3.151 \(\int \sinh ^2(c+d x) (a+b \sinh ^3(c+d x))^2 \, dx\)

Optimal. Leaf size=180 \[ \frac {a^2 \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {a^2 x}{2}+\frac {2 a b \cosh ^5(c+d x)}{5 d}-\frac {4 a b \cosh ^3(c+d x)}{3 d}+\frac {2 a b \cosh (c+d x)}{d}+\frac {b^2 \sinh ^7(c+d x) \cosh (c+d x)}{8 d}-\frac {7 b^2 \sinh ^5(c+d x) \cosh (c+d x)}{48 d}+\frac {35 b^2 \sinh ^3(c+d x) \cosh (c+d x)}{192 d}-\frac {35 b^2 \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac {35 b^2 x}{128} \]

[Out]

-1/2*a^2*x+35/128*b^2*x+2*a*b*cosh(d*x+c)/d-4/3*a*b*cosh(d*x+c)^3/d+2/5*a*b*cosh(d*x+c)^5/d+1/2*a^2*cosh(d*x+c
)*sinh(d*x+c)/d-35/128*b^2*cosh(d*x+c)*sinh(d*x+c)/d+35/192*b^2*cosh(d*x+c)*sinh(d*x+c)^3/d-7/48*b^2*cosh(d*x+
c)*sinh(d*x+c)^5/d+1/8*b^2*cosh(d*x+c)*sinh(d*x+c)^7/d

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Rubi [A]  time = 0.15, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3220, 2635, 8, 2633} \[ \frac {a^2 \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {a^2 x}{2}+\frac {2 a b \cosh ^5(c+d x)}{5 d}-\frac {4 a b \cosh ^3(c+d x)}{3 d}+\frac {2 a b \cosh (c+d x)}{d}+\frac {b^2 \sinh ^7(c+d x) \cosh (c+d x)}{8 d}-\frac {7 b^2 \sinh ^5(c+d x) \cosh (c+d x)}{48 d}+\frac {35 b^2 \sinh ^3(c+d x) \cosh (c+d x)}{192 d}-\frac {35 b^2 \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac {35 b^2 x}{128} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*x)/2 + (35*b^2*x)/128 + (2*a*b*Cosh[c + d*x])/d - (4*a*b*Cosh[c + d*x]^3)/(3*d) + (2*a*b*Cosh[c + d*x]^5
)/(5*d) + (a^2*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) - (35*b^2*Cosh[c + d*x]*Sinh[c + d*x])/(128*d) + (35*b^2*Cos
h[c + d*x]*Sinh[c + d*x]^3)/(192*d) - (7*b^2*Cosh[c + d*x]*Sinh[c + d*x]^5)/(48*d) + (b^2*Cosh[c + d*x]*Sinh[c
+ d*x]^7)/(8*d)

