Integrand size = 24, antiderivative size = 73 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{a-6 d-5 b x}}{320 b}-\frac {3 e^{a-2 d-b x}}{64 b}-\frac {e^{a+2 d+3 b x}}{64 b}+\frac {e^{a+6 d+7 b x}}{448 b} \] Output:
1/320*exp(-5*b*x+a-6*d)/b-3/64*exp(-b*x+a-2*d)/b-1/64*exp(3*b*x+a+2*d)/b+1 /448*exp(7*b*x+a+6*d)/b
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{a-6 d-5 b x} \left (7-105 e^{4 (d+b x)}-35 e^{8 (d+b x)}+5 e^{12 (d+b x)}\right )}{2240 b} \] Input:
Integrate[E^(a + b*x)*Cosh[d + b*x]^3*Sinh[d + b*x]^3,x]
Output:
(E^(a - 6*d - 5*b*x)*(7 - 105*E^(4*(d + b*x)) - 35*E^(8*(d + b*x)) + 5*E^( 12*(d + b*x))))/(2240*b)
Time = 0.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.62, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2720, 27, 802, 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.
\(\displaystyle \int e^{a+b x} \sinh ^3(b x+d) \cosh ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {1}{64} e^{a-6 b x} \left (1-e^{4 b x}\right )^3de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^a \int e^{-6 b x} \left (1-e^{4 b x}\right )^3de^{b x}}{64 b}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle -\frac {e^a \int \left (e^{-6 b x}-3 e^{-2 b x}+3 e^{2 b x}-e^{6 b x}\right )de^{b x}}{64 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^a \left (-\frac {1}{5} e^{-5 b x}+3 e^{-b x}+e^{3 b x}-\frac {1}{7} e^{7 b x}\right )}{64 b}\) |
Input:
Int[E^(a + b*x)*Cosh[d + b*x]^3*Sinh[d + b*x]^3,x]
Output:
-1/64*(E^a*(-1/5*1/E^(5*b*x) + 3/E^(b*x) + E^(3*b*x) - E^(7*b*x)/7))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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 = 0.04 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.67
\[\frac {\sinh \left (-5 b x +a -6 d \right )}{320 b}-\frac {3 \sinh \left (-b x +a -2 d \right )}{64 b}-\frac {\sinh \left (3 b x +a +2 d \right )}{64 b}+\frac {\sinh \left (7 b x +a +6 d \right )}{448 b}+\frac {\cosh \left (-5 b x +a -6 d \right )}{320 b}-\frac {3 \cosh \left (-b x +a -2 d \right )}{64 b}-\frac {\cosh \left (3 b x +a +2 d \right )}{64 b}+\frac {\cosh \left (7 b x +a +6 d \right )}{448 b}\]
Input:
int(exp(b*x+a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x)
Output:
1/320/b*sinh(-5*b*x+a-6*d)-3/64/b*sinh(-b*x+a-2*d)-1/64/b*sinh(3*b*x+a+2*d )+1/448/b*sinh(7*b*x+a+6*d)+1/320*cosh(-5*b*x+a-6*d)/b-3/64*cosh(-b*x+a-2* d)/b-1/64*cosh(3*b*x+a+2*d)/b+1/448*cosh(7*b*x+a+6*d)/b
Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (61) = 122\).
Time = 0.08 (sec) , antiderivative size = 352, normalized size of antiderivative = 4.82 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {3 \, \cosh \left (b x + d\right )^{6} \cosh \left (-a + d\right ) + 3 \, {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{6} - 3 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{5} + 45 \, {\left (\cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) - \cosh \left (b x + d\right )^{2} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{4} - 10 \, {\left (\cosh \left (b x + d\right )^{3} \cosh \left (-a + d\right ) - \cosh \left (b x + d\right )^{3} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{3} - 35 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + 5 \, {\left (9 \, \cosh \left (b x + d\right )^{4} \cosh \left (-a + d\right ) - {\left (9 \, \cosh \left (b x + d\right )^{4} - 7\right )} \sinh \left (-a + d\right ) - 7 \, \cosh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} - {\left (3 \, \cosh \left (b x + d\right )^{5} \cosh \left (-a + d\right ) - 35 \, \cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - {\left (3 \, \cosh \left (b x + d\right )^{5} - 35 \, \cosh \left (b x + d\right )\right )} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (3 \, \cosh \left (b x + d\right )^{6} - 35 \, \cosh \left (b x + d\right )^{2}\right )} \sinh \left (-a + d\right )}{560 \, {\left (b \cosh \left (b x + d\right ) - b \sinh \left (b x + d\right )\right )}} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x, algorithm="fricas")
Output:
1/560*(3*cosh(b*x + d)^6*cosh(-a + d) + 3*(cosh(-a + d) - sinh(-a + d))*si nh(b*x + d)^6 - 3*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d) )*sinh(b*x + d)^5 + 45*(cosh(b*x + d)^2*cosh(-a + d) - cosh(b*x + d)^2*sin h(-a + d))*sinh(b*x + d)^4 - 10*(cosh(b*x + d)^3*cosh(-a + d) - cosh(b*x + d)^3*sinh(-a + d))*sinh(b*x + d)^3 - 35*cosh(b*x + d)^2*cosh(-a + d) + 5* (9*cosh(b*x + d)^4*cosh(-a + d) - (9*cosh(b*x + d)^4 - 7)*sinh(-a + d) - 7 *cosh(-a + d))*sinh(b*x + d)^2 - (3*cosh(b*x + d)^5*cosh(-a + d) - 35*cosh (b*x + d)*cosh(-a + d) - (3*cosh(b*x + d)^5 - 35*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - (3*cosh(b*x + d)^6 - 35*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. 202 vs. \(2 (60) = 120\).
