Integrand size = 18, antiderivative size = 94 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=-\frac {x \cosh (a+b x)}{8 b}-\frac {x \cosh (3 a+3 b x)}{48 b}+\frac {x \cosh (5 a+5 b x)}{80 b}+\frac {\sinh (a+b x)}{8 b^2}+\frac {\sinh (3 a+3 b x)}{144 b^2}-\frac {\sinh (5 a+5 b x)}{400 b^2} \] Output:
-1/8*x*cosh(b*x+a)/b-1/48*x*cosh(3*b*x+3*a)/b+1/80*x*cosh(5*b*x+5*a)/b+1/8 *sinh(b*x+a)/b^2+1/144*sinh(3*b*x+3*a)/b^2-1/400*sinh(5*b*x+5*a)/b^2
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {-450 b x \cosh (a+b x)-75 b x \cosh (3 (a+b x))+45 b x \cosh (5 (a+b x))+450 \sinh (a+b x)+25 \sinh (3 (a+b x))-9 \sinh (5 (a+b x))}{3600 b^2} \] Input:
Integrate[x*Cosh[a + b*x]^2*Sinh[a + b*x]^3,x]
Output:
(-450*b*x*Cosh[a + b*x] - 75*b*x*Cosh[3*(a + b*x)] + 45*b*x*Cosh[5*(a + b* x)] + 450*Sinh[a + b*x] + 25*Sinh[3*(a + b*x)] - 9*Sinh[5*(a + b*x)])/(360 0*b^2)
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int x \sinh ^3(a+b x) \cosh ^2(a+b x) \, dx\) |
\(\Big \downarrow \) 5971 |
\(\displaystyle \int \left (-\frac {1}{8} x \sinh (a+b x)-\frac {1}{16} x \sinh (3 a+3 b x)+\frac {1}{16} x \sinh (5 a+5 b x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sinh (a+b x)}{8 b^2}+\frac {\sinh (3 a+3 b x)}{144 b^2}-\frac {\sinh (5 a+5 b x)}{400 b^2}-\frac {x \cosh (a+b x)}{8 b}-\frac {x \cosh (3 a+3 b x)}{48 b}+\frac {x \cosh (5 a+5 b x)}{80 b}\) |
Input:
Int[x*Cosh[a + b*x]^2*Sinh[a + b*x]^3,x]
Output:
-1/8*(x*Cosh[a + b*x])/b - (x*Cosh[3*a + 3*b*x])/(48*b) + (x*Cosh[5*a + 5* b*x])/(80*b) + Sinh[a + b*x]/(8*b^2) + Sinh[3*a + 3*b*x]/(144*b^2) - Sinh[ 5*a + 5*b*x]/(400*b^2)
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 = 21.57 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23
method | result | size |
derivativedivides | \(\frac {\frac {\left (b x +a \right ) \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{5}-\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right )^{3}}{15}-\frac {\sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{25}+\frac {26 \sinh \left (b x +a \right )}{225}+\frac {13 \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{225}-a \left (\frac {\cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2}}{5}-\frac {2 \cosh \left (b x +a \right )^{3}}{15}\right )}{b^{2}}\) | \(116\) |
default | \(\frac {\frac {\left (b x +a \right ) \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}}{5}-\frac {2 \left (b x +a \right ) \cosh \left (b x +a \right )^{3}}{15}-\frac {\sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{25}+\frac {26 \sinh \left (b x +a \right )}{225}+\frac {13 \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )}{225}-a \left (\frac {\cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2}}{5}-\frac {2 \cosh \left (b x +a \right )^{3}}{15}\right )}{b^{2}}\) | \(116\) |
risch | \(\frac {\left (5 b x -1\right ) {\mathrm e}^{5 b x +5 a}}{800 b^{2}}-\frac {\left (3 b x -1\right ) {\mathrm e}^{3 b x +3 a}}{288 b^{2}}-\frac {\left (b x -1\right ) {\mathrm e}^{b x +a}}{16 b^{2}}-\frac {\left (b x +1\right ) {\mathrm e}^{-b x -a}}{16 b^{2}}-\frac {\left (3 b x +1\right ) {\mathrm e}^{-3 b x -3 a}}{288 b^{2}}+\frac {\left (5 b x +1\right ) {\mathrm e}^{-5 b x -5 a}}{800 b^{2}}\) | \(117\) |