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rubi steps

\begin {align*} \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right )^2 \, dx &=-\int \left (-a^2 \sinh ^2(c+d x)-2 a b \sinh ^5(c+d x)-b^2 \sinh ^8(c+d x)\right ) \, dx\\ &=a^2 \int \sinh ^2(c+d x) \, dx+(2 a b) \int \sinh ^5(c+d x) \, dx+b^2 \int \sinh ^8(c+d x) \, dx\\ &=\frac {a^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {b^2 \cosh (c+d x) \sinh ^7(c+d x)}{8 d}-\frac {1}{2} a^2 \int 1 \, dx-\frac {1}{8} \left (7 b^2\right ) \int \sinh ^6(c+d x) \, dx+\frac {(2 a b) \operatorname {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac {a^2 x}{2}+\frac {2 a b \cosh (c+d x)}{d}-\frac {4 a b \cosh ^3(c+d x)}{3 d}+\frac {2 a b \cosh ^5(c+d x)}{5 d}+\frac {a^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {7 b^2 \cosh (c+d x) \sinh ^5(c+d x)}{48 d}+\frac {b^2 \cosh (c+d x) \sinh ^7(c+d x)}{8 d}+\frac {1}{48} \left (35 b^2\right ) \int \sinh ^4(c+d x) \, dx\\ &=-\frac {a^2 x}{2}+\frac {2 a b \cosh (c+d x)}{d}-\frac {4 a b \cosh ^3(c+d x)}{3 d}+\frac {2 a b \cosh ^5(c+d x)}{5 d}+\frac {a^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {35 b^2 \cosh (c+d x) \sinh ^3(c+d x)}{192 d}-\frac {7 b^2 \cosh (c+d x) \sinh ^5(c+d x)}{48 d}+\frac {b^2 \cosh (c+d x) \sinh ^7(c+d x)}{8 d}-\frac {1}{64} \left (35 b^2\right ) \int \sinh ^2(c+d x) \, dx\\ &=-\frac {a^2 x}{2}+\frac {2 a b \cosh (c+d x)}{d}-\frac {4 a b \cosh ^3(c+d x)}{3 d}+\frac {2 a b \cosh ^5(c+d x)}{5 d}+\frac {a^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {35 b^2 \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {35 b^2 \cosh (c+d x) \sinh ^3(c+d x)}{192 d}-\frac {7 b^2 \cosh (c+d x) \sinh ^5(c+d x)}{48 d}+\frac {b^2 \cosh (c+d x) \sinh ^7(c+d x)}{8 d}+\frac {1}{128} \left (35 b^2\right ) \int 1 \, dx\\ &=-\frac {a^2 x}{2}+\frac {35 b^2 x}{128}+\frac {2 a b \cosh (c+d x)}{d}-\frac {4 a b \cosh ^3(c+d x)}{3 d}+\frac {2 a b \cosh ^5(c+d x)}{5 d}+\frac {a^2 \cosh (c+d x) \sinh (c+d x)}{2 d}-\frac {35 b^2 \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {35 b^2 \cosh (c+d x) \sinh ^3(c+d x)}{192 d}-\frac {7 b^2 \cosh (c+d x) \sinh ^5(c+d x)}{48 d}+\frac {b^2 \cosh (c+d x) \sinh ^7(c+d x)}{8 d}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 133, normalized size = 0.74 \[ \frac {3840 a^2 \sinh (2 (c+d x))-7680 a^2 c-7680 a^2 d x+19200 a b \cosh (c+d x)-3200 a b \cosh (3 (c+d x))+384 a b \cosh (5 (c+d x))-3360 b^2 \sinh (2 (c+d x))+840 b^2 \sinh (4 (c+d x))-160 b^2 \sinh (6 (c+d x))+15 b^2 \sinh (8 (c+d x))+4200 b^2 c+4200 b^2 d x}{15360 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7680*a^2*c + 4200*b^2*c - 7680*a^2*d*x + 4200*b^2*d*x + 19200*a*b*Cosh[c + d*x] - 3200*a*b*Cosh[3*(c + d*x)]
+ 384*a*b*Cosh[5*(c + d*x)] + 3840*a^2*Sinh[2*(c + d*x)] - 3360*b^2*Sinh[2*(c + d*x)] + 840*b^2*Sinh[4*(c + d
*x)] - 160*b^2*Sinh[6*(c + d*x)] + 15*b^2*Sinh[8*(c + d*x)])/(15360*d)

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fricas [A]  time = 1.53, size = 274, normalized size = 1.52 \[ \frac {15 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + 48 \, a b \cosh \left (d x + c\right )^{5} + 240 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 15 \, {\left (7 \, b^{2} \cosh \left (d x + c\right )^{3} - 8 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} - 400 \, a b \cosh \left (d x + c\right )^{3} + 5 \, {\left (21 \, b^{2} \cosh \left (d x + c\right )^{5} - 80 \, b^{2} \cosh \left (d x + c\right )^{3} + 84 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 15 \, {\left (64 \, a^{2} - 35 \, b^{2}\right )} d x + 2400 \, a b \cosh \left (d x + c\right ) + 240 \, {\left (2 \, a b \cosh \left (d x + c\right )^{3} - 5 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 15 \, {\left (b^{2} \cosh \left (d x + c\right )^{7} - 8 \, b^{2} \cosh \left (d x + c\right )^{5} + 28 \, b^{2} \cosh \left (d x + c\right )^{3} + 8 \, {\left (8 \, a^{2} - 7 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{1920 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*(15*b^2*cosh(d*x + c)*sinh(d*x + c)^7 + 48*a*b*cosh(d*x + c)^5 + 240*a*b*cosh(d*x + c)*sinh(d*x + c)^4
+ 15*(7*b^2*cosh(d*x + c)^3 - 8*b^2*cosh(d*x + c))*sinh(d*x + c)^5 - 400*a*b*cosh(d*x + c)^3 + 5*(21*b^2*cosh(
d*x + c)^5 - 80*b^2*cosh(d*x + c)^3 + 84*b^2*cosh(d*x + c))*sinh(d*x + c)^3 - 15*(64*a^2 - 35*b^2)*d*x + 2400*
a*b*cosh(d*x + c) + 240*(2*a*b*cosh(d*x + c)^3 - 5*a*b*cosh(d*x + c))*sinh(d*x + c)^2 + 15*(b^2*cosh(d*x + c)^
7 - 8*b^2*cosh(d*x + c)^5 + 28*b^2*cosh(d*x + c)^3 + 8*(8*a^2 - 7*b^2)*cosh(d*x + c))*sinh(d*x + c))/d

