\(\int e^{a+b x} \text {sech}^5(a+b x) \, dx\) [273]

Optimal result

Integrand size = 16, antiderivative size = 60 \[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=-\frac {4}{b \left (1+e^{2 a+2 b x}\right )^4}+\frac {32}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {8}{b \left (1+e^{2 a+2 b x}\right )^2} \] Output:

-4/b/(1+exp(2*b*x+2*a))^4+32/3/b/(1+exp(2*b*x+2*a))^3-8/b/(1+exp(2*b*x+2*a 
))^2
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=-\frac {4 \left (1+4 e^{2 (a+b x)}+6 e^{4 (a+b x)}\right )}{3 b \left (1+e^{2 (a+b x)}\right )^4} \] Input:

Integrate[E^(a + b*x)*Sech[a + b*x]^5,x]
 

Output:

(-4*(1 + 4*E^(2*(a + b*x)) + 6*E^(4*(a + b*x))))/(3*b*(1 + E^(2*(a + b*x)) 
)^4)
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2720, 27, 243, 53, 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.

\(\displaystyle \int e^{a+b x} \text {sech}^5(a+b x) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\int \frac {32 e^{5 a+5 b x}}{\left (1+e^{2 a+2 b x}\right )^5}de^{a+b x}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {32 \int \frac {e^{5 a+5 b x}}{\left (1+e^{2 a+2 b x}\right )^5}de^{a+b x}}{b}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {16 \int \frac {e^{2 a+2 b x}}{\left (1+e^{2 a+2 b x}\right )^5}de^{2 a+2 b x}}{b}\)

\(\Big \downarrow \) 53

\(\displaystyle \frac {16 \int \left (\frac {1}{\left (1+e^{2 a+2 b x}\right )^3}-\frac {2}{\left (1+e^{2 a+2 b x}\right )^4}+\frac {1}{\left (1+e^{2 a+2 b x}\right )^5}\right )de^{2 a+2 b x}}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {16 \left (-\frac {1}{2 \left (e^{2 a+2 b x}+1\right )^2}+\frac {2}{3 \left (e^{2 a+2 b x}+1\right )^3}-\frac {1}{4 \left (e^{2 a+2 b x}+1\right )^4}\right )}{b}\)

Input:

Int[E^(a + b*x)*Sech[a + b*x]^5,x]
 

Output:

(16*(-1/4*1/(1 + E^(2*a + 2*b*x))^4 + 2/(3*(1 + E^(2*a + 2*b*x))^3) - 1/(2 
*(1 + E^(2*a + 2*b*x))^2)))/b
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 53

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, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2720

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]]]
 

Maple [A] (verified)

Time = 3.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(\frac {-\frac {1}{4 \cosh \left (b x +a \right )^{4}}+\left (\frac {2}{3}+\frac {\operatorname {sech}\left (b x +a \right )^{2}}{3}\right ) \tanh \left (b x +a \right )}{b}\)	\(35\)
default	\(\frac {-\frac {1}{4 \cosh \left (b x +a \right )^{4}}+\left (\frac {2}{3}+\frac {\operatorname {sech}\left (b x +a \right )^{2}}{3}\right ) \tanh \left (b x +a \right )}{b}\)	\(35\)
risch	\(-\frac {4 \left (6 \,{\mathrm e}^{4 b x +4 a}+4 \,{\mathrm e}^{2 b x +2 a}+1\right )}{3 b \left (1+{\mathrm e}^{2 b x +2 a}\right )^{4}}\)	\(43\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{b x +a} \left (\cosh \left (3 b x +3 a \right )+\sinh \left (3 b x +3 a \right )+4 \cosh \left (b x +a \right )+4 \sinh \left (b x +a \right )\right )}{3 b \left (\cosh \left (4 b x +4 a \right )+4 \cosh \left (2 b x +2 a \right )+3\right )}\)	\(71\)

Input:

int(exp(b*x+a)*sech(b*x+a)^5,x,method=_RETURNVERBOSE)
 

Output:

1/b*(-1/4/cosh(b*x+a)^4+(2/3+1/3*sech(b*x+a)^2)*tanh(b*x+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (55) = 110\).

