Integrand size = 20, antiderivative size = 148 \[ \int x^2 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=-\frac {\cosh (a+b x)}{4 b^3}-\frac {x^2 \cosh (a+b x)}{8 b}-\frac {\cosh (3 a+3 b x)}{216 b^3}-\frac {x^2 \cosh (3 a+3 b x)}{48 b}+\frac {\cosh (5 a+5 b x)}{1000 b^3}+\frac {x^2 \cosh (5 a+5 b x)}{80 b}+\frac {x \sinh (a+b x)}{4 b^2}+\frac {x \sinh (3 a+3 b x)}{72 b^2}-\frac {x \sinh (5 a+5 b x)}{200 b^2} \]
-1/4*cosh(b*x+a)/b^3-1/8*x^2*cosh(b*x+a)/b-1/216*cosh(3*b*x+3*a)/b^3-1/48* x^2*cosh(3*b*x+3*a)/b+1/1000*cosh(5*b*x+5*a)/b^3+1/80*x^2*cosh(5*b*x+5*a)/ b+1/4*x*sinh(b*x+a)/b^2+1/72*x*sinh(3*b*x+3*a)/b^2-1/200*x*sinh(5*b*x+5*a) /b^2
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.66 \[ \int x^2 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {-6750 \left (2+b^2 x^2\right ) \cosh (a+b x)-125 \left (2+9 b^2 x^2\right ) \cosh (3 (a+b x))+27 \left (2+25 b^2 x^2\right ) \cosh (5 (a+b x))+30 b x (450 \sinh (a+b x)+25 \sinh (3 (a+b x))-9 \sinh (5 (a+b x)))}{54000 b^3} \]
(-6750*(2 + b^2*x^2)*Cosh[a + b*x] - 125*(2 + 9*b^2*x^2)*Cosh[3*(a + b*x)] + 27*(2 + 25*b^2*x^2)*Cosh[5*(a + b*x)] + 30*b*x*(450*Sinh[a + b*x] + 25* Sinh[3*(a + b*x)] - 9*Sinh[5*(a + b*x)]))/(54000*b^3)
Time = 0.39 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {5971, 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.
| \(\displaystyle \int x^2 \sinh ^3(a+b x) \cosh ^2(a+b x) \, dx\) |
| \(\Big \downarrow \) 5971 |
| \(\displaystyle \int \left (-\frac {1}{8} x^2 \sinh (a+b x)-\frac {1}{16} x^2 \sinh (3 a+3 b x)+\frac {1}{16} x^2 \sinh (5 a+5 b x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\cosh (a+b x)}{4 b^3}-\frac {\cosh (3 a+3 b x)}{216 b^3}+\frac {\cosh (5 a+5 b x)}{1000 b^3}+\frac {x \sinh (a+b x)}{4 b^2}+\frac {x \sinh (3 a+3 b x)}{72 b^2}-\frac {x \sinh (5 a+5 b x)}{200 b^2}-\frac {x^2 \cosh (a+b x)}{8 b}-\frac {x^2 \cosh (3 a+3 b x)}{48 b}+\frac {x^2 \cosh (5 a+5 b x)}{80 b}\) |
-1/4*Cosh[a + b*x]/b^3 - (x^2*Cosh[a + b*x])/(8*b) - Cosh[3*a + 3*b*x]/(21 6*b^3) - (x^2*Cosh[3*a + 3*b*x])/(48*b) + Cosh[5*a + 5*b*x]/(1000*b^3) + ( x^2*Cosh[5*a + 5*b*x])/(80*b) + (x*Sinh[a + b*x])/(4*b^2) + (x*Sinh[3*a + 3*b*x])/(72*b^2) - (x*Sinh[5*a + 5*b*x])/(200*b^2)
3.4.18.3.1 Defintions of rubi rules used
Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] & & IGtQ[p, 0]
Time = 26.87 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.11
| method | result | size |
| risch | \(\frac {\left (25 x^{2} b^{2}-10 b x +2\right ) {\mathrm e}^{5 b x +5 a}}{4000 b^{3}}-\frac {\left (9 x^{2} b^{2}-6 b x +2\right ) {\mathrm e}^{3 b x +3 a}}{864 b^{3}}-\frac {\left (x^{2} b^{2}-2 b x +2\right ) {\mathrm e}^{b x +a}}{16 b^{3}}-\frac {\left (x^{2} b^{2}+2 b x +2\right ) {\mathrm e}^{-b x -a}}{16 b^{3}}-\frac {\left (9 x^{2} b^{2}+6 b x +2\right ) {\mathrm e}^{-3 b x -3 a}}{864 b^{3}}+\frac {\left (25 x^{2} b^{2}+10 b x +2\right ) {\mathrm e}^{-5 b x -5 a}}{4000 b^{3}}\) | \(165\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {\cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2}}{5}-\frac {2 \cosh \left (b x +a \right )^{3}}{15}\right )-2 a \left (\frac {\left (b x +a \right ) \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{5}-\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right )^{3}}{15}-\frac {\sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{25}+\frac {26 \sinh \left (b x +a \right )}{225}+\frac {13 \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{225}\right )+\frac {\left (b x +a \right )^{2} \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{5}-\frac {2 \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{15}-\frac {2 \left (b x +a \right ) \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{25}+\frac {52 \left (b x +a \right ) \sinh \left (b x +a \right )}{225}+\frac {26 \left (b x +a \right ) \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{2}}{225}+\frac {2 \cosh \left (b x +a \right )^{5}}{125}-\frac {52 \cosh \left (b x +a \right )}{225}-\frac {26 \cosh \left (b x +a \right )^{3}}{675}}{b^{3}}\) | \(246\) |
