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

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

[Out]

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

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Rubi [A]  time = 0.0543121, 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, 6275} \[ \frac{e^{a c+b c x} \tanh ^{-1}(\coth (c (a+b x)))}{b c}-\frac{e^{a c+b c x}}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(a*c + b*c*x)/(b*c)) + (E^(a*c + b*c*x)*ArcTanh[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 6275

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

Rubi steps

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

Mathematica [A]  time = 0.0832905, size = 46, normalized size = 1.02 \[ \frac{e^{c (a+b x)} \left (\tanh ^{-1}\left (\frac{e^{2 c (a+b x)}+1}{e^{2 c (a+b x)}-1}\right )-1\right )}{b c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.257, size = 351, normalized size = 7.8 \begin{align*}{\frac{{{\rm e}^{c \left ( bx+a \right ) }}\ln \left ({{\rm e}^{c \left ( bx+a \right ) }} \right ) }{bc}}+{\frac{{\frac{i}{4}}{{\rm e}^{c \left ( bx+a \right ) }}}{bc} \left ( 2\,\pi \, \left ({\it csgn} \left ({\frac{i}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ) \right ) ^{2}-2\,\pi \, \left ({\it csgn} \left ({\frac{i}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ) \right ) ^{3}-\pi \,{\it csgn} \left ({\frac{i}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ){\it csgn} \left ( i{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ){\it csgn} \left ({\frac{i{{\rm e}^{2\,c \left ( bx+a \right ) }}}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ) +\pi \,{\it csgn} \left ({\frac{i}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ) \left ({\it csgn} \left ({\frac{i{{\rm e}^{2\,c \left ( bx+a \right ) }}}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ) \right ) ^{2}-\pi \, \left ({\it csgn} \left ( i{{\rm e}^{c \left ( bx+a \right ) }} \right ) \right ) ^{2}{\it csgn} \left ( i{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) +2\,\pi \,{\it csgn} \left ( i{{\rm e}^{c \left ( bx+a \right ) }} \right ) \left ({\it csgn} \left ( i{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) \right ) ^{2}-\pi \, \left ({\it csgn} \left ( i{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) \right ) ^{3}+\pi \,{\it csgn} \left ( i{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) \left ({\it csgn} \left ({\frac{i{{\rm e}^{2\,c \left ( bx+a \right ) }}}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ) \right ) ^{2}-\pi \, \left ({\it csgn} \left ({\frac{i{{\rm e}^{2\,c \left ( bx+a \right ) }}}{{{\rm e}^{2\,c \left ( bx+a \right ) }}-1}} \right ) \right ) ^{3}+4\,i-2\,\pi \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.986189, size = 58, normalized size = 1.29 \begin{align*} \frac{\operatorname{artanh}\left (\coth \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac{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))*arctanh(coth(b*c*x+a*c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.93226, 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))*arctanh(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))*atanh(coth(b*c*x+a*c)),x)

[Out]

Timed out

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Giac [A]  time = 1.19302, size = 54, normalized size = 1.2 \begin{align*} \frac{{\left (e^{\left (b c x\right )} \log \left (-e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - 2 \, e^{\left (b c x\right )}\right )} e^{\left (a c\right )}}{2 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^(b*c*x)*log(-e^(2*b*c*x + 2*a*c)) - 2*e^(b*c*x))*e^(a*c)/(b*c)