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giac [A]  time = 0.17, size = 260, normalized size = 1.44 \[ -\frac {1}{128} \, {\left (64 \, a^{2} - 35 \, b^{2}\right )} x + \frac {b^{2} e^{\left (8 \, d x + 8 \, c\right )}}{2048 \, d} - \frac {b^{2} e^{\left (6 \, d x + 6 \, c\right )}}{192 \, d} + \frac {a b e^{\left (5 \, d x + 5 \, c\right )}}{80 \, d} + \frac {7 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )}}{256 \, d} - \frac {5 \, a b e^{\left (3 \, d x + 3 \, c\right )}}{48 \, d} + \frac {5 \, a b e^{\left (d x + c\right )}}{8 \, d} + \frac {5 \, a b e^{\left (-d x - c\right )}}{8 \, d} - \frac {5 \, a b e^{\left (-3 \, d x - 3 \, c\right )}}{48 \, d} - \frac {7 \, b^{2} e^{\left (-4 \, d x - 4 \, c\right )}}{256 \, d} + \frac {a b e^{\left (-5 \, d x - 5 \, c\right )}}{80 \, d} + \frac {b^{2} e^{\left (-6 \, d x - 6 \, c\right )}}{192 \, d} - \frac {b^{2} e^{\left (-8 \, d x - 8 \, c\right )}}{2048 \, d} + \frac {{\left (8 \, a^{2} - 7 \, b^{2}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d} - \frac {{\left (8 \, a^{2} - 7 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(64*a^2 - 35*b^2)*x + 1/2048*b^2*e^(8*d*x + 8*c)/d - 1/192*b^2*e^(6*d*x + 6*c)/d + 1/80*a*b*e^(5*d*x +
5*c)/d + 7/256*b^2*e^(4*d*x + 4*c)/d - 5/48*a*b*e^(3*d*x + 3*c)/d + 5/8*a*b*e^(d*x + c)/d + 5/8*a*b*e^(-d*x -
c)/d - 5/48*a*b*e^(-3*d*x - 3*c)/d - 7/256*b^2*e^(-4*d*x - 4*c)/d + 1/80*a*b*e^(-5*d*x - 5*c)/d + 1/192*b^2*e^
(-6*d*x - 6*c)/d - 1/2048*b^2*e^(-8*d*x - 8*c)/d + 1/64*(8*a^2 - 7*b^2)*e^(2*d*x + 2*c)/d - 1/64*(8*a^2 - 7*b^
2)*e^(-2*d*x - 2*c)/d

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maple [A]  time = 0.05, size = 122, normalized size = 0.68 \[ \frac {b^{2} \left (\left (\frac {\left (\sinh ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sinh ^{5}\left (d x +c \right )\right )}{48}+\frac {35 \left (\sinh ^{3}\left (d x +c \right )\right )}{192}-\frac {35 \sinh \left (d x +c \right )}{128}\right ) \cosh \left (d x +c \right )+\frac {35 d x}{128}+\frac {35 c}{128}\right )+2 a b \left (\frac {8}{15}+\frac {\left (\sinh ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sinh ^{2}\left (d x +c \right )\right )}{15}\right ) \cosh \left (d x +c \right )+a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^2*((1/8*sinh(d*x+c)^7-7/48*sinh(d*x+c)^5+35/192*sinh(d*x+c)^3-35/128*sinh(d*x+c))*cosh(d*x+c)+35/128*d*
x+35/128*c)+2*a*b*(8/15+1/5*sinh(d*x+c)^4-4/15*sinh(d*x+c)^2)*cosh(d*x+c)+a^2*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2
*d*x-1/2*c))