Time = 12.59 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.77 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\begin {cases} - \frac {2 e^{a} e^{b x} \sinh ^{6}{\left (b x + d \right )}}{35 b} + \frac {2 e^{a} e^{b x} \sinh ^{5}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{35 b} + \frac {e^{a} e^{b x} \sinh ^{4}{\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{7 b} - \frac {e^{a} e^{b x} \sinh ^{3}{\left (b x + d \right )} \cosh ^{3}{\left (b x + d \right )}}{7 b} + \frac {e^{a} e^{b x} \sinh ^{2}{\left (b x + d \right )} \cosh ^{4}{\left (b x + d \right )}}{7 b} + \frac {2 e^{a} e^{b x} \sinh {\left (b x + d \right )} \cosh ^{5}{\left (b x + d \right )}}{35 b} - \frac {2 e^{a} e^{b x} \cosh ^{6}{\left (b x + d \right )}}{35 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{3}{\left (d \right )} \cosh ^{3}{\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)**3*sinh(b*x+d)**3,x)
Output:
Piecewise((-2*exp(a)*exp(b*x)*sinh(b*x + d)**6/(35*b) + 2*exp(a)*exp(b*x)* sinh(b*x + d)**5*cosh(b*x + d)/(35*b) + exp(a)*exp(b*x)*sinh(b*x + d)**4*c osh(b*x + d)**2/(7*b) - exp(a)*exp(b*x)*sinh(b*x + d)**3*cosh(b*x + d)**3/ (7*b) + exp(a)*exp(b*x)*sinh(b*x + d)**2*cosh(b*x + d)**4/(7*b) + 2*exp(a) *exp(b*x)*sinh(b*x + d)*cosh(b*x + d)**5/(35*b) - 2*exp(a)*exp(b*x)*cosh(b *x + d)**6/(35*b), Ne(b, 0)), (x*exp(a)*sinh(d)**3*cosh(d)**3, True))
Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=-\frac {{\left (15 \, e^{\left (4 \, b x + 6 \, a + 4 \, d\right )} - e^{\left (6 \, a\right )}\right )} e^{\left (-5 \, b x - 5 \, a - 6 \, d\right )}}{320 \, b} + \frac {{\left (e^{\left (7 \, b x + 7 \, a + 6 \, d\right )} - 7 \, e^{\left (3 \, b x + 7 \, a + 2 \, d\right )}\right )} e^{\left (-6 \, a\right )}}{448 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x, algorithm="maxima")
Output:
-1/320*(15*e^(4*b*x + 6*a + 4*d) - e^(6*a))*e^(-5*b*x - 5*a - 6*d)/b + 1/4 48*(e^(7*b*x + 7*a + 6*d) - 7*e^(3*b*x + 7*a + 2*d))*e^(-6*a)/b
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=-\frac {{\left (7 \, {\left (15 \, e^{\left (4 \, b x + a + 4 \, d\right )} - e^{a}\right )} e^{\left (-5 \, b x\right )} - 5 \, e^{\left (7 \, b x + a + 12 \, d\right )} + 35 \, e^{\left (3 \, b x + a + 8 \, d\right )}\right )} e^{\left (-6 \, d\right )}}{2240 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x, algorithm="giac")
Output:
-1/2240*(7*(15*e^(4*b*x + a + 4*d) - e^a)*e^(-5*b*x) - 5*e^(7*b*x + a + 12 *d) + 35*e^(3*b*x + a + 8*d))*e^(-6*d)/b
Time = 3.14 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {{\mathrm {e}}^{-6\,d}\,{\mathrm {e}}^{-5\,b\,x}\,{\mathrm {e}}^a}{320\,b}-\frac {{\mathrm {e}}^{2\,d}\,{\mathrm {e}}^{3\,b\,x}\,{\mathrm {e}}^a}{64\,b}-\frac {3\,{\mathrm {e}}^{-2\,d}\,{\mathrm {e}}^{-b\,x}\,{\mathrm {e}}^a}{64\,b}+\frac {{\mathrm {e}}^{6\,d}\,{\mathrm {e}}^{7\,b\,x}\,{\mathrm {e}}^a}{448\,b} \] Input:
int(cosh(d + b*x)^3*exp(a + b*x)*sinh(d + b*x)^3,x)
Output:
(exp(-6*d)*exp(-5*b*x)*exp(a))/(320*b) - (exp(2*d)*exp(3*b*x)*exp(a))/(64* b) - (3*exp(-2*d)*exp(-b*x)*exp(a))/(64*b) + (exp(6*d)*exp(7*b*x)*exp(a))/ (448*b)
Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int e^{a+b x} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{a} \left (5 e^{12 b x +12 d}-35 e^{8 b x +8 d}-105 e^{4 b x +4 d}+7\right )}{2240 e^{5 b x +6 d} b} \] Input:
int(exp(b*x+a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x)
Output:
(e**a*(5*e**(12*b*x + 12*d) - 35*e**(8*b*x + 8*d) - 105*e**(4*b*x + 4*d) + 7))/(2240*e**(5*b*x + 6*d)*b)