Integrand size = 20, antiderivative size = 135 \[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}+\frac {\arctan \left (1+\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}+\frac {e^{a c+b c x} \arctan (\coth (c (a+b x)))}{b c}-\frac {\text {arctanh}\left (\frac {\sqrt {2} e^{a c+b c x}}{1+e^{2 c (a+b x)}}\right )}{\sqrt {2} b c} \] Output:
1/2*arctan(-1+2^(1/2)*exp(b*c*x+a*c))*2^(1/2)/b/c+1/2*arctan(1+2^(1/2)*exp (b*c*x+a*c))*2^(1/2)/b/c+exp(b*c*x+a*c)*arctan(coth(c*(b*x+a)))/b/c-1/2*ar ctanh(2^(1/2)*exp(b*c*x+a*c)/(1+exp(2*c*(b*x+a))))*2^(1/2)/b/c
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.66 \[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=\frac {2 e^{c (a+b x)} \arctan \left (\frac {1+e^{2 c (a+b x)}}{-1+e^{2 c (a+b x)}}\right )+\text {RootSum}\left [1+\text {$\#$1}^4\&,\frac {-a c-b c x+\log \left (e^{c (a+b x)}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ]}{2 b c} \] Input:
Integrate[E^(c*(a + b*x))*ArcTan[Coth[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)
Time = 0.52 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.30, 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.
\(\displaystyle \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {\int e^{a c+b x c} \arctan (\coth (a c+b x c))d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 5730 |
\(\displaystyle \frac {e^{a c+b c x} \arctan (\coth (a c+b c x))-\int -\frac {2 e^{3 (a c+b x c)}}{1+e^{4 (a c+b x c)}}d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {e^{3 (a c+b x c)}}{1+e^{4 (a c+b x c)}}d(a c+b x c)+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 2679 |
\(\displaystyle \frac {2 \int \frac {e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \int \frac {1+e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}-\frac {1}{2} \int \frac {1-e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}+\frac {1}{2} \int \frac {1}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}\right )-\frac {1}{2} \int \frac {1-e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-1-e^{2 a c+2 b x c}}d\left (1-\sqrt {2} e^{a c+b x c}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-1-e^{2 a c+2 b x c}}d\left (1+\sqrt {2} e^{a c+b x c}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )-\frac {1}{2} \int \frac {1-e^{2 a c+2 b x c}}{1+e^{4 a c+4 b x c}}de^{a c+b x c}\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2}-2 e^{a c+b x c}}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}+\frac {\int -\frac {\sqrt {2} \left (1+\sqrt {2} e^{a c+b x c}\right )}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^{a c+b x c}}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2} \left (1+\sqrt {2} e^{a c+b x c}\right )}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2}-2 e^{a c+b x c}}{1-\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}}{2 \sqrt {2}}-\frac {1}{2} \int \frac {1+\sqrt {2} e^{a c+b x c}}{1+\sqrt {2} e^{a c+b x c}+e^{2 a c+2 b x c}}de^{a c+b x c}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} e^{a c+b c x}+e^{2 a c+2 b c x}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\sqrt {2} e^{a c+b c x}+e^{2 a c+2 b c x}+1\right )}{2 \sqrt {2}}\right )\right )+e^{a c+b c x} \arctan (\coth (a c+b c x))}{b c}\) |
Input:
Int[E^(c*(a + b*x))*ArcTan[Coth[a*c + b*c*x]],x]
Output:
(E^(a*c + b*c*x)*ArcTan[Coth[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)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]]))
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]
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]
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]
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]
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]
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]]
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]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.84 (sec) , antiderivative size = 1355, normalized size of antiderivative = 10.04
Input:
int(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x,method=_RETURNVERBOSE)
Output:
-1/2*I/b/c*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))+I)+1/4/b/c*exp(c*(b*x+a))*Pi +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))))^3*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))))^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*I/b/c*ln(exp(c*(b*x+a)) +1/2*I*2^(1/2)-1/2*2^(1/2))*2^(1/2)-1/4*I/b/c*ln(exp(c*(b*x+a))+1/2*2^(1/2 )-1/2*I*2^(1/2))*2^(1/2)+1/2*I/b/c*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))-I)+1 /4*I/b/c*ln(exp(c*(b*x+a))+1/2*2^(1/2)+1/2*I*2^(1/2))*2^(1/2)-1/4*I/b/c*ln (exp(c*(b*x+a))-1/2*I*2^(1/2)-1/2*2^(1/2))*2^(1/2)+1/4/b/c*Pi*csgn(I*(exp( 2*c*(b*x+a))-I))*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))))*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*(exp(2*c*(b*x+a)) +I))*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))))*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*cs gn(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)/(-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(...
