3.150 \(\int e^{c (a+b x)} \tan ^{-1}(\coth (a c+b c x)) \, dx\)

Optimal. Leaf size=180 \[ \frac{\log \left (e^{2 c (a+b x)}-\sqrt{2} e^{a c+b c x}+1\right )}{2 \sqrt{2} b c}-\frac{\log \left (e^{2 c (a+b x)}+\sqrt{2} e^{a c+b c x}+1\right )}{2 \sqrt{2} b c}-\frac{\tan ^{-1}\left (1-\sqrt{2} e^{a c+b c x}\right )}{\sqrt{2} b c}+\frac{\tan ^{-1}\left (\sqrt{2} e^{a c+b c x}+1\right )}{\sqrt{2} b c}+\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c} \]

[Out]

-(ArcTan[1 - Sqrt[2]*E^(a*c + b*c*x)]/(Sqrt[2]*b*c)) + ArcTan[1 + Sqrt[2]*E^(a*c + b*c*x)]/(Sqrt[2]*b*c) + (E^
(a*c + b*c*x)*ArcTan[Coth[c*(a + b*x)]])/(b*c) + Log[1 + E^(2*c*(a + b*x)) - Sqrt[2]*E^(a*c + b*c*x)]/(2*Sqrt[
2]*b*c) - Log[1 + E^(2*c*(a + b*x)) + Sqrt[2]*E^(a*c + b*c*x)]/(2*Sqrt[2]*b*c)

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Rubi [A]  time = 0.179136, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {2194, 5207, 12, 2249, 297, 1162, 617, 204, 1165, 628} \[ \frac{\log \left (e^{2 c (a+b x)}-\sqrt{2} e^{a c+b c x}+1\right )}{2 \sqrt{2} b c}-\frac{\log \left (e^{2 c (a+b x)}+\sqrt{2} e^{a c+b c x}+1\right )}{2 \sqrt{2} b c}-\frac{\tan ^{-1}\left (1-\sqrt{2} e^{a c+b c x}\right )}{\sqrt{2} b c}+\frac{\tan ^{-1}\left (\sqrt{2} e^{a c+b c x}+1\right )}{\sqrt{2} b c}+\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*ArcTan[Coth[a*c + b*c*x]],x]

[Out]

-(ArcTan[1 - Sqrt[2]*E^(a*c + b*c*x)]/(Sqrt[2]*b*c)) + ArcTan[1 + Sqrt[2]*E^(a*c + b*c*x)]/(Sqrt[2]*b*c) + (E^
(a*c + b*c*x)*ArcTan[Coth[c*(a + b*x)]])/(b*c) + Log[1 + E^(2*c*(a + b*x)) - Sqrt[2]*E^(a*c + b*c*x)]/(2*Sqrt[
2]*b*c) - Log[1 + E^(2*c*(a + b*x)) + Sqrt[2]*E^(a*c + b*c*x)]/(2*Sqrt[2]*b*c)

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 5207

Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTan[u], w, x] - Dist
[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_.) /; FreeQ[{c, d, m}, x]] && FalseQ[Functi
onOfLinear[v*(a + b*ArcTan[u]), x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2249

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

Rule 297

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]

Rubi steps

\begin{align*} \int e^{c (a+b x)} \tan ^{-1}(\coth (a c+b c x)) \, dx &=\frac{\operatorname{Subst}\left (\int e^x \tan ^{-1}(\coth (x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c}-\frac{\operatorname{Subst}\left (\int \frac{2 e^{3 x}}{-1-e^{4 x}} \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{3 x}}{-1-e^{4 x}} \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c}-\frac{2 \operatorname{Subst}\left (\int \frac{x^2}{-1-x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int \frac{1-x^2}{-1-x^4} \, dx,x,e^{a c+b c x}\right )}{b c}-\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{-1-x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,e^{a c+b c x}\right )}{2 b c}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,e^{a c+b c x}\right )}{2 b c}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,e^{a c+b c x}\right )}{2 \sqrt{2} b c}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,e^{a c+b c x}\right )}{2 \sqrt{2} b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c}+\frac{\log \left (1-\sqrt{2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt{2} b c}-\frac{\log \left (1+\sqrt{2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt{2} b c}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} e^{a c+b c x}\right )}{\sqrt{2} b c}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} e^{a c+b c x}\right )}{\sqrt{2} b c}\\ &=-\frac{\tan ^{-1}\left (1-\sqrt{2} e^{a c+b c x}\right )}{\sqrt{2} b c}+\frac{\tan ^{-1}\left (1+\sqrt{2} e^{a c+b c x}\right )}{\sqrt{2} b c}+\frac{e^{a c+b c x} \tan ^{-1}(\coth (c (a+b x)))}{b c}+\frac{\log \left (1-\sqrt{2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt{2} b c}-\frac{\log \left (1+\sqrt{2} e^{a c+b c x}+e^{2 a c+2 b c x}\right )}{2 \sqrt{2} b c}\\ \end{align*}

