Integrand size = 26, antiderivative size = 118 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=-\frac {e^{2 a-5 d-3 b x}}{96 b}-\frac {e^{2 a-3 d-b x}}{32 b}-\frac {e^{2 a-d+b x}}{16 b}-\frac {e^{2 a+d+3 b x}}{48 b}+\frac {e^{2 a+3 d+5 b x}}{160 b}+\frac {e^{2 a+5 d+7 b x}}{224 b} \] Output:
-1/96*exp(-3*b*x+2*a-5*d)/b-1/32*exp(-b*x+2*a-3*d)/b-1/16*exp(b*x+2*a-d)/b -1/48*exp(3*b*x+2*a+d)/b+1/160*exp(5*b*x+2*a+3*d)/b+1/224*exp(7*b*x+2*a+5* d)/b
Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.99 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=\frac {e^{2 a-3 b x} \left (-70 e^{4 b x} \left (\left (3+e^{2 b x}\right ) \cosh (d)+\left (-3+e^{2 b x}\right ) \sinh (d)\right )+21 e^{2 b x} \left (\left (-5+e^{6 b x}\right ) \cosh (3 d)+\left (5+e^{6 b x}\right ) \sinh (3 d)\right )+5 \left (\left (-7+3 e^{10 b x}\right ) \cosh (5 d)+\left (7+3 e^{10 b x}\right ) \sinh (5 d)\right )\right )}{3360 b} \] Input:
Integrate[E^(2*(a + b*x))*Cosh[d + b*x]^3*Sinh[d + b*x]^2,x]
Output:
(E^(2*a - 3*b*x)*(-70*E^(4*b*x)*((3 + E^(2*b*x))*Cosh[d] + (-3 + E^(2*b*x) )*Sinh[d]) + 21*E^(2*b*x)*((-5 + E^(6*b*x))*Cosh[3*d] + (5 + E^(6*b*x))*Si nh[3*d]) + 5*((-7 + 3*E^(10*b*x))*Cosh[5*d] + (7 + 3*E^(10*b*x))*Sinh[5*d] )))/(3360*b)
Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.58, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, 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.
\(\displaystyle \int e^{2 (a+b x)} \sinh ^2(b x+d) \cosh ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {1}{32} e^{2 a-4 b x} \left (1-e^{2 b x}\right )^2 \left (1+e^{2 b x}\right )^3de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{2 a} \int e^{-4 b x} \left (1-e^{2 b x}\right )^2 \left (1+e^{2 b x}\right )^3de^{b x}}{32 b}\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \frac {e^{2 a} \int \left (-2+e^{-4 b x}+e^{-2 b x}-2 e^{2 b x}+e^{4 b x}+e^{6 b x}\right )de^{b x}}{32 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{2 a} \left (-\frac {1}{3} e^{-3 b x}-e^{-b x}-2 e^{b x}-\frac {2}{3} e^{3 b x}+\frac {1}{5} e^{5 b x}+\frac {1}{7} e^{7 b x}\right )}{32 b}\) |
Input:
Int[E^(2*(a + b*x))*Cosh[d + b*x]^3*Sinh[d + b*x]^2,x]
Output:
(E^(2*a)*(-1/3*1/E^(3*b*x) - E^(-(b*x)) - 2*E^(b*x) - (2*E^(3*b*x))/3 + E^ (5*b*x)/5 + E^(7*b*x)/7))/(32*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 = 0.24 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.69
\[-\frac {\sinh \left (-3 b x +2 a -5 d \right )}{96 b}-\frac {\sinh \left (-b x +2 a -3 d \right )}{32 b}-\frac {\sinh \left (b x +2 a -d \right )}{16 b}-\frac {\sinh \left (3 b x +2 a +d \right )}{48 b}+\frac {\sinh \left (5 b x +2 a +3 d \right )}{160 b}+\frac {\sinh \left (7 b x +2 a +5 d \right )}{224 b}-\frac {\cosh \left (-3 b x +2 a -5 d \right )}{96 b}-\frac {\cosh \left (-b x +2 a -3 d \right )}{32 b}-\frac {\cosh \left (b x +2 a -d \right )}{16 b}-\frac {\cosh \left (3 b x +2 a +d \right )}{48 b}+\frac {\cosh \left (5 b x +2 a +3 d \right )}{160 b}+\frac {\cosh \left (7 b x +2 a +5 d \right )}{224 b}\]
Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^2,x)
Output:
-1/96/b*sinh(-3*b*x+2*a-5*d)-1/32/b*sinh(-b*x+2*a-3*d)-1/16/b*sinh(b*x+2*a -d)-1/48/b*sinh(3*b*x+2*a+d)+1/160/b*sinh(5*b*x+2*a+3*d)+1/224/b*sinh(7*b* x+2*a+5*d)-1/96*cosh(-3*b*x+2*a-5*d)/b-1/32*cosh(-b*x+2*a-3*d)/b-1/16*cosh (b*x+2*a-d)/b-1/48*cosh(3*b*x+2*a+d)/b+1/160*cosh(5*b*x+2*a+3*d)/b+1/224*c osh(7*b*x+2*a+5*d)/b
Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (100) = 200\).
