Integrand size = 16, antiderivative size = 87 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=-\frac {e^{a-4 d-3 b x}}{48 b}-\frac {e^{a-2 d-b x}}{4 b}+\frac {3 e^{a+b x}}{8 b}+\frac {e^{a+2 d+3 b x}}{12 b}+\frac {e^{a+4 d+5 b x}}{80 b} \] Output:
-1/48*exp(-3*b*x+a-4*d)/b-1/4*exp(-b*x+a-2*d)/b+3/8*exp(b*x+a)/b+1/12*exp( 3*b*x+a+2*d)/b+1/80*exp(5*b*x+a+4*d)/b
Time = 0.13 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.16 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=\frac {e^{a-3 b x} \left (90 e^{4 b x}+20 e^{2 b x} \left (-3+e^{4 b x}\right ) \cosh (2 d)+\left (-5+3 e^{8 b x}\right ) \cosh (4 d)+60 e^{2 b x} \sinh (2 d)+20 e^{6 b x} \sinh (2 d)+5 \sinh (4 d)+3 e^{8 b x} \sinh (4 d)\right )}{240 b} \] Input:
Integrate[E^(a + b*x)*Cosh[d + b*x]^4,x]
Output:
(E^(a - 3*b*x)*(90*E^(4*b*x) + 20*E^(2*b*x)*(-3 + E^(4*b*x))*Cosh[2*d] + ( -5 + 3*E^(8*b*x))*Cosh[4*d] + 60*E^(2*b*x)*Sinh[2*d] + 20*E^(6*b*x)*Sinh[2 *d] + 5*Sinh[4*d] + 3*E^(8*b*x)*Sinh[4*d]))/(240*b)
Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.64, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.
\(\displaystyle \int e^{a+b x} \cosh ^4(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {1}{16} e^{a-4 b x} \left (1+e^{2 b x}\right )^4de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^a \int e^{-4 b x} \left (1+e^{2 b x}\right )^4de^{b x}}{16 b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {e^a \int \left (6+e^{-4 b x}+4 e^{-2 b x}+4 e^{2 b x}+e^{4 b x}\right )de^{b x}}{16 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^a \left (-\frac {1}{3} e^{-3 b x}-4 e^{-b x}+6 e^{b x}+\frac {4}{3} e^{3 b x}+\frac {1}{5} e^{5 b x}\right )}{16 b}\) |
Input:
Int[E^(a + b*x)*Cosh[d + b*x]^4,x]
Output:
(E^a*(-1/3*1/E^(3*b*x) - 4/E^(b*x) + 6*E^(b*x) + (4*E^(3*b*x))/3 + E^(5*b* x)/5))/(16*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 = 1.79 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.84
method | result | size |
risch | \(-\frac {{\mathrm e}^{-3 b x +a -4 d}}{48 b}-\frac {{\mathrm e}^{-b x +a -2 d}}{4 b}+\frac {3 \,{\mathrm e}^{b x +a}}{8 b}+\frac {{\mathrm e}^{3 b x +a +2 d}}{12 b}+\frac {{\mathrm e}^{5 b x +a +4 d}}{80 b}\) | \(73\) |
parallelrisch | \(\frac {2 \,{\mathrm e}^{b x +a} \left (15 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{7}-15 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{6}-5 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{5}+25 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{4}+13 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{3}-21 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )^{2}+9 \tanh \left (\frac {b x}{2}+\frac {d}{2}\right )+3\right )}{15 b \left (-1+\tanh \left (\frac {b x}{2}+\frac {d}{2}\right )\right )^{4} \left (1+\tanh \left (\frac {b x}{2}+\frac {d}{2}\right )\right )^{4}}\) | \(129\) |
default | \(\frac {3 \sinh \left (b x +a \right )}{8 b}-\frac {\sinh \left (-3 b x +a -4 d \right )}{48 b}-\frac {\sinh \left (-b x +a -2 d \right )}{4 b}+\frac {\sinh \left (3 b x +a +2 d \right )}{12 b}+\frac {\sinh \left (5 b x +a +4 d \right )}{80 b}+\frac {3 \cosh \left (b x +a \right )}{8 b}-\frac {\cosh \left (-3 b x +a -4 d \right )}{48 b}-\frac {\cosh \left (-b x +a -2 d \right )}{4 b}+\frac {\cosh \left (3 b x +a +2 d \right )}{12 b}+\frac {\cosh \left (5 b x +a +4 d \right )}{80 b}\) | \(144\) |
