3.359 \(\int e^{c (a+b x)} \tanh ^{-1}(\text {sech}(a c+b c x)) \, dx\)

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

[Out]

exp(b*c*x+a*c)*arctanh(sech(c*(b*x+a)))/b/c+ln(1-exp(2*c*(b*x+a)))/b/c

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Rubi [A]  time = 0.07, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2194, 6275, 2282, 12, 260} \[ \frac {\log \left (1-e^{2 c (a+b x)}\right )}{b c}+\frac {e^{a c+b c x} \tanh ^{-1}(\text {sech}(c (a+b x)))}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

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Mathematica [A]  time = 0.08, size = 59, normalized size = 1.20 \[ \frac {\log \left (1-e^{2 c (a+b x)}\right )+e^{c (a+b x)} \tanh ^{-1}\left (\frac {2 e^{c (a+b x)}}{e^{2 c (a+b x)}+1}\right )}{b c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 92, normalized size = 1.88 \[ \frac {{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \log \left (\frac {\cosh \left (b c x + a c\right ) + 1}{\cosh \left (b c x + a c\right ) - 1}\right ) + 2 \, \log \left (\frac {2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{2 \, b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctanh(sech(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*si
nh(b*c*x + a*c)/(cosh(b*c*x + a*c) - sinh(b*c*x + a*c))))/(b*c)

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giac [B]  time = 0.32, size = 98, normalized size = 2.00 \[ \frac {e^{\left ({\left (b x + a\right )} c\right )} \log \left (-\frac {\frac {2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}} + 1}{\frac {2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}} - 1}\right )}{2 \, b c} + \frac {\log \left ({\left | e^{\left (2 \, b c x + 2 \, a c\right )} - 1 \right |}\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctanh(sech(b*c*x+a*c)),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.38, size = 872, normalized size = 17.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I/b/c*Pi*csgn(I*(exp(c*(b*x+a))+1)^2)*csgn(I/(exp(2*c*(b*x+a))+1))*csgn(I*(exp(c*(b*x+a))+1)^2/(exp(2*c*(
b*x+a))+1))*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*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/(exp(2*c*(b*x+a))+1))^3*exp(c*(b*x+a))+1/4*I/b/c*Pi*csgn(I/(exp(2*c*(b*x+a))+1))*csgn(I*(e
xp(c*(b*x+a))+1)^2/(exp(2*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)^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))-1)^2)*csgn(I/(exp(2*c*(b*x+a))+1
))*csgn(I*(exp(c*(b*x+a))-1)^2/(exp(2*c*(b*x+a))+1))*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(2*c*(b*x+a))+1))^2*exp(c*(b*x+a))-1/4*I/b/c*Pi*csgn(I/(exp(2*c*(b*x+a))+1))*c
sgn(I*(exp(c*(b*x+a))-1)^2/(exp(2*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))+1)^
2/(exp(2*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/(exp(2*c*(b*x+a))+1))^3*exp(
c*(b*x+a))+1/b/c*exp(c*(b*x+a))*ln(exp(c*(b*x+a))+1)-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 = 0.33, size = 64, normalized size = 1.31 \[ \frac {\operatorname {artanh}\left (\operatorname {sech}\left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctanh(sech(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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mupad [B]  time = 1.71, size = 119, normalized size = 2.43 \[ \frac {\ln \left ({\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}-1\right )}{b\,c}-\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\ln \left (1-\frac {1}{\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}}\right )}{2\,b\,c}+\frac {\ln \left (\frac {1}{\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}}+1\right )\,{\mathrm {e}}^{a\,c+b\,c\,x}}{2\,b\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(1/cosh(a*c + b*c*x))*exp(c*(a + b*x)),x)

[Out]

log(exp(2*b*c*x)*exp(2*a*c) - 1)/(b*c) - (exp(a*c + b*c*x)*log(1 - 1/((exp(b*c*x)*exp(a*c))/2 + (exp(-b*c*x)*e
xp(-a*c))/2)))/(2*b*c) + (log(1/((exp(b*c*x)*exp(a*c))/2 + (exp(-b*c*x)*exp(-a*c))/2) + 1)*exp(a*c + b*c*x))/(
2*b*c)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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