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

Optimal. Leaf size=203 \[ \frac{b \left (88 a^2-68 a b+15 b^2\right ) \sinh (c+d x) \cosh ^3(c+d x)}{192 d}+\frac{\left (-48 a^2 b+64 a^3+24 a b^2-5 b^3\right ) \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac{1}{128} x \left (-48 a^2 b+64 a^3+24 a b^2-5 b^3\right )+\frac{b \sinh (c+d x) \cosh ^7(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}+\frac{b \sinh (c+d x) \cosh ^5(c+d x) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{48 d} \]

[Out]

((64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*x)/128 + ((64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*Cosh[c + d*x]*Sinh[c
+ d*x])/(128*d) + (b*(88*a^2 - 68*a*b + 15*b^2)*Cosh[c + d*x]^3*Sinh[c + d*x])/(192*d) + (b*Cosh[c + d*x]^7*Si
nh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2)^2)/(8*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x]*(a*(8*a - b) - (8*a - 5*
b)*(a - b)*Tanh[c + d*x]^2))/(48*d)

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Rubi [A]  time = 0.268032, antiderivative size = 203, 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 = {3191, 413, 526, 385, 199, 206} \[ \frac{b \left (88 a^2-68 a b+15 b^2\right ) \sinh (c+d x) \cosh ^3(c+d x)}{192 d}+\frac{\left (-48 a^2 b+64 a^3+24 a b^2-5 b^3\right ) \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac{1}{128} x \left (-48 a^2 b+64 a^3+24 a b^2-5 b^3\right )+\frac{b \sinh (c+d x) \cosh ^7(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}+\frac{b \sinh (c+d x) \cosh ^5(c+d x) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*x)/128 + ((64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*Cosh[c + d*x]*Sinh[c
+ d*x])/(128*d) + (b*(88*a^2 - 68*a*b + 15*b^2)*Cosh[c + d*x]^3*Sinh[c + d*x])/(192*d) + (b*Cosh[c + d*x]^7*Si
nh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2)^2)/(8*d) + (b*Cosh[c + d*x]^5*Sinh[c + d*x]*(a*(8*a - b) - (8*a - 5*
b)*(a - b)*Tanh[c + d*x]^2))/(48*d)

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 413

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

Rule 526

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)/(a*b*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n
)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))*x
^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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])

Rubi steps

\begin{align*} \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-(a-b) x^2\right )^3}{\left (1-x^2\right )^5} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b \cosh ^7(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}-\frac{\operatorname{Subst}\left (\int \frac{\left (-a (8 a-b)+(8 a-5 b) (a-b) x^2\right ) \left (a+(-a+b) x^2\right )}{\left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{b \cosh ^7(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{48 d}-\frac{\operatorname{Subst}\left (\int \frac{-a (6 a-b) (8 a-b)+3 (8 a-5 b) (a-b) (2 a-b) x^2}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=\frac{b \left (88 a^2-68 a b+15 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac{b \cosh ^7(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{48 d}+\frac{\left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{64 d}\\ &=\frac{\left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac{b \left (88 a^2-68 a b+15 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac{b \cosh ^7(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{48 d}+\frac{\left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{128 d}\\ &=\frac{1}{128} \left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) x+\frac{\left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac{b \left (88 a^2-68 a b+15 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac{b \cosh ^7(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}+\frac{b \cosh ^5(c+d x) \sinh (c+d x) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{48 d}\\ \end{align*}

Mathematica [A]  time = 0.302241, size = 120, normalized size = 0.59 \[ \frac{24 \left (-48 a^2 b+64 a^3+24 a b^2-5 b^3\right ) (c+d x)+24 b \left (12 a^2-6 a b+b^2\right ) \sinh (4 (c+d x))+48 \left (16 a^3-3 a b^2+b^3\right ) \sinh (2 (c+d x))+16 b^2 (3 a-b) \sinh (6 (c+d x))+3 b^3 \sinh (8 (c+d x))}{3072 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(24*(64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*(c + d*x) + 48*(16*a^3 - 3*a*b^2 + b^3)*Sinh[2*(c + d*x)] + 24*b*(1
2*a^2 - 6*a*b + b^2)*Sinh[4*(c + d*x)] + 16*(3*a - b)*b^2*Sinh[6*(c + d*x)] + 3*b^3*Sinh[8*(c + d*x)])/(3072*d
)