Mathematica [C]  time = 0.111277, size = 89, normalized size = 0.49 \[ \frac{\text{RootSum}\left [\text{$\#$1}^4+1\& ,\frac{\log \left (e^{c (a+b x)}-\text{$\#$1}\right )-a c-b c x}{\text{$\#$1}}\& \right ]+2 e^{c (a+b x)} \tan ^{-1}\left (\frac{e^{2 c (a+b x)}+1}{e^{2 c (a+b x)}-1}\right )}{2 b c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(c*(a + b*x))*ArcTan[Coth[a*c + b*c*x]],x]

[Out]

(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)

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Maple [C]  time = 0.591, size = 1355, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x)

[Out]

-1/4*I/c/b*ln(exp(c*(b*x+a))+1/2*2^(1/2)-1/2*I*2^(1/2))*2^(1/2)-1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))+I)/(exp(2*
c*(b*x+a))-1))^3*exp(c*(b*x+a))+1/4/c/b*Pi*csgn((1-I)*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+a))-1))^3*exp(c*(b*x+
a))+1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))-I)/(exp(2*c*(b*x+a))-1))^3*exp(c*(b*x+a))+1/4/c/b*Pi*csgn((1+I)*(exp(2
*c*(b*x+a))-I)/(exp(2*c*(b*x+a))-1))^3*exp(c*(b*x+a))+1/4/c/b*ln(exp(c*(b*x+a))-1/2*2^(1/2)+1/2*I*2^(1/2))*2^(
1/2)+1/4/c/b*ln(exp(c*(b*x+a))-1/2*2^(1/2)-1/2*I*2^(1/2))*2^(1/2)-1/4/c/b*ln(exp(c*(b*x+a))+1/2*2^(1/2)+1/2*I*
2^(1/2))*2^(1/2)-1/4/c/b*ln(exp(c*(b*x+a))+1/2*2^(1/2)-1/2*I*2^(1/2))*2^(1/2)+1/4/c/b*exp(c*(b*x+a))*Pi+1/4*I/
c/b*ln(exp(c*(b*x+a))-1/2*2^(1/2)+1/2*I*2^(1/2))*2^(1/2)+1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+
a))-1))*csgn((1-I)*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+a))-1))^2*exp(c*(b*x+a))-1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x
+a))-I)/(exp(2*c*(b*x+a))-1))*csgn((1+I)*(exp(2*c*(b*x+a))-I)/(exp(2*c*(b*x+a))-1))^2*exp(c*(b*x+a))-1/4/c/b*P
i*csgn(I*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+a))-1))*csgn((1-I)*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+a))-1))*exp(
c*(b*x+a))+1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))-I)/(exp(2*c*(b*x+a))-1))*csgn((1+I)*(exp(2*c*(b*x+a))-I)/(exp(2
*c*(b*x+a))-1))*exp(c*(b*x+a))-1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))-I))*csgn(I*(exp(2*c*(b*x+a))-I)/(exp(2*c*(b
*x+a))-1))^2*exp(c*(b*x+a))+1/4/c/b*Pi*csgn(I/(exp(2*c*(b*x+a))-1))*csgn(I*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+
a))-1))^2*exp(c*(b*x+a))-1/4/c/b*Pi*csgn(I/(exp(2*c*(b*x+a))-1))*csgn(I*(exp(2*c*(b*x+a))-I)/(exp(2*c*(b*x+a))
-1))^2*exp(c*(b*x+a))-1/2*I/c/b*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))+I)+1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))+I))*
csgn(I*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+a))-1))^2*exp(c*(b*x+a))-1/4/c/b*Pi*csgn((1-I)*(exp(2*c*(b*x+a))+I)/
(exp(2*c*(b*x+a))-1))^2*exp(c*(b*x+a))-1/4/c/b*Pi*csgn((1+I)*(exp(2*c*(b*x+a))-I)/(exp(2*c*(b*x+a))-1))^2*exp(
c*(b*x+a))+1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))-I))*csgn(I/(exp(2*c*(b*x+a))-1))*csgn(I*(exp(2*c*(b*x+a))-I)/(e
xp(2*c*(b*x+a))-1))*exp(c*(b*x+a))-1/4/c/b*Pi*csgn(I*(exp(2*c*(b*x+a))+I))*csgn(I/(exp(2*c*(b*x+a))-1))*csgn(I
*(exp(2*c*(b*x+a))+I)/(exp(2*c*(b*x+a))-1))*exp(c*(b*x+a))+1/2*I/c/b*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))-I)-1/4
*I/c/b*ln(exp(c*(b*x+a))-1/2*2^(1/2)-1/2*I*2^(1/2))*2^(1/2)+1/4*I/c/b*ln(exp(c*(b*x+a))+1/2*2^(1/2)+1/2*I*2^(1
/2))*2^(1/2)

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Maxima [A]  time = 1.54574, size = 225, normalized size = 1.25 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x, algorithm="maxima")

[Out]

arctan(coth(b*c*x + a*c))*e^((b*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) + 1/2*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)

