Integrand size = 20, antiderivative size = 49 \[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=\frac {e^{a c+b c x} \text {arctanh}(\text {sech}(c (a+b x)))}{b c}+\frac {\log \left (1-e^{2 c (a+b x)}\right )}{b c} \] Output:
exp(b*c*x+a*c)*arctanh(sech(c*(b*x+a)))/b/c+ln(1-exp(2*c*(b*x+a)))/b/c
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.20 \[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=\frac {e^{c (a+b x)} \text {arctanh}\left (\frac {2 e^{c (a+b x)}}{1+e^{2 c (a+b x)}}\right )+\log \left (1-e^{2 c (a+b x)}\right )}{b c} \] Input:
Integrate[E^(c*(a + b*x))*ArcTanh[Sech[a*c + b*c*x]],x]
Output:
(E^(c*(a + b*x))*ArcTanh[(2*E^(c*(a + b*x)))/(1 + E^(2*c*(a + b*x)))] + Lo g[1 - E^(2*c*(a + b*x))])/(b*c)
Time = 0.33 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {7281, 6829, 25, 2720, 27, 240}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {\int e^{a c+b x c} \text {arctanh}(\text {sech}(a c+b x c))d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 6829 |
\(\displaystyle \frac {e^{a c+b c x} \text {arctanh}(\text {sech}(a c+b c x))-\int -e^{a c+b x c} \text {csch}(a c+b x c)d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int e^{a c+b x c} \text {csch}(a c+b x c)d(a c+b x c)+e^{a c+b c x} \text {arctanh}(\text {sech}(a c+b c x))}{b c}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {2 e^{a c+b x c}}{1-e^{2 a c+2 b x c}}de^{a c+b x c}+e^{a c+b c x} \text {arctanh}(\text {sech}(a c+b c x))}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{a c+b c x} \text {arctanh}(\text {sech}(a c+b c x))-2 \int \frac {e^{a c+b x c}}{1-e^{2 a c+2 b x c}}de^{a c+b x c}}{b c}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {e^{a c+b c x} \text {arctanh}(\text {sech}(a c+b c x))+\log \left (1-e^{2 a c+2 b c x}\right )}{b c}\) |
Input:
Int[E^(c*(a + b*x))*ArcTanh[Sech[a*c + b*c*x]],x]
Output:
(E^(a*c + b*c*x)*ArcTanh[Sech[a*c + b*c*x]] + Log[1 - E^(2*a*c + 2*b*c*x)] )/(b*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[((a_.) + ArcTanh[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]} , Simp[(a + b*ArcTanh[u]) w, x] - Simp[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_.) /; F reeQ[{c, d, m}, x]] && FalseQ[FunctionOfLinear[v*(a + b*ArcTanh[u]), x]]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.72 (sec) , antiderivative size = 872, normalized size of antiderivative = 17.80
Input:
int(exp(c*(b*x+a))*arctanh(sech(b*c*x+a*c)),x,method=_RETURNVERBOSE)
Output:
-1/4*I/b/c*Pi*csgn(I*(exp(c*(b*x+a))+1))^2*csgn(I*(exp(c*(b*x+a))+1)^2)*ex p(c*(b*x+a))-1/4*I/b/c*Pi*csgn(I/(1+exp(2*c*(b*x+a))))*csgn(I*(exp(c*(b*x+ a))-1)^2/(1+exp(2*c*(b*x+a))))^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*c sgn(I*(exp(c*(b*x+a))+1)^2)*csgn(I*(exp(c*(b*x+a))+1)^2/(1+exp(2*c*(b*x+a) )))^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) ^3*exp(c*(b*x+a))+1/4*I/b/c*Pi*csgn(I*(exp(c*(b*x+a))-1)^2)*csgn(I/(1+exp( 2*c*(b*x+a))))*csgn(I*(exp(c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a))))*exp(c*(b* x+a))+1/4*I/b/c*Pi*csgn(I/(1+exp(2*c*(b*x+a))))*csgn(I*(exp(c*(b*x+a))+1)^ 2/(1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))+1/4*I/b/c*Pi*csgn(I*(exp(c*(b*x+a ))-1)^2/(1+exp(2*c*(b*x+a))))^3*exp(c*(b*x+a))-1/4*I/b/c*Pi*csgn(I*(exp(c* (b*x+a))+1)^2/(1+exp(2*c*(b*x+a))))^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/(1+exp(2*c*(b*x+a))))^2*e xp(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 /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/(1+exp(2*c*(b* x+a))))*csgn(I*(exp(c*(b*x+a))+1)^2/(1+exp(2*c*(b*x+a))))*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(-1+exp(2*c*(b*x+a)))
Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.88 \[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=\frac {{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \log \left (\frac {\cosh \left (b c x + a c\right ) + 1}{\cosh \left (b c x + a c\right ) - 1}\right ) + 2 \, \log \left (\frac {2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{2 \, b c} \] Input:
integrate(exp(c*(b*x+a))*arctanh(sech(b*c*x+a*c)),x, algorithm="fricas")
Output:
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*sinh(b*c*x + a*c)/(cosh(b*c*x + a*c) - s inh(b*c*x + a*c))))/(b*c)
Timed out. \[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=\text {Timed out} \] Input:
integrate(exp(c*(b*x+a))*atanh(sech(b*c*x+a*c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.31 \[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=\frac {\operatorname {artanh}\left (\operatorname {sech}\left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \] Input:
integrate(exp(c*(b*x+a))*arctanh(sech(b*c*x+a*c)),x, algorithm="maxima")
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (47) = 94\).
Time = 0.20 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.00 \[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=\frac {e^{\left ({\left (b x + a\right )} c\right )} \log \left (-\frac {\frac {2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}} + 1}{\frac {2}{e^{\left (b c x + a c\right )} + e^{\left (-b c x - a c\right )}} - 1}\right )}{2 \, b c} + \frac {\log \left ({\left | e^{\left (2 \, b c x + 2 \, a c\right )} - 1 \right |}\right )}{b c} \] Input:
integrate(exp(c*(b*x+a))*arctanh(sech(b*c*x+a*c)),x, algorithm="giac")
Output:
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)
Time = 4.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.43 \[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}-1\right )}{b\,c}-\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\ln \left (1-\frac {1}{\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}}\right )}{2\,b\,c}+\frac {\ln \left (\frac {1}{\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}}+1\right )\,{\mathrm {e}}^{a\,c+b\,c\,x}}{2\,b\,c} \] Input:
int(atanh(1/cosh(a*c + b*c*x))*exp(c*(a + b*x)),x)
Output:
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)*exp(-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)
\[ \int e^{c (a+b x)} \text {arctanh}(\text {sech}(a c+b c x)) \, dx=e^{a c} \left (\int e^{b c x} \mathit {atanh} \left (\mathrm {sech}\left (b c x +a c \right )\right )d x \right ) \] Input:
int(exp(c*(b*x+a))*atanh(sech(b*c*x+a*c)),x)
Output:
e**(a*c)*int(e**(b*c*x)*atanh(sech(a*c + b*c*x)),x)