Integrand size = 23, antiderivative size = 238 \[ \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {3}{256} (4 a-b) \left (8 a^2-2 a b+b^2\right ) x+\frac {3 (4 a-b) \left (8 a^2-2 a b+b^2\right ) \cosh (c+d x) \sinh (c+d x)}{256 d}+\frac {(4 a-b) \left (8 a^2-2 a b+b^2\right ) \cosh ^3(c+d x) \sinh (c+d x)}{128 d}+\frac {b \left (44 a^2-28 a b+5 b^2\right ) \cosh ^5(c+d x) \sinh (c+d x)}{160 d}+\frac {b \cosh ^9(c+d x) \sinh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 d}+\frac {b \cosh ^7(c+d x) \sinh (c+d x) \left (a (10 a-b)-5 (a-b) (2 a-b) \tanh ^2(c+d x)\right )}{80 d} \]
3/256*(4*a-b)*(8*a^2-2*a*b+b^2)*x+3/256*(4*a-b)*(8*a^2-2*a*b+b^2)*cosh(d*x +c)*sinh(d*x+c)/d+1/128*(4*a-b)*(8*a^2-2*a*b+b^2)*cosh(d*x+c)^3*sinh(d*x+c )/d+1/160*b*(44*a^2-28*a*b+5*b^2)*cosh(d*x+c)^5*sinh(d*x+c)/d+1/10*b*cosh( d*x+c)^9*sinh(d*x+c)*(a-(a-b)*tanh(d*x+c)^2)^2/d+1/80*b*cosh(d*x+c)^7*sinh (d*x+c)*(a*(10*a-b)-5*(a-b)*(2*a-b)*tanh(d*x+c)^2)/d
Time = 2.63 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.61 \[ \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {120 (4 a-b) \left (8 a^2-2 a b+b^2\right ) (c+d x)+20 \left (128 a^3-24 a^2 b+b^3\right ) \sinh (2 (c+d x))+40 \left (8 a^3+12 a^2 b-6 a b^2+b^3\right ) \sinh (4 (c+d x))-10 b \left (-16 a^2+b^2\right ) \sinh (6 (c+d x))+5 (6 a-b) b^2 \sinh (8 (c+d x))+2 b^3 \sinh (10 (c+d x))}{10240 d} \]
(120*(4*a - b)*(8*a^2 - 2*a*b + b^2)*(c + d*x) + 20*(128*a^3 - 24*a^2*b + b^3)*Sinh[2*(c + d*x)] + 40*(8*a^3 + 12*a^2*b - 6*a*b^2 + b^3)*Sinh[4*(c + d*x)] - 10*b*(-16*a^2 + b^2)*Sinh[6*(c + d*x)] + 5*(6*a - b)*b^2*Sinh[8*( c + d*x)] + 2*b^3*Sinh[10*(c + d*x)])/(10240*d)
Time = 0.42 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3670, 315, 25, 401, 298, 215, 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.
| \(\displaystyle \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (i c+i d x)^4 \left (a-b \sin (i c+i d x)^2\right )^3dx\) |
| \(\Big \downarrow \) 3670 |
| \(\displaystyle \frac {\int \frac {\left (a-(a-b) \tanh ^2(c+d x)\right )^3}{\left (1-\tanh ^2(c+d x)\right )^6}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 \left (1-\tanh ^2(c+d x)\right )^5}-\frac {1}{10} \int -\frac {\left (a-(a-b) \tanh ^2(c+d x)\right ) \left (a (10 a-b)-5 (a-b) (2 a-b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right )^5}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{10} \int \frac {\left (a-(a-b) \tanh ^2(c+d x)\right ) \left (a (10 a-b)-5 (a-b) (2 a-b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right )^5}d\tanh (c+d x)+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 401 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \int \frac {a \left (80 a^2-22 b a+5 b^2\right )-5 (a-b) \left (16 a^2-10 b a+3 b^2\right ) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^4}d\tanh (c+d x)+\frac {b (14 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} (4 a-b) \left (8 a^2-2 a b+b^2\right ) \int \frac {1}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)+\frac {b \left (36 a^2-20 a b+5 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b (14 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} (4 a-b) \left (8 a^2-2 a b+b^2\right ) \left (\frac {3}{4} \int \frac {1}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {\tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b \left (36 a^2-20 a b+5 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b (14 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} (4 a-b) \left (8 a^2-2 a b+b^2\right ) \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b \left (36 a^2-20 a b+5 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b (14 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{10} \left (\frac {1}{8} \left (\frac {5}{2} (4 a-b) \left (8 a^2-2 a b+b^2\right ) \left (\frac {3}{4} \left (\frac {1}{2} \text {arctanh}(\tanh (c+d x))+\frac {\tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {\tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}\right )+\frac {b \left (36 a^2-20 a b+5 b^2\right ) \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )^3}\right )+\frac {b (14 a-5 b) \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )}{8 \left (1-\tanh ^2(c+d x)\right )^4}\right )+\frac {b \tanh (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{10 \left (1-\tanh ^2(c+d x)\right )^5}}{d}\) |
((b*Tanh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2)^2)/(10*(1 - Tanh[c + d*x]^ 2)^5) + (((14*a - 5*b)*b*Tanh[c + d*x]*(a - (a - b)*Tanh[c + d*x]^2))/(8*( 1 - Tanh[c + d*x]^2)^4) + ((b*(36*a^2 - 20*a*b + 5*b^2)*Tanh[c + d*x])/(2* (1 - Tanh[c + d*x]^2)^3) + (5*(4*a - b)*(8*a^2 - 2*a*b + b^2)*(Tanh[c + d* x]/(4*(1 - Tanh[c + d*x]^2)^2) + (3*(ArcTanh[Tanh[c + d*x]]/2 + Tanh[c + d *x]/(2*(1 - Tanh[c + d*x]^2))))/4))/2)/8)/10)/d
3.4.4.3.1 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 = 0.07 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.12
\[\frac {a^{3} \left (\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{2} b \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{5}}{6}-\frac {\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )}{6}-\frac {d x}{16}-\frac {c}{16}\right )+3 a \,b^{2} \left (\frac {\sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )^{5}}{8}-\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{5}}{16}+\frac {\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )}{16}+\frac {3 d x}{128}+\frac {3 c}{128}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5} \cosh \left (d x +c \right )^{5}}{10}-\frac {\sinh \left (d x +c \right )^{3} \cosh \left (d x +c \right )^{5}}{16}+\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )^{5}}{32}-\frac {\left (\frac {\cosh \left (d x +c \right )^{3}}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )}{32}-\frac {3 d x}{256}-\frac {3 c}{256}\right )}{d}\]
1/d*(a^3*((1/4*cosh(d*x+c)^3+3/8*cosh(d*x+c))*sinh(d*x+c)+3/8*d*x+3/8*c)+3 *a^2*b*(1/6*sinh(d*x+c)*cosh(d*x+c)^5-1/6*(1/4*cosh(d*x+c)^3+3/8*cosh(d*x+ c))*sinh(d*x+c)-1/16*d*x-1/16*c)+3*a*b^2*(1/8*sinh(d*x+c)^3*cosh(d*x+c)^5- 1/16*sinh(d*x+c)*cosh(d*x+c)^5+1/16*(1/4*cosh(d*x+c)^3+3/8*cosh(d*x+c))*si nh(d*x+c)+3/128*d*x+3/128*c)+b^3*(1/10*sinh(d*x+c)^5*cosh(d*x+c)^5-1/16*si nh(d*x+c)^3*cosh(d*x+c)^5+1/32*sinh(d*x+c)*cosh(d*x+c)^5-1/32*(1/4*cosh(d* x+c)^3+3/8*cosh(d*x+c))*sinh(d*x+c)-3/256*d*x-3/256*c))
Time = 0.26 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.58 \[ \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {5 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{9} + 10 \, {\left (6 \, b^{3} \cosh \left (d x + c\right )^{3} + {\left (6 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{7} + {\left (126 \, b^{3} \cosh \left (d x + c\right )^{5} + 70 \, {\left (6 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{3} + 15 \, {\left (16 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{5} + 10 \, {\left (6 \, b^{3} \cosh \left (d x + c\right )^{7} + 7 \, {\left (6 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (16 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{3} + 4 \, {\left (8 \, a^{3} + 