\(\int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx\) [86]

Optimal result

Integrand size = 18, antiderivative size = 144 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx=\frac {3 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{b}+\frac {\sqrt {3} e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1-2 e^{\frac {2}{3} (d+b x)}}{\sqrt {3}}\right )}{b}-\frac {e^{\frac {5 (a-d)}{3}} \log \left (1+e^{\frac {2}{3} (d+b x)}\right )}{b}+\frac {e^{\frac {5 (a-d)}{3}} \log \left (1-e^{\frac {2}{3} (d+b x)}+e^{\frac {4}{3} (d+b x)}\right )}{2 b} \] Output:

3*exp(5/3*a-d+2/3*b*x)/b+3^(1/2)*exp(5/3*a-5/3*d)*arctan(1/3*(1-2*exp(2/3* 
b*x+2/3*d))*3^(1/2))/b-exp(5/3*a-5/3*d)*ln(1+exp(2/3*b*x+2/3*d))/b+1/2*exp 
(5/3*a-5/3*d)*ln(1-exp(2/3*b*x+2/3*d)+exp(4/3*b*x+4/3*d))/b
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.29 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx=\frac {3 e^{d+b x+\frac {5}{3} (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {7}{3},-e^{2 (d+b x)}\right )}{4 b} \] Input:

Integrate[E^((5*(a + b*x))/3)*Sech[d + b*x],x]
 

Output:

(3*E^(d + b*x + (5*(a + b*x))/3)*Hypergeometric2F1[1, 4/3, 7/3, -E^(2*(d + 
 b*x))])/(4*b)
 

Rubi [A] (warning: unable to verify)

Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.43, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {2720, 27, 807, 843, 750, 16, 1142, 25, 1083, 217, 1103}

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.

\(\displaystyle \int e^{\frac {5}{3} (a+b x)} \text {sech}(b x+d) \, dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {3 \int \frac {2 e^{\frac {5 a}{3}+\frac {7 b x}{3}}}{1+e^{2 b x}}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6 e^{5 a/3} \int \frac {e^{\frac {7 b x}{3}}}{1+e^{2 b x}}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3 e^{5 a/3} \int \frac {e^{b x}}{1+e^{b x}}de^{\frac {2 b x}{3}}}{b}\)

\(\Big \downarrow \) 843

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

\(\Big \downarrow \) 750

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 e^{5 a/3} \left (\frac {1}{3} \left (-\frac {1}{2} \int \left (1-2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}-\sqrt {3} \arctan \left (\frac {2 e^{\frac {2 b x}{3}}-1}{\sqrt {3}}\right )\right )+e^{\frac {2 b x}{3}}-\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 e^{5 a/3} \left (-\frac {\arctan \left (\frac {2 e^{\frac {2 b x}{3}}-1}{\sqrt {3}}\right )}{\sqrt {3}}+e^{\frac {2 b x}{3}}-\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )}{b}\)

Input:

Int[E^((5*(a + b*x))/3)*Sech[d + b*x],x]
 

Output:

(3*E^((5*a)/3)*(E^((2*b*x)/3) - ArcTan[(-1 + 2*E^((2*b*x)/3))/Sqrt[3]]/Sqr 
t[3] - Log[1 + E^((2*b*x)/3)]/3))/b
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*Rt[a, 3] - 
 Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; 
 FreeQ[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

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 [C] (verified)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}}}{b}+\frac {\ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}+\frac {i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}+\frac {\ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{2 b}-\frac {i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{2 b}-\frac {{\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \ln \left (1+{\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}\right )}{b}\)	\(176\)

Input:

int(exp(5/3*b*x+5/3*a)*sech(b*x+d),x,method=_RETURNVERBOSE)
 

Output:

3*exp(5/3*a-d+2/3*b*x)/b+1/2*ln(exp(2/3*b*x+2/3*d)-1/2-1/2*I*3^(1/2))/b*ex 
p(5/3*a-5/3*d)+1/2*I*ln(exp(2/3*b*x+2/3*d)-1/2-1/2*I*3^(1/2))/b*exp(5/3*a- 
5/3*d)*3^(1/2)+1/2*ln(exp(2/3*b*x+2/3*d)-1/2+1/2*I*3^(1/2))/b*exp(5/3*a-5/ 
3*d)-1/2*I*ln(exp(2/3*b*x+2/3*d)-1/2+1/2*I*3^(1/2))/b*exp(5/3*a-5/3*d)*3^( 
1/2)-exp(5/3*a-5/3*d)*ln(1+exp(2/3*b*x+2/3*d))/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (113) = 226\).

Time = 0.09 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.51 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx =\text {Too large to display} \] Input:

integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d),x, algorithm="fricas")
 

Output:

1/2*(6*cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) + 6*(cosh(-5/3*a + 5/3 
*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 - 6*cosh(1/3*b*x + 1/3 
*d)^2*sinh(-5/3*a + 5/3*d) + 2*(sqrt(3)*cosh(-5/3*a + 5/3*d) - sqrt(3)*sin 
h(-5/3*a + 5/3*d))*arctan(-1/3*(sqrt(3)*cosh(1/3*b*x + 1/3*d) + 3*sqrt(3)* 
sinh(1/3*b*x + 1/3*d))/(cosh(1/3*b*x + 1/3*d) - sinh(1/3*b*x + 1/3*d))) + 
(cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*log((2*cosh(1/3*b*x + 1/3*d) 
^2 + 2*sinh(1/3*b*x + 1/3*d)^2 - 1)/(cosh(1/3*b*x + 1/3*d)^2 - 2*cosh(1/3* 
b*x + 1/3*d)*sinh(1/3*b*x + 1/3*d) + sinh(1/3*b*x + 1/3*d)^2)) - 2*(cosh(- 
5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*log(2*cosh(1/3*b*x + 1/3*d)/(cosh(1 
/3*b*x + 1/3*d) - sinh(1/3*b*x + 1/3*d))) + 12*(cosh(1/3*b*x + 1/3*d)*cosh 
(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b* 
x + 1/3*d))/b
 