| default | \(\frac {a^{2} \left (\frac {\cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2}}{5}-\frac {2 \cosh \left (b x +a \right )^{3}}{15}\right )-2 a \left (\frac {\left (b x +a \right ) \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{5}-\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right )^{3}}{15}-\frac {\sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{25}+\frac {26 \sinh \left (b x +a \right )}{225}+\frac {13 \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{225}\right )+\frac {\left (b x +a \right )^{2} \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{5}-\frac {2 \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{15}-\frac {2 \left (b x +a \right ) \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{25}+\frac {52 \left (b x +a \right ) \sinh \left (b x +a \right )}{225}+\frac {26 \left (b x +a \right ) \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{2}}{225}+\frac {2 \cosh \left (b x +a \right )^{5}}{125}-\frac {52 \cosh \left (b x +a \right )}{225}-\frac {26 \cosh \left (b x +a \right )^{3}}{675}}{b^{3}}\) | \(246\) |
1/4000*(25*b^2*x^2-10*b*x+2)/b^3*exp(5*b*x+5*a)-1/864*(9*b^2*x^2-6*b*x+2)/ b^3*exp(3*b*x+3*a)-1/16*(b^2*x^2-2*b*x+2)/b^3*exp(b*x+a)-1/16*(b^2*x^2+2*b *x+2)/b^3*exp(-b*x-a)-1/864*(9*b^2*x^2+6*b*x+2)/b^3*exp(-3*b*x-3*a)+1/4000 *(25*b^2*x^2+10*b*x+2)/b^3*exp(-5*b*x-5*a)
Time = 0.27 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.45 \[ \int x^2 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=-\frac {270 \, b x \sinh \left (b x + a\right )^{5} - 27 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{5} - 135 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + 125 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} + 150 \, {\left (18 \, b x \cosh \left (b x + a\right )^{2} - 5 \, b x\right )} \sinh \left (b x + a\right )^{3} - 15 \, {\left (18 \, {\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{3} - 25 \, {\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 6750 \, {\left (b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right ) + 450 \, {\left (3 \, b x \cosh \left (b x + a\right )^{4} - 5 \, b x \cosh \left (b x + a\right )^{2} - 30 \, b x\right )} \sinh \left (b x + a\right )}{54000 \, b^{3}} \]
-1/54000*(270*b*x*sinh(b*x + a)^5 - 27*(25*b^2*x^2 + 2)*cosh(b*x + a)^5 - 135*(25*b^2*x^2 + 2)*cosh(b*x + a)*sinh(b*x + a)^4 + 125*(9*b^2*x^2 + 2)*c osh(b*x + a)^3 + 150*(18*b*x*cosh(b*x + a)^2 - 5*b*x)*sinh(b*x + a)^3 - 15 *(18*(25*b^2*x^2 + 2)*cosh(b*x + a)^3 - 25*(9*b^2*x^2 + 2)*cosh(b*x + a))* sinh(b*x + a)^2 + 6750*(b^2*x^2 + 2)*cosh(b*x + a) + 450*(3*b*x*cosh(b*x + a)^4 - 5*b*x*cosh(b*x + a)^2 - 30*b*x)*sinh(b*x + a))/b^3
