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]

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

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Rubi [A]  time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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*}

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Mathematica [A]  time = 0.09, 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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fricas [A]  time = 0.60, size = 25, normalized size = 0.56 \[ \frac {{\left (b c x + a c - 1\right )} e^{\left (b c x + a c\right )}}{b c} \]

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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giac [A]  time = 0.13, size = 40, normalized size = 0.89 \[ \frac {{\left (e^{\left (b c x\right )} \log \left (-e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - 2 \, e^{\left (b c x\right )}\right )} e^{\left (a c\right )}}{2 \, b c} \]

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)

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maple [C]  time = 0.31, size = 351, normalized size = 7.80 \[ \frac {{\mathrm e}^{c \left (b x +a \right )} \ln \left ({\mathrm e}^{c \left (b x +a \right )}\right )}{b c}+\frac {i \left (2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{2}-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )+2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}-\pi \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{3}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{3}+4 i-2 \pi \right ) {\mathrm e}^{c \left (b x +a \right )}}{4 b c} \]

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 = 0.33, size = 43, normalized size = 0.96 \[ \frac {\operatorname {artanh}\left (\coth \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {e^{\left (b c x + a c\right )}}{b c} \]

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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mupad [B]  time = 0.07, size = 28, normalized size = 0.62 \[ \frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\left (\mathrm {atanh}\left (\mathrm {coth}\left (a\,c+b\,c\,x\right )\right )-1\right )}{b\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int e^{b c x} \operatorname {atanh}{\left (\coth {\left (a c + b c x \right )} \right )}\, dx \]

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]

exp(a*c)*Integral(exp(b*c*x)*atanh(coth(a*c + b*c*x)), x)

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