12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 30 \, {\left (32 \, a^{3} - 16 \, a^{2} b + 6 \, a b^{2} - b^{3}\right )} d x + 5 \, {\left (b^{3} \cosh \left (d x + c\right )^{9} + 2 \, {\left (6 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right )^{7} + 3 \, {\left (16 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right )^{5} + 8 \, {\left (8 \, a^{3} + 12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{3} + 2 \, {\left (128 \, a^{3} - 24 \, a^{2} b + b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{2560 \, d} \]
1/2560*(5*b^3*cosh(d*x + c)*sinh(d*x + c)^9 + 10*(6*b^3*cosh(d*x + c)^3 + (6*a*b^2 - b^3)*cosh(d*x + c))*sinh(d*x + c)^7 + (126*b^3*cosh(d*x + c)^5 + 70*(6*a*b^2 - b^3)*cosh(d*x + c)^3 + 15*(16*a^2*b - b^3)*cosh(d*x + c))* sinh(d*x + c)^5 + 10*(6*b^3*cosh(d*x + c)^7 + 7*(6*a*b^2 - b^3)*cosh(d*x + c)^5 + 5*(16*a^2*b - b^3)*cosh(d*x + c)^3 + 4*(8*a^3 + 12*a^2*b - 6*a*b^2 + b^3)*cosh(d*x + c))*sinh(d*x + c)^3 + 30*(32*a^3 - 16*a^2*b + 6*a*b^2 - b^3)*d*x + 5*(b^3*cosh(d*x + c)^9 + 2*(6*a*b^2 - b^3)*cosh(d*x + c)^7 + 3 *(16*a^2*b - b^3)*cosh(d*x + c)^5 + 8*(8*a^3 + 12*a^2*b - 6*a*b^2 + b^3)*c osh(d*x + c)^3 + 2*(128*a^3 - 24*a^2*b + b^3)*cosh(d*x + c))*sinh(d*x + c) )/d
Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (219) = 438\).
Time = 1.36 (sec) , antiderivative size = 774, normalized size of antiderivative = 3.25 \[ \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\begin {cases} \frac {3 a^{3} x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac {3 a^{3} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {3 a^{3} x \cosh ^{4}{\left (c + d x \right )}}{8} - \frac {3 a^{3} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} + \frac {5 a^{3} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac {3 a^{2} b x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac {9 a^{2} b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac {9 a^{2} b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac {3 a^{2} b x \cosh ^{6}{\left (c + d x \right )}}{16} - \frac {3 a^{2} b \sinh ^{5}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{16 d} + \frac {a^{2} b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{2 d} + \frac {3 a^{2} b \sinh {\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{16 d} + \frac {9 a b^{2} x \sinh ^{8}{\left (c + d x \right )}}{128} - \frac {9 a b^{2} x \sinh ^{6}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{32} + \frac {27 a b^{2} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{64} - \frac {9 a b^{2} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{6}{\left (c + d x \right )}}{32} + \frac {9 a b^{2} x \cosh ^{8}{\left (c + d x \right )}}{128} - \frac {9 a b^{2} \sinh ^{7}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{128 d} + \frac {33 a b^{2} \sinh ^{5}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{128 d} + \frac {33 a b^{2} \sinh ^{3}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{128 d} - \frac {9 a b^{2} \sinh {\left (c + d x \right )} \cosh ^{7}{\left (c + d x \right )}}{128 d} + \frac {3 b^{3} x \sinh ^{10}{\left (c + d x \right )}}{256} - \frac {15 b^{3} x \sinh ^{8}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{256} + \frac {15 b^{3} x \sinh ^{6}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{128} - \frac {15 b^{3} x \sinh ^{4}{\left (c + d x \right )} \cosh ^{6}{\left (c + d x \right )}}{128} + \frac {15 b^{3} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{8}{\left (c + d x \right )}}{256} - \frac {3 b^{3} x \cosh ^{10}{\left (c + d x \right )}}{256} - \frac {3 b^{3} \sinh ^{9}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{256 d} + \frac {7 b^{3} \sinh ^{7}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{128 d} + \frac {b^{3} \sinh ^{5}{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{10 d} - \frac {7 b^{3} \sinh ^{3}{\left (c + d x \right )} \cosh ^{7}{\left (c + d x \right )}}{128 d} + \frac {3 b^{3} \sinh {\left (c + d x \right )} \cosh ^{9}{\left (c + d x \right )}}{256 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{2}{\left (c \right )}\right )^{3} \cosh ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((3*a**3*x*sinh(c + d*x)**4/8 - 3*a**3*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 + 3*a**3*x*cosh(c + d*x)**4/8 - 3*a**3*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) + 5*a**3*sinh(c + d*x)*cosh(c + d*x)**3/(8*d) + 3*a**2*b*x*s inh(c + d*x)**6/16 - 9*a**2*b*x*sinh(c + d*x)**4*cosh(c + d*x)**2/16 + 9*a **2*b*x*sinh(c + d*x)**2*cosh(c + d*x)**4/16 - 3*a**2*b*x*cosh(c + d*x)**6 /16 - 3*a**2*b*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) + a**2*b*sinh(c + d*x )**3*cosh(c + d*x)**3/(2*d) + 3*a**2*b*sinh(c + d*x)*cosh(c + d*x)**5/(16* d) + 9*a*b**2*x*sinh(c + d*x)**8/128 - 9*a*b**2*x*sinh(c + d*x)**6*cosh(c + d*x)**2/32 + 27*a*b**2*x*sinh(c + d*x)**4*cosh(c + d*x)**4/64 - 9*a*b**2 *x*sinh(c + d*x)**2*cosh(c + d*x)**6/32 + 9*a*b**2*x*cosh(c + d*x)**8/128 - 9*a*b**2*sinh(c + d*x)**7*cosh(c + d*x)/(128*d) + 33*a*b**2*sinh(c + d*x )**5*cosh(c + d*x)**3/(128*d) + 33*a*b**2*sinh(c + d*x)**3*cosh(c + d*x)** 5/(128*d) - 9*a*b**2*sinh(c + d*x)*cosh(c + d*x)**7/(128*d) + 3*b**3*x*sin h(c + d*x)**10/256 - 15*b**3*x*sinh(c + d*x)**8*cosh(c + d*x)**2/256 + 15* b**3*x*sinh(c + d*x)**6*cosh(c + d*x)**4/128 - 15*b**3*x*sinh(c + d*x)**4* cosh(c + d*x)**6/128 + 15*b**3*x*sinh(c + d*x)**2*cosh(c + d*x)**8/256 - 3 *b**3*x*cosh(c + d*x)**10/256 - 3*b**3*sinh(c + d*x)**9*cosh(c + d*x)/(256 *d) + 7*b**3*sinh(c + d*x)**7*cosh(c + d*x)**3/(128*d) + b**3*sinh(c + d*x )**5*cosh(c + d*x)**5/(10*d) - 7*b**3*sinh(c + d*x)**3*cosh(c + d*x)**7/(1 28*d) + 3*b**3*sinh(c + d*x)*cosh(c + d*x)**9/(256*d), Ne(d, 0)), (x*(a...
Time = 0.20 (sec) , antiderivative size = 363, normalized size of antiderivative = 1.53 \[ \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {1}{64} \, a^{3} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac {1}{20480} \, b^{3} {\left (\frac {{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} - 40 \, e^{\left (-6 \, d x - 6 \, c\right )} - 20 \, e^{\left (-8 \, d x - 8 \, c\right )} - 2\right )} e^{\left (10 \, d x + 10 \, c\right )}}{d} + \frac {240 \, {\left (d x + c\right )}}{d} + \frac {20 \, e^{\left (-2 \, d x - 2 \, c\right )} + 40 \, e^{\left (-4 \, d x - 4 \, c\right )} - 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + 2 \, e^{\left (-10 \, d x - 10 \, c\right )}}{d}\right )} - \frac {3}{2048} \, a b^{2} {\left (\frac {{\left (8 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (8 \, d x + 8 \, c\right )}}{d} - \frac {48 \, {\left (d x + c\right )}}{d} - \frac {8 \, e^{\left (-4 \, d x - 4 \, c\right )} - e^{\left (-8 \, d x - 8 \, c\right )}}{d}\right )} + \frac {1}{128} \, a^{2} b {\left (\frac {{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} - \frac {24 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} \]