Time = 0.08 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.60 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=-\frac {10 \, \cosh \left (b x + d\right )^{5} \cosh \left (-2 \, a + 2 \, d\right ) - 25 \, {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{5} + 50 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - \cosh \left (b x + d\right ) \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{4} + 42 \, \cosh \left (b x + d\right )^{3} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (250 \, \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (250 \, \cosh \left (b x + d\right )^{2} + 63\right )} \sinh \left (-2 \, a + 2 \, d\right ) + 63 \, \cosh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{3} + 2 \, {\left (50 \, \cosh \left (b x + d\right )^{3} \cosh \left (-2 \, a + 2 \, d\right ) + 63 \, \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - {\left (50 \, \cosh \left (b x + d\right )^{3} + 63 \, \cosh \left (b x + d\right )\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{2} + 140 \, \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - {\left (125 \, \cosh \left (b x + d\right )^{4} \cosh \left (-2 \, a + 2 \, d\right ) + 189 \, \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (125 \, \cosh \left (b x + d\right )^{4} + 189 \, \cosh \left (b x + d\right )^{2} + 70\right )} \sinh \left (-2 \, a + 2 \, d\right ) + 70 \, \cosh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right ) - 2 \, {\left (5 \, \cosh \left (b x + d\right )^{5} + 21 \, \cosh \left (b x + d\right )^{3} + 70 \, \cosh \left (b x + d\right )\right )} \sinh \left (-2 \, a + 2 \, d\right )}{1680 \, {\left (b \cosh \left (b x + d\right )^{2} - 2 \, b \cosh \left (b x + d\right ) \sinh \left (b x + d\right ) + b \sinh \left (b x + d\right )^{2}\right )}} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^2,x, algorithm="fricas" )
Output:
-1/1680*(10*cosh(b*x + d)^5*cosh(-2*a + 2*d) - 25*(cosh(-2*a + 2*d) - sinh (-2*a + 2*d))*sinh(b*x + d)^5 + 50*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh( b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^4 + 42*cosh(b*x + d)^3*cosh(-2*a + 2*d) - (250*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (250*cosh(b*x + d)^2 + 63 )*sinh(-2*a + 2*d) + 63*cosh(-2*a + 2*d))*sinh(b*x + d)^3 + 2*(50*cosh(b*x + d)^3*cosh(-2*a + 2*d) + 63*cosh(b*x + d)*cosh(-2*a + 2*d) - (50*cosh(b* x + d)^3 + 63*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d)^2 + 140*cosh( b*x + d)*cosh(-2*a + 2*d) - (125*cosh(b*x + d)^4*cosh(-2*a + 2*d) + 189*co sh(b*x + d)^2*cosh(-2*a + 2*d) - (125*cosh(b*x + d)^4 + 189*cosh(b*x + d)^ 2 + 70)*sinh(-2*a + 2*d) + 70*cosh(-2*a + 2*d))*sinh(b*x + d) - 2*(5*cosh( b*x + d)^5 + 21*cosh(b*x + d)^3 + 70*cosh(b*x + d))*sinh(-2*a + 2*d))/(b*c osh(b*x + d)^2 - 2*b*cosh(b*x + d)*sinh(b*x + d) + b*sinh(b*x + d)^2)
Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (94) = 188\).
