Integrand size = 21, antiderivative size = 70 \[ \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx=-\frac {a x}{2}+\frac {b \cosh (c+d x)}{d}-\frac {2 b \cosh ^3(c+d x)}{3 d}+\frac {b \cosh ^5(c+d x)}{5 d}+\frac {a \cosh (c+d x) \sinh (c+d x)}{2 d} \]
-1/2*a*x+b*cosh(d*x+c)/d-2/3*b*cosh(d*x+c)^3/d+1/5*b*cosh(d*x+c)^5/d+1/2*a *cosh(d*x+c)*sinh(d*x+c)/d
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.13 \[ \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx=\frac {a (-c-d x)}{2 d}+\frac {5 b \cosh (c+d x)}{8 d}-\frac {5 b \cosh (3 (c+d x))}{48 d}+\frac {b \cosh (5 (c+d x))}{80 d}+\frac {a \sinh (2 (c+d x))}{4 d} \]
(a*(-c - d*x))/(2*d) + (5*b*Cosh[c + d*x])/(8*d) - (5*b*Cosh[3*(c + d*x)]) /(48*d) + (b*Cosh[5*(c + d*x)])/(80*d) + (a*Sinh[2*(c + d*x)])/(4*d)
Time = 0.27 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 25, 3699, 2009}
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.
\(\displaystyle \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin (i c+i d x)^2 \left (a+i b \sin (i c+i d x)^3\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin (i c+i d x)^2 \left (i b \sin (i c+i d x)^3+a\right )dx\) |
\(\Big \downarrow \) 3699 |
\(\displaystyle -\int \left (-b \sinh ^5(c+d x)-a \sinh ^2(c+d x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac {a x}{2}+\frac {b \cosh ^5(c+d x)}{5 d}-\frac {2 b \cosh ^3(c+d x)}{3 d}+\frac {b \cosh (c+d x)}{d}\) |
-1/2*(a*x) + (b*Cosh[c + d*x])/d - (2*b*Cosh[c + d*x]^3)/(3*d) + (b*Cosh[c + d*x]^5)/(5*d) + (a*Cosh[c + d*x]*Sinh[c + d*x])/(2*d)
3.2.43.3.1 Defintions of rubi rules used
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> Int[ExpandTrig[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] || Gt Q[p, 0] || (EqQ[p, -1] && IntegerQ[n]))
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.86
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b \left (\frac {8}{15}+\frac {\sinh \left (d x +c \right )^{4}}{5}-\frac {4 \sinh \left (d x +c \right )^{2}}{15}\right ) \cosh \left (d x +c \right )}{d}\) | \(60\) |
default | \(\frac {a \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b \left (\frac {8}{15}+\frac {\sinh \left (d x +c \right )^{4}}{5}-\frac {4 \sinh \left (d x +c \right )^{2}}{15}\right ) \cosh \left (d x +c \right )}{d}\) | \(60\) |
parallelrisch | \(\frac {-120 a x d +150 b \cosh \left (d x +c \right )+3 b \cosh \left (5 d x +5 c \right )-25 b \cosh \left (3 d x +3 c \right )+60 a \sinh \left (2 d x +2 c \right )+128 b}{240 d}\) | \(60\) |
parts | \(\frac {a \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )}{d}+\frac {b \left (\frac {8}{15}+\frac {\sinh \left (d x +c \right )^{4}}{5}-\frac {4 \sinh \left (d x +c \right )^{2}}{15}\right ) \cosh \left (d x +c \right )}{d}\) | \(62\) |
risch | \(-\frac {a x}{2}+\frac {b \,{\mathrm e}^{5 d x +5 c}}{160 d}-\frac {5 b \,{\mathrm e}^{3 d x +3 c}}{96 d}+\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 d}+\frac {5 b \,{\mathrm e}^{d x +c}}{16 d}+\frac {5 b \,{\mathrm e}^{-d x -c}}{16 d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {5 b \,{\mathrm e}^{-3 d x -3 c}}{96 d}+\frac {b \,{\mathrm e}^{-5 d x -5 c}}{160 d}\) | \(123\) |
1/d*(a*(1/2*sinh(d*x+c)*cosh(d*x+c)-1/2*d*x-1/2*c)+b*(8/15+1/5*sinh(d*x+c) ^4-4/15*sinh(d*x+c)^2)*cosh(d*x+c))
