Integrand size = 23, antiderivative size = 203 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {1}{128} \left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) x+\frac {\left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac {b \left (88 a^2-68 a b+15 b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}+\frac {b \cosh ^7(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 d}+\frac {b \cosh ^5(c+d x) \sinh (c+d x) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{48 d} \]
1/128*(64*a^3-48*a^2*b+24*a*b^2-5*b^3)*x+1/128*(64*a^3-48*a^2*b+24*a*b^2-5 *b^3)*cosh(d*x+c)*sinh(d*x+c)/d+1/192*b*(88*a^2-68*a*b+15*b^2)*cosh(d*x+c) ^3*sinh(d*x+c)/d+1/8*b*cosh(d*x+c)^7*sinh(d*x+c)*(a-(a-b)*tanh(d*x+c)^2)^2 /d+1/48*b*cosh(d*x+c)^5*sinh(d*x+c)*(a*(8*a-b)-(8*a-5*b)*(a-b)*tanh(d*x+c) ^2)/d
Time = 2.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.59 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {24 \left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) (c+d x)+48 \left (16 a^3-3 a b^2+b^3\right ) \sinh (2 (c+d x))+24 b \left (12 a^2-6 a b+b^2\right ) \sinh (4 (c+d x))+16 (3 a-b) b^2 \sinh (6 (c+d x))+3 b^3 \sinh (8 (c+d x))}{3072 d} \]
(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*(12*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)
Time = 0.41 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3670, 315, 25, 401, 298, 215, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (i c+i d x)^2 \left (a-b \sin (i c+i d x)^2\right )^3dx\) |
| \(\Big \downarrow \) 3670 |
| \(\displaystyle \frac {\int \frac {\left (a-(a-b) \tanh ^2(c+d x)\right )^3}{\left (1-\tanh ^2(c+d x)\right )^5}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 \left (1-\tanh ^2(c+d x)\right )^4}-\frac {1}{8} \int -\frac {\left (a-(a-b) \tanh ^2(c+d x)\right ) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right )^4}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{8} \int \frac {\left (a-(a-b) \tanh ^2(c+d x)\right ) \left (a (8 a-b)-(8 a-5 b) (a-b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right )^4}d\tanh (c+d x)+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 \left (1-\tanh ^2(c+d x)\right )^4}}{d}\) |
| \(\Big \downarrow \) 401 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \int \frac {a \left (48 a^2-18 b a+5 b^2\right )-3 (a-b) \left (16 a^2-14 b a+5 b^2\right ) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)+\frac {b (12 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 \left (1-\tanh ^2(c+d x)\right )^4}}{d}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} \left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \int \frac {1}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {b \left (72 a^2-52 a b+15 b^2\right ) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b (12 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 \left (1-\tanh ^2(c+d x)\right )^4}}{d}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} \left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \left (\frac {1}{2} \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {b \left (72 a^2-52 a b+15 b^2\right ) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b (12 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 \left (1-\tanh ^2(c+d x)\right )^4}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {b \left (72 a^2-52 a b+15 b^2\right ) \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}+\frac {3}{4} \left (64 a^3-48 a^2 b+24 a b^2-5 b^3\right ) \left (\frac {1}{2} \text {arctanh}(\tanh (c+d x))+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )\right )+\frac {b (12 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{6 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{8 \left (1-\tanh ^2(c+d x)\right )^4}}{d}\) |
((b*Tanh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2)^2)/(8*(1 - Tanh[c + d*x]^2 )^4) + (((12*a - 5*b)*b*Tanh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2))/(6*(1 - Tanh[c + d*x]^2)^3) + ((b*(72*a^2 - 52*a*b + 15*b^2)*Tanh[c + d*x])/(4* (1 - Tanh[c + d*x]^2)^2) + (3*(64*a^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*(ArcT anh[Tanh[c + d*x]]/2 + Tanh[c + d*x]/(2*(1 - Tanh[c + d*x]^2))))/4)/6)/8)/ d