orering | \(-\frac {2 \left (259 b^{4} x^{4}-140 b^{2} x^{2}+72\right ) \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )^{3}}{225 b^{6} x^{4}}+\frac {\left (259 b^{4} x^{4}-280 b^{2} x^{2}+144\right ) \left (\cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )^{3}+2 x \cosh \left (b x +a \right ) \sinh \left (b x +a \right )^{4} b +3 x \cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2} b \right )}{225 b^{6} x^{4}}+\frac {4 \left (35 b^{2} x^{2}-18\right ) \left (4 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )^{4} b +6 \cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2} b +2 x \,b^{2} \sinh \left (b x +a \right )^{5}+17 x \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )^{3} b^{2}+6 x \cosh \left (b x +a \right )^{4} \sinh \left (b x +a \right ) b^{2}\right )}{225 b^{6} x^{3}}-\frac {\left (35 b^{2} x^{2}-24\right ) \left (6 b^{2} \sinh \left (b x +a \right )^{5}+51 \cosh \left (b x +a \right )^{2} \sinh \left (b x +a \right )^{3} b^{2}+18 \cosh \left (b x +a \right )^{4} \sinh \left (b x +a \right ) b^{2}+44 x \,b^{3} \sinh \left (b x +a \right )^{4} \cosh \left (b x +a \right )+75 x \cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2} b^{3}+6 x \cosh \left (b x +a \right )^{5} b^{3}\right )}{225 b^{6} x^{2}}-\frac {2 \left (176 b^{3} \sinh \left (b x +a \right )^{4} \cosh \left (b x +a \right )+300 \cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )^{2} b^{3}+24 \cosh \left (b x +a \right )^{5} b^{3}+401 x \,b^{4} \sinh \left (b x +a \right )^{3} \cosh \left (b x +a \right )^{2}+44 x \,b^{4} \sinh \left (b x +a \right )^{5}+180 x \cosh \left (b x +a \right )^{4} \sinh \left (b x +a \right ) b^{4}\right )}{75 b^{6} x}+\frac {2005 b^{4} \sinh \left (b x +a \right )^{3} \cosh \left (b x +a \right )^{2}+220 b^{4} \sinh \left (b x +a \right )^{5}+900 \cosh \left (b x +a \right )^{4} \sinh \left (b x +a \right ) b^{4}+1923 x \,b^{5} \sinh \left (b x +a \right )^{2} \cosh \left (b x +a \right )^{3}+1022 x \,b^{5} \sinh \left (b x +a \right )^{4} \cosh \left (b x +a \right )+180 x \cosh \left (b x +a \right )^{5} b^{5}}{225 b^{6}}\) | \(598\) |
Input:
int(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b^2*(1/5*(b*x+a)*sinh(b*x+a)^2*cosh(b*x+a)^3-2/15*(b*x+a)*cosh(b*x+a)^3- 1/25*sinh(b*x+a)*cosh(b*x+a)^4+26/225*sinh(b*x+a)+13/225*cosh(b*x+a)^2*sin h(b*x+a)-a*(1/5*cosh(b*x+a)^3*sinh(b*x+a)^2-2/15*cosh(b*x+a)^3))
Time = 0.10 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.63 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {45 \, b x \cosh \left (b x + a\right )^{5} + 225 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 75 \, b x \cosh \left (b x + a\right )^{3} - 9 \, \sinh \left (b x + a\right )^{5} - 5 \, {\left (18 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )^{3} - 450 \, b x \cosh \left (b x + a\right ) + 225 \, {\left (2 \, b x \cosh \left (b x + a\right )^{3} - b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 15 \, {\left (3 \, \cosh \left (b x + a\right )^{4} - 5 \, \cosh \left (b x + a\right )^{2} - 30\right )} \sinh \left (b x + a\right )}{3600 \, b^{2}} \] Input:
integrate(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="fricas")
Output:
1/3600*(45*b*x*cosh(b*x + a)^5 + 225*b*x*cosh(b*x + a)*sinh(b*x + a)^4 - 7 5*b*x*cosh(b*x + a)^3 - 9*sinh(b*x + a)^5 - 5*(18*cosh(b*x + a)^2 - 5)*sin h(b*x + a)^3 - 450*b*x*cosh(b*x + a) + 225*(2*b*x*cosh(b*x + a)^3 - b*x*co sh(b*x + a))*sinh(b*x + a)^2 - 15*(3*cosh(b*x + a)^4 - 5*cosh(b*x + a)^2 - 30)*sinh(b*x + a))/b^2
Time = 0.44 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.19 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\begin {cases} \frac {x \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{3 b} - \frac {2 x \cosh ^{5}{\left (a + b x \right )}}{15 b} + \frac {26 \sinh ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac {13 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{45 b^{2}} + \frac {2 \sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{15 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh ^{3}{\left (a \right )} \cosh ^{2}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*cosh(b*x+a)**2*sinh(b*x+a)**3,x)