orering | \(\frac {{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{4}}{5 b}+\frac {\frac {10 b \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{4}}{9}+\frac {40 \,{\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} b \sinh \left (b x +d \right )}{9}}{b^{2}}-\frac {2 \left (5 b^{2} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{4}+8 b^{2} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} \sinh \left (b x +d \right )+12 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{2} b^{2} \cosh \left (b x +d \right )^{2}\right )}{9 b^{3}}-\frac {13 b^{3} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{4}+52 b^{3} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} \sinh \left (b x +d \right )+36 b^{3} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )^{2}+24 \,{\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{3} b^{3} \cosh \left (b x +d \right )}{9 b^{4}}+\frac {65 b^{4} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{4}+176 b^{4} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{3} \sinh \left (b x +d \right )+264 b^{4} {\mathrm e}^{b x +a} \cosh \left (b x +d \right )^{2} \sinh \left (b x +d \right )^{2}+96 b^{4} {\mathrm e}^{b x +a} \cosh \left (b x +d \right ) \sinh \left (b x +d \right )^{3}+24 b^{4} {\mathrm e}^{b x +a} \sinh \left (b x +d \right )^{4}}{45 b^{5}}\) | \(366\) |
Input:
int(exp(b*x+a)*cosh(b*x+d)^4,x,method=_RETURNVERBOSE)
Output:
-1/48*exp(-3*b*x+a-4*d)/b-1/4*exp(-b*x+a-2*d)/b+3/8*exp(b*x+a)/b+1/12*exp( 3*b*x+a+2*d)/b+1/80*exp(5*b*x+a+4*d)/b
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (72) = 144\).
Time = 0.09 (sec) , antiderivative size = 270, normalized size of antiderivative = 3.10 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=-\frac {\cosh \left (b x + d\right )^{4} \cosh \left (-a + d\right ) + {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{4} - 16 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{3} + 20 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + 2 \, {\left (3 \, \cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) - {\left (3 \, \cosh \left (b x + d\right )^{2} + 10\right )} \sinh \left (-a + d\right ) + 10 \, \cosh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} - 16 \, {\left (\cosh \left (b x + d\right )^{3} \cosh \left (-a + d\right ) + 5 \, \cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - {\left (\cosh \left (b x + d\right )^{3} + 5 \, \cosh \left (b x + d\right )\right )} \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{4} + 20 \, \cosh \left (b x + d\right )^{2} - 45\right )} \sinh \left (-a + d\right ) - 45 \, \cosh \left (-a + d\right )}{120 \, {\left (b \cosh \left (b x + d\right ) - b \sinh \left (b x + d\right )\right )}} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^4,x, algorithm="fricas")
Output:
-1/120*(cosh(b*x + d)^4*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh( b*x + d)^4 - 16*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))* sinh(b*x + d)^3 + 20*cosh(b*x + d)^2*cosh(-a + d) + 2*(3*cosh(b*x + d)^2*c osh(-a + d) - (3*cosh(b*x + d)^2 + 10)*sinh(-a + d) + 10*cosh(-a + d))*sin h(b*x + d)^2 - 16*(cosh(b*x + d)^3*cosh(-a + d) + 5*cosh(b*x + d)*cosh(-a + d) - (cosh(b*x + d)^3 + 5*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d) - ( cosh(b*x + d)^4 + 20*cosh(b*x + d)^2 - 45)*sinh(-a + d) - 45*cosh(-a + d)) /(b*cosh(b*x + d) - b*sinh(b*x + d))
