Integrand size = 26, antiderivative size = 76 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{2 (a-3 d)-4 b x}}{256 b}-\frac {3 e^{2 (a+d)+4 b x}}{256 b}+\frac {e^{2 (a+3 d)+8 b x}}{512 b}+\frac {3}{64} e^{2 a-2 d} x \] Output:
1/256*exp(-4*b*x+2*a-6*d)/b-3/256*exp(4*b*x+2*a+2*d)/b+1/512*exp(8*b*x+2*a +6*d)/b+3/64*exp(2*a-2*d)*x
Time = 0.26 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.75 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{2 a-2 d} \left (e^{-4 (d+b x)}-3 e^{4 (d+b x)}+\frac {1}{2} e^{8 (d+b x)}+12 (d+b x)\right )}{256 b} \] Input:
Integrate[E^(2*(a + b*x))*Cosh[d + b*x]^3*Sinh[d + b*x]^3,x]
Output:
(E^(2*a - 2*d)*(E^(-4*(d + b*x)) - 3*E^(4*(d + b*x)) + E^(8*(d + b*x))/2 + 12*(d + b*x)))/(256*b)
Time = 0.41 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.63, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {2720, 27, 798, 49, 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.
\(\displaystyle \int e^{2 (a+b x)} \sinh ^3(b x+d) \cosh ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {1}{64} e^{2 a-5 b x} \left (1-e^{4 b x}\right )^3de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^{2 a} \int e^{-5 b x} \left (1-e^{4 b x}\right )^3de^{b x}}{64 b}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -\frac {e^{2 a} \int e^{-2 b x} \left (1-e^{4 b x}\right )^3de^{4 b x}}{256 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle -\frac {e^{2 a} \int \left (3+e^{-2 b x}-3 e^{-b x}-e^{4 b x}\right )de^{4 b x}}{256 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 a} \left (-e^{-b x}-\frac {1}{2} e^{2 b x}+3 e^{4 b x}-3 \log \left (e^{4 b x}\right )\right )}{256 b}\) |
Input:
Int[E^(2*(a + b*x))*Cosh[d + b*x]^3*Sinh[d + b*x]^3,x]
Output:
-1/256*(E^(2*a)*(-E^(-(b*x)) - E^(2*b*x)/2 + 3*E^(4*b*x) - 3*Log[E^(4*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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
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 = 0.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.66
\[\frac {3 x \cosh \left (2 a -2 d \right )}{64}+\frac {\sinh \left (-4 b x +2 a -6 d \right )}{256 b}-\frac {3 \sinh \left (4 b x +2 a +2 d \right )}{256 b}+\frac {\sinh \left (8 b x +2 a +6 d \right )}{512 b}+\frac {3 x \sinh \left (2 a -2 d \right )}{64}+\frac {\cosh \left (-4 b x +2 a -6 d \right )}{256 b}-\frac {3 \cosh \left (4 b x +2 a +2 d \right )}{256 b}+\frac {\cosh \left (8 b x +2 a +6 d \right )}{512 b}\]
Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x)
Output:
3/64*x*cosh(2*a-2*d)+1/256/b*sinh(-4*b*x+2*a-6*d)-3/256/b*sinh(4*b*x+2*a+2 *d)+1/512/b*sinh(8*b*x+2*a+6*d)+3/64*x*sinh(2*a-2*d)+1/256*cosh(-4*b*x+2*a -6*d)/b-3/256*cosh(4*b*x+2*a+2*d)/b+1/512*cosh(8*b*x+2*a+6*d)/b
Leaf count of result is larger than twice the leaf count of optimal. 433 vs. \(2 (63) = 126\).
