\(\int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx\) [297]

Optimal result

Integrand size = 20, antiderivative size = 45 \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=-\frac {e^{a c+b c x}}{b c}+\frac {e^{a c+b c x} \coth ^{-1}(\tanh (c (a+b x)))}{b c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2225, 6411} \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=\frac {e^{a c+b c x} \coth ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {e^{a c+b c x}}{b c} \]

[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 2225

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 6411

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*} \text {integral}& = \frac {\text {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 {\text {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*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02 \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=\frac {e^{c (a+b x)} \left (-1+\coth ^{-1}\left (\frac {-1+e^{2 c (a+b x)}}{1+e^{2 c (a+b x)}}\right )\right )}{b c} \]

[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)

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.82

method	result	size
parallelrisch	\(-\frac {-{\mathrm e}^{c \left (b x +a \right )} \operatorname {arccoth}\left (\tanh \left (c \left (b x +a \right )\right )\right )+{\mathrm e}^{c \left (b x +a \right )}}{b c}\)	\(37\)
risch	\(\frac {{\mathrm e}^{c \left (b x +a \right )} \ln \left ({\mathrm e}^{c \left (b x +a \right )}\right )}{b c}-\frac {i \left (\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{3}-2 \pi \operatorname {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}+\pi \operatorname {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{3}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{3}-4 i+2 \pi \right ) {\mathrm e}^{c \left (b x +a \right )}}{4 b c}\)	\(349\)

[In]

int(exp(c*(b*x+a))*arccoth(tanh(b*c*x+a*c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.27 \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=\frac {{\left (i \, \pi + 2 \, b c x + 2 \, a c - 2\right )} \cosh \left (b c x + a c\right ) + {\left (i \, \pi + 2 \, b c x + 2 \, a c - 2\right )} \sinh \left (b c x + a c\right )}{2 \, b c} \]

[In]

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

[Out]

1/2*((I*pi + 2*b*c*x + 2*a*c - 2)*cosh(b*c*x + a*c) + (I*pi + 2*b*c*x + 2*a*c - 2)*sinh(b*c*x + a*c))/(b*c)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.71 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47 \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=\begin {cases} \frac {i \pi x}{2} & \text {for}\: b = 0 \wedge c = 0 \\x e^{a c} \operatorname {acoth}{\left (\tanh {\left (a c \right )} \right )} & \text {for}\: b = 0 \\\frac {i \pi x}{2} & \text {for}\: c = 0 \\\frac {e^{a c} e^{b c x} \operatorname {acoth}{\left (\tanh {\left (a c + b c x \right )} \right )}}{b c} - \frac {e^{a c} e^{b c x}}{b c} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.96 \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=\frac {\operatorname {arcoth}\left (\tanh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {e^{\left (b c x + a c\right )}}{b c} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.89 \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=\frac {{\left (e^{\left (b c x\right )} \log \left (-e^{\left (2 \, b c x + 2 \, a c\right )}\right ) - 2 \, e^{\left (b c x\right )}\right )} e^{\left (a c\right )}}{2 \, b c} \]

[In]

integrate(exp(c*(b*x+a))*arccoth(tanh(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)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.62 \[ \int e^{c (a+b x)} \coth ^{-1}(\tanh (a c+b c x)) \, dx=\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\left (\mathrm {acoth}\left (\mathrm {tanh}\left (a\,c+b\,c\,x\right )\right )-1\right )}{b\,c} \]

[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)