2*b^3)/d*cosh(3*d*x+3*c)+1/2*(-5/2*a*b^4+5*a^3*b^2)*sinh(2*d*x+2*c)/d+(5/8*b^5-15/2
*a^2*b^3+5*a^4*b)/d*cosh(d*x+c)+15/8*x*a*b^4-5*x*a^3*b^2+1/80*b^5/d*cosh(5*d*x+5*c)+5/32*a*b^4*sinh(4*d*x+4*c)
/d

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Maxima [A]
time = 0.29, size = 272, normalized size = 1.49 \begin {gather*} \frac {5}{64} \, a b^{4} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {5}{4} \, a^{3} b^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + a^{5} x + \frac {1}{480} \, b^{5} {\left (\frac {3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} - \frac {25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {150 \, e^{\left (d x + c\right )}}{d} + \frac {150 \, e^{\left (-d x - c\right )}}{d} - \frac {25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} + \frac {5}{12} \, a^{2} b^{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 )} + \frac {5 \, a^{4} b \cosh \left (d x + c\right )}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c))^5,x, algorithm="maxima")

[Out]

5/64*a*b^4*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) - 5/4*
a^3*b^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + a^5*x + 1/480*b^5*(3*e^(5*d*x + 5*c)/d - 25*e^(3*d*x
+ 3*c)/d + 150*e^(d*x + c)/d + 150*e^(-d*x - c)/d - 25*e^(-3*d*x - 3*c)/d + 3*e^(-5*d*x - 5*c)/d) + 5/12*a^2*b
^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) + 5*a^4*b*cosh(d*x + c)/d

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Fricas [A]
time = 0.47, size = 223, normalized size = 1.22 \begin {gather*} \frac {3 \, b^{5} \cosh \left (d x + c\right )^{5} + 15 \, b^{5} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 150 \, a b^{4} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 25 \, {\left (8 \, a^{2} b^{3} - b^{5}\right )} \cosh \left (d x + c\right )^{3} + 30 \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 15 \, {\left (2 \, b^{5} \cosh \left (d x + c\right )^{3} + 5 \, {\left (8 \, a^{2} b^{3} - b^{5}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 150 \, {\left (8 \, a^{4} b - 12 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (d x + c\right ) + 150 \, {\left (a b^{4} \cosh \left (d x + c\right )^{3} + 4 \, {\left (2 \, a^{3} b^{2} - a b^{4}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c))^5,x, algorithm="fricas")

[Out]

1/240*(3*b^5*cosh(d*x + c)^5 + 15*b^5*cosh(d*x + c)*sinh(d*x + c)^4 + 150*a*b^4*cosh(d*x + c)*sinh(d*x + c)^3
+ 25*(8*a^2*b^3 - b^5)*cosh(d*x + c)^3 + 30*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*d*x + 15*(2*b^5*cosh(d*x + c)^3 +
5*(8*a^2*b^3 - b^5)*cosh(d*x + c))*sinh(d*x + c)^2 + 150*(8*a^4*b - 12*a^2*b^3 + b^5)*cosh(d*x + c) + 150*(a*b
^4*cosh(d*x + c)^3 + 4*(2*a^3*b^2 - a*b^4)*cosh(d*x + c))*sinh(d*x + c))/d

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Sympy [A]
time = 0.32, size = 314, normalized size = 1.72 \begin {gather*} \begin {cases} a^{5} x + \frac {5 a^{4} b \cosh {\left (c + d x \right )}}{d} + 5 a^{3} b^{2} x \sinh ^{2}{\left (c + d x \right )} - 5 a^{3} b^{2} x \cosh ^{2}{\left (c + d x \right )} + \frac {5 a^{3} b^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} + \frac {10 a^{2} b^{3} \sinh ^{2}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {20 a^{2} b^{3} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {15 a b^{4} x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac {15 a b^{4} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {15 a b^{4} x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {25 a b^{4} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} - \frac {15 a b^{4} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac {b^{5} \sinh ^{4}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {4 b^{5} \sinh ^{2}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {8 b^{5} \cosh ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh {\left (c \right )}\right )^{5} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c))**5,x)

[Out]