Time = 5.22 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.67 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=\begin {cases} \frac {2 e^{2 a} e^{2 b x} \sinh ^{5}{\left (b x + d \right )}}{105 b} - \frac {4 e^{2 a} e^{2 b x} \sinh ^{4}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{105 b} - \frac {e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{105 b} + \frac {2 e^{2 a} e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \cosh ^{3}{\left (b x + d \right )}}{35 b} + \frac {8 e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{4}{\left (b x + d \right )}}{35 b} - \frac {4 e^{2 a} e^{2 b x} \cosh ^{5}{\left (b x + d \right )}}{35 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\left (d \right )} \cosh ^{3}{\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)**3*sinh(b*x+d)**2,x)
Output:
Piecewise((2*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**5/(105*b) - 4*exp(2*a)*exp (2*b*x)*sinh(b*x + d)**4*cosh(b*x + d)/(105*b) - exp(2*a)*exp(2*b*x)*sinh( b*x + d)**3*cosh(b*x + d)**2/(105*b) + 2*exp(2*a)*exp(2*b*x)*sinh(b*x + d) **2*cosh(b*x + d)**3/(35*b) + 8*exp(2*a)*exp(2*b*x)*sinh(b*x + d)*cosh(b*x + d)**4/(35*b) - 4*exp(2*a)*exp(2*b*x)*cosh(b*x + d)**5/(35*b), Ne(b, 0)) , (x*exp(2*a)*sinh(d)**2*cosh(d)**3, True))
Time = 0.04 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=\frac {{\left (21 \, e^{\left (-2 \, b x - 2 \, d\right )} - 70 \, e^{\left (-4 \, b x - 4 \, d\right )} - 210 \, e^{\left (-6 \, b x - 6 \, d\right )} + 15\right )} e^{\left (7 \, b x + 2 \, a + 5 \, d\right )}}{3360 \, b} - \frac {{\left (3 \, e^{\left (-b x - d\right )} + e^{\left (-3 \, b x - 3 \, d\right )}\right )} e^{\left (2 \, a - 2 \, d\right )}}{96 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^2,x, algorithm="maxima" )
Output:
1/3360*(21*e^(-2*b*x - 2*d) - 70*e^(-4*b*x - 4*d) - 210*e^(-6*b*x - 6*d) + 15)*e^(7*b*x + 2*a + 5*d)/b - 1/96*(3*e^(-b*x - d) + e^(-3*b*x - 3*d))*e^ (2*a - 2*d)/b
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=-\frac {{\left (35 \, {\left (3 \, e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (-3 \, b x - 3 \, d\right )} - 15 \, e^{\left (7 \, b x + 2 \, a + 7 \, d\right )} - 21 \, e^{\left (5 \, b x + 2 \, a + 5 \, d\right )} + 70 \, e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} + 210 \, e^{\left (b x + 2 \, a + d\right )}\right )} e^{\left (-2 \, d\right )}}{3360 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^2,x, algorithm="giac")
Output:
-1/3360*(35*(3*e^(2*b*x + 2*a + 2*d) + e^(2*a))*e^(-3*b*x - 3*d) - 15*e^(7 *b*x + 2*a + 7*d) - 21*e^(5*b*x + 2*a + 5*d) + 70*e^(3*b*x + 2*a + 3*d) + 210*e^(b*x + 2*a + d))*e^(-2*d)/b
Time = 0.62 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=\frac {{\mathrm {e}}^{2\,a+3\,d+5\,b\,x}}{160\,b}-\frac {{\mathrm {e}}^{2\,a-d+b\,x}}{16\,b}-\frac {{\mathrm {e}}^{2\,a-3\,d-b\,x}}{32\,b}-\frac {{\mathrm {e}}^{2\,a+d+3\,b\,x}}{48\,b}-\frac {{\mathrm {e}}^{2\,a-5\,d-3\,b\,x}}{96\,b}+\frac {{\mathrm {e}}^{2\,a+5\,d+7\,b\,x}}{224\,b} \] Input:
int(cosh(d + b*x)^3*exp(2*a + 2*b*x)*sinh(d + b*x)^2,x)
Output:
exp(2*a + 3*d + 5*b*x)/(160*b) - exp(2*a - d + b*x)/(16*b) - exp(2*a - 3*d - b*x)/(32*b) - exp(2*a + d + 3*b*x)/(48*b) - exp(2*a - 5*d - 3*b*x)/(96* b) + exp(2*a + 5*d + 7*b*x)/(224*b)
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^2(d+b x) \, dx=\frac {e^{2 a} \left (15 e^{10 b x +10 d}+21 e^{8 b x +8 d}-70 e^{6 b x +6 d}-210 e^{4 b x +4 d}-105 e^{2 b x +2 d}-35\right )}{3360 e^{3 b x +5 d} b} \] Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^2,x)
Output:
(e**(2*a)*(15*e**(10*b*x + 10*d) + 21*e**(8*b*x + 8*d) - 70*e**(6*b*x + 6* d) - 210*e**(4*b*x + 4*d) - 105*e**(2*b*x + 2*d) - 35))/(3360*e**(3*b*x + 5*d)*b)