3.2.49 \(\int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx\) [149]

3.2.49.1 Optimal result
3.2.49.2 Mathematica [C] (verified)
3.2.49.3 Rubi [A] (verified)
3.2.49.4 Maple [C] (warning: unable to verify)
3.2.49.5 Fricas [C] (verification not implemented)
3.2.49.6 Sympy [F]
3.2.49.7 Maxima [A] (verification not implemented)
3.2.49.8 Giac [F]
3.2.49.9 Mupad [B] (verification not implemented)

3.2.49.1 Optimal result

Integrand size = 20, antiderivative size = 180 \[ \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx=\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}-\frac {\arctan \left (1+\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}+\frac {e^{a c+b c x} \arctan (\tanh (c (a+b x)))}{b c}-\frac {\log \left (1+e^{2 c (a+b x)}-\sqrt {2} e^{a c+b c x}\right )}{2 \sqrt {2} b c}+\frac {\log \left (1+e^{2 c (a+b x)}+\sqrt {2} e^{a c+b c x}\right )}{2 \sqrt {2} b c} \]

output
exp(b*c*x+a*c)*arctan(tanh(c*(b*x+a)))/b/c-1/2*arctan(-1+exp(b*c*x+a*c)*2^ 
(1/2))/b/c*2^(1/2)-1/2*arctan(1+exp(b*c*x+a*c)*2^(1/2))/b/c*2^(1/2)-1/4*ln 
(1+exp(2*c*(b*x+a))-exp(b*c*x+a*c)*2^(1/2))/b/c*2^(1/2)+1/4*ln(1+exp(2*c*( 
b*x+a))+exp(b*c*x+a*c)*2^(1/2))/b/c*2^(1/2)
 
3.2.49.2 Mathematica [C] (verified)

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

Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.49 \[ \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx=\frac {2 e^{c (a+b x)} \arctan \left (\frac {-1+e^{2 c (a+b x)}}{1+e^{2 c (a+b x)}}\right )+\text {RootSum}\left [1+\text {$\#$1}^4\&,\frac {a c+b c x-\log \left (e^{c (a+b x)}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ]}{2 b c} \]

input
Integrate[E^(c*(a + b*x))*ArcTan[Tanh[a*c + b*c*x]],x]
 
output
(2*E^(c*(a + b*x))*ArcTan[(-1 + E^(2*c*(a + b*x)))/(1 + E^(2*c*(a + b*x))) 
] + RootSum[1 + #1^4 & , (a*c + b*c*x - Log[E^(c*(a + b*x)) - #1])/#1 & ]) 
/(2*b*c)
 
3.2.49.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.97, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {7281, 5730, 27, 2679, 826, 1476, 1082, 217, 1479, 25, 27, 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.

\(\displaystyle \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx\)

\(\Big \downarrow \) 7281

\(\displaystyle \frac {\int e^{a c+b x c} \arctan (\tanh (a c+b x c))d(a c+b x c)}{b c}\)

\(\Big \downarrow \) 5730

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-\int \frac {2 e^{3 (a c+b x c)}}{1+e^{4 (a c+b x c)}}d(a c+b x c)}{b c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \int \frac {e^{3 (a c+b x c)}}{1+e^{4 (a c+b x c)}}d(a c+b x c)}{b c}\)

\(\Big \downarrow \) 2679

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \int \frac {e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}}{b c}\)

\(\Big \downarrow \) 826

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

\(\Big \downarrow \) 1476

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}+\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}\right )-\frac {1}{2} \int \frac {1-e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}\right )}{b c}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 a c+2 b x c}}d\left (1-\sqrt {2} e^{a c+b x c}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 a c+2 b x c}}d\left (1+\sqrt {2} e^{a c+b x c}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}\right )}{b c}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}\right )}{b c}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 e^{a c+b x c}}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^{a c+b x c}\right )}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )\right )}{b c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^{a c+b x c}}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^{a c+b x c}\right )}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )\right )}{b c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^{a c+b c x} \arctan (\tanh (a c+b c x))-2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^{a c+b x c}}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {1+\sqrt {2} e^{a c+b x c}}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )\right )}{b c}\)

\(\Big \downarrow \) 1103

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

input
Int[E^(c*(a + b*x))*ArcTan[Tanh[a*c + b*c*x]],x]
 
output
(E^(a*c + b*c*x)*ArcTan[Tanh[a*c + b*c*x]] - 2*((-(ArcTan[1 - Sqrt[2]*E^(a 
*c + b*c*x)]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*E^(a*c + b*c*x)]/Sqrt[2])/2 + ( 
Log[1 - Sqrt[2]*E^(a*c + b*c*x) + E^(2*a*c + 2*b*c*x)]/(2*Sqrt[2]) - Log[1 
 + Sqrt[2]*E^(a*c + b*c*x) + E^(2*a*c + 2*b*c*x)]/(2*Sqrt[2]))/2))/(b*c)
 