Piecewise((a**5*x + 5*a**4*b*cosh(c + d*x)/d + 5*a**3*b**2*x*sinh(c + d*x)**2 - 5*a**3*b**2*x*cosh(c + d*x)**2
+ 5*a**3*b**2*sinh(c + d*x)*cosh(c + d*x)/d + 10*a**2*b**3*sinh(c + d*x)**2*cosh(c + d*x)/d - 20*a**2*b**3*co
sh(c + d*x)**3/(3*d) + 15*a*b**4*x*sinh(c + d*x)**4/8 - 15*a*b**4*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 + 15*a
*b**4*x*cosh(c + d*x)**4/8 + 25*a*b**4*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) - 15*a*b**4*sinh(c + d*x)*cosh(c +
d*x)**3/(8*d) + b**5*sinh(c + d*x)**4*cosh(c + d*x)/d - 4*b**5*sinh(c + d*x)**2*cosh(c + d*x)**3/(3*d) + 8*b*
*5*cosh(c + d*x)**5/(15*d), Ne(d, 0)), (x*(a + b*sinh(c))**5, True))

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Giac [A]
time = 0.43, size = 269, normalized size = 1.47 \begin {gather*} \frac {b^{5} e^{\left (5 \, d x + 5 \, c\right )}}{160 \, d} + \frac {5 \, a b^{4} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, d} - \frac {5 \, a b^{4} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac {b^{5} e^{\left (-5 \, d x - 5 \, c\right )}}{160 \, d} + \frac {1}{8} \, {\left (8 \, a^{5} - 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x + \frac {5 \, {\left (8 \, a^{2} b^{3} - b^{5}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{96 \, d} + \frac {5 \, {\left (2 \, a^{3} b^{2} - a b^{4}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac {5 \, {\left (8 \, a^{4} b - 12 \, a^{2} b^{3} + b^{5}\right )} e^{\left (d x + c\right )}}{16 \, d} + \frac {5 \, {\left (8 \, a^{4} b - 12 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-d x - c\right )}}{16 \, d} - \frac {5 \, {\left (2 \, a^{3} b^{2} - a b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} + \frac {5 \, {\left (8 \, a^{2} b^{3} - b^{5}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sinh(d*x+c))^5,x, algorithm="giac")

[Out]

1/160*b^5*e^(5*d*x + 5*c)/d + 5/64*a*b^4*e^(4*d*x + 4*c)/d - 5/64*a*b^4*e^(-4*d*x - 4*c)/d + 1/160*b^5*e^(-5*d
*x - 5*c)/d + 1/8*(8*a^5 - 40*a^3*b^2 + 15*a*b^4)*x + 5/96*(8*a^2*b^3 - b^5)*e^(3*d*x + 3*c)/d + 5/8*(2*a^3*b^
2 - a*b^4)*e^(2*d*x + 2*c)/d + 5/16*(8*a^4*b - 12*a^2*b^3 + b^5)*e^(d*x + c)/d + 5/16*(8*a^4*b - 12*a^2*b^3 +
b^5)*e^(-d*x - c)/d - 5/8*(2*a^3*b^2 - a*b^4)*e^(-2*d*x - 2*c)/d + 5/96*(8*a^2*b^3 - b^5)*e^(-3*d*x - 3*c)/d

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Mupad [B]
time = 0.60, size = 160, normalized size = 0.87 \begin {gather*} \frac {75\,b^5\,\mathrm {cosh}\left (c+d\,x\right )-\frac {25\,b^5\,\mathrm {cosh}\left (3\,c+3\,d\,x\right )}{2}+\frac {3\,b^5\,\mathrm {cosh}\left (5\,c+5\,d\,x\right )}{2}-900\,a^2\,b^3\,\mathrm {cosh}\left (c+d\,x\right )-150\,a\,b^4\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+\frac {75\,a\,b^4\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+100\,a^2\,b^3\,\mathrm {cosh}\left (3\,c+3\,d\,x\right )+300\,a^3\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+600\,a^4\,b\,\mathrm {cosh}\left (c+d\,x\right )+120\,a^5\,d\,x+225\,a\,b^4\,d\,x-600\,a^3\,b^2\,d\,x}{120\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x))^5,x)

[Out]

(75*b^5*cosh(c + d*x) - (25*b^5*cosh(3*c + 3*d*x))/2 + (3*b^5*cosh(5*c + 5*d*x))/2 - 900*a^2*b^3*cosh(c + d*x)
- 150*a*b^4*sinh(2*c + 2*d*x) + (75*a*b^4*sinh(4*c + 4*d*x))/4 + 100*a^2*b^3*cosh(3*c + 3*d*x) + 300*a^3*b^2*
sinh(2*c + 2*d*x) + 600*a^4*b*cosh(c + d*x) + 120*a^5*d*x + 225*a*b^4*d*x - 600*a^3*b^2*d*x)/(120*d)

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