3.4.6.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ q/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1)) Int[(a + b*x^2)^(p + 1)*( c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + (b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L tQ[p, -1] && GtQ[q, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Su bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
Time = 0.05 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.06
\[\frac {a^{3} \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{4}-\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{8}-\frac {d x}{8}-\frac {c}{8}\right )+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )^{3}}{6}-\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{8}+\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{16}+\frac {d x}{16}+\frac {c}{16}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5} \cosh \left (d x +c \right )^{3}}{8}-\frac {5 \sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )^{3}}{48}+\frac {5 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{3}}{64}-\frac {5 \sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{128}-\frac {5 d x}{128}-\frac {5 c}{128}\right )}{d}\]
1/d*(a^3*(1/2*sinh(d*x+c)*cosh(d*x+c)+1/2*d*x+1/2*c)+3*a^2*b*(1/4*sinh(d*x +c)*cosh(d*x+c)^3-1/8*sinh(d*x+c)*cosh(d*x+c)-1/8*d*x-1/8*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*sinh(d*x+c) *cosh(d*x+c)+1/16*d*x+1/16*c)+b^3*(1/8*sinh(d*x+c)^5*cosh(d*x+c)^3-5/48*si nh(d*x+c)^3*cosh(d*x+c)^3+5/64*sinh(d*x+c)*cosh(d*x+c)^3-5/128*sinh(d*x+c) *cosh(d*x+c)-5/128*d*x-5/128*c))
Time = 0.28 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.27 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {3 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{7} + 3 \, {\left (7 \, b^{3} \cosh \left (d x + c\right )^{3} + 4 \, {\left (3 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + {\left (21 \, b^{3} \cosh \left (d x + c\right )^{5} + 40 \, {\left (3 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{3} + 12 \, {\left (12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (64 \, a^{3} - 48 \, a^{2} b + 24 \, a b^{2} - 5 \, b^{3}\right )} d x + 3 \, {\left (b^{3} \cosh \left (d x + c\right )^{7} + 4 \, {\left (3 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{5} + 4 \, {\left (12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{3} + 4 \, {\left (16 \, a^{3} - 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{384 \, d} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 559 vs. \(2 (190) = 380\).
Time = 0.70 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.75 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\begin {cases} - \frac {a^{3} x \sinh ^{2}{\left (c + d x \right )}}{2} + \frac {a^{3} x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {a^{3} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} - \frac {3 a^{2} b x \sinh ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} - \frac {3 a^{2} b x \cosh ^{4}{\left (c + d x \right )}}{8} + \frac {3 a^{2} b \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} + \frac {3 a^{2} b \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} - \frac {3 a b^{2} x \sinh ^{6}{\left (c + d x \right )}}{16} + \frac {9 a b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} - \frac {9 a b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac {3 a b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{16 d} + \frac {a b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} \sinh {\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{16 d} - \frac {5 b^{3} x \sinh ^{8}{\left (c + d x \right )}}{128} + \frac {5 b^{3} x \sinh ^{6}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{32} - \frac {15 b^{3} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{64} + \frac {5 b^{3} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{6}{\left (c + d x \right )}}{32} - \frac {5 b^{3} x \cosh ^{8}{\left (c + d x \right )}}{128} + \frac {5 b^{3} \sinh ^{7}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{128 d} + \frac {73 b^{3} \sinh ^{5}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{384 