Time = 0.09 (sec) , antiderivative size = 433, normalized size of antiderivative = 5.70 \[ \int e^{2 (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 (-2 \, a + 2 \, d\right ) + 3 \, {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{6} - 6 \, {\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 )^{5} + 45 \, {\left (\cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) - \cosh \left (b x + d\right )^{2} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{4} + 6 \, {\left (4 \, b x - 1\right )} \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) - 20 \, {\left (\cosh \left (b x + d\right )^{3} \cosh \left (-2 \, a + 2 \, d\right ) - \cosh \left (b x + d\right )^{3} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{3} + 3 \, {\left (15 \, \cosh \left (b x + d\right )^{4} \cosh \left (-2 \, a + 2 \, d\right ) + 2 \, {\left (4 \, b x - 1\right )} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (15 \, \cosh \left (b x + d\right )^{4} + 8 \, b x - 2\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{2} - 6 \, {\left (\cosh \left (b x + d\right )^{5} \cosh \left (-2 \, a + 2 \, d\right ) + 2 \, {\left (4 \, b x + 1\right )} \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - {\left (\cosh \left (b x + d\right )^{5} + 2 \, {\left (4 \, b x + 1\right )} \cosh \left (b x + d\right )\right )} \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right ) - 3 \, {\left (\cosh \left (b x + d\right )^{6} + 2 \, {\left (4 \, b x - 1\right )} \cosh \left (b x + d\right )^{2}\right )} \sinh \left (-2 \, a + 2 \, d\right )}{512 \, {\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)^3,x, algorithm="fricas" )
Output:
1/512*(3*cosh(b*x + d)^6*cosh(-2*a + 2*d) + 3*(cosh(-2*a + 2*d) - sinh(-2* a + 2*d))*sinh(b*x + d)^6 - 6*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^5 + 45*(cosh(b*x + d)^2*cosh(-2*a + 2* d) - cosh(b*x + d)^2*sinh(-2*a + 2*d))*sinh(b*x + d)^4 + 6*(4*b*x - 1)*cos h(b*x + d)^2*cosh(-2*a + 2*d) - 20*(cosh(b*x + d)^3*cosh(-2*a + 2*d) - cos h(b*x + d)^3*sinh(-2*a + 2*d))*sinh(b*x + d)^3 + 3*(15*cosh(b*x + d)^4*cos h(-2*a + 2*d) + 2*(4*b*x - 1)*cosh(-2*a + 2*d) - (15*cosh(b*x + d)^4 + 8*b *x - 2)*sinh(-2*a + 2*d))*sinh(b*x + d)^2 - 6*(cosh(b*x + d)^5*cosh(-2*a + 2*d) + 2*(4*b*x + 1)*cosh(b*x + d)*cosh(-2*a + 2*d) - (cosh(b*x + d)^5 + 2*(4*b*x + 1)*cosh(b*x + d))*sinh(-2*a + 2*d))*sinh(b*x + d) - 3*(cosh(b*x + d)^6 + 2*(4*b*x - 1)*cosh(b*x + d)^2)*sinh(-2*a + 2*d))/(b*cosh(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. 382 vs. \(2 (65) = 130\).
Time = 12.15 (sec) , antiderivative size = 382, normalized size of antiderivative = 5.03 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\begin {cases} \frac {3 x e^{2 a} e^{2 b x} \sinh ^{6}{\left (b x + d \right )}}{64} - \frac {3 x e^{2 a} e^{2 b x} \sinh ^{5}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{32} - \frac {3 x e^{2 a} e^{2 b x} \sinh ^{4}{\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{64} + \frac {3 x e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )} \cosh ^{3}{\left (b x + d \right )}}{16} - \frac {3 x e^{2 a} e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \cosh ^{4}{\left (b x + d \right )}}{64} - \frac {3 x e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{5}{\left (b x + d \right )}}{32} + \frac {3 x e^{2 a} e^{2 b x} \cosh ^{6}{\left (b x + d \right )}}{64} + \frac {3 e^{2 a} e^{2 b x} \sinh ^{6}{\left (b x + d \right )}}{32 b} - \frac {15 e^{2 a} e^{2 b x} \sinh ^{5}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{64 b} + \frac {13 e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )} \cosh ^{3}{\left (b x + d \right )}}{32 b} - \frac {15 e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{5}{\left (b x + d \right )}}{64 b} + \frac {3 e^{2 a} e^{2 b x} \cosh ^{6}{\left (b x + d \right )}}{32 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh ^{3}{\left (d \right )} \cosh ^{3}{\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)**3*sinh(b*x+d)**3,x)