3.2.49.3.1 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 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 826
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{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 1476
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 2679
Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_ 
.) + (g_.)*(x_))), x_Symbol] :> With[{m = FullSimplify[d*e*(Log[F]/(g*h*Log 
[G]))]}, Simp[Denominator[m]/(g*h*Log[G])   Subst[Int[x^(Denominator[m] - 1 
)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Deno 
minator[m]))], x] /; LtQ[m, -1] || GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e 
, f, g, h, p}, x]
 

rule 5730
Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, 
 Simp[(a + b*ArcTan[u])   w, x] - Simp[b   Int[SimplifyIntegrand[w*(D[u, x] 
/(1 + u^2)), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b}, x] 
&& InverseFunctionFreeQ[u, x] &&  !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; Fre 
eQ[{c, d, m}, x]] && FalseQ[FunctionOfLinear[v*(a + b*ArcTan[u]), x]]
 

rule 7281
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] 
 Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /;  !FalseQ[lst]]
 
3.2.49.4 Maple [C] (warning: unable to verify)

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

Time = 1.38 (sec) , antiderivative size = 1355, normalized size of antiderivative = 7.53

method result size
risch \(\text {Expression too large to display}\) \(1355\)

input
int(exp(c*(b*x+a))*arctan(tanh(b*c*x+a*c)),x,method=_RETURNVERBOSE)
 
output
1/4/b/c*Pi*csgn(I*(exp(2*c*(b*x+a))-I))*csgn(I*(exp(2*c*(b*x+a))-I)/(1+exp 
(2*c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/b/c*Pi*csgn(I/(1+exp(2*c*(b*x+a))))*c 
sgn(I*(exp(2*c*(b*x+a))-I)/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/b/c* 
Pi*csgn(I*(exp(2*c*(b*x+a))-I)/(1+exp(2*c*(b*x+a))))*csgn((1-I)*(exp(2*c*( 
b*x+a))-I)/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I/(1+exp 
(2*c*(b*x+a))))*csgn(I*(exp(2*c*(b*x+a))+I)/(1+exp(2*c*(b*x+a))))^2*exp(c* 
(b*x+a))-1/4/b/c*Pi*csgn(I*(exp(2*c*(b*x+a))+I))*csgn(I*(exp(2*c*(b*x+a))+ 
I)/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*(exp(2*c*(b*x+ 
a))+I)/(1+exp(2*c*(b*x+a))))*csgn((1+I)*(exp(2*c*(b*x+a))+I)/(1+exp(2*c*(b 
*x+a))))^2*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*(exp(2*c*(b*x+a))-I)/(1+exp(2* 
c*(b*x+a))))*csgn((1-I)*(exp(2*c*(b*x+a))-I)/(1+exp(2*c*(b*x+a))))*exp(c*( 
b*x+a))+1/4/b/c*Pi*csgn(I*(exp(2*c*(b*x+a))+I)/(1+exp(2*c*(b*x+a))))*csgn( 
(1+I)*(exp(2*c*(b*x+a))+I)/(1+exp(2*c*(b*x+a))))*exp(c*(b*x+a))+1/4/b/c*Pi 
*csgn(I*(exp(2*c*(b*x+a))+I)/(1+exp(2*c*(b*x+a))))^3*exp(c*(b*x+a))+1/4/b/ 
c*Pi*csgn((1+I)*(exp(2*c*(b*x+a))+I)/(1+exp(2*c*(b*x+a))))^3*exp(c*(b*x+a) 
)-1/4/b/c*Pi*csgn((1-I)*(exp(2*c*(b*x+a))-I)/(1+exp(2*c*(b*x+a))))^2*exp(c 
*(b*x+a))-1/4/b/c*Pi*csgn((1+I)*(exp(2*c*(b*x+a))+I)/(1+exp(2*c*(b*x+a)))) 
^2*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*(exp(2*c*(b*x+a))-I)/(1+exp(2*c*(b*x+a 
))))^3*exp(c*(b*x+a))+1/4/b/c*Pi*csgn((1-I)*(exp(2*c*(b*x+a))-I)/(1+exp(2* 
c*(b*x+a))))^3*exp(c*(b*x+a))-1/2*I/b/c*exp(c*(b*x+a))*ln(exp(2*c*(b*x+...
 