Time = 0.26 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.50 \[ \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx=\frac {3 \, b \cosh \left (d x + c\right )^{5} + 15 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - 25 \, b \cosh \left (d x + c\right )^{3} - 120 \, a d x + 120 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + 15 \, {\left (2 \, b \cosh \left (d x + c\right )^{3} - 5 \, b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 150 \, b \cosh \left (d x + c\right )}{240 \, d} \]
1/240*(3*b*cosh(d*x + c)^5 + 15*b*cosh(d*x + c)*sinh(d*x + c)^4 - 25*b*cos h(d*x + c)^3 - 120*a*d*x + 120*a*cosh(d*x + c)*sinh(d*x + c) + 15*(2*b*cos h(d*x + c)^3 - 5*b*cosh(d*x + c))*sinh(d*x + c)^2 + 150*b*cosh(d*x + c))/d
Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.67 \[ \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx=\begin {cases} \frac {a x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac {a x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {a \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d} + \frac {b \sinh ^{4}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {4 b \sinh ^{2}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac {8 b \cosh ^{5}{\left (c + d x \right )}}{15 d} & \text {for}\: d \neq 0 \\x \left (a + b \sinh ^{3}{\left (c \right )}\right ) \sinh ^{2}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a*x*sinh(c + d*x)**2/2 - a*x*cosh(c + d*x)**2/2 + a*sinh(c + d* x)*cosh(c + d*x)/(2*d) + b*sinh(c + d*x)**4*cosh(c + d*x)/d - 4*b*sinh(c + d*x)**2*cosh(c + d*x)**3/(3*d) + 8*b*cosh(c + d*x)**5/(15*d), Ne(d, 0)), (x*(a + b*sinh(c)**3)*sinh(c)**2, True))
Time = 0.21 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.71 \[ \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx=-\frac {1}{8} \, a {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + \frac {1}{480} \, b {\left (\frac {3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} - \frac {25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {150 \, e^{\left (d x + c\right )}}{d} + \frac {150 \, e^{\left (-d x - c\right )}}{d} - \frac {25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac {3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} \]
-1/8*a*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + 1/480*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)
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.74 \[ \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx=-\frac {1}{2} \, a x + \frac {b e^{\left (5 \, d x + 5 \, c\right )}}{160 \, d} - \frac {5 \, b e^{\left (3 \, d x + 3 \, c\right )}}{96 \, d} + \frac {a e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac {5 \, b e^{\left (d x + c\right )}}{16 \, d} + \frac {5 \, b e^{\left (-d x - c\right )}}{16 \, d} - \frac {a e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac {5 \, b e^{\left (-3 \, d x - 3 \, c\right )}}{96 \, d} + \frac {b e^{\left (-5 \, d x - 5 \, c\right )}}{160 \, d} \]
-1/2*a*x + 1/160*b*e^(5*d*x + 5*c)/d - 5/96*b*e^(3*d*x + 3*c)/d + 1/8*a*e^ (2*d*x + 2*c)/d + 5/16*b*e^(d*x + c)/d + 5/16*b*e^(-d*x - c)/d - 1/8*a*e^( -2*d*x - 2*c)/d - 5/96*b*e^(-3*d*x - 3*c)/d + 1/160*b*e^(-5*d*x - 5*c)/d
Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79 \[ \int \sinh ^2(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx=\frac {b\,\mathrm {cosh}\left (c+d\,x\right )-\frac {2\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{3}+\frac {b\,{\mathrm {cosh}\left (c+d\,x\right )}^5}{5}+\frac {a\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )}{2}}{d}-\frac {a\,x}{2} \]