Integrand size = 22, antiderivative size = 73 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{a-4 d-3 b x}}{48 b}-\frac {e^{a-2 d-b x}}{8 b}-\frac {e^{a+2 d+3 b x}}{24 b}+\frac {e^{a+4 d+5 b x}}{80 b} \] Output:
1/48*exp(-3*b*x+a-4*d)/b-1/8*exp(-b*x+a-2*d)/b-1/24*exp(3*b*x+a+2*d)/b+1/8 0*exp(5*b*x+a+4*d)/b
Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.22 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=-\frac {e^{a-b x} \left (\left (3+e^{4 b x}\right ) \cosh (2 d)+\left (-3+e^{4 b x}\right ) \sinh (2 d)\right )}{24 b}+\frac {e^{a-3 b x} \left (\left (5+3 e^{8 b x}\right ) \cosh (4 d)+\left (-5+3 e^{8 b x}\right ) \sinh (4 d)\right )}{240 b} \] Input:
Integrate[E^(a + b*x)*Cosh[d + b*x]*Sinh[d + b*x]^3,x]
Output:
-1/24*(E^(a - b*x)*((3 + E^(4*b*x))*Cosh[2*d] + (-3 + E^(4*b*x))*Sinh[2*d] ))/b + (E^(a - 3*b*x)*((5 + 3*E^(8*b*x))*Cosh[4*d] + (-5 + 3*E^(8*b*x))*Si nh[4*d]))/(240*b)
Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.67, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2720, 27, 355, 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.
\(\displaystyle \int e^{a+b x} \sinh ^3(b x+d) \cosh (b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {1}{16} e^{a-4 b x} \left (1-e^{2 b x}\right )^3 \left (1+e^{2 b x}\right )de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^a \int e^{-4 b x} \left (1-e^{2 b x}\right )^3 \left (1+e^{2 b x}\right )de^{b x}}{16 b}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle -\frac {e^a \int \left (e^{-4 b x}-2 e^{-2 b x}+2 e^{2 b x}-e^{4 b x}\right )de^{b x}}{16 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^a \left (-\frac {1}{3} e^{-3 b x}+2 e^{-b x}+\frac {2}{3} e^{3 b x}-\frac {1}{5} e^{5 b x}\right )}{16 b}\) |
Input:
Int[E^(a + b*x)*Cosh[d + b*x]*Sinh[d + b*x]^3,x]
Output:
-1/16*(E^a*(-1/3*1/E^(3*b*x) + 2/E^(b*x) + (2*E^(3*b*x))/3 - E^(5*b*x)/5)) /b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 279.98 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.85
method | result | size |
risch | \(\frac {{\mathrm e}^{-3 b x +a -4 d}}{48 b}-\frac {{\mathrm e}^{-b x +a -2 d}}{8 b}-\frac {{\mathrm e}^{3 b x +a +2 d}}{24 b}+\frac {{\mathrm e}^{5 b x +a +4 d}}{80 b}\) | \(62\) |
default | \(\frac {\sinh \left (-3 b x +a -4 d \right )}{48 b}-\frac {\sinh \left (-b x +a -2 d \right )}{8 b}-\frac {\sinh \left (3 b x +a +2 d \right )}{24 b}+\frac {\sinh \left (5 b x +a +4 d \right )}{80 b}+\frac {\cosh \left (-3 b x +a -4 d \right )}{48 b}-\frac {\cosh \left (-b x +a -2 d \right )}{8 b}-\frac {\cosh \left (3 b x +a +2 d \right )}{24 b}+\frac {\cosh \left (5 b x +a +4 d \right )}{80 b}\) | \(122\) |
orering | \(-\frac {4 \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{3}}{5 b}+\frac {\frac {14 b \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{3}}{45}+\frac {14 \,{\mathrm e}^{b x +a} b \sinh \left (b x +d \right )^{4}}{45}+\frac {14 \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )^{2} b}{15}}{b^{2}}+\frac {\frac {44 b^{2} {\mathrm e}^{b x +a} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{3}}{45}+\frac {8 b^{2} {\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{4}}{45}+\frac {8 b^{2} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )^{2}}{15}+\frac {8 \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} \sinh \left (b x +d \right ) b^{2}}{15}}{b^{3}}-\frac {31 b^{3} {\mathrm e}^{b x +a} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{3}+13 b^{3} {\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{4}+57 b^{3} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )^{2}+18 b^{3} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} \sinh \left (b x +d \right )+6 \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{4} b^{3}}{45 b^{4}}\) | \(319\) |
Input:
int(exp(b*x+a)*cosh(b*x+d)*sinh(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
1/48*exp(-3*b*x+a-4*d)/b-1/8*exp(-b*x+a-2*d)/b-1/24*exp(3*b*x+a+2*d)/b+1/8 0*exp(5*b*x+a+4*d)/b
Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (61) = 122\).