Output:
Piecewise((x*sinh(a + b*x)**2*cosh(a + b*x)**3/(3*b) - 2*x*cosh(a + b*x)** 5/(15*b) + 26*sinh(a + b*x)**5/(225*b**2) - 13*sinh(a + b*x)**3*cosh(a + b *x)**2/(45*b**2) + 2*sinh(a + b*x)*cosh(a + b*x)**4/(15*b**2), Ne(b, 0)), (x**2*sinh(a)**3*cosh(a)**2/2, True))
Time = 0.05 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.37 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {{\left (5 \, b x e^{\left (5 \, a\right )} - e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{800 \, b^{2}} - \frac {{\left (3 \, b x e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{288 \, b^{2}} - \frac {{\left (b x e^{a} - e^{a}\right )} e^{\left (b x\right )}}{16 \, b^{2}} - \frac {{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{16 \, b^{2}} - \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{288 \, b^{2}} + \frac {{\left (5 \, b x + 1\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{800 \, b^{2}} \] Input:
integrate(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="maxima")
Output:
1/800*(5*b*x*e^(5*a) - e^(5*a))*e^(5*b*x)/b^2 - 1/288*(3*b*x*e^(3*a) - e^( 3*a))*e^(3*b*x)/b^2 - 1/16*(b*x*e^a - e^a)*e^(b*x)/b^2 - 1/16*(b*x + 1)*e^ (-b*x - a)/b^2 - 1/288*(3*b*x + 1)*e^(-3*b*x - 3*a)/b^2 + 1/800*(5*b*x + 1 )*e^(-5*b*x - 5*a)/b^2
Time = 0.11 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {{\left (5 \, b x - 1\right )} e^{\left (5 \, b x + 5 \, a\right )}}{800 \, b^{2}} - \frac {{\left (3 \, b x - 1\right )} e^{\left (3 \, b x + 3 \, a\right )}}{288 \, b^{2}} - \frac {{\left (b x - 1\right )} e^{\left (b x + a\right )}}{16 \, b^{2}} - \frac {{\left (b x + 1\right )} e^{\left (-b x - a\right )}}{16 \, b^{2}} - \frac {{\left (3 \, b x + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{288 \, b^{2}} + \frac {{\left (5 \, b x + 1\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{800 \, b^{2}} \] Input:
integrate(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x, algorithm="giac")
Output:
1/800*(5*b*x - 1)*e^(5*b*x + 5*a)/b^2 - 1/288*(3*b*x - 1)*e^(3*b*x + 3*a)/ b^2 - 1/16*(b*x - 1)*e^(b*x + a)/b^2 - 1/16*(b*x + 1)*e^(-b*x - a)/b^2 - 1 /288*(3*b*x + 1)*e^(-3*b*x - 3*a)/b^2 + 1/800*(5*b*x + 1)*e^(-5*b*x - 5*a) /b^2
Time = 0.95 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.76 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {\frac {26\,\mathrm {sinh}\left (a+b\,x\right )}{225}-b\,\left (\frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{3}-\frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^5}{5}\right )+\frac {13\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,\mathrm {sinh}\left (a+b\,x\right )}{225}-\frac {{\mathrm {cosh}\left (a+b\,x\right )}^4\,\mathrm {sinh}\left (a+b\,x\right )}{25}}{b^2} \] Input:
int(x*cosh(a + b*x)^2*sinh(a + b*x)^3,x)
Output:
((26*sinh(a + b*x))/225 - b*((x*cosh(a + b*x)^3)/3 - (x*cosh(a + b*x)^5)/5 ) + (13*cosh(a + b*x)^2*sinh(a + b*x))/225 - (cosh(a + b*x)^4*sinh(a + b*x ))/25)/b^2
Time = 0.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.63 \[ \int x \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx=\frac {45 e^{10 b x +10 a} b x -9 e^{10 b x +10 a}-75 e^{8 b x +8 a} b x +25 e^{8 b x +8 a}-450 e^{6 b x +6 a} b x +450 e^{6 b x +6 a}-450 e^{4 b x +4 a} b x -450 e^{4 b x +4 a}-75 e^{2 b x +2 a} b x -25 e^{2 b x +2 a}+45 b x +9}{7200 e^{5 b x +5 a} b^{2}} \] Input:
int(x*cosh(b*x+a)^2*sinh(b*x+a)^3,x)
Output:
(45*e**(10*a + 10*b*x)*b*x - 9*e**(10*a + 10*b*x) - 75*e**(8*a + 8*b*x)*b* x + 25*e**(8*a + 8*b*x) - 450*e**(6*a + 6*b*x)*b*x + 450*e**(6*a + 6*b*x) - 450*e**(4*a + 4*b*x)*b*x - 450*e**(4*a + 4*b*x) - 75*e**(2*a + 2*b*x)*b* x - 25*e**(2*a + 2*b*x) + 45*b*x + 9)/(7200*e**(5*a + 5*b*x)*b**2)