1/64*a^3*(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) - 1/20480*b^3*((5*e^(-2*d*x - 2*c) + 10*e^(-4* d*x - 4*c) - 40*e^(-6*d*x - 6*c) - 20*e^(-8*d*x - 8*c) - 2)*e^(10*d*x + 10 *c)/d + 240*(d*x + c)/d + (20*e^(-2*d*x - 2*c) + 40*e^(-4*d*x - 4*c) - 10* e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + 2*e^(-10*d*x - 10*c))/d) - 3/2048* a*b^2*((8*e^(-4*d*x - 4*c) - 1)*e^(8*d*x + 8*c)/d - 48*(d*x + c)/d - (8*e^ (-4*d*x - 4*c) - e^(-8*d*x - 8*c))/d) + 1/128*a^2*b*((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)
Time = 0.31 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.23 \[ \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {b^{3} e^{\left (10 \, d x + 10 \, c\right )}}{10240 \, d} - \frac {b^{3} e^{\left (-10 \, d x - 10 \, c\right )}}{10240 \, d} + \frac {3}{256} \, {\left (32 \, a^{3} - 16 \, a^{2} b + 6 \, a b^{2} - b^{3}\right )} x + \frac {{\left (6 \, a b^{2} - b^{3}\right )} e^{\left (8 \, d x + 8 \, c\right )}}{4096 \, d} + \frac {{\left (16 \, a^{2} b - b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )}}{2048 \, d} + \frac {{\left (8 \, a^{3} + 12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} e^{\left (4 \, d x + 4 \, c\right )}}{512 \, d} + \frac {{\left (128 \, a^{3} - 24 \, a^{2} b + b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{1024 \, d} - \frac {{\left (128 \, a^{3} - 24 \, a^{2} b + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{1024 \, d} - \frac {{\left (8 \, a^{3} + 12 \, a^{2} b - 6 \, a b^{2} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{512 \, d} - \frac {{\left (16 \, a^{2} b - b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{2048 \, d} - \frac {{\left (6 \, a b^{2} - b^{3}\right )} e^{\left (-8 \, d x - 8 \, c\right )}}{4096 \, d} \]
1/10240*b^3*e^(10*d*x + 10*c)/d - 1/10240*b^3*e^(-10*d*x - 10*c)/d + 3/256 *(32*a^3 - 16*a^2*b + 6*a*b^2 - b^3)*x + 1/4096*(6*a*b^2 - b^3)*e^(8*d*x + 8*c)/d + 1/2048*(16*a^2*b - b^3)*e^(6*d*x + 6*c)/d + 1/512*(8*a^3 + 12*a^ 2*b - 6*a*b^2 + b^3)*e^(4*d*x + 4*c)/d + 1/1024*(128*a^3 - 24*a^2*b + b^3) *e^(2*d*x + 2*c)/d - 1/1024*(128*a^3 - 24*a^2*b + b^3)*e^(-2*d*x - 2*c)/d - 1/512*(8*a^3 + 12*a^2*b - 6*a*b^2 + b^3)*e^(-4*d*x - 4*c)/d - 1/2048*(16 *a^2*b - b^3)*e^(-6*d*x - 6*c)/d - 1/4096*(6*a*b^2 - b^3)*e^(-8*d*x - 8*c) /d
Time = 0.61 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.88 \[ \int \cosh ^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {320\,a^3\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+40\,a^3\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+\frac {5\,b^3\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )}{2}+5\,b^3\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )-\frac {5\,b^3\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )}{4}-\frac {5\,b^3\,\mathrm {sinh}\left (8\,c+8\,d\,x\right )}{8}+\frac {b^3\,\mathrm {sinh}\left (10\,c+10\,d\,x\right )}{4}-60\,a^2\,b\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )-30\,a\,b^2\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+60\,a^2\,b\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )+20\,a^2\,b\,\mathrm {sinh}\left (6\,c+6\,d\,x\right )+\frac {15\,a\,b^2\,\mathrm {sinh}\left (8\,c+8\,d\,x\right )}{4}+480\,a^3\,d\,x-15\,b^3\,d\,x+90\,a\,b^2\,d\,x-240\,a^2\,b\,d\,x}{1280\,d} \]
(320*a^3*sinh(2*c + 2*d*x) + 40*a^3*sinh(4*c + 4*d*x) + (5*b^3*sinh(2*c + 2*d*x))/2 + 5*b^3*sinh(4*c + 4*d*x) - (5*b^3*sinh(6*c + 6*d*x))/4 - (5*b^3 *sinh(8*c + 8*d*x))/8 + (b^3*sinh(10*c + 10*d*x))/4 - 60*a^2*b*sinh(2*c + 2*d*x) - 30*a*b^2*sinh(4*c + 4*d*x) + 60*a^2*b*sinh(4*c + 4*d*x) + 20*a^2* b*sinh(6*c + 6*d*x) + (15*a*b^2*sinh(8*c + 8*d*x))/4 + 480*a^3*d*x - 15*b^ 3*d*x + 90*a*b^2*d*x - 240*a^2*b*d*x)/(1280*d)