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

Optimal. Leaf size=45 \[ \frac{e^{a c+b c x} \coth ^{-1}(\coth (c (a+b x)))}{b c}-\frac{e^{a c+b c x}}{b c} \]

[Out]

-(E^(a*c + b*c*x)/(b*c)) + (E^(a*c + b*c*x)*ArcCoth[Coth[c*(a + b*x)]])/(b*c)

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Rubi [A]  time = 0.0586381, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {2194, 6276} \[ \frac{e^{a c+b c x} \coth ^{-1}(\coth (c (a+b x)))}{b c}-\frac{e^{a c+b c x}}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(a*c + b*c*x)/(b*c)) + (E^(a*c + b*c*x)*ArcCoth[Coth[c*(a + b*x)]])/(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 6276

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

Rubi steps

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

Mathematica [A]  time = 0.0797835, size = 46, normalized size = 1.02 \[ \frac{e^{c (a+b x)} \left (\coth ^{-1}\left (\frac{e^{2 c (a+b x)}+1}{e^{2 c (a+b x)}-1}\right )-1\right )}{b c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(E^(c*(a + b*x))*(-1 + ArcCoth[(1 + E^(2*c*(a + b*x)))/(-1 + E^(2*c*(a + b*x)))]))/(b*c)

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Maple [A]  time = 0.069, size = 68, normalized size = 1.5 \begin{align*}{\frac{ \left ( xbc+ac \right ){{\rm e}^{xbc+ac}}-{{\rm e}^{xbc+ac}}+{{\rm e}^{xbc+ac}} \left ({\rm arccoth} \left ({\rm coth} \left (xbc+ac\right )\right )-xbc-ac \right ) }{bc}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b/c*((b*c*x+a*c)*exp(b*c*x+a*c)-exp(b*c*x+a*c)+exp(b*c*x+a*c)*(arccoth(coth(b*c*x+a*c))-x*b*c-a*c))

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Maxima [A]  time = 1.1811, size = 57, normalized size = 1.27 \begin{align*} \frac{a e^{\left (b c x + a c\right )}}{b} + \frac{{\left (b c x e^{\left (a c\right )} - e^{\left (a c\right )}\right )} e^{\left (b c x\right )}}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e^(b*c*x + a*c)/b + (b*c*x*e^(a*c) - e^(a*c))*e^(b*c*x)/(b*c)

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Fricas [A]  time = 1.51155, size = 55, normalized size = 1.22 \begin{align*} \frac{{\left (b c x + a c - 1\right )} e^{\left (b c x + a c\right )}}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c*x + 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))*acoth(coth(b*c*x+a*c)),x)

[Out]

Timed out

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Giac [A]  time = 1.18341, size = 47, normalized size = 1.04 \begin{align*} \frac{{\left (b^{2} c^{2} x + a b c^{2} - b c\right )} e^{\left (b c x + a c\right )}}{b^{2} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^2*c^2*x + a*b*c^2 - b*c)*e^(b*c*x + a*c)/(b^2*c^2)