\(\int e^{\frac {5}{3} (a+b x)} \text {sech}^3(d+b x) \, dx\) [88]

Optimal result

Integrand size = 20, antiderivative size = 203 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^3(d+b x) \, dx=-\frac {2 e^{\frac {5 (a-d)}{3}+\frac {8}{3} (d+b x)}}{b \left (1+e^{2 (d+b x)}\right )^2}-\frac {8 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)}}{3 b \left (1+e^{2 (d+b x)}\right )}-\frac {8 e^{\frac {5 (a-d)}{3}} \arctan \left (\frac {1-2 e^{\frac {2}{3} (d+b x)}}{\sqrt {3}}\right )}{3 \sqrt {3} b}+\frac {8 e^{\frac {5 (a-d)}{3}} \log \left (1+e^{\frac {2}{3} (d+b x)}\right )}{9 b}-\frac {4 e^{\frac {5 (a-d)}{3}} \log \left (1-e^{\frac {2}{3} (d+b x)}+e^{\frac {4}{3} (d+b x)}\right )}{9 b} \] Output:

-2*exp(5/3*a+d+8/3*b*x)/b/(1+exp(2*b*x+2*d))^2-8/3*exp(5/3*a-d+2/3*b*x)/b/ 
(1+exp(2*b*x+2*d))-8/9*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+8/9*exp(5/3*a-5/3*d)*ln(1+exp(2/3*b*x+2/3*d))/b-4/9*e 
xp(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 = 0.09 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.32 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^3(d+b x) \, dx=\frac {e^{\frac {5}{3} (a+b x)} \left (-4 e^{d+b x} \operatorname {Hypergeometric2F1}\left (1,\frac {4}{3},\frac {7}{3},-e^{2 (d+b x)}\right )+\text {sech}(d+b x) (5+3 \tanh (d+b x))\right )}{6 b} \] Input:

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

Output:

(E^((5*(a + b*x))/3)*(-4*E^(d + b*x)*Hypergeometric2F1[1, 4/3, 7/3, -E^(2* 
(d + b*x))] + Sech[d + b*x]*(5 + 3*Tanh[d + b*x])))/(6*b)
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.52, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2720, 27, 807, 817, 817, 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.

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

\(\Big \downarrow \) 2720

\(\displaystyle \frac {3 \int \frac {8 e^{\frac {5 a}{3}+\frac {13 b x}{3}}}{\left (1+e^{2 b x}\right )^3}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {24 e^{5 a/3} \int \frac {e^{\frac {13 b x}{3}}}{\left (1+e^{2 b x}\right )^3}de^{\frac {b x}{3}}}{b}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {12 e^{5 a/3} \int \frac {e^{2 b x}}{\left (1+e^{b x}\right )^3}de^{\frac {2 b x}{3}}}{b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \int \frac {e^{b x}}{\left (1+e^{b x}\right )^2}de^{\frac {2 b x}{3}}-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \int \frac {1}{1+e^{b x}}de^{\frac {2 b x}{3}}-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \int \left (2-e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}+\frac {1}{3} \int \frac {1}{1+e^{\frac {2 b x}{3}}}de^{\frac {2 b x}{3}}\right )-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \int \left (2-e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {3}{2} \int 1de^{\frac {2 b x}{3}}-\frac {1}{2} \int \left (-1+2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {3}{2} \int 1de^{\frac {2 b x}{3}}+\frac {1}{2} \int \left (1-2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}-3 \int \frac {1}{-2-2 e^{\frac {2 b x}{3}}}d\left (-1+2 e^{\frac {2 b x}{3}}\right )\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (1-2 e^{\frac {2 b x}{3}}\right )de^{\frac {2 b x}{3}}+\sqrt {3} \arctan \left (\frac {2 e^{\frac {2 b x}{3}}-1}{\sqrt {3}}\right )\right )+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {12 e^{5 a/3} \left (\frac {2}{3} \left (\frac {1}{3} \left (\frac {\arctan \left (\frac {2 e^{\frac {2 b x}{3}}-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (e^{\frac {2 b x}{3}}+1\right )\right )-\frac {e^{\frac {2 b x}{3}}}{3 \left (e^{b x}+1\right )}\right )-\frac {e^{\frac {4 b x}{3}}}{6 \left (e^{b x}+1\right )^2}\right )}{b}\)