Time = 0.54 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.23 \[ \int x^2 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\begin {cases} \frac {x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 x^{2} \cosh ^{5}{\left (a + b x \right )}}{15 b} + \frac {52 x \sinh ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac {26 x \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac {4 x \sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{15 b^{2}} - \frac {52 \sinh ^{4}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{225 b^{3}} + \frac {338 \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{675 b^{3}} - \frac {856 \cosh ^{5}{\left (a + b x \right )}}{3375 b^{3}} & \text {for}\: b \neq 0 \\\frac {x^{3} \sinh ^{3}{\left (a \right )} \cosh ^{2}{\left (a \right )}}{3} & \text {otherwise} \end {cases} \]
Piecewise((x**2*sinh(a + b*x)**2*cosh(a + b*x)**3/(3*b) - 2*x**2*cosh(a + b*x)**5/(15*b) + 52*x*sinh(a + b*x)**5/(225*b**2) - 26*x*sinh(a + b*x)**3* cosh(a + b*x)**2/(45*b**2) + 4*x*sinh(a + b*x)*cosh(a + b*x)**4/(15*b**2) - 52*sinh(a + b*x)**4*cosh(a + b*x)/(225*b**3) + 338*sinh(a + b*x)**2*cosh (a + b*x)**3/(675*b**3) - 856*cosh(a + b*x)**5/(3375*b**3), Ne(b, 0)), (x* *3*sinh(a)**3*cosh(a)**2/3, True))
Time = 0.24 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.26 \[ \int x^2 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {{\left (25 \, b^{2} x^{2} e^{\left (5 \, a\right )} - 10 \, b x e^{\left (5 \, a\right )} + 2 \, e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{4000 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} e^{\left (3 \, a\right )} - 6 \, b x e^{\left (3 \, a\right )} + 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{864 \, b^{3}} - \frac {{\left (b^{2} x^{2} e^{a} - 2 \, b x e^{a} + 2 \, e^{a}\right )} e^{\left (b x\right )}}{16 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} + \frac {{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \]
1/4000*(25*b^2*x^2*e^(5*a) - 10*b*x*e^(5*a) + 2*e^(5*a))*e^(5*b*x)/b^3 - 1 /864*(9*b^2*x^2*e^(3*a) - 6*b*x*e^(3*a) + 2*e^(3*a))*e^(3*b*x)/b^3 - 1/16* (b^2*x^2*e^a - 2*b*x*e^a + 2*e^a)*e^(b*x)/b^3 - 1/16*(b^2*x^2 + 2*b*x + 2) *e^(-b*x - a)/b^3 - 1/864*(9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^3 + 1 /4000*(25*b^2*x^2 + 10*b*x + 2)*e^(-5*b*x - 5*a)/b^3
Time = 0.28 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.11 \[ \int x^2 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {{\left (25 \, b^{2} x^{2} - 10 \, b x + 2\right )} e^{\left (5 \, b x + 5 \, a\right )}}{4000 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} - 6 \, b x + 2\right )} e^{\left (3 \, b x + 3 \, a\right )}}{864 \, b^{3}} - \frac {{\left (b^{2} x^{2} - 2 \, b x + 2\right )} e^{\left (b x + a\right )}}{16 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac {{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} + \frac {{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \]
1/4000*(25*b^2*x^2 - 10*b*x + 2)*e^(5*b*x + 5*a)/b^3 - 1/864*(9*b^2*x^2 - 6*b*x + 2)*e^(3*b*x + 3*a)/b^3 - 1/16*(b^2*x^2 - 2*b*x + 2)*e^(b*x + a)/b^ 3 - 1/16*(b^2*x^2 + 2*b*x + 2)*e^(-b*x - a)/b^3 - 1/864*(9*b^2*x^2 + 6*b*x + 2)*e^(-3*b*x - 3*a)/b^3 + 1/4000*(25*b^2*x^2 + 10*b*x + 2)*e^(-5*b*x - 5*a)/b^3
Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.76 \[ \int x^2 \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=-\frac {780\,\mathrm {cosh}\left (a+b\,x\right )+130\,{\mathrm {cosh}\left (a+b\,x\right )}^3-54\,{\mathrm {cosh}\left (a+b\,x\right )}^5-780\,b\,x\,\mathrm {sinh}\left (a+b\,x\right )+1125\,b^2\,x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^3-675\,b^2\,x^2\,{\mathrm {cosh}\left (a+b\,x\right )}^5-390\,b\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )+270\,b\,x\,{\mathrm {cosh}\left (a+b\,x\right )}^4\,\mathrm {sinh}\left (a+b\,x\right )}{3375\,b^3} \]