d} - \frac {55 b^{3} \sinh ^{3}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{384 d} + \frac {5 b^{3} \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 ^{2}{\left (c \right )}\right )^{3} \cosh ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((-a**3*x*sinh(c + d*x)**2/2 + a**3*x*cosh(c + d*x)**2/2 + a**3*s inh(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)*cos h(c + d*x)**3/(8*d) - 3*a*b**2*x*sinh(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*s inh(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*co sh(c + d*x)**4/64 + 5*b**3*x*sinh(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)/(128*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))
Time = 0.23 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.41 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {1}{8} \, a^{3} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{6144} \, b^{3} {\left (\frac {{\left (16 \, e^{\left (-2 \, d x - 2 \, c\right )} - 24 \, e^{\left (-4 \, d x - 4 \, c\right )} - 48 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3\right )} e^{\left (8 \, d x + 8 \, c\right )}}{d} + \frac {240 \, {\left (d x + c\right )}}{d} + \frac {48 \, e^{\left (-2 \, d x - 2 \, c\right )} + 24 \, e^{\left (-4 \, d x - 4 \, c\right )} - 16 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d}\right )} - \frac {1}{128} \, a b^{2} {\left (\frac {{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} - \frac {24 \, {\left (d x + c\right )}}{d} - \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} - \frac {3}{64} \, a^{2} b {\left (\frac {8 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} \]
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)
Time = 0.30 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.14 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {b^{3} e^{\left (8 \, d x + 8 \, c\right )}}{2048 \, d} - \frac {b^{3} e^{\left (-8 \, d x - 8 \, c\right )}}{2048 \, d} + \frac {1}{128} \, {\left (64 \, a^{3} - 48 \, a^{2} b + 24 \, a b^{2} - 5 \, b^{3}\right )} x + \frac {{\left (3 \, a b^{2} - b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )}}{384 \, d} + \frac {{\left (12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{256 \, d} + \frac {{\left (16 \, a^{3} - 3 \, a b^{2} + b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{128 \, d} - \frac {{\left (16 \, a^{3} - 3 \, a b^{2} + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{128 \, d} - \frac {{\left (12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{256 \, d} - \frac {{\left (3 \, a b^{2} - b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} \]
1/2048*b^3*e^(8*d*x + 8*c)/d - 1/2048*b^3*e^(-8*d*x - 8*c)/d + 1/128*(64*a ^3 - 48*a^2*b + 24*a*b^2 - 5*b^3)*x + 1/384*(3*a*b^2 - b^3)*e^(6*d*x + 6*c )/d + 1/256*(12*a^2*b - 6*a*b^2 + b^3)*e^(4*d*x + 4*c)/d + 1/128*(16*a^3 - 3*a*b^2 + b^3)*e^(2*d*x + 2*c)/d - 1/128*(16*a^3 - 3*a*b^2 + b^3)*e^(-2*d *x - 2*c)/d - 1/256*(12*a^2*b - 6*a*b^2 + b^3)*e^(-4*d*x - 4*c)/d - 1/384* (3*a*b^2 - b^3)*e^(-6*d*x - 6*c)/d
Time = 0.48 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.82 \[ \int \cosh ^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {96\,a^3\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+6\,b^3\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+3\,b^3\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )-2\,b^3\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )+\frac {3\,b^3\,\mathrm {sinh}\left (8\,c+8\,d\,x\right )}{8}-18\,a\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-18\,a\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+36\,a^2\,b\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+6\,a\,b^2\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )+192\,a^3\,d\,x-15\,b^3\,d\,x+72\,a\,b^2\,d\,x-144\,a^2\,b\,d\,x}{384\,d} \]
(96*a^3*sinh(2*c + 2*d*x) + 6*b^3*sinh(2*c + 2*d*x) + 3*b^3*sinh(4*c + 4*d *x) - 2*b^3*sinh(6*c + 6*d*x) + (3*b^3*sinh(8*c + 8*d*x))/8 - 18*a*b^2*sin h(2*c + 2*d*x) - 18*a*b^2*sinh(4*c + 4*d*x) + 36*a^2*b*sinh(4*c + 4*d*x) + 6*a*b^2*sinh(6*c + 6*d*x) + 192*a^3*d*x - 15*b^3*d*x + 72*a*b^2*d*x - 144 *a^2*b*d*x)/(384*d)