Sympy [F]

\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx=e^{\frac {5 a}{3}} \int e^{\frac {5 b x}{3}} \operatorname {sech}{\left (b x + d \right )}\, dx \] Input:

integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d),x)
 

Output:

exp(5*a/3)*Integral(exp(5*b*x/3)*sech(b*x + d), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.28 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx=-\frac {\sqrt {3} \arctan \left (\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{b} + \frac {\sqrt {3} \arctan \left (-\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{b} + \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{2 \, b} + \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{2 \, b} - \frac {e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )}{b} + \frac {3 \, e^{\left (\frac {2}{3} \, b x + \frac {5}{3} \, a - d\right )}}{b} \] Input:

integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d),x, algorithm="maxima")
 

Output:

-sqrt(3)*arctan(sqrt(3) + 2*e^(-1/3*b*x - 1/3*d))*e^(5/3*a - 5/3*d)/b + sq 
rt(3)*arctan(-sqrt(3) + 2*e^(-1/3*b*x - 1/3*d))*e^(5/3*a - 5/3*d)/b + 1/2* 
e^(5/3*a - 5/3*d)*log(sqrt(3)*e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x - 2/3*d) 
+ 1)/b + 1/2*e^(5/3*a - 5/3*d)*log(-sqrt(3)*e^(-1/3*b*x - 1/3*d) + e^(-2/3 
*b*x - 2/3*d) + 1)/b - e^(5/3*a - 5/3*d)*log(e^(-2/3*b*x - 2/3*d) + 1)/b + 
 3*e^(2/3*b*x + 5/3*a - d)/b
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.70 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx=-\frac {{\left (2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {2}{3} \, b x\right )} - e^{\left (-\frac {2}{3} \, d\right )}\right )} e^{\left (\frac {2}{3} \, d\right )}\right ) e^{\left (-\frac {8}{3} \, d\right )} - e^{\left (-\frac {8}{3} \, d\right )} \log \left (e^{\left (\frac {4}{3} \, b x\right )} - e^{\left (\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + e^{\left (-\frac {4}{3} \, d\right )}\right ) + 2 \, e^{\left (-\frac {8}{3} \, d\right )} \log \left (e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 6 \, e^{\left (\frac {2}{3} \, b x - 2 \, d\right )}\right )} e^{\left (\frac {5}{3} \, a + d\right )}}{2 \, b} \] Input:

integrate(exp(5/3*b*x+5/3*a)*sech(b*x+d),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-1/2*(2*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(2/3*b*x) - e^(-2/3*d))*e^(2/3*d)) 
*e^(-8/3*d) - e^(-8/3*d)*log(e^(4/3*b*x) - e^(2/3*b*x - 2/3*d) + e^(-4/3*d 
)) + 2*e^(-8/3*d)*log(e^(2/3*b*x) + e^(-2/3*d)) - 6*e^(2/3*b*x - 2*d))*e^( 
5/3*a + d)/b
 

Mupad [B] (verification not implemented)

Time = 0.99 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.42 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx=\frac {3\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}}{b}+\frac {{\left (-{\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (2\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}-2\,{\left (-{\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}\right )}{b}+\frac {{\left (-{\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (2\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}-2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (-{\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b}-\frac {{\left (-{\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left (-{\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}+2\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{b} \] Input:

int(exp((5*a)/3 + (5*b*x)/3)/cosh(d + b*x),x)
 

Output:

(3*exp((5*a)/3 - d + (2*b*x)/3))/b + ((-exp(5*a - 5*d))^(1/3)*log(2*exp((5 
*a)/3)*exp((2*d)/3)*exp(-(5*d)/3)*exp((2*b*x)/3) - 2*(-exp(5*a)*exp(-5*d)) 
^(1/3)))/b + ((-exp(5*a - 5*d))^(1/3)*log(2*exp((5*a)/3)*exp((2*d)/3)*exp( 
-(5*d)/3)*exp((2*b*x)/3) - 2*((3^(1/2)*1i)/2 - 1/2)*(-exp(5*a)*exp(-5*d))^ 
(1/3))*((3^(1/2)*1i)/2 - 1/2))/b - ((-exp(5*a - 5*d))^(1/3)*log(2*((3^(1/2 
)*1i)/2 + 1/2)*(-exp(5*a)*exp(-5*d))^(1/3) + 2*exp((5*a)/3)*exp((2*d)/3)*e 
xp(-(5*d)/3)*exp((2*b*x)/3))*((3^(1/2)*1i)/2 + 1/2))/b
 

Reduce [F]

\[ \int e^{\frac {5}{3} (a+b x)} \text {sech}(d+b x) \, dx=\int e^{\frac {5 b x}{3}+\frac {5 a}{3}} \mathrm {sech}\left (b x +d \right )d x \] Input:

int(exp(5/3*b*x+5/3*a)*sech(b*x+d),x)
 

Output:

int(e**((5*a + 5*b*x)/3)*sech(b*x + d),x)