3.231 \(\int e^{c (a+b x)} \cot ^{-1}(\tanh (a c+b c x)) \, dx\)

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

[Out]

exp(b*c*x+a*c)*arccot(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)

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Rubi [A]  time = 0.18, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2194, 5208, 12, 2249, 297, 1162, 617, 204, 1165, 628} \[ \frac {\log \left (e^{2 c (a+b x)}-\sqrt {2} e^{a c+b c x}+1\right )}{2 \sqrt {2} b c}-\frac {\log \left (e^{2 c (a+b x)}+\sqrt {2} e^{a c+b c x}+1\right )}{2 \sqrt {2} b c}-\frac {\tan ^{-1}\left (1-\sqrt {2} e^{a c+b c x}\right )}{\sqrt {2} b c}+\frac {\tan ^{-1}\left (\sqrt {2} e^{a c+b c x}+1\right )}{\sqrt {2} b c}+\frac {e^{a c+b c x} \cot ^{-1}(\tanh (c (a+b x)))}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*ArcCot[Tanh[a*c + b*c*x]],x]

[Out]

(E^(a*c + b*c*x)*ArcCot[Tanh[c*(a + b*x)]])/(b*c) - 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) + 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 12

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

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 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 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 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]

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 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 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 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 5208

Int[((a_.) + ArcCot[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcCot[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*ArcCot[u]), x]]

Rubi steps

\begin {align*} \int e^{c (a+b x)} \cot ^{-1}(\tanh (a c+b c x)) \, dx &=\frac {\operatorname {Subst}\left (\int e^x \cot ^{-1}(\tanh (x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac {e^{a c+b c x} \cot ^{-1}(\tanh (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} \cot ^{-1}(\tanh (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} \cot ^{-1}(\tanh (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} \cot ^{-1}(\tanh (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} \cot ^{-1}(\tanh (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} \cot ^{-1}(\tanh (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 {e^{a c+b c x} \cot ^{-1}(\tanh (c (a+b x)))}{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 {\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*}

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Mathematica [C]  time = 0.12, size = 89, normalized size = 0.49 \[ \frac {\text {RootSum}\left [\text {$\#$1}^4+1\& ,\frac {\log \left (e^{c (a+b x)}-\text {$\#$1}\right )-a c-b c x}{\text {$\#$1}}\& \right ]+2 e^{c (a+b x)} \cot ^{-1}\left (\frac {e^{2 c (a+b x)}-1}{e^{2 c (a+b x)}+1}\right )}{2 b c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(c*(a + b*x))*ArcCot[Tanh[a*c + b*c*x]],x]

[Out]

(2*E^(c*(a + b*x))*ArcCot[(-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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fricas [B]  time = 0.71, size = 431, normalized size = 2.39 \[ -\frac {4 \, \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} e^{\left (b c x + a c\right )} + \sqrt {2} \sqrt {\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} - 1\right ) + 4 \, \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \arctan \left (-\sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} e^{\left (b c x + a c\right )} + \sqrt {2} \sqrt {-\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} + 1\right ) + \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - \sqrt {2} b c \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} b^{3} c^{3} \left (\frac {1}{b^{4} c^{4}}\right )^{\frac {3}{4}} e^{\left (b c x + a c\right )} + b^{2} c^{2} \sqrt {\frac {1}{b^{4} c^{4}}} + e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - 4 \, \arctan \left (\frac {e^{\left (2 \, b c x + 2 \, a c\right )} + 1}{e^{\left (2 \, b c x + 2 \, a c\right )} - 1}\right ) e^{\left (b c x + a c\right )}}{4 \, b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arccot(tanh(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arccot(tanh(b*c*x+a*c)),x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 1.74, size = 1323, normalized size = 7.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*arccot(tanh(b*c*x+a*c)),x)

[Out]

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

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maxima [A]  time = 0.43, size = 167, normalized size = 0.93 \[ \frac {\operatorname {arccot}\left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac {\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} + \frac {\sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, e^{\left (b c x + a c\right )}\right )}\right )}{2 \, b c} - \frac {\sqrt {2} \log \left (\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} + \frac {\sqrt {2} \log \left (-\sqrt {2} e^{\left (b c x + a c\right )} + e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{4 \, b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arccot(tanh(b*c*x+a*c)),x, algorithm="maxima")

[Out]

arccot(tanh(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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mupad [B]  time = 2.56, size = 164, normalized size = 0.91 \[ \frac {4\,{\mathrm {e}}^{a\,c+b\,c\,x}\,\mathrm {acot}\left (\frac {{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}-1}{{\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}+1}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4-4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (-4+4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (-1+1{}\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4-4{}\mathrm {i}\right )+{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1-\mathrm {i}\right )+\sqrt {2}\,\ln \left (\sqrt {2}\,\left (4+4{}\mathrm {i}\right )-{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,8{}\mathrm {i}\right )\,\left (1+1{}\mathrm {i}\right )}{4\,b\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(a + b*x))*acot(tanh(a*c + b*c*x)),x)

[Out]

(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)*acot((exp(2*b*c*x)*exp(2*a*c) - 1)/(exp(2*b*
c*x)*exp(2*a*c) + 1)))/(4*b*c)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*acot(tanh(b*c*x+a*c)),x)

[Out]

Timed out

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