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maxima [A]  time = 0.37, size = 237, normalized size = 1.32 \[ -\frac {1}{8} \, a^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{6144} \, b^{2} {\left (\frac {{\left (32 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 672 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3\right )} e^{\left (8 \, d x + 8 \, c\right )}}{d} - \frac {1680 \, {\left (d x + c\right )}}{d} - \frac {672 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 32 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d}\right )} + \frac {1}{240} \, a b {\left (\frac {3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} - \frac {25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {150 \, e^{\left (d x + c\right )}}{d} + \frac {150 \, e^{\left (-d x - c\right )}}{d} - \frac {25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/6144*b^2*((32*e^(-2*d*x - 2*c) - 168*e^(-4*d*x - 4
*c) + 672*e^(-6*d*x - 6*c) - 3)*e^(8*d*x + 8*c)/d - 1680*(d*x + c)/d - (672*e^(-2*d*x - 2*c) - 168*e^(-4*d*x -
4*c) + 32*e^(-6*d*x - 6*c) - 3*e^(-8*d*x - 8*c))/d) + 1/240*a*b*(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)

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mupad [B]  time = 1.63, size = 126, normalized size = 0.70 \[ \frac {480\,a^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-420\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+105\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )-20\,b^2\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )+\frac {15\,b^2\,\mathrm {sinh}\left (8\,c+8\,d\,x\right )}{8}+2400\,a\,b\,\mathrm {cosh}\left (c+d\,x\right )-400\,a\,b\,\mathrm {cosh}\left (3\,c+3\,d\,x\right )+48\,a\,b\,\mathrm {cosh}\left (5\,c+5\,d\,x\right )-960\,a^2\,d\,x+525\,b^2\,d\,x}{1920\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(480*a^2*sinh(2*c + 2*d*x) - 420*b^2*sinh(2*c + 2*d*x) + 105*b^2*sinh(4*c + 4*d*x) - 20*b^2*sinh(6*c + 6*d*x)
+ (15*b^2*sinh(8*c + 8*d*x))/8 + 2400*a*b*cosh(c + d*x) - 400*a*b*cosh(3*c + 3*d*x) + 48*a*b*cosh(5*c + 5*d*x)
- 960*a^2*d*x + 525*b^2*d*x)/(1920*d)

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sympy [A]  time = 8.65, size = 340, normalized size = 1.89 \[ \begin {cases} \frac {a^{2} x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac {a^{2} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {a^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} + \frac {2 a b \sinh ^{4}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {8 a b \sinh ^{2}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {16 a b \cosh ^{5}{\left (c + d x \right )}}{15 d} + \frac {35 b^{2} x \sinh ^{8}{\left (c + d x \right )}}{128} - \frac {35 b^{2} x \sinh ^{6}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{32} + \frac {105 b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{64} - \frac {35 b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{6}{\left (c + d x \right )}}{32} + \frac {35 b^{2} x \cosh ^{8}{\left (c + d x \right )}}{128} + \frac {93 b^{2} \sinh ^{7}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{128 d} - \frac {511 b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{384 d} + \frac {385 b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{384 d} - \frac {35 b^{2} \sinh {\left (c + d x \right )} \cosh ^{7}{\left (c + d x \right )}}{128 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{3}{\relax (c )}\right )^{2} \sinh ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)**2*(a+b*sinh(d*x+c)**3)**2,x)

[Out]

Piecewise((a**2*x*sinh(c + d*x)**2/2 - a**2*x*cosh(c + d*x)**2/2 + a**2*sinh(c + d*x)*cosh(c + d*x)/(2*d) + 2*
a*b*sinh(c + d*x)**4*cosh(c + d*x)/d - 8*a*b*sinh(c + d*x)**2*cosh(c + d*x)**3/(3*d) + 16*a*b*cosh(c + d*x)**5
/(15*d) + 35*b**2*x*sinh(c + d*x)**8/128 - 35*b**2*x*sinh(c + d*x)**6*cosh(c + d*x)**2/32 + 105*b**2*x*sinh(c
+ d*x)**4*cosh(c + d*x)**4/64 - 35*b**2*x*sinh(c + d*x)**2*cosh(c + d*x)**6/32 + 35*b**2*x*cosh(c + d*x)**8/12
8 + 93*b**2*sinh(c + d*x)**7*cosh(c + d*x)/(128*d) - 511*b**2*sinh(c + d*x)**5*cosh(c + d*x)**3/(384*d) + 385*
b**2*sinh(c + d*x)**3*cosh(c + d*x)**5/(384*d) - 35*b**2*sinh(c + d*x)*cosh(c + d*x)**7/(128*d), Ne(d, 0)), (x
*(a + b*sinh(c)**3)**2*sinh(c)**2, True))

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