Time = 0.09 (sec) , antiderivative size = 260, normalized size of antiderivative = 3.56 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=\frac {\cosh \left (b x + d\right )^{4} \cosh \left (-a + d\right ) + {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{4} - {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{3} - 5 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + {\left (6 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) - {\left (6 \, \cosh \left (b x + d\right )^{2} - 5\right )} \sinh \left (-a + d\right ) - 5 \, \cosh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} - {\left (\cosh \left (b x + d\right )^{3} \cosh \left (-a + d\right ) - 5 \, \cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - {\left (\cosh \left (b x + d\right )^{3} - 5 \, \cosh \left (b x + d\right )\right )} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{4} - 5 \, \cosh \left (b x + d\right )^{2}\right )} \sinh \left (-a + d\right )}{30 \, {\left (b \cosh \left (b x + d\right ) - b \sinh \left (b x + d\right )\right )}} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)*sinh(b*x+d)^3,x, algorithm="fricas")
Output:
1/30*(cosh(b*x + d)^4*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b* x + d)^4 - (cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh( b*x + d)^3 - 5*cosh(b*x + d)^2*cosh(-a + d) + (6*cosh(b*x + d)^2*cosh(-a + d) - (6*cosh(b*x + d)^2 - 5)*sinh(-a + d) - 5*cosh(-a + d))*sinh(b*x + d) ^2 - (cosh(b*x + d)^3*cosh(-a + d) - 5*cosh(b*x + d)*cosh(-a + d) - (cosh( b*x + d)^3 - 5*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - (cosh(b*x + d) ^4 - 5*cosh(b*x + d)^2)*sinh(-a + d))/(b*cosh(b*x + d) - b*sinh(b*x + d))
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (58) = 116\).
Time = 2.19 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.90 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=\begin {cases} \frac {e^{a} e^{b x} \sinh ^{4}{\left (b x + d \right )}}{5 b} - \frac {e^{a} e^{b x} \sinh ^{3}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{5 b} + \frac {e^{a} e^{b x} \sinh ^{2}{\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{5 b} + \frac {2 e^{a} e^{b x} \sinh {\left (b x + d \right )} \cosh ^{3}{\left (b x + d \right )}}{15 b} - \frac {2 e^{a} e^{b x} \cosh ^{4}{\left (b x + d \right )}}{15 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{3}{\left (d \right )} \cosh {\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)*sinh(b*x+d)**3,x)
Output:
Piecewise((exp(a)*exp(b*x)*sinh(b*x + d)**4/(5*b) - exp(a)*exp(b*x)*sinh(b *x + d)**3*cosh(b*x + d)/(5*b) + exp(a)*exp(b*x)*sinh(b*x + d)**2*cosh(b*x + d)**2/(5*b) + 2*exp(a)*exp(b*x)*sinh(b*x + d)*cosh(b*x + d)**3/(15*b) - 2*exp(a)*exp(b*x)*cosh(b*x + d)**4/(15*b), Ne(b, 0)), (x*exp(a)*sinh(d)** 3*cosh(d), True))
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.05 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=-\frac {{\left (6 \, e^{\left (2 \, b x + 4 \, a + 2 \, d\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (-3 \, b x - 3 \, a - 4 \, d\right )}}{48 \, b} + \frac {{\left (3 \, e^{\left (5 \, b x + 5 \, a + 4 \, d\right )} - 10 \, e^{\left (3 \, b x + 5 \, a + 2 \, d\right )}\right )} e^{\left (-4 \, a\right )}}{240 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)*sinh(b*x+d)^3,x, algorithm="maxima")
Output:
-1/48*(6*e^(2*b*x + 4*a + 2*d) - e^(4*a))*e^(-3*b*x - 3*a - 4*d)/b + 1/240 *(3*e^(5*b*x + 5*a + 4*d) - 10*e^(3*b*x + 5*a + 2*d))*e^(-4*a)/b
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=-\frac {{\left (5 \, {\left (6 \, e^{\left (2 \, b x + a + 2 \, d\right )} - e^{a}\right )} e^{\left (-3 \, b x\right )} - 3 \, e^{\left (5 \, b x + a + 8 \, d\right )} + 10 \, e^{\left (3 \, b x + a + 6 \, d\right )}\right )} e^{\left (-4 \, d\right )}}{240 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)*sinh(b*x+d)^3,x, algorithm="giac")
Output:
-1/240*(5*(6*e^(2*b*x + a + 2*d) - e^a)*e^(-3*b*x) - 3*e^(5*b*x + a + 8*d) + 10*e^(3*b*x + a + 6*d))*e^(-4*d)/b
Time = 0.62 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=\frac {{\mathrm {e}}^{-4\,d}\,{\mathrm {e}}^{-3\,b\,x}\,{\mathrm {e}}^a}{48\,b}-\frac {{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{3\,b\,x}\,{\mathrm {e}}^a}{24\,b}-\frac {{\mathrm {e}}^{-2\,d}\,{\mathrm {e}}^{-b\,x}\,{\mathrm {e}}^a}{8\,b}+\frac {{\mathrm {e}}^{4\,d}\,{\mathrm {e}}^{5\,b\,x}\,{\mathrm {e}}^a}{80\,b} \] Input:
int(cosh(d + b*x)*exp(a + b*x)*sinh(d + b*x)^3,x)
Output:
(exp(-4*d)*exp(-3*b*x)*exp(a))/(48*b) - (exp(2*d)*exp(3*b*x)*exp(a))/(24*b ) - (exp(-2*d)*exp(-b*x)*exp(a))/(8*b) + (exp(4*d)*exp(5*b*x)*exp(a))/(80* b)
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int e^{a+b x} \cosh (d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{a} \left (3 e^{8 b x +8 d}-10 e^{6 b x +6 d}-30 e^{2 b x +2 d}+5\right )}{240 e^{3 b x +4 d} b} \] Input:
int(exp(b*x+a)*cosh(b*x+d)*sinh(b*x+d)^3,x)
Output:
(e**a*(3*e**(8*b*x + 8*d) - 10*e**(6*b*x + 6*d) - 30*e**(2*b*x + 2*d) + 5) )/(240*e**(3*b*x + 4*d)*b)