Input:

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

Output:

(12*E^((5*a)/3)*(-1/6*E^((4*b*x)/3)/(1 + E^(b*x))^2 + (2*(-1/3*E^((2*b*x)/ 
3)/(1 + E^(b*x)) + (ArcTan[(-1 + 2*E^((2*b*x)/3))/Sqrt[3]]/Sqrt[3] + Log[1 
 + E^((2*b*x)/3)]/3)/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 817

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*n*(p + 1))), x] - Simp[c^n 
*((m - n + 1)/(b*n*(p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x 
] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  ! 
ILtQ[(m + n*(p + 1) + 1)/n, 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 = 1.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {2 \left (7 \,{\mathrm e}^{2 b x +2 d}+4\right ) {\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}}}{3 \left (1+{\mathrm e}^{2 b x +2 d}\right )^{2} b}-\frac {4 \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{9 b}+\frac {4 i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{9 b}-\frac {4 \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{9 b}-\frac {4 i \ln \left ({\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \sqrt {3}}{9 b}+\frac {8 \,{\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}} \ln \left (1+{\mathrm e}^{\frac {2 b x}{3}+\frac {2 d}{3}}\right )}{9 b}\)	\(202\)

Input:

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

Output:

-2/3/(1+exp(2*b*x+2*d))^2/b*(7*exp(2*b*x+2*d)+4)*exp(5/3*a-d+2/3*b*x)-4/9* 
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)+4/9*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)-4/9*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)-4/9*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)+8/9*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. 3955 vs. \(2 (155) = 310\).

Time = 0.12 (sec) , antiderivative size = 3955, normalized size of antiderivative = 19.48 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^3(d+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.93 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^3(d+b x) \, dx=\frac {4 \, {\left (2 \, \sqrt {3} \arctan \left (\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) - 2 \, \sqrt {3} \arctan \left (-\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) - \log \left (\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) - \log \left (-\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) + 2 \, \log \left (e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right )\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{9 \, b} - \frac {2 \, {\left (7 \, e^{\left (-\frac {4}{3} \, b x - \frac {4}{3} \, d\right )} + 4 \, e^{\left (-\frac {10}{3} \, b x - \frac {10}{3} \, d\right )}\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{3 \, b {\left (2 \, e^{\left (-2 \, b x - 2 \, d\right )} + e^{\left (-4 \, b x - 4 \, d\right )} + 1\right )}} \] Input:

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

Output:

4/9*(2*sqrt(3)*arctan(sqrt(3) + 2*e^(-1/3*b*x - 1/3*d)) - 2*sqrt(3)*arctan 
(-sqrt(3) + 2*e^(-1/3*b*x - 1/3*d)) - log(sqrt(3)*e^(-1/3*b*x - 1/3*d) + e 
^(-2/3*b*x - 2/3*d) + 1) - log(-sqrt(3)*e^(-1/3*b*x - 1/3*d) + e^(-2/3*b*x 
 - 2/3*d) + 1) + 2*log(e^(-2/3*b*x - 2/3*d) + 1))*e^(5/3*a - 5/3*d)/b - 2/ 
3*(7*e^(-4/3*b*x - 4/3*d) + 4*e^(-10/3*b*x - 10/3*d))*e^(5/3*a - 5/3*d)/(b 
*(2*e^(-2*b*x - 2*d) + e^(-4*b*x - 4*d) + 1))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.64 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^3(d+b x) \, dx=\frac {2 \, {\left (4 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, e^{\left (\frac {2}{3} \, b x\right )} - e^{\left (-\frac {2}{3} \, d\right )}\right )} e^{\left (\frac {2}{3} \, d\right )}\right ) e^{\left (-\frac {14}{3} \, d\right )} - 2 \, e^{\left (-\frac {14}{3} \, d\right )} \log \left (e^{\left (\frac {4}{3} \, b x\right )} - e^{\left (\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + e^{\left (-\frac {4}{3} \, d\right )}\right ) + 4 \, e^{\left (-\frac {14}{3} \, d\right )} \log \left (e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - \frac {3 \, {\left (7 \, e^{\left (\frac {8}{3} \, b x + 2 \, d\right )} + 4 \, e^{\left (\frac {2}{3} \, b x\right )}\right )} e^{\left (-4 \, d\right )}}{{\left (e^{\left (2 \, b x + 2 \, d\right )} + 1\right )}^{2}}\right )} e^{\left (\frac {5}{3} \, a + 3 \, d\right )}}{9 \, b} \] Input:

