Integrand size = 16, antiderivative size = 65 \[ \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx=-\frac {3 x}{32 b}+\frac {x \cosh ^4(a+b x)}{4 b}-\frac {3 \cosh (a+b x) \sinh (a+b x)}{32 b^2}-\frac {\cosh ^3(a+b x) \sinh (a+b x)}{16 b^2} \]
-3/32*x/b+1/4*x*cosh(b*x+a)^4/b-3/32*cosh(b*x+a)*sinh(b*x+a)/b^2-1/16*cosh (b*x+a)^3*sinh(b*x+a)/b^2
Time = 0.09 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.77 \[ \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx=-\frac {-16 b x \cosh (2 (a+b x))-4 b x \cosh (4 (a+b x))+8 \sinh (2 (a+b x))+\sinh (4 (a+b x))}{128 b^2} \]
-1/128*(-16*b*x*Cosh[2*(a + b*x)] - 4*b*x*Cosh[4*(a + b*x)] + 8*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)])/b^2
Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5896, 3042, 3115, 3042, 3115, 24}
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.
\(\displaystyle \int x \sinh (a+b x) \cosh ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 5896 |
\(\displaystyle \frac {x \cosh ^4(a+b x)}{4 b}-\frac {\int \cosh ^4(a+b x)dx}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {x \cosh ^4(a+b x)}{4 b}-\frac {\int \sin \left (i a+i b x+\frac {\pi }{2}\right )^4dx}{4 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {x \cosh ^4(a+b x)}{4 b}-\frac {\frac {3}{4} \int \cosh ^2(a+b x)dx+\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}}{4 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {x \cosh ^4(a+b x)}{4 b}-\frac {\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac {3}{4} \int \sin \left (i a+i b x+\frac {\pi }{2}\right )^2dx}{4 b}\) |
\(\Big \downarrow \) 3115 |
\(\displaystyle \frac {x \cosh ^4(a+b x)}{4 b}-\frac {\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sinh (a+b x) \cosh (a+b x)}{2 b}\right )+\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}}{4 b}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {x \cosh ^4(a+b x)}{4 b}-\frac {\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac {3}{4} \left (\frac {\sinh (a+b x) \cosh (a+b x)}{2 b}+\frac {x}{2}\right )}{4 b}\) |
(x*Cosh[a + b*x]^4)/(4*b) - ((Cosh[a + b*x]^3*Sinh[a + b*x])/(4*b) + (3*(x /2 + (Cosh[a + b*x]*Sinh[a + b*x])/(2*b)))/4)/(4*b)
3.3.71.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_) ^(n_.)], x_Symbol] :> Simp[x^(m - n + 1)*(Cosh[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Simp[(m - n + 1)/(b*n*(p + 1)) Int[x^(m - n)*Cosh[a + b*x^n]^ (p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]
Time = 3.92 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06
method | result | size |
derivativedivides | \(\frac {\frac {\left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{4}-\frac {\cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )}{16}-\frac {3 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{32}-\frac {3 b x}{32}-\frac {3 a}{32}-\frac {a \cosh \left (b x +a \right )^{4}}{4}}{b^{2}}\) | \(69\) |
default | \(\frac {\frac {\left (b x +a \right ) \cosh \left (b x +a \right )^{4}}{4}-\frac {\cosh \left (b x +a \right )^{3} \sinh \left (b x +a \right )}{16}-\frac {3 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{32}-\frac {3 b x}{32}-\frac {3 a}{32}-\frac {a \cosh \left (b x +a \right )^{4}}{4}}{b^{2}}\) | \(69\) |
risch | \(\frac {\left (4 b x -1\right ) {\mathrm e}^{4 b x +4 a}}{256 b^{2}}+\frac {\left (2 b x -1\right ) {\mathrm e}^{2 b x +2 a}}{32 b^{2}}+\frac {\left (2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{32 b^{2}}+\frac {\left (4 b x +1\right ) {\mathrm e}^{-4 b x -4 a}}{256 b^{2}}\) | \(82\) |
1/b^2*(1/4*(b*x+a)*cosh(b*x+a)^4-1/16*cosh(b*x+a)^3*sinh(b*x+a)-3/32*cosh( b*x+a)*sinh(b*x+a)-3/32*b*x-3/32*a-1/4*a*cosh(b*x+a)^4)
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.66 \[ \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx=\frac {b x \cosh \left (b x + a\right )^{4} + b x \sinh \left (b x + a\right )^{4} + 4 \, b x \cosh \left (b x + a\right )^{2} - \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 2 \, {\left (3 \, b x \cosh \left (b x + a\right )^{2} + 2 \, b x\right )} \sinh \left (b x + a\right )^{2} - {\left (\cosh \left (b x + a\right )^{3} + 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{32 \, b^{2}} \]
1/32*(b*x*cosh(b*x + a)^4 + b*x*sinh(b*x + a)^4 + 4*b*x*cosh(b*x + a)^2 - cosh(b*x + a)*sinh(b*x + a)^3 + 2*(3*b*x*cosh(b*x + a)^2 + 2*b*x)*sinh(b*x + a)^2 - (cosh(b*x + a)^3 + 4*cosh(b*x + a))*sinh(b*x + a))/b^2
Time = 0.30 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.69 \[ \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx=\begin {cases} - \frac {3 x \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} + \frac {5 x \cosh ^{4}{\left (a + b x \right )}}{32 b} + \frac {3 \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{32 b^{2}} - \frac {5 \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{32 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh {\left (a \right )} \cosh ^{3}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \]
Piecewise((-3*x*sinh(a + b*x)**4/(32*b) + 3*x*sinh(a + b*x)**2*cosh(a + b* x)**2/(16*b) + 5*x*cosh(a + b*x)**4/(32*b) + 3*sinh(a + b*x)**3*cosh(a + b *x)/(32*b**2) - 5*sinh(a + b*x)*cosh(a + b*x)**3/(32*b**2), Ne(b, 0)), (x* *2*sinh(a)*cosh(a)**3/2, True))
Time = 0.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.40 \[ \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx=\frac {{\left (4 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{256 \, b^{2}} + \frac {{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{2}} + \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{2}} + \frac {{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \]
1/256*(4*b*x*e^(4*a) - e^(4*a))*e^(4*b*x)/b^2 + 1/32*(2*b*x*e^(2*a) - e^(2 *a))*e^(2*b*x)/b^2 + 1/32*(2*b*x + 1)*e^(-2*b*x - 2*a)/b^2 + 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx=\frac {{\left (4 \, b x - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{256 \, b^{2}} + \frac {{\left (2 \, b x - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{2}} + \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{2}} + \frac {{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \]
1/256*(4*b*x - 1)*e^(4*b*x + 4*a)/b^2 + 1/32*(2*b*x - 1)*e^(2*b*x + 2*a)/b ^2 + 1/32*(2*b*x + 1)*e^(-2*b*x - 2*a)/b^2 + 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int x \cosh ^3(a+b x) \sinh (a+b x) \, dx=-\frac {\frac {3\,x}{32}-\frac {x\,{\mathrm {cosh}\left (a+b\,x\right )}^4}{4}}{b}-\frac {{\mathrm {cosh}\left (a+b\,x\right )}^3\,\mathrm {sinh}\left (a+b\,x\right )}{16\,b^2}-\frac {3\,\mathrm {cosh}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right )}{32\,b^2} \]