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Maple [A]  time = 0.032, size = 216, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5} \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{8}}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3} \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{48}}+{\frac{5\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}{64}}-{\frac{5\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{128}}-{\frac{5\,dx}{128}}-{\frac{5\,c}{128}} \right ) +3\,a{b}^{2} \left ( 1/6\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3} \left ( \cosh \left ( dx+c \right ) \right ) ^{3}-1/8\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}+1/16\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) +1/16\,dx+c/16 \right ) +3\,{a}^{2}b \left ( 1/4\,\sinh \left ( dx+c \right ) \left ( \cosh \left ( dx+c \right ) \right ) ^{3}-1/8\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/8\,dx-c/8 \right ) +{a}^{3} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^3*(1/8*sinh(d*x+c)^5*cosh(d*x+c)^3-5/48*sinh(d*x+c)^3*cosh(d*x+c)^3+5/64*sinh(d*x+c)*cosh(d*x+c)^3-5/12
8*cosh(d*x+c)*sinh(d*x+c)-5/128*d*x-5/128*c)+3*a*b^2*(1/6*sinh(d*x+c)^3*cosh(d*x+c)^3-1/8*sinh(d*x+c)*cosh(d*x
+c)^3+1/16*cosh(d*x+c)*sinh(d*x+c)+1/16*d*x+1/16*c)+3*a^2*b*(1/4*sinh(d*x+c)*cosh(d*x+c)^3-1/8*cosh(d*x+c)*sin
h(d*x+c)-1/8*d*x-1/8*c)+a^3*(1/2*cosh(d*x+c)*sinh(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 1.11971, size = 387, normalized size = 1.91 \begin{align*} \frac{1}{8} \, a^{3}{\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^{3}{\left (\frac{{\left (16 \, e^{\left (-2 \, d x - 2 \, c\right )} - 24 \, e^{\left (-4 \, d x - 4 \, c\right )} - 48 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3\right )} e^{\left (8 \, d x + 8 \, c\right )}}{d} + \frac{240 \,{\left (d x + c\right )}}{d} + \frac{48 \, e^{\left (-2 \, d x - 2 \, c\right )} + 24 \, e^{\left (-4 \, d x - 4 \, c\right )} - 16 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d}\right )} - \frac{1}{128} \, a b^{2}{\left (\frac{{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} - \frac{24 \,{\left (d x + c\right )}}{d} - \frac{3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} - \frac{3}{64} \, a^{2} b{\left (\frac{8 \,{\left (d x + c\right )}}{d} - \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^3*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) - 1/6144*b^3*((16*e^(-2*d*x - 2*c) - 24*e^(-4*d*x - 4*c
) - 48*e^(-6*d*x - 6*c) - 3)*e^(8*d*x + 8*c)/d + 240*(d*x + c)/d + (48*e^(-2*d*x - 2*c) + 24*e^(-4*d*x - 4*c)
- 16*e^(-6*d*x - 6*c) + 3*e^(-8*d*x - 8*c))/d) - 1/128*a*b^2*((3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) - 1)*e^
(6*d*x + 6*c)/d - 24*(d*x + c)/d - (3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) - e^(-6*d*x - 6*c))/d) - 3/64*a^2*
b*(8*(d*x + c)/d - e^(4*d*x + 4*c)/d + e^(-4*d*x - 4*c)/d)

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Fricas [A]  time = 1.48108, size = 626, normalized size = 3.08 \begin{align*} \frac{3 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + 3 \,{\left (7 \, b^{3} \cosh \left (d x + c\right )^{3} + 4 \,{\left (3 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} +{\left (21 \, b^{3} \cosh \left (d x + c\right )^{5} + 40 \,{\left (3 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{3} + 12 \,{\left (12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \,{\left (64 \, a^{3} - 48 \, a^{2} b + 24 \, a b^{2} - 5 \, b^{3}\right )} d x + 3 \,{\left (b^{3} \cosh \left (d x + c\right )^{7} + 4 \,{\left (3 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{5} + 4 \,{\left (12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{3} + 4 \,{\left (16 \, a^{3} - 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(3*b^3*cosh(d*x + c)*sinh(d*x + c)^7 + 3*(7*b^3*cosh(d*x + c)^3 + 4*(3*a*b^2 - b^3)*cosh(d*x + c))*sinh(
d*x + c)^5 + (21*b^3*cosh(d*x + c)^5 + 40*(3*a*b^2 - b^3)*cosh(d*x + c)^3 + 12*(12*a^2*b - 6*a*b^2 + b^3)*cosh
(d*x + c))*sinh(d*x + c)^3 + 3*(64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*d*x + 3*(b^3*cosh(d*x + c)^7 + 4*(3*a*b^
2 - b^3)*cosh(d*x + c)^5 + 4*(12*a^2*b - 6*a*b^2 + b^3)*cosh(d*x + c)^3 + 4*(16*a^3 - 3*a*b^2 + b^3)*cosh(d*x
+ c))*sinh(d*x + c))/d