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

Output:

2/9*(4*sqrt(3)*arctan(1/3*sqrt(3)*(2*e^(2/3*b*x) - e^(-2/3*d))*e^(2/3*d))* 
e^(-14/3*d) - 2*e^(-14/3*d)*log(e^(4/3*b*x) - e^(2/3*b*x - 2/3*d) + e^(-4/ 
3*d)) + 4*e^(-14/3*d)*log(e^(2/3*b*x) + e^(-2/3*d)) - 3*(7*e^(8/3*b*x + 2* 
d) + 4*e^(2/3*b*x))*e^(-4*d)/(e^(2*b*x + 2*d) + 1)^2)*e^(5/3*a + 3*d)/b
 

Mupad [B] (verification not implemented)

Time = 4.93 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.23 \[ \int e^{\frac {5}{3} (a+b x)} \text {sech}^3(d+b x) \, dx=\frac {8\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-\frac {16\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{9}-\frac {16\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{9}\right )}{9\,b}-\frac {8\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}}{3\,b\,\left ({\mathrm {e}}^{2\,d+2\,b\,x}+1\right )}-\frac {2\,{\mathrm {e}}^{\frac {5\,a}{3}+d+\frac {8\,b\,x}{3}}}{b\,\left (2\,{\mathrm {e}}^{2\,d+2\,b\,x}+{\mathrm {e}}^{4\,d+4\,b\,x}+1\right )}+\frac {8\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (-\frac {16\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{9}-\frac {16\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{9}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,b}-\frac {8\,{\left ({\mathrm {e}}^{5\,a-5\,d}\right )}^{1/3}\,\ln \left (\frac {16\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,{\left ({\mathrm {e}}^{5\,a}\,{\mathrm {e}}^{-5\,d}\right )}^{1/3}}{9}-\frac {16\,{\mathrm {e}}^{\frac {5\,a}{3}}\,{\mathrm {e}}^{\frac {2\,d}{3}}\,{\mathrm {e}}^{-\frac {5\,d}{3}}\,{\mathrm {e}}^{\frac {2\,b\,x}{3}}}{9}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{9\,b} \] Input:

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

Output:

(8*exp(5*a - 5*d)^(1/3)*log(- (16*(exp(5*a)*exp(-5*d))^(1/3))/9 - (16*exp( 
(5*a)/3)*exp((2*d)/3)*exp(-(5*d)/3)*exp((2*b*x)/3))/9))/(9*b) - (8*exp((5* 
a)/3 - d + (2*b*x)/3))/(3*b*(exp(2*d + 2*b*x) + 1)) - (2*exp((5*a)/3 + d + 
 (8*b*x)/3))/(b*(2*exp(2*d + 2*b*x) + exp(4*d + 4*b*x) + 1)) + (8*exp(5*a 
- 5*d)^(1/3)*log(- (16*((3^(1/2)*1i)/2 - 1/2)*(exp(5*a)*exp(-5*d))^(1/3))/ 
9 - (16*exp((5*a)/3)*exp((2*d)/3)*exp(-(5*d)/3)*exp((2*b*x)/3))/9)*((3^(1/ 
2)*1i)/2 - 1/2))/(9*b) - (8*exp(5*a - 5*d)^(1/3)*log((16*((3^(1/2)*1i)/2 + 
 1/2)*(exp(5*a)*exp(-5*d))^(1/3))/9 - (16*exp((5*a)/3)*exp((2*d)/3)*exp(-( 
5*d)/3)*exp((2*b*x)/3))/9)*((3^(1/2)*1i)/2 + 1/2))/(9*b)
 

Reduce [F]

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

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

Output:

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