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

Optimal. Leaf size=49 \[ \frac{\log \left (1-e^{2 c (a+b x)}\right )}{b c}+\frac{e^{a c+b c x} \tanh ^{-1}(\cosh (c (a+b x)))}{b c} \]

[Out]

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

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Rubi [A]  time = 0.0746668, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2194, 6275, 2282, 12, 260} \[ \frac{\log \left (1-e^{2 c (a+b x)}\right )}{b c}+\frac{e^{a c+b c x} \tanh ^{-1}(\cosh (c (a+b x)))}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int e^{c (a+b x)} \tanh ^{-1}(\cosh (a c+b c x)) \, dx &=\frac{\operatorname{Subst}\left (\int e^x \tanh ^{-1}(\cosh (x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\cosh (c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int e^x \text{csch}(x) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\cosh (c (a+b x)))}{b c}+\frac{\operatorname{Subst}\left (\int \frac{2 x}{-1+x^2} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\cosh (c (a+b x)))}{b c}+\frac{2 \operatorname{Subst}\left (\int \frac{x}{-1+x^2} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \tanh ^{-1}(\cosh (c (a+b x)))}{b c}+\frac{\log \left (1-e^{2 c (a+b x)}\right )}{b c}\\ \end{align*}

Mathematica [A]  time = 0.0879195, size = 60, normalized size = 1.22 \[ \frac{\log \left (1-e^{2 c (a+b x)}\right )+e^{c (a+b x)} \tanh ^{-1}\left (\frac{1}{2} e^{-c (a+b x)} \left (e^{2 c (a+b x)}+1\right )\right )}{b c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.375, size = 887, normalized size = 18.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00298, size = 86, normalized size = 1.76 \begin{align*} \frac{\operatorname{artanh}\left (\cosh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac{\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac{\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.76039, size = 238, normalized size = 4.86 \begin{align*} \frac{{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \log \left (-\frac{\cosh \left (b c x + a c\right ) + 1}{\cosh \left (b c x + a c\right ) - 1}\right ) + 2 \, \log \left (\frac{2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{2 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*log(-(cosh(b*c*x + a*c) + 1)/(cosh(b*c*x + a*c) - 1)) + 2*log(2*s
inh(b*c*x + a*c)/(cosh(b*c*x + a*c) - sinh(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(cosh(b*c*x+a*c)),x)

[Out]

Timed out

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Giac [B]  time = 1.23512, size = 198, normalized size = 4.04 \begin{align*} \frac{{\left (e^{\left (b c x\right )} \log \left (-\frac{e^{\left (2 \, b c x + 2 \, a c\right )}}{e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} + 1} - \frac{2 \, e^{\left (b c x + a c\right )}}{e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} + 1} - \frac{1}{e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} + 1}\right ) + 2 \, e^{\left (-a c\right )} \log \left ({\left | e^{\left (2 \, b c x + 2 \, a c\right )} - 1 \right |}\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(cosh(b*c*x+a*c)),x, algorithm="giac")

[Out]

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