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

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

[Out]

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

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Rubi [A]  time = 0.06, 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, 6276} \[ \frac {e^{a c+b c x} \coth ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {e^{a c+b c x}}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, 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[Tanh[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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fricas [A]  time = 0.58, 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))*arccoth(tanh(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 [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(tanh(b*c*x + a*c))*e^((b*x + a)*c), x)

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maple [C]  time = 0.38, size = 349, normalized size = 7.76 \[ \frac {{\mathrm e}^{c \left (b x +a \right )} \ln \left ({\mathrm e}^{c \left (b x +a \right )}\right )}{b c}-\frac {i \left (-2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{3}+\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{3}-4 i+2 \pi \right ) {\mathrm e}^{c \left (b x +a \right )}}{4 b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*arccoth(tanh(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+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))+1))*csgn(I*exp(2*c*(b*x+a))/(exp(2*c*(b*x+a))+1))-Pi*cs
gn(I*exp(2*c*(b*x+a)))*csgn(I*exp(2*c*(b*x+a))/(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))/(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 \[ \frac {\operatorname {arcoth}\left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {e^{\left (b c x + a c\right )}}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arccoth(tanh(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 = 0.10, size = 28, normalized size = 0.62 \[ \frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\left (\mathrm {acoth}\left (\mathrm {tanh}\left (a\,c+b\,c\,x\right )\right )-1\right )}{b\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(a*c + b*c*x)*(acoth(tanh(a*c + b*c*x)) - 1))/(b*c)

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sympy [A]  time = 3.12, size = 66, normalized size = 1.47 \[ \begin {cases} \frac {i \pi x}{2} & \text {for}\: b = 0 \wedge c = 0 \\x e^{a c} \operatorname {acoth}{\left (\tanh {\left (a c \right )} \right )} & \text {for}\: b = 0 \\\frac {i \pi x}{2} & \text {for}\: c = 0 \\\frac {e^{a c} e^{b c x} \operatorname {acoth}{\left (\tanh {\left (a c + b c x \right )} \right )}}{b c} - \frac {e^{a c} e^{b c x}}{b c} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*pi*x/2, Eq(b, 0) & Eq(c, 0)), (x*exp(a*c)*acoth(tanh(a*c)), Eq(b, 0)), (I*pi*x/2, Eq(c, 0)), (exp
(a*c)*exp(b*c*x)*acoth(tanh(a*c + b*c*x))/(b*c) - exp(a*c)*exp(b*c*x)/(b*c), True))

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