Integrand size = 18, antiderivative size = 78 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=-\frac {e^{2 a-3 d-b x}}{8 b}+\frac {3 e^{2 a-d+b x}}{8 b}+\frac {e^{2 a+d+3 b x}}{8 b}+\frac {e^{2 a+3 d+5 b x}}{40 b} \] Output:
-1/8*exp(-b*x+2*a-3*d)/b+3/8*exp(b*x+2*a-d)/b+1/8*exp(3*b*x+2*a+d)/b+1/40* exp(5*b*x+2*a+3*d)/b
Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=\frac {e^{2 a-b x} \left (5 e^{2 b x} \left (3+e^{2 b x}\right ) \cosh (d)+\left (-5+e^{6 b x}\right ) \cosh (3 d)-15 e^{2 b x} \sinh (d)+5 e^{4 b x} \sinh (d)+5 \sinh (3 d)+e^{6 b x} \sinh (3 d)\right )}{40 b} \] Input:
Integrate[E^(2*(a + b*x))*Cosh[d + b*x]^3,x]
Output:
(E^(2*a - b*x)*(5*E^(2*b*x)*(3 + E^(2*b*x))*Cosh[d] + (-5 + E^(6*b*x))*Cos h[3*d] - 15*E^(2*b*x)*Sinh[d] + 5*E^(4*b*x)*Sinh[d] + 5*Sinh[3*d] + E^(6*b *x)*Sinh[3*d]))/(40*b)
Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.56, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2720, 27, 244, 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.
\(\displaystyle \int e^{2 (a+b x)} \cosh ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {1}{8} e^{2 a-2 b x} \left (1+e^{2 b x}\right )^3de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{2 a} \int e^{-2 b x} \left (1+e^{2 b x}\right )^3de^{b x}}{8 b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {e^{2 a} \int \left (3+e^{-2 b x}+3 e^{2 b x}+e^{4 b x}\right )de^{b x}}{8 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{2 a} \left (-e^{-b x}+3 e^{b x}+e^{3 b x}+\frac {1}{5} e^{5 b x}\right )}{8 b}\) |
Input:
Int[E^(2*(a + b*x))*Cosh[d + b*x]^3,x]
Output:
(E^(2*a)*(-E^(-(b*x)) + 3*E^(b*x) + E^(3*b*x) + E^(5*b*x)/5))/(8*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_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, 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 = 0.77 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.86
method | result | size |
risch | \(-\frac {{\mathrm e}^{-b x +2 a -3 d}}{8 b}+\frac {3 \,{\mathrm e}^{b x +2 a -d}}{8 b}+\frac {{\mathrm e}^{3 b x +2 a +d}}{8 b}+\frac {{\mathrm e}^{5 b x +2 a +3 d}}{40 b}\) | \(67\) |
parallelrisch | \(-\frac {2 \,{\mathrm e}^{2 b x +2 a} \left (5 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}-10 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}+10 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}-3 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )+2\right )}{5 b \left (-1+\tanh \left (\frac {b x}{2}+\frac {d}{2}\right )\right )^{3} \left (1+\tanh \left (\frac {b x}{2}+\frac {d}{2}\right )\right )^{3}}\) | \(93\) |
default | \(-\frac {\sinh \left (-b x +2 a -3 d \right )}{8 b}+\frac {3 \sinh \left (b x +2 a -d \right )}{8 b}+\frac {\sinh \left (3 b x +2 a +d \right )}{8 b}+\frac {\sinh \left (5 b x +2 a +3 d \right )}{40 b}-\frac {\cosh \left (-b x +2 a -3 d \right )}{8 b}+\frac {3 \cosh \left (b x +2 a -d \right )}{8 b}+\frac {\cosh \left (3 b x +2 a +d \right )}{8 b}+\frac {\cosh \left (5 b x +2 a +3 d \right )}{40 b}\) | \(132\) |
orering | \(\frac {8 \,{\mathrm e}^{2 b x +2 a} \cosh \left (b x +d \right )^{3}}{15 b}+\frac {\frac {28 \,{\mathrm e}^{2 b x +2 a} \cosh \left (b x +d \right )^{3} b}{15}+\frac {14 \,{\mathrm e}^{2 b x +2 a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right ) b}{5}}{b^{2}}-\frac {8 \left (7 \,{\mathrm e}^{2 b x +2 a} \cosh \left (b x +d \right )^{3} b^{2}+12 \,{\mathrm e}^{2 b x +2 a} b^{2} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )+6 \,{\mathrm e}^{2 b x +2 a} b^{2} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{2}\right )}{15 b^{3}}+\frac {26 \,{\mathrm e}^{2 b x +2 a} b^{3} \cosh \left (b x +d \right )^{3}+57 \,{\mathrm e}^{2 b x +2 a} b^{3} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )+36 \,{\mathrm e}^{2 b x +2 a} b^{3} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{2}+6 \,{\mathrm e}^{2 b x +2 a} b^{3} \sinh \left (b x +d \right )^{3}}{15 b^{4}}\) | \(266\) |
Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/8*exp(-b*x+2*a-3*d)/b+3/8*exp(b*x+2*a-d)/b+1/8*exp(3*b*x+2*a+d)/b+1/40* exp(5*b*x+2*a+3*d)/b
Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (66) = 132\).