Time = 2.30 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.60 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=\begin {cases} \frac {8 e^{a} e^{b x} \sinh ^{4}{\left (b x + d \right )}}{15 b} - \frac {8 e^{a} e^{b x} \sinh ^{3}{\left (b x + d \right )} \cosh {\left (b x + d \right )}}{15 b} - \frac {4 e^{a} e^{b x} \sinh ^{2}{\left (b x + d \right )} \cosh ^{2}{\left (b x + d \right )}}{5 b} + \frac {4 e^{a} e^{b x} \sinh {\left (b x + d \right )} \cosh ^{3}{\left (b x + d \right )}}{5 b} + \frac {e^{a} e^{b x} \cosh ^{4}{\left (b x + d \right )}}{5 b} & \text {for}\: b \neq 0 \\x e^{a} \cosh ^{4}{\left (d \right )} & \text {otherwise} \end {cases} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)**4,x)
Output:
Piecewise((8*exp(a)*exp(b*x)*sinh(b*x + d)**4/(15*b) - 8*exp(a)*exp(b*x)*s inh(b*x + d)**3*cosh(b*x + d)/(15*b) - 4*exp(a)*exp(b*x)*sinh(b*x + d)**2* cosh(b*x + d)**2/(5*b) + 4*exp(a)*exp(b*x)*sinh(b*x + d)*cosh(b*x + d)**3/ (5*b) + exp(a)*exp(b*x)*cosh(b*x + d)**4/(5*b), Ne(b, 0)), (x*exp(a)*cosh( d)**4, True))
Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=\frac {e^{\left (5 \, b x + a + 4 \, d\right )}}{80 \, b} + \frac {e^{\left (3 \, b x + a + 2 \, d\right )}}{12 \, b} + \frac {3 \, e^{\left (b x + a\right )}}{8 \, b} - \frac {e^{\left (-b x + a - 2 \, d\right )}}{4 \, b} - \frac {e^{\left (-3 \, b x + a - 4 \, d\right )}}{48 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^4,x, algorithm="maxima")
Output:
1/80*e^(5*b*x + a + 4*d)/b + 1/12*e^(3*b*x + a + 2*d)/b + 3/8*e^(b*x + a)/ b - 1/4*e^(-b*x + a - 2*d)/b - 1/48*e^(-3*b*x + a - 4*d)/b
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=-\frac {{\left (5 \, {\left (12 \, e^{\left (2 \, b x + a + 2 \, d\right )} + e^{a}\right )} e^{\left (-3 \, b x\right )} - 3 \, e^{\left (5 \, b x + a + 8 \, d\right )} - 20 \, e^{\left (3 \, b x + a + 6 \, d\right )} - 90 \, e^{\left (b x + a + 4 \, d\right )}\right )} e^{\left (-4 \, d\right )}}{240 \, b} \] Input:
integrate(exp(b*x+a)*cosh(b*x+d)^4,x, algorithm="giac")
Output:
-1/240*(5*(12*e^(2*b*x + a + 2*d) + e^a)*e^(-3*b*x) - 3*e^(5*b*x + a + 8*d ) - 20*e^(3*b*x + a + 6*d) - 90*e^(b*x + a + 4*d))*e^(-4*d)/b
Time = 0.55 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.83 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=\frac {{\mathrm {e}}^{a+2\,d+3\,b\,x}}{12\,b}-\frac {{\mathrm {e}}^{a-2\,d-b\,x}}{4\,b}-\frac {{\mathrm {e}}^{a-4\,d-3\,b\,x}}{48\,b}+\frac {{\mathrm {e}}^{a+4\,d+5\,b\,x}}{80\,b}+\frac {3\,{\mathrm {e}}^{a+b\,x}}{8\,b} \] Input:
int(cosh(d + b*x)^4*exp(a + b*x),x)
Output:
exp(a + 2*d + 3*b*x)/(12*b) - exp(a - 2*d - b*x)/(4*b) - exp(a - 4*d - 3*b *x)/(48*b) + exp(a + 4*d + 5*b*x)/(80*b) + (3*exp(a + b*x))/(8*b)
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int e^{a+b x} \cosh ^4(d+b x) \, dx=\frac {e^{a} \left (3 e^{8 b x +8 d}+20 e^{6 b x +6 d}+90 e^{4 b x +4 d}-60 e^{2 b x +2 d}-5\right )}{240 e^{3 b x +4 d} b} \] Input:
int(exp(b*x+a)*cosh(b*x+d)^4,x)
Output:
(e**a*(3*e**(8*b*x + 8*d) + 20*e**(6*b*x + 6*d) + 90*e**(4*b*x + 4*d) - 60 *e**(2*b*x + 2*d) - 5))/(240*e**(3*b*x + 4*d)*b)