\(\int e^{\frac {5}{3} (a+b x)} \tanh (d+b x) \, dx\) [50]

Optimal result

Integrand size = 18, antiderivative size = 181 \[ \int e^{\frac {5}{3} (a+b x)} \tanh (d+b x) \, dx=\frac {3 e^{\frac {5 (a-d)}{3}+\frac {5}{3} (d+b x)}}{5 b}-\frac {2 e^{\frac {5 (a-d)}{3}} \arctan \left (e^{\frac {1}{3} (d+b x)}\right )}{b}+\frac {e^{\frac {5 (a-d)}{3}} \arctan \left (\sqrt {3}-2 e^{\frac {1}{3} (d+b x)}\right )}{b}-\frac {e^{\frac {5 (a-d)}{3}} \arctan \left (\sqrt {3}+2 e^{\frac {1}{3} (d+b x)}\right )}{b}+\frac {\sqrt {3} e^{\frac {5 (a-d)}{3}} \text {arctanh}\left (\frac {\sqrt {3} e^{\frac {1}{3} (d+b x)}}{1+e^{\frac {2}{3} (d+b x)}}\right )}{b} \] Output:

3/5*exp(5/3*b*x+5/3*a)/b-2*exp(5/3*a-5/3*d)*arctan(exp(1/3*b*x+1/3*d))/b-e 
xp(5/3*a-5/3*d)*arctan(-3^(1/2)+2*exp(1/3*b*x+1/3*d))/b-exp(5/3*a-5/3*d)*a 
rctan(3^(1/2)+2*exp(1/3*b*x+1/3*d))/b+3^(1/2)*exp(5/3*a-5/3*d)*arctanh(3^( 
1/2)*exp(1/3*b*x+1/3*d)/(1+exp(2/3*b*x+2/3*d)))/b
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.15 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.48 \[ \int e^{\frac {5}{3} (a+b x)} \tanh (d+b x) \, dx=\frac {e^{5 a/3} \left (9 e^{\frac {5 b x}{3}}+5 \text {RootSum}\left [\cosh (d)-\sinh (d)+\cosh (d) \text {$\#$1}^6+\sinh (d) \text {$\#$1}^6\&,\frac {b x-3 \log \left (e^{\frac {b x}{3}}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ] (\cosh (2 d)-\sinh (2 d))\right )}{15 b} \] Input:

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

Output:

(E^((5*a)/3)*(9*E^((5*b*x)/3) + 5*RootSum[Cosh[d] - Sinh[d] + Cosh[d]*#1^6 
 + Sinh[d]*#1^6 & , (b*x - 3*Log[E^((b*x)/3) - #1])/#1 & ]*(Cosh[2*d] - Si 
nh[2*d])))/(15*b)
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.86, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {2720, 25, 27, 959, 824, 27, 216, 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.

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 824

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

Defintions of rubi rules used

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 216

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

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 824

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator 
[Rt[a/b, n]], s = Denominator[Rt[a/b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k 
- 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 
 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k 
- 1)*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] 
; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))   Int[1/(r^2 + s^2*x^2), x] + 2*(r^(m 
+ 1)/(a*n*s^m))   Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGt 
Q[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]
 

rule 959

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

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 higher order function than in optimal. Order 9 vs. order 3.

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}}}{5 b}+\frac {i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}-i\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{b}-\frac {i \ln \left ({\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}+i\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}}{b}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{4} \textit {\_Z}^{4}-b^{2} \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-b^{3} \textit {\_R}^{3}+b \textit {\_R} +{\mathrm e}^{\frac {b x}{3}+\frac {d}{3}}\right )\right ) {\mathrm e}^{\frac {5 a}{3}-\frac {5 d}{3}}\)	\(124\)

Input:

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

Output:

3/5*exp(5/3*b*x+5/3*a)/b+I/b*ln(exp(1/3*b*x+1/3*d)-I)*exp(5/3*a-5/3*d)-I/b 
*ln(exp(1/3*b*x+1/3*d)+I)*exp(5/3*a-5/3*d)+sum(_R*ln(-b^3*_R^3+b*_R+exp(1/ 
3*b*x+1/3*d)),_R=RootOf(_Z^4*b^4-_Z^2*b^2+1))*exp(5/3*a-5/3*d)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1200 vs. \(2 (140) = 280\).