Output:
Piecewise((3*x*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**6/64 - 3*x*exp(2*a)*exp( 2*b*x)*sinh(b*x + d)**5*cosh(b*x + d)/32 - 3*x*exp(2*a)*exp(2*b*x)*sinh(b* x + d)**4*cosh(b*x + d)**2/64 + 3*x*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**3*c osh(b*x + d)**3/16 - 3*x*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**2*cosh(b*x + d )**4/64 - 3*x*exp(2*a)*exp(2*b*x)*sinh(b*x + d)*cosh(b*x + d)**5/32 + 3*x* exp(2*a)*exp(2*b*x)*cosh(b*x + d)**6/64 + 3*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**6/(32*b) - 15*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**5*cosh(b*x + d)/(64* b) + 13*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**3*cosh(b*x + d)**3/(32*b) - 15* exp(2*a)*exp(2*b*x)*sinh(b*x + d)*cosh(b*x + d)**5/(64*b) + 3*exp(2*a)*exp (2*b*x)*cosh(b*x + d)**6/(32*b), Ne(b, 0)), (x*exp(2*a)*sinh(d)**3*cosh(d) **3, True))
Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=-\frac {{\left (6 \, e^{\left (-4 \, b x - 4 \, d\right )} - 1\right )} e^{\left (8 \, b x + 2 \, a + 6 \, d\right )}}{512 \, b} + \frac {3 \, {\left (b x + d\right )} e^{\left (2 \, a - 2 \, d\right )}}{64 \, b} + \frac {e^{\left (-4 \, b x + 2 \, a - 6 \, d\right )}}{256 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x, algorithm="maxima" )
Output:
-1/512*(6*e^(-4*b*x - 4*d) - 1)*e^(8*b*x + 2*a + 6*d)/b + 3/64*(b*x + d)*e ^(2*a - 2*d)/b + 1/256*e^(-4*b*x + 2*a - 6*d)/b
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.07 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=-\frac {{\left (2 \, {\left (3 \, e^{\left (4 \, b x + 2 \, a + 4 \, d\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (-4 \, b x - 4 \, d\right )} - 24 \, {\left (b x + d\right )} e^{\left (2 \, a\right )} - e^{\left (8 \, b x + 2 \, a + 8 \, d\right )} + 6 \, e^{\left (4 \, b x + 2 \, a + 4 \, d\right )}\right )} e^{\left (-2 \, d\right )}}{512 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x, algorithm="giac")
Output:
-1/512*(2*(3*e^(4*b*x + 2*a + 4*d) - e^(2*a))*e^(-4*b*x - 4*d) - 24*(b*x + d)*e^(2*a) - e^(8*b*x + 2*a + 8*d) + 6*e^(4*b*x + 2*a + 4*d))*e^(-2*d)/b
Time = 0.42 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.83 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {3\,x\,{\mathrm {e}}^{2\,a-2\,d}}{64}-\frac {3\,{\mathrm {e}}^{2\,a+2\,d+4\,b\,x}}{256\,b}+\frac {{\mathrm {e}}^{2\,a-6\,d-4\,b\,x}}{256\,b}+\frac {{\mathrm {e}}^{2\,a+6\,d+8\,b\,x}}{512\,b} \] Input:
int(cosh(d + b*x)^3*exp(2*a + 2*b*x)*sinh(d + b*x)^3,x)
Output:
(3*x*exp(2*a - 2*d))/64 - (3*exp(2*a + 2*d + 4*b*x))/(256*b) + exp(2*a - 6 *d - 4*b*x)/(256*b) + exp(2*a + 6*d + 8*b*x)/(512*b)
Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.79 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \sinh ^3(d+b x) \, dx=\frac {e^{2 a} \left (e^{12 b x +12 d}-6 e^{8 b x +8 d}+24 e^{4 b x +4 d} b x +2\right )}{512 e^{4 b x +6 d} b} \] Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)^3*sinh(b*x+d)^3,x)
Output:
(e**(2*a)*(e**(12*b*x + 12*d) - 6*e**(8*b*x + 8*d) + 24*e**(4*b*x + 4*d)*b *x + 2))/(512*e**(4*b*x + 6*d)*b)