Integrand size = 22, antiderivative size = 65 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=\frac {e^{a-3 d-2 b x}}{16 b}+\frac {e^{a+d+2 b x}}{16 b}+\frac {e^{a+3 d+4 b x}}{32 b}-\frac {1}{8} e^{a-d} x \] Output:
1/16*exp(-2*b*x+a-3*d)/b+1/16*exp(2*b*x+a+d)/b+1/32*exp(4*b*x+a+3*d)/b-1/8 *exp(a-d)*x
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.17 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=\frac {e^a \left (2 \left (\left (e^{2 b x}-2 b x\right ) \cosh (d)+\left (e^{2 b x}+2 b x\right ) \sinh (d)\right )+e^{-2 b x} \left (\left (2+e^{6 b x}\right ) \cosh (3 d)+\left (-2+e^{6 b x}\right ) \sinh (3 d)\right )\right )}{32 b} \] Input:
Integrate[E^(a + b*x)*Cosh[d + b*x]^2*Sinh[d + b*x],x]
Output:
(E^a*(2*((E^(2*b*x) - 2*b*x)*Cosh[d] + (E^(2*b*x) + 2*b*x)*Sinh[d]) + ((2 + E^(6*b*x))*Cosh[3*d] + (-2 + E^(6*b*x))*Sinh[3*d])/E^(2*b*x)))/(32*b)
Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.55, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2720, 27, 354, 84, 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.
\(\displaystyle \int e^{a+b x} \sinh (b x+d) \cosh ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {1}{8} e^{a-3 b x} \left (1-e^{2 b x}\right ) \left (1+e^{2 b x}\right )^2de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^a \int e^{-3 b x} \left (1-e^{2 b x}\right ) \left (1+e^{2 b x}\right )^2de^{b x}}{8 b}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {e^a \int e^{-2 b x} \left (1-e^{2 b x}\right ) \left (1+e^{2 b x}\right )^2de^{2 b x}}{16 b}\) |
\(\Big \downarrow \) 84 |
\(\displaystyle -\frac {e^a \int \left (-1+e^{-2 b x}+e^{-b x}-e^{2 b x}\right )de^{2 b x}}{16 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^a \left (-e^{-b x}-\frac {3}{2} e^{2 b x}+\log \left (e^{2 b x}\right )\right )}{16 b}\) |
Input:
Int[E^(a + b*x)*Cosh[d + b*x]^2*Sinh[d + b*x],x]
Output:
-1/16*(E^a*(-E^(-(b*x)) - (3*E^(2*b*x))/2 + Log[E^(2*b*x)]))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] && !(ILtQ[n + p + 2, 0 ] && GtQ[n + 2*p, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
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 = 11.46 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {{\mathrm e}^{-2 b x +a -3 d}}{16 b}+\frac {{\mathrm e}^{2 b x +a +d}}{16 b}+\frac {{\mathrm e}^{4 b x +a +3 d}}{32 b}-\frac {{\mathrm e}^{a -d} x}{8}\) | \(54\) |
default | \(-\frac {\cosh \left (a -d \right ) x}{8}+\frac {\sinh \left (-2 b x +a -3 d \right )}{16 b}+\frac {\sinh \left (2 b x +a +d \right )}{16 b}+\frac {\sinh \left (4 b x +a +3 d \right )}{32 b}-\frac {\sinh \left (a -d \right ) x}{8}+\frac {\cosh \left (-2 b x +a -3 d \right )}{16 b}+\frac {\cosh \left (2 b x +a +d \right )}{16 b}+\frac {\cosh \left (4 b x +a +3 d \right )}{32 b}\) | \(106\) |
orering | \(\frac {\left (4 b x +1\right ) {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )}{4 b}-\frac {\left (b x -1\right ) \left ({\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right ) b +2 b \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{2}+{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} b \right )}{4 b^{2}}-\frac {\left (4 b x +1\right ) \left (8 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right ) \cosh \left (b x +d \right )^{2} b^{2}+4 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{2} \cosh \left (b x +d \right ) b^{2}+2 \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} b^{2}+2 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{3} b^{2}\right )}{16 b^{3}}+\frac {x \left (22 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right ) \cosh \left (b x +d \right )^{2} b^{3}+10 \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} b^{3}+26 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{2} \cosh \left (b x +d \right ) b^{3}+6 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{3} b^{3}\right )}{16 b^{3}}\) | \(300\) |
Input:
int(exp(b*x+a)*cosh(b*x+d)^2*sinh(b*x+d),x,method=_RETURNVERBOSE)
Output:
1/16*exp(-2*b*x+a-3*d)/b+1/16*exp(2*b*x+a+d)/b+1/32*exp(4*b*x+a+3*d)/b-1/8 *exp(a-d)*x
Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (53) = 106\).