Time = 0.13 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.56 \[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=\frac {4 \, {\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \arctan \left (\frac {\cosh \left (b c x + a c\right )}{\sinh \left (b c x + a c\right )}\right ) + 2 \, \sqrt {2} \arctan \left (\sqrt {2} \cosh \left (b c x + a c\right ) + \sqrt {2} \sinh \left (b c x + a c\right ) + 1\right ) + 2 \, \sqrt {2} \arctan \left (\sqrt {2} \cosh \left (b c x + a c\right ) + \sqrt {2} \sinh \left (b c x + a c\right ) - 1\right ) - \sqrt {2} \log \left (\frac {\sqrt {2} + 2 \, \cosh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right ) + \sqrt {2} \log \left (-\frac {\sqrt {2} - 2 \, \cosh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{4 \, b c} \] Input:
integrate(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x, algorithm="fricas")
Output:
1/4*(4*(cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*arctan(cosh(b*c*x + a*c)/si nh(b*c*x + a*c)) + 2*sqrt(2)*arctan(sqrt(2)*cosh(b*c*x + a*c) + sqrt(2)*si nh(b*c*x + a*c) + 1) + 2*sqrt(2)*arctan(sqrt(2)*cosh(b*c*x + a*c) + sqrt(2 )*sinh(b*c*x + a*c) - 1) - sqrt(2)*log((sqrt(2) + 2*cosh(b*c*x + a*c))/(co sh(b*c*x + a*c) - sinh(b*c*x + a*c))) + sqrt(2)*log(-(sqrt(2) - 2*cosh(b*c *x + a*c))/(cosh(b*c*x + a*c) - sinh(b*c*x + a*c))))/(b*c)
\[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=e^{a c} \int e^{b c x} \operatorname {atan}{\left (\coth {\left (a c + b c x \right )} \right )}\, dx \] Input:
integrate(exp(c*(b*x+a))*atan(coth(b*c*x+a*c)),x)
Output:
exp(a*c)*Integral(exp(b*c*x)*atan(coth(a*c + b*c*x)), x)
Time = 0.18 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.24 \[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=\frac {\arctan \left (\coth \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac {\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} + \frac {\sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} - \frac {\sqrt {2} \log \left (\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} + \frac {\sqrt {2} \log \left (-\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} \] Input:
integrate(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x, algorithm="maxima")
Output:
arctan(coth(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)
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (118) = 236\).