Time = 0.08 (sec) , antiderivative size = 232, normalized size of antiderivative = 2.97 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=-\frac {2 \, \cosh \left (b x + d\right )^{3} \cosh \left (-2 \, a + 2 \, d\right ) - 3 \, {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{3} + 6 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - \cosh \left (b x + d\right ) \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{2} - 10 \, \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - {\left (9 \, \cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (9 \, \cosh \left (b x + d\right )^{2} - 5\right )} \sinh \left (-2 \, a + 2 \, d\right ) - 5 \, \cosh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right ) - 2 \, {\left (\cosh \left (b x + d\right )^{3} - 5 \, \cosh \left (b x + d\right )\right )} \sinh \left (-2 \, a + 2 \, d\right )}{20 \, {\left (b \cosh \left (b x + d\right )^{2} - 2 \, b \cosh \left (b x + d\right ) \sinh \left (b x + d\right ) + b \sinh \left (b x + d\right )^{2}\right )}} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3,x, algorithm="fricas")
Output:
-1/20*(2*cosh(b*x + d)^3*cosh(-2*a + 2*d) - 3*(cosh(-2*a + 2*d) - sinh(-2* a + 2*d))*sinh(b*x + d)^3 + 6*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*sinh(-2*a + 2*d))*sinh(b*x + d)^2 - 10*cosh(b*x + d)*cosh(-2*a + 2*d) - (9*cosh(b*x + d)^2*cosh(-2*a + 2*d) - (9*cosh(b*x + d)^2 - 5)*sinh(-2*a + 2*d) - 5*cosh(-2*a + 2*d))*sinh(b*x + d) - 2*(cosh(b*x + d)^3 - 5*cosh(b *x + d))*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. 124 vs. \(2 (61) = 122\).
Time = 0.92 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.59 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=\begin {cases} \frac {2 e^{2 a} e^{2 b x} \sinh ^{3}{\left (b x + d \right )}}{5 b} - \frac {4 e^{2 a} e^{2 b x} \sinh ^{2}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{5 b} + \frac {e^{2 a} e^{2 b x} \sinh {\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{5 b} + \frac {2 e^{2 a} e^{2 b x} \cosh ^{3}{\left (b x + d \right )}}{5 b} & \text {for}\: b \neq 0 \\x e^{2 a} \cosh ^{3}{\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)**3,x)
Output:
Piecewise((2*exp(2*a)*exp(2*b*x)*sinh(b*x + d)**3/(5*b) - 4*exp(2*a)*exp(2 *b*x)*sinh(b*x + d)**2*cosh(b*x + d)/(5*b) + exp(2*a)*exp(2*b*x)*sinh(b*x + d)*cosh(b*x + d)**2/(5*b) + 2*exp(2*a)*exp(2*b*x)*cosh(b*x + d)**3/(5*b) , Ne(b, 0)), (x*exp(2*a)*cosh(d)**3, True))
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.76 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=\frac {{\left (5 \, e^{\left (-2 \, b x - 2 \, d\right )} + 15 \, e^{\left (-4 \, b x - 4 \, d\right )} + 1\right )} e^{\left (5 \, b x + 2 \, a + 3 \, d\right )}}{40 \, b} - \frac {e^{\left (-b x + 2 \, a - 3 \, d\right )}}{8 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3,x, algorithm="maxima")
Output:
1/40*(5*e^(-2*b*x - 2*d) + 15*e^(-4*b*x - 4*d) + 1)*e^(5*b*x + 2*a + 3*d)/ b - 1/8*e^(-b*x + 2*a - 3*d)/b
Time = 0.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.78 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=\frac {{\left (e^{\left (5 \, b x + 2 \, a + 5 \, d\right )} + 5 \, e^{\left (3 \, b x + 2 \, a + 3 \, d\right )} + 15 \, e^{\left (b x + 2 \, a + d\right )} - 5 \, e^{\left (-b x + 2 \, a - d\right )}\right )} e^{\left (-2 \, d\right )}}{40 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)^3,x, algorithm="giac")
Output:
1/40*(e^(5*b*x + 2*a + 5*d) + 5*e^(3*b*x + 2*a + 3*d) + 15*e^(b*x + 2*a + d) - 5*e^(-b*x + 2*a - d))*e^(-2*d)/b
Time = 0.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=\frac {{\mathrm {e}}^{2\,a+d+3\,b\,x}}{8\,b}+\frac {3\,{\mathrm {e}}^{2\,a-d+b\,x}}{8\,b}-\frac {{\mathrm {e}}^{2\,a-3\,d-b\,x}}{8\,b}+\frac {{\mathrm {e}}^{2\,a+3\,d+5\,b\,x}}{40\,b} \] Input:
int(cosh(d + b*x)^3*exp(2*a + 2*b*x),x)
Output:
exp(2*a + d + 3*b*x)/(8*b) + (3*exp(2*a - d + b*x))/(8*b) - exp(2*a - 3*d - b*x)/(8*b) + exp(2*a + 3*d + 5*b*x)/(40*b)
Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.73 \[ \int e^{2 (a+b x)} \cosh ^3(d+b x) \, dx=\frac {e^{2 a} \left (e^{6 b x +6 d}+5 e^{4 b x +4 d}+15 e^{2 b x +2 d}-5\right )}{40 e^{b x +3 d} b} \] Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)^3,x)
Output:
(e**(2*a)*(e**(6*b*x + 6*d) + 5*e**(4*b*x + 4*d) + 15*e**(2*b*x + 2*d) - 5 ))/(40*e**(b*x + 3*d)*b)