3.4.57 \(\int e^{c (a+b x)} \tanh ^{-1}(\tanh (a c+b c x)) \, dx\) [357]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {2225, 6410} \begin {gather*} \frac {e^{a c+b c x} \tanh ^{-1}(\tanh (c (a+b x)))}{b c}-\frac {e^{a c+b c x}}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(E^(a*c + b*c*x)/(b*c)) + (E^(a*c + b*c*x)*ArcTanh[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 6410

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

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Mathematica [A]
time = 0.04, size = 46, normalized size = 1.02 \begin {gather*} \frac {e^{c (a+b x)} \left (-1+\tanh ^{-1}\left (\frac {-1+e^{2 c (a+b x)}}{1+e^{2 c (a+b x)}}\right )\right )}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))*ArcTanh[Tanh[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 [A]
time = 0.16, size = 68, normalized size = 1.51

method	result	size
default	\(\frac {{\mathrm e}^{b c x +a c} \left (b c x +a c \right )-{\mathrm e}^{b c x +a c}+{\mathrm e}^{b c x +a c} \left (\arctanh \left (\tanh \left (b c x +a c \right )\right )-b c x -a c \right )}{c b}\)	\(68\)
risch	\(\frac {{\mathrm e}^{c \left (b x +a \right )} \ln \left ({\mathrm e}^{c \left (b x +a \right )}\right )}{c b}-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )-2 \pi \,\mathrm {csgn}\left (i {\mathrm e}^{c \left (b x +a \right )}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}+\pi \mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right )^{3}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{2 c \left (b x +a \right )}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{2 c \left (b x +a \right )}}{{\mathrm e}^{2 c \left (b x +a \right )}+1}\right )^{3}-4 i\right ) {\mathrm e}^{c \left (b x +a \right )}}{4 c b}\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 43, normalized size = 0.96 \begin {gather*} \frac {\operatorname {artanh}\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctanh(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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Fricas [A]
time = 0.33, size = 46, normalized size = 1.02 \begin {gather*} \frac {{\left (b c x + a c - 1\right )} \cosh \left (b c x + a c\right ) + {\left (b c x + a c - 1\right )} \sinh \left (b c x + a c\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.78, size = 58, normalized size = 1.29 \begin {gather*} \begin {cases} 0 & \text {for}\: c = 0 \wedge \left (b = 0 \vee c = 0\right ) \\x e^{a c} \operatorname {atanh}{\left (\tanh {\left (a c \right )} \right )} & \text {for}\: b = 0 \\\frac {e^{a c} e^{b c x} \operatorname {atanh}{\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((0, Eq(c, 0) & (Eq(b, 0) | Eq(c, 0))), (x*exp(a*c)*atanh(tanh(a*c)), Eq(b, 0)), (exp(a*c)*exp(b*c*x)
*atanh(tanh(a*c + b*c*x))/(b*c) - exp(a*c)*exp(b*c*x)/(b*c), True))

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Giac [A]
time = 0.39, size = 35, normalized size = 0.78 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 0.16, size = 28, normalized size = 0.62 \begin {gather*} \frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\left (\mathrm {atanh}\left (\mathrm {tanh}\left (a\,c+b\,c\,x\right )\right )-1\right )}{b\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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