Time = 0.13 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.77 \[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=\frac {4 \, \pi e^{\left (b c x + a c\right )} \left \lfloor \frac {3 \, \pi + 4 \, \arctan \left (e^{\left (2 \, b c x + 2 \, a c\right )}\right )}{4 \, \pi } \right \rfloor - \pi e^{\left (b c x + a c\right )} - {\left (4 \, \arctan \left (e^{\left (2 \, b c x + 2 \, a c\right )}\right ) e^{\left (b c x\right )} - {\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-a c\right )} + 2 \, e^{\left (b c x\right )}\right )} e^{\left (a c\right )}\right ) e^{\left (-3 \, a c\right )} + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} e^{\left (-a c\right )} - 2 \, e^{\left (b c x\right )}\right )} e^{\left (a c\right )}\right ) e^{\left (-3 \, a c\right )} - \sqrt {2} e^{\left (-3 \, a c\right )} \log \left (\sqrt {2} e^{\left (b c x - a c\right )} + e^{\left (2 \, b c x\right )} + e^{\left (-2 \, a c\right )}\right ) + \sqrt {2} e^{\left (-3 \, a c\right )} \log \left (-\sqrt {2} e^{\left (b c x - a c\right )} + e^{\left (2 \, b c x\right )} + e^{\left (-2 \, a c\right )}\right )\right )} e^{\left (2 \, a c\right )}\right )} e^{\left (a c\right )}}{4 \, b c} \] Input:
integrate(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x, algorithm="giac")
Output:
1/4*(4*pi*e^(b*c*x + a*c)*floor(1/4*(3*pi + 4*arctan(e^(2*b*c*x + 2*a*c))) /pi) - pi*e^(b*c*x + a*c) - (4*arctan(e^(2*b*c*x + 2*a*c))*e^(b*c*x) - (2* sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-a*c) + 2*e^(b*c*x))*e^(a*c))*e^(-3 *a*c) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*e^(-a*c) - 2*e^(b*c*x))*e^( a*c))*e^(-3*a*c) - sqrt(2)*e^(-3*a*c)*log(sqrt(2)*e^(b*c*x - a*c) + e^(2*b *c*x) + e^(-2*a*c)) + sqrt(2)*e^(-3*a*c)*log(-sqrt(2)*e^(b*c*x - a*c) + e^ (2*b*c*x) + e^(-2*a*c)))*e^(2*a*c))*e^(a*c))/(b*c)
Time = 1.13 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.21 \[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=\frac {4\,{\mathrm {e}}^{a\,c+b\,c\,x}\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}+1}{{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}-1}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4-4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4+4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1+1{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4-4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4+4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1+1{}\mathrm {i}\right )}{4\,b\,c} \] Input:
int(exp(c*(a + b*x))*atan(coth(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(exp(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) + 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)
Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.25 \[ \int e^{c (a+b x)} \arctan (\coth (a c+b c x)) \, dx=\frac {4 e^{b c x +a c} \mathit {atan} \left (\frac {e^{2 b c x +2 a c}+1}{e^{2 b c x +2 a c}-1}\right )+2 \sqrt {2}\, \mathit {atan} \left (\frac {2 e^{b c x +a c}-\sqrt {2}}{\sqrt {2}}\right )+2 \sqrt {2}\, \mathit {atan} \left (\frac {2 e^{b c x +a c}+\sqrt {2}}{\sqrt {2}}\right )+\sqrt {2}\, \mathrm {log}\left (e^{2 b c x +2 a c}-e^{b c x +a c} \sqrt {2}+1\right )-\sqrt {2}\, \mathrm {log}\left (e^{2 b c x +2 a c}+e^{b c x +a c} \sqrt {2}+1\right )}{4 b c} \] Input:
int(exp(c*(b*x+a))*atan(coth(b*c*x+a*c)),x)
Output:
(4*e**(a*c + b*c*x)*atan((e**(2*a*c + 2*b*c*x) + 1)/(e**(2*a*c + 2*b*c*x) - 1)) + 2*sqrt(2)*atan((2*e**(a*c + b*c*x) - sqrt(2))/sqrt(2)) + 2*sqrt(2) *atan((2*e**(a*c + b*c*x) + sqrt(2))/sqrt(2)) + sqrt(2)*log(e**(2*a*c + 2* b*c*x) - e**(a*c + b*c*x)*sqrt(2) + 1) - sqrt(2)*log(e**(2*a*c + 2*b*c*x) + e**(a*c + b*c*x)*sqrt(2) + 1))/(4*b*c)