Time = 0.11 (sec) , antiderivative size = 1200, normalized size of antiderivative = 6.63 \[ \int e^{\frac {5}{3} (a+b x)} \tanh (d+b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/10*(6*cosh(1/3*b*x + 1/3*d)^5*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)^5 - 6*cosh(1/3*b*x + 1/ 
3*d)^5*sinh(-5/3*a + 5/3*d) + 30*(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)^4 + 
 60*(cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) - cosh(1/3*b*x + 1/3*d)^ 
2*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^3 + 5*sqrt(3)*b*sqrt((cosh(- 
5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))/(b^2*cosh(-5/3*a + 5/3*d) + b^2*sin 
h(-5/3*a + 5/3*d)))*log(cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) + (co 
sh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 + sqrt( 
3)*(b*cosh(1/3*b*x + 1/3*d) + b*sinh(1/3*b*x + 1/3*d))*sqrt((cosh(-5/3*a + 
 5/3*d) - sinh(-5/3*a + 5/3*d))/(b^2*cosh(-5/3*a + 5/3*d) + b^2*sinh(-5/3* 
a + 5/3*d))) + 2*(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) - (cosh(1/3*b*x + 1 
/3*d)^2 + 1)*sinh(-5/3*a + 5/3*d) + cosh(-5/3*a + 5/3*d)) - 5*sqrt(3)*b*sq 
rt((cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))/(b^2*cosh(-5/3*a + 5/3*d) 
 + b^2*sinh(-5/3*a + 5/3*d)))*log(cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/ 
3*d) + (cosh(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d) 
^2 - sqrt(3)*(b*cosh(1/3*b*x + 1/3*d) + b*sinh(1/3*b*x + 1/3*d))*sqrt((cos 
h(-5/3*a + 5/3*d) - sinh(-5/3*a + 5/3*d))/(b^2*cosh(-5/3*a + 5/3*d) + b^2* 
sinh(-5/3*a + 5/3*d))) + 2*(cosh(1/3*b*x + 1/3*d)*cosh(-5/3*a + 5/3*d) ...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.77 \[ \int e^{\frac {5}{3} (a+b x)} \tanh (d+b x) \, dx=\frac {{\left (\sqrt {3} \log \left (\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) - \sqrt {3} \log \left (-\sqrt {3} e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (-\frac {2}{3} \, b x - \frac {2}{3} \, d\right )} + 1\right ) + 2 \, \arctan \left (\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) + 2 \, \arctan \left (-\sqrt {3} + 2 \, e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right ) + 4 \, \arctan \left (e^{\left (-\frac {1}{3} \, b x - \frac {1}{3} \, d\right )}\right )\right )} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )}}{2 \, b} + \frac {3 \, e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a\right )}}{5 \, b} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.98 \[ \int e^{\frac {5}{3} (a+b x)} \tanh (d+b x) \, dx=\frac {5 \, \sqrt {3} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (\sqrt {3} e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 5 \, \sqrt {3} e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} \log \left (-\sqrt {3} e^{\left (\frac {1}{3} \, b x - \frac {1}{3} \, d\right )} + e^{\left (\frac {2}{3} \, b x\right )} + e^{\left (-\frac {2}{3} \, d\right )}\right ) - 10 \, \arctan \left ({\left (\sqrt {3} e^{\left (-\frac {1}{3} \, d\right )} + 2 \, e^{\left (\frac {1}{3} \, b x\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} - 10 \, \arctan \left (-{\left (\sqrt {3} e^{\left (-\frac {1}{3} \, d\right )} - 2 \, e^{\left (\frac {1}{3} \, b x\right )}\right )} e^{\left (\frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} - 20 \, \arctan \left (e^{\left (\frac {1}{3} \, b x + \frac {1}{3} \, d\right )}\right ) e^{\left (\frac {5}{3} \, a - \frac {5}{3} \, d\right )} + 6 \, e^{\left (\frac {5}{3} \, b x + \frac {5}{3} \, a\right )}}{10 \, b} \] Input:

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

Output:

1/10*(5*sqrt(3)*e^(5/3*a - 5/3*d)*log(sqrt(3)*e^(1/3*b*x - 1/3*d) + e^(2/3 
*b*x) + e^(-2/3*d)) - 5*sqrt(3)*e^(5/3*a - 5/3*d)*log(-sqrt(3)*e^(1/3*b*x 
- 1/3*d) + e^(2/3*b*x) + e^(-2/3*d)) - 10*arctan((sqrt(3)*e^(-1/3*d) + 2*e 
^(1/3*b*x))*e^(1/3*d))*e^(5/3*a - 5/3*d) - 10*arctan(-(sqrt(3)*e^(-1/3*d) 
- 2*e^(1/3*b*x))*e^(1/3*d))*e^(5/3*a - 5/3*d) - 20*arctan(e^(1/3*b*x + 1/3 
*d))*e^(5/3*a - 5/3*d) + 6*e^(5/3*b*x + 5/3*a))/b
 

Mupad [B] (verification not implemented)

Time = 4.66 (sec) , antiderivative size = 436, normalized size of antiderivative = 2.41 \[ \int e^{\frac {5}{3} (a+b x)} \tanh (d+b x) \, dx =\text {Too large to display} \] Input:

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

Output:

(3*exp((5*a)/3 + (5*b*x)/3))/(5*b) - ((-exp(10*a - 10*d))^(1/6)*log(4*exp( 
(10*a)/3)*exp(-(10*d)/3) - 4*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x) 
/3)*(-exp(10*a)*exp(-10*d))^(1/6)))/b + ((-exp(10*a - 10*d))^(1/6)*log(4*e 
xp((10*a)/3)*exp(-(10*d)/3) + 4*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b 
*x)/3)*(-exp(10*a)*exp(-10*d))^(1/6)))/b - (log(4*exp((10*a)/3)*exp(-(10*d 
)/3) - 4*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i)/2 
- 1/2)*(-exp(10*a)*exp(-10*d))^(1/6))*(-exp(10*a - 10*d))^(1/6)*((3^(1/2)* 
1i)/2 - 1/2))/b + (log(4*exp((10*a)/3)*exp(-(10*d)/3) + 4*exp((5*a)/3)*exp 
(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i)/2 - 1/2)*(-exp(10*a)*exp(-1 
0*d))^(1/6))*(-exp(10*a - 10*d))^(1/6)*((3^(1/2)*1i)/2 - 1/2))/b - (log(4* 
exp((10*a)/3)*exp(-(10*d)/3) - 4*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp(( 
b*x)/3)*((3^(1/2)*1i)/2 + 1/2)*(-exp(10*a)*exp(-10*d))^(1/6))*(-exp(10*a - 
 10*d))^(1/6)*((3^(1/2)*1i)/2 + 1/2))/b + (log(4*exp((10*a)/3)*exp(-(10*d) 
/3) + 4*exp((5*a)/3)*exp(d/3)*exp(-(5*d)/3)*exp((b*x)/3)*((3^(1/2)*1i)/2 + 
 1/2)*(-exp(10*a)*exp(-10*d))^(1/6))*(-exp(10*a - 10*d))^(1/6)*((3^(1/2)*1 
i)/2 + 1/2))/b
 

Reduce [F]

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

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

Output:

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