Time = 0.09 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.34 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=\frac {3 \, \cosh \left (b x + d\right )^{3} \cosh \left (-a + d\right ) - {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{3} - 2 \, {\left (2 \, b x - 1\right )} \cosh \left (b x + d\right ) \cosh \left (-a + d\right ) + 9 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} - {\left (3 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) - 2 \, {\left (2 \, b x + 1\right )} \cosh \left (-a + d\right ) + {\left (4 \, b x - 3 \, \cosh \left (b x + d\right )^{2} + 2\right )} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (3 \, \cosh \left (b x + d\right )^{3} - 2 \, {\left (2 \, b x - 1\right )} \cosh \left (b x + d\right )\right )} \sinh \left (-a + d\right )}{32 \, {\left (b \cosh \left (b x + d\right ) - b \sinh \left (b x + d\right )\right )}} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^2*sinh(b*x+d),x, algorithm="fricas")
Output:
1/32*(3*cosh(b*x + d)^3*cosh(-a + d) - (cosh(-a + d) - sinh(-a + d))*sinh( b*x + d)^3 - 2*(2*b*x - 1)*cosh(b*x + d)*cosh(-a + d) + 9*(cosh(b*x + d)*c osh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^2 - (3*cosh(b*x + d)^2*cosh(-a + d) - 2*(2*b*x + 1)*cosh(-a + d) + (4*b*x - 3*cosh(b*x + d)^ 2 + 2)*sinh(-a + d))*sinh(b*x + d) - (3*cosh(b*x + d)^3 - 2*(2*b*x - 1)*co sh(b*x + d))*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. 175 vs. \(2 (51) = 102\).
Time = 0.90 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.69 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=\begin {cases} - \frac {x e^{a} e^{b x} \sinh ^{3}{\left (b x + d \right )}}{8} + \frac {x e^{a} e^{b x} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{8} + \frac {x e^{a} e^{b x} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{8} - \frac {x e^{a} e^{b x} \cosh ^{3}{\left (b x + d \right )}}{8} - \frac {e^{a} e^{b x} \sinh ^{3}{\left (b x + d \right )}}{8 b} + \frac {e^{a} e^{b x} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{4 b} + \frac {e^{a} e^{b x} \cosh ^{3}{\left (b x + d \right )}}{8 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh {\left (d \right )} \cosh ^{2}{\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)**2*sinh(b*x+d),x)
Output:
Piecewise((-x*exp(a)*exp(b*x)*sinh(b*x + d)**3/8 + x*exp(a)*exp(b*x)*sinh( b*x + d)**2*cosh(b*x + d)/8 + x*exp(a)*exp(b*x)*sinh(b*x + d)*cosh(b*x + d )**2/8 - x*exp(a)*exp(b*x)*cosh(b*x + d)**3/8 - exp(a)*exp(b*x)*sinh(b*x + d)**3/(8*b) + exp(a)*exp(b*x)*sinh(b*x + d)**2*cosh(b*x + d)/(4*b) + exp( a)*exp(b*x)*cosh(b*x + d)**3/(8*b), Ne(b, 0)), (x*exp(a)*sinh(d)*cosh(d)** 2, True))
Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=\frac {{\left (e^{\left (4 \, b x + 4 \, a + 3 \, d\right )} + 2 \, e^{\left (2 \, b x + 4 \, a + d\right )}\right )} e^{\left (-3 \, a\right )}}{32 \, b} - \frac {{\left (b x + a\right )} e^{\left (a - d\right )}}{8 \, b} + \frac {e^{\left (-2 \, b x + a - 3 \, d\right )}}{16 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^2*sinh(b*x+d),x, algorithm="maxima")
Output:
1/32*(e^(4*b*x + 4*a + 3*d) + 2*e^(2*b*x + 4*a + d))*e^(-3*a)/b - 1/8*(b*x + a)*e^(a - d)/b + 1/16*e^(-2*b*x + a - 3*d)/b
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.98 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=-\frac {{\left (4 \, b x e^{\left (a + 2 \, d\right )} - 2 \, {\left (e^{\left (2 \, b x + a + 2 \, d\right )} + e^{a}\right )} e^{\left (-2 \, b x\right )} - e^{\left (4 \, b x + a + 6 \, d\right )} - 2 \, e^{\left (2 \, b x + a + 4 \, d\right )}\right )} e^{\left (-3 \, d\right )}}{32 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^2*sinh(b*x+d),x, algorithm="giac")
Output:
-1/32*(4*b*x*e^(a + 2*d) - 2*(e^(2*b*x + a + 2*d) + e^a)*e^(-2*b*x) - e^(4 *b*x + a + 6*d) - 2*e^(2*b*x + a + 4*d))*e^(-3*d)/b
Time = 3.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\mathrm {cosh}\left (d+b\,x\right )+{\mathrm {cosh}\left (d+b\,x\right )}^2\,\mathrm {sinh}\left (d+b\,x\right )-3\,{\mathrm {cosh}\left (d+b\,x\right )}^3+b\,x\,\mathrm {cosh}\left (d+b\,x\right )-b\,x\,\mathrm {sinh}\left (d+b\,x\right )\right )}{8\,b} \] Input:
int(cosh(d + b*x)^2*exp(a + b*x)*sinh(d + b*x),x)
Output:
-(exp(a + b*x)*(cosh(d + b*x) + cosh(d + b*x)^2*sinh(d + b*x) - 3*cosh(d + b*x)^3 + b*x*cosh(d + b*x) - b*x*sinh(d + b*x)))/(8*b)
Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \cosh ^2(d+b x) \sinh (d+b x) \, dx=\frac {e^{a} \left (e^{6 b x +6 d}+2 e^{4 b x +4 d}-4 e^{2 b x +2 d} b x +2\right )}{32 e^{2 b x +3 d} b} \] Input:
int(exp(b*x+a)*cosh(b*x+d)^2*sinh(b*x+d),x)
Output:
(e**a*(e**(6*b*x + 6*d) + 2*e**(4*b*x + 4*d) - 4*e**(2*b*x + 2*d)*b*x + 2) )/(32*e**(2*b*x + 3*d)*b)