Time = 0.09 (sec) , antiderivative size = 233, normalized size of antiderivative = 3.88 \[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=-\frac {4 \, {\left (7 \, \cosh \left (b x + a\right )^{2} + 10 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + 7 \, \sinh \left (b x + a\right )^{2} + 4\right )}}{3 \, {\left (b \cosh \left (b x + a\right )^{6} + 6 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + b \sinh \left (b x + a\right )^{6} + 4 \, b \cosh \left (b x + a\right )^{4} + {\left (15 \, b \cosh \left (b x + a\right )^{2} + 4 \, b\right )} \sinh \left (b x + a\right )^{4} + 4 \, {\left (5 \, b \cosh \left (b x + a\right )^{3} + 4 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{3} + 7 \, b \cosh \left (b x + a\right )^{2} + {\left (15 \, b \cosh \left (b x + a\right )^{4} + 24 \, b \cosh \left (b x + a\right )^{2} + 7 \, b\right )} \sinh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{5} + 8 \, b \cosh \left (b x + a\right )^{3} + 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 4 \, b\right )}} \] Input:

integrate(exp(b*x+a)*sech(b*x+a)^5,x, algorithm="fricas")
 

Output:

-4/3*(7*cosh(b*x + a)^2 + 10*cosh(b*x + a)*sinh(b*x + a) + 7*sinh(b*x + a) 
^2 + 4)/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5 + b*sinh(b* 
x + a)^6 + 4*b*cosh(b*x + a)^4 + (15*b*cosh(b*x + a)^2 + 4*b)*sinh(b*x + a 
)^4 + 4*(5*b*cosh(b*x + a)^3 + 4*b*cosh(b*x + a))*sinh(b*x + a)^3 + 7*b*co 
sh(b*x + a)^2 + (15*b*cosh(b*x + a)^4 + 24*b*cosh(b*x + a)^2 + 7*b)*sinh(b 
*x + a)^2 + 2*(3*b*cosh(b*x + a)^5 + 8*b*cosh(b*x + a)^3 + 5*b*cosh(b*x + 
a))*sinh(b*x + a) + 4*b)
 

Sympy [F]

\[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=e^{a} \int e^{b x} \operatorname {sech}^{5}{\left (a + b x \right )}\, dx \] Input:

integrate(exp(b*x+a)*sech(b*x+a)**5,x)
 

Output:

exp(a)*Integral(exp(b*x)*sech(a + b*x)**5, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (55) = 110\).

Time = 0.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.87 \[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=-\frac {8 \, e^{\left (4 \, b x + 4 \, a\right )}}{b {\left (e^{\left (8 \, b x + 8 \, a\right )} + 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} - \frac {16 \, e^{\left (2 \, b x + 2 \, a\right )}}{3 \, b {\left (e^{\left (8 \, b x + 8 \, a\right )} + 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} - \frac {4}{3 \, b {\left (e^{\left (8 \, b x + 8 \, a\right )} + 4 \, e^{\left (6 \, b x + 6 \, a\right )} + 6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \] Input:

integrate(exp(b*x+a)*sech(b*x+a)^5,x, algorithm="maxima")
 

Output:

-8*e^(4*b*x + 4*a)/(b*(e^(8*b*x + 8*a) + 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 
4*a) + 4*e^(2*b*x + 2*a) + 1)) - 16/3*e^(2*b*x + 2*a)/(b*(e^(8*b*x + 8*a) 
+ 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) + 4*e^(2*b*x + 2*a) + 1)) - 4/3/(b 
*(e^(8*b*x + 8*a) + 4*e^(6*b*x + 6*a) + 6*e^(4*b*x + 4*a) + 4*e^(2*b*x + 2 
*a) + 1))
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=-\frac {4 \, {\left (6 \, e^{\left (4 \, b x + 4 \, a\right )} + 4 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{3 \, b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{4}} \] Input:

integrate(exp(b*x+a)*sech(b*x+a)^5,x, algorithm="giac")
 

Output:

-4/3*(6*e^(4*b*x + 4*a) + 4*e^(2*b*x + 2*a) + 1)/(b*(e^(2*b*x + 2*a) + 1)^ 
4)
 

Mupad [B] (verification not implemented)

Time = 2.00 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=-\frac {4\,\left (4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}{3\,b\,{\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}^4} \] Input:

int(exp(a + b*x)/cosh(a + b*x)^5,x)
 

Output:

-(4*(4*exp(2*a + 2*b*x) + 6*exp(4*a + 4*b*x) + 1))/(3*b*(exp(2*a + 2*b*x) 
+ 1)^4)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.35 \[ \int e^{a+b x} \text {sech}^5(a+b x) \, dx=\frac {-8 e^{4 b x +4 a}-\frac {16 e^{2 b x +2 a}}{3}-\frac {4}{3}}{b \left (e^{8 b x +8 a}+4 e^{6 b x +6 a}+6 e^{4 b x +4 a}+4 e^{2 b x +2 a}+1\right )} \] Input:

int(exp(b*x+a)*sech(b*x+a)^5,x)
 

Output:

(4*( - 6*e**(4*a + 4*b*x) - 4*e**(2*a + 2*b*x) - 1))/(3*b*(e**(8*a + 8*b*x 
) + 4*e**(6*a + 6*b*x) + 6*e**(4*a + 4*b*x) + 4*e**(2*a + 2*b*x) + 1))