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Fricas [B]  time = 2.2317, size = 1153, normalized size = 6.41 \begin{align*} -\frac{4 \, \sqrt{2} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} \arctan \left (-\sqrt{2} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} e^{\left (b c x + a c\right )} + \sqrt{2} \sqrt{\sqrt{2} b^{3} c^{3} \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt{\frac{1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} - 1\right ) + 4 \, \sqrt{2} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} \arctan \left (-\sqrt{2} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} e^{\left (b c x + a c\right )} + \sqrt{2} \sqrt{-\sqrt{2} b^{3} c^{3} \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt{\frac{1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} + 1\right ) + \sqrt{2} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} \log \left (\sqrt{2} b^{3} c^{3} \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt{\frac{1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - \sqrt{2} b c \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{1}{4}} \log \left (-\sqrt{2} b^{3} c^{3} \left (\frac{1}{b^{4} c^{4}}\right )^{\frac{3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt{\frac{1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - 4 \, \arctan \left (\frac{e^{\left (2 \, b c x + 2 \, a c\right )} + 1}{e^{\left (2 \, b c x + 2 \, a c\right )} - 1}\right ) e^{\left (b c x + a c\right )}}{4 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x, algorithm="fricas")

[Out]

-1/4*(4*sqrt(2)*b*c*(1/(b^4*c^4))^(1/4)*arctan(-sqrt(2)*b*c*(1/(b^4*c^4))^(1/4)*e^(b*c*x + a*c) + sqrt(2)*sqrt
(sqrt(2)*b^3*c^3*(1/(b^4*c^4))^(3/4)*e^(b*c*x + a*c) + b^2*c^2*sqrt(1/(b^4*c^4)) + e^(2*b*c*x + 2*a*c))*b*c*(1
/(b^4*c^4))^(1/4) - 1) + 4*sqrt(2)*b*c*(1/(b^4*c^4))^(1/4)*arctan(-sqrt(2)*b*c*(1/(b^4*c^4))^(1/4)*e^(b*c*x +
a*c) + sqrt(2)*sqrt(-sqrt(2)*b^3*c^3*(1/(b^4*c^4))^(3/4)*e^(b*c*x + a*c) + b^2*c^2*sqrt(1/(b^4*c^4)) + e^(2*b*
c*x + 2*a*c))*b*c*(1/(b^4*c^4))^(1/4) + 1) + sqrt(2)*b*c*(1/(b^4*c^4))^(1/4)*log(sqrt(2)*b^3*c^3*(1/(b^4*c^4))
^(3/4)*e^(b*c*x + a*c) + b^2*c^2*sqrt(1/(b^4*c^4)) + e^(2*b*c*x + 2*a*c)) - sqrt(2)*b*c*(1/(b^4*c^4))^(1/4)*lo
g(-sqrt(2)*b^3*c^3*(1/(b^4*c^4))^(3/4)*e^(b*c*x + a*c) + b^2*c^2*sqrt(1/(b^4*c^4)) + e^(2*b*c*x + 2*a*c)) - 4*
arctan((e^(2*b*c*x + 2*a*c) + 1)/(e^(2*b*c*x + 2*a*c) - 1))*e^(b*c*x + a*c))/(b*c)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*atan(coth(b*c*x+a*c)),x)

[Out]

Timed out

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Giac [A]  time = 1.36402, size = 344, normalized size = 1.91 \begin{align*} \frac{1}{4} \,{\left (\frac{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 (-11 \, a c\right )}}{b c} + \frac{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 (-11 \, a c\right )}}{b c} - \frac{\sqrt{2} e^{\left (-11 \, 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 )}{b c} + \frac{\sqrt{2} e^{\left (-11 \, 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 )}{b c}\right )} e^{\left (11 \, a c\right )} + \frac{4 \, \pi e^{\left (b c x + a c\right )} \left \lfloor \frac{5 \, \pi - 4 \, \arctan \left (e^{\left (-2 \, a c\right )}\right )}{4 \, \pi } \right \rfloor - 3 \, \pi e^{\left (b c x + a c\right )} + 4 \, \arctan \left (e^{\left (-2 \, b c x - 2 \, a c\right )}\right ) e^{\left (b c x + a c\right )}}{4 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(coth(b*c*x+a*c)),x, algorithm="giac")

[Out]

1/4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*e^(-a*c) + 2*e^(b*c*x))*e^(a*c))*e^(-11*a*c)/(b*c) + 2*sqrt(2)*arct
an(-1/2*sqrt(2)*(sqrt(2)*e^(-a*c) - 2*e^(b*c*x))*e^(a*c))*e^(-11*a*c)/(b*c) - sqrt(2)*e^(-11*a*c)*log(sqrt(2)*
e^(b*c*x - a*c) + e^(2*b*c*x) + e^(-2*a*c))/(b*c) + sqrt(2)*e^(-11*a*c)*log(-sqrt(2)*e^(b*c*x - a*c) + e^(2*b*
c*x) + e^(-2*a*c))/(b*c))*e^(11*a*c) + 1/4*(4*pi*e^(b*c*x + a*c)*floor(1/4*(5*pi - 4*arctan(e^(-2*a*c)))/pi) -
3*pi*e^(b*c*x + a*c) + 4*arctan(e^(-2*b*c*x - 2*a*c))*e^(b*c*x + a*c))/(b*c)