3.2.49.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.43 \[ \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx=-\frac {b c \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (b^{3} c^{3} \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} + \cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) - i \, b c \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (i \, b^{3} c^{3} \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} + \cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) + i \, b c \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (-i \, b^{3} c^{3} \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} + \cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) - b c \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (-b^{3} c^{3} \left (-\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} + \cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) - 2 \, {\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \arctan \left (\frac {\sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right )}\right )}{2 \, b c} \]

input
integrate(exp(c*(b*x+a))*arctan(tanh(b*c*x+a*c)),x, algorithm="fricas")
 
output
-1/2*(b*c*(-1/(b^4*c^4))^(1/4)*log(b^3*c^3*(-1/(b^4*c^4))^(3/4) + cosh(b*c 
*x + a*c) + sinh(b*c*x + a*c)) - I*b*c*(-1/(b^4*c^4))^(1/4)*log(I*b^3*c^3* 
(-1/(b^4*c^4))^(3/4) + cosh(b*c*x + a*c) + sinh(b*c*x + a*c)) + I*b*c*(-1/ 
(b^4*c^4))^(1/4)*log(-I*b^3*c^3*(-1/(b^4*c^4))^(3/4) + cosh(b*c*x + a*c) + 
 sinh(b*c*x + a*c)) - b*c*(-1/(b^4*c^4))^(1/4)*log(-b^3*c^3*(-1/(b^4*c^4)) 
^(3/4) + cosh(b*c*x + a*c) + sinh(b*c*x + a*c)) - 2*(cosh(b*c*x + a*c) + s 
inh(b*c*x + a*c))*arctan(sinh(b*c*x + a*c)/cosh(b*c*x + a*c)))/(b*c)
 
3.2.49.6 Sympy [F]

\[ \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx=e^{a c} \int e^{b c x} \operatorname {atan}{\left (\tanh {\left (a c + b c x \right )} \right )}\, dx \]

input
integrate(exp(c*(b*x+a))*atan(tanh(b*c*x+a*c)),x)
 
output
exp(a*c)*Integral(exp(b*c*x)*atan(tanh(a*c + b*c*x)), x)
 
3.2.49.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.93 \[ \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx=\frac {\arctan \left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} - \frac {\sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} + \frac {\sqrt {2} \log \left (\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} - \frac {\sqrt {2} \log \left (-\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} \]

input
integrate(exp(c*(b*x+a))*arctan(tanh(b*c*x+a*c)),x, algorithm="maxima")
 
output
arctan(tanh(b*c*x + a*c))*e^((b*x + a)*c)/(b*c) - 1/2*sqrt(2)*arctan(1/2*s 
qrt(2)*(sqrt(2) + 2*e^(b*c*x + a*c)))/(b*c) - 1/2*sqrt(2)*arctan(-1/2*sqrt 
(2)*(sqrt(2) - 2*e^(b*c*x + a*c)))/(b*c) + 1/4*sqrt(2)*log(sqrt(2)*e^(b*c* 
x + a*c) + e^(2*b*c*x + 2*a*c) + 1)/(b*c) - 1/4*sqrt(2)*log(-sqrt(2)*e^(b* 
c*x + a*c) + e^(2*b*c*x + 2*a*c) + 1)/(b*c)
 
3.2.49.8 Giac [F]

\[ \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx=\int { \arctan \left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )} \,d x } \]

input
integrate(exp(c*(b*x+a))*arctan(tanh(b*c*x+a*c)),x, algorithm="giac")
 
output
sage0*x
 
3.2.49.9 Mupad [B] (verification not implemented)

Time = 1.59 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.91 \[ \int e^{c (a+b x)} \arctan (\tanh (a c+b c x)) \, dx=\frac {4\,{\mathrm {e}}^{a\,c+b\,c\,x}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}-1}{{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}+1}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4-4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4+4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1+1{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4-4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4+4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1+1{}\mathrm {i}\right )}{4\,b\,c} \]

input
int(exp(c*(a + b*x))*atan(tanh(a*c + b*c*x)),x)
 
output
(2^(1/2)*log(2^(1/2)*(4 - 4i) - exp(b*c*x)*exp(a*c)*8i)*(1 - 1i) - 2^(1/2) 
*log(- 2^(1/2)*(4 - 4i) - exp(b*c*x)*exp(a*c)*8i)*(1 - 1i) - 2^(1/2)*log(e 
xp(b*c*x)*exp(a*c)*8i - 2^(1/2)*(4 + 4i))*(1 + 1i) + 2^(1/2)*log(2^(1/2)*( 
4 + 4i) + exp(b*c*x)*exp(a*c)*8i)*(1 + 1i) + 4*exp(a*c + b*c*x)*atan((exp( 
2*b*c*x)*exp(2*a*c) - 1)/(exp(2*b*c*x)*exp(2*a*c) + 1)))/(4*b*c)