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Sympy [A]  time = 13.7037, size = 559, normalized size = 2.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**3*x*sinh(c + d*x)**2/2 + a**3*x*cosh(c + d*x)**2/2 + a**3*sinh(c + d*x)*cosh(c + d*x)/(2*d) - 3
*a**2*b*x*sinh(c + d*x)**4/8 + 3*a**2*b*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 - 3*a**2*b*x*cosh(c + d*x)**4/8
+ 3*a**2*b*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) + 3*a**2*b*sinh(c + d*x)*cosh(c + d*x)**3/(8*d) - 3*a*b**2*x*s
inh(c + d*x)**6/16 + 9*a*b**2*x*sinh(c + d*x)**4*cosh(c + d*x)**2/16 - 9*a*b**2*x*sinh(c + d*x)**2*cosh(c + d*
x)**4/16 + 3*a*b**2*x*cosh(c + d*x)**6/16 + 3*a*b**2*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) + a*b**2*sinh(c + d
*x)**3*cosh(c + d*x)**3/(2*d) - 3*a*b**2*sinh(c + d*x)*cosh(c + d*x)**5/(16*d) - 5*b**3*x*sinh(c + d*x)**8/128
+ 5*b**3*x*sinh(c + d*x)**6*cosh(c + d*x)**2/32 - 15*b**3*x*sinh(c + d*x)**4*cosh(c + d*x)**4/64 + 5*b**3*x*s
inh(c + d*x)**2*cosh(c + d*x)**6/32 - 5*b**3*x*cosh(c + d*x)**8/128 + 5*b**3*sinh(c + d*x)**7*cosh(c + d*x)/(1
28*d) + 73*b**3*sinh(c + d*x)**5*cosh(c + d*x)**3/(384*d) - 55*b**3*sinh(c + d*x)**3*cosh(c + d*x)**5/(384*d)
+ 5*b**3*sinh(c + d*x)*cosh(c + d*x)**7/(128*d), Ne(d, 0)), (x*(a + b*sinh(c)**2)**3*cosh(c)**2, True))

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Giac [A]  time = 1.27872, size = 482, normalized size = 2.37 \begin{align*} \frac{3 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 48 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 16 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 288 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 768 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 144 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 48 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 48 \,{\left (64 \, a^{3} - 48 \, a^{2} b + 24 \, a b^{2} - 5 \, b^{3}\right )}{\left (d x + c\right )} -{\left (3200 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} - 2400 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 1200 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 250 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 768 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} - 144 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 48 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 288 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 144 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 48 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 16 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{3}\right )} e^{\left (-8 \, d x - 8 \, c\right )}}{6144 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6144*(3*b^3*e^(8*d*x + 8*c) + 48*a*b^2*e^(6*d*x + 6*c) - 16*b^3*e^(6*d*x + 6*c) + 288*a^2*b*e^(4*d*x + 4*c)
- 144*a*b^2*e^(4*d*x + 4*c) + 24*b^3*e^(4*d*x + 4*c) + 768*a^3*e^(2*d*x + 2*c) - 144*a*b^2*e^(2*d*x + 2*c) + 4
8*b^3*e^(2*d*x + 2*c) + 48*(64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*(d*x + c) - (3200*a^3*e^(8*d*x + 8*c) - 2400
*a^2*b*e^(8*d*x + 8*c) + 1200*a*b^2*e^(8*d*x + 8*c) - 250*b^3*e^(8*d*x + 8*c) + 768*a^3*e^(6*d*x + 6*c) - 144*
a*b^2*e^(6*d*x + 6*c) + 48*b^3*e^(6*d*x + 6*c) + 288*a^2*b*e^(4*d*x + 4*c) - 144*a*b^2*e^(4*d*x + 4*c) + 24*b^
3*e^(4*d*x + 4*c) + 48*a*b^2*e^(2*d*x + 2*c) - 16*b^3*e^(2*d*x + 2*c) + 3*b^3)*e^(-8*d*x - 8*c))/d