Integrand size = 20, antiderivative size = 48 \[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=\frac {e^{a c+b c x} \cot ^{-1}(\text {csch}(c (a+b x)))}{b c}-\frac {\log \left (1+e^{2 c (a+b x)}\right )}{b c} \] Output:
exp(b*c*x+a*c)*arccot(csch(c*(b*x+a)))/b/c-ln(1+exp(2*c*(b*x+a)))/b/c
Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=\frac {e^{c (a+b x)} \cot ^{-1}\left (\frac {2 e^{c (a+b x)}}{-1+e^{2 c (a+b x)}}\right )-\log \left (1+e^{2 c (a+b x)}\right )}{b c} \] Input:
Integrate[E^(c*(a + b*x))*ArcCot[Csch[a*c + b*c*x]],x]
Output:
(E^(c*(a + b*x))*ArcCot[(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 = 0.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {7281, 5731, 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.
\(\displaystyle \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {\int e^{a c+b x c} \cot ^{-1}(\text {csch}(a c+b x c))d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 5731 |
\(\displaystyle \frac {\int -e^{a c+b x c} \text {sech}(a c+b x c)d(a c+b x c)+e^{a c+b c x} \cot ^{-1}(\text {csch}(a c+b c x))}{b c}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {e^{a c+b c x} \cot ^{-1}(\text {csch}(a c+b c x))-\int e^{a c+b x c} \text {sech}(a c+b x c)d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {e^{a c+b c x} \cot ^{-1}(\text {csch}(a c+b c x))-\int \frac {2 e^{a c+b x c}}{1+e^{2 a c+2 b x c}}de^{a c+b x c}}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^{a c+b c x} \cot ^{-1}(\text {csch}(a c+b c x))-2 \int \frac {e^{a c+b x c}}{1+e^{2 a c+2 b x c}}de^{a c+b x c}}{b c}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {e^{a c+b c x} \cot ^{-1}(\text {csch}(a c+b c x))-\log \left (e^{2 a c+2 b c x}+1\right )}{b c}\) |
Input:
Int[E^(c*(a + b*x))*ArcCot[Csch[a*c + b*c*x]],x]
Output:
(E^(a*c + b*c*x)*ArcCot[Csch[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_.) + ArcCot[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[(a + b*ArcCot[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_.) /; FreeQ [{c, d, m}, x]] && FalseQ[FunctionOfLinear[v*(a + b*ArcCot[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 = 1.24 (sec) , antiderivative size = 903, normalized size of antiderivative = 18.81
Input:
int(exp(c*(b*x+a))*arccot(csch(b*c*x+a*c)),x,method=_RETURNVERBOSE)
Output:
I/b/c*exp(c*(b*x+a))*ln(exp(c*(b*x+a))+I)-1/4/b/c*Pi*csgn(I*(exp(c*(b*x+a) )-I))^2*csgn(I*(exp(c*(b*x+a))-I)^2)*exp(c*(b*x+a))+1/2/b/c*Pi*csgn(I*(exp (c*(b*x+a))-I))*csgn(I*(exp(c*(b*x+a))-I)^2)^2*exp(c*(b*x+a))-1/4/b/c*Pi*c sgn(I*(exp(c*(b*x+a))-I)^2)^3*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*(exp(c*(b*x +a))-I)^2)*csgn(I/(-1+exp(2*c*(b*x+a))))*csgn(I*(exp(c*(b*x+a))-I)^2/(-1+e xp(2*c*(b*x+a))))*exp(c*(b*x+a))+1/4/b/c*Pi*csgn(I*(exp(c*(b*x+a))-I)^2)*c sgn(I*(exp(c*(b*x+a))-I)^2/(-1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))+1/4/b/c *Pi*csgn(I/(-1+exp(2*c*(b*x+a))))*csgn(I*(exp(c*(b*x+a))-I)^2/(-1+exp(2*c* (b*x+a))))^2*exp(c*(b*x+a))+1/4/b/c*Pi*csgn(I*(exp(c*(b*x+a))+I)^2)*csgn(I /(-1+exp(2*c*(b*x+a))))*csgn(I*(exp(c*(b*x+a))+I)^2/(-1+exp(2*c*(b*x+a)))) *exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I/(-1+exp(2*c*(b*x+a))))*csgn(I*(exp(c*(b* x+a))+I)^2/(-1+exp(2*c*(b*x+a))))^2*exp(c*(b*x+a))-1/4/b/c*Pi*csgn(I*(exp( c*(b*x+a))-I)^2/(-1+exp(2*c*(b*x+a))))^3*exp(c*(b*x+a))+1/4/b/c*Pi*csgn(I* (exp(c*(b*x+a))+I))^2*csgn(I*(exp(c*(b*x+a))+I)^2)*exp(c*(b*x+a))-1/2/b/c* Pi*csgn(I*(exp(c*(b*x+a))+I))*csgn(I*(exp(c*(b*x+a))+I)^2)^2*exp(c*(b*x+a) )+1/4/b/c*Pi*csgn(I*(exp(c*(b*x+a))+I)^2)^3*exp(c*(b*x+a))-1/4/b/c*Pi*csgn (I*(exp(c*(b*x+a))+I)^2)*csgn(I*(exp(c*(b*x+a))+I)^2/(-1+exp(2*c*(b*x+a))) )^2*exp(c*(b*x+a))+1/4/b/c*Pi*csgn(I*(exp(c*(b*x+a))+I)^2/(-1+exp(2*c*(b*x +a))))^3*exp(c*(b*x+a))-I/b/c*exp(c*(b*x+a))*ln(exp(c*(b*x+a))-I)+1/2*Pi/b /c*exp(c*(b*x+a))+2*a/b-ln(1+exp(2*c*(b*x+a)))/b/c
Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.56 \[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=\frac {{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \arctan \left (\sinh \left (b c x + a c\right )\right ) - \log \left (\frac {2 \, \cosh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{b c} \] Input:
integrate(exp(c*(b*x+a))*arccot(csch(b*c*x+a*c)),x, algorithm="fricas")
Output:
((cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*arctan(sinh(b*c*x + a*c)) - log(2 *cosh(b*c*x + a*c)/(cosh(b*c*x + a*c) - sinh(b*c*x + a*c))))/(b*c)
\[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=e^{a c} \int e^{b c x} \operatorname {acot}{\left (\operatorname {csch}{\left (a c + b c x \right )} \right )}\, dx \] Input:
integrate(exp(c*(b*x+a))*acot(csch(b*c*x+a*c)),x)
Output:
exp(a*c)*Integral(exp(b*c*x)*acot(csch(a*c + b*c*x)), x)
Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=\frac {\operatorname {arccot}\left (\operatorname {csch}\left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} - \frac {\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{b c} \] Input:
integrate(exp(c*(b*x+a))*arccot(csch(b*c*x+a*c)),x, algorithm="maxima")
Output:
arccot(csch(b*c*x + a*c))*e^((b*x + a)*c)/(b*c) - log(e^(2*b*c*x + 2*a*c) + 1)/(b*c)
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=\frac {{\left (\arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} e^{\left (-b c x - a c\right )}\right ) e^{\left (b c x\right )} - e^{\left (-a c\right )} \log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )\right )} e^{\left (a c\right )}}{b c} \] Input:
integrate(exp(c*(b*x+a))*arccot(csch(b*c*x+a*c)),x, algorithm="giac")
Output:
(arctan(1/2*(e^(2*b*c*x + 2*a*c) - 1)*e^(-b*c*x - a*c))*e^(b*c*x) - e^(-a* c)*log(e^(2*b*c*x + 2*a*c) + 1))*e^(a*c)/(b*c)
Time = 0.88 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.42 \[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,\mathrm {acot}\left (\frac {1}{\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}-\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}}\right )}{b\,c}-\frac {\ln \left ({\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}+1\right )}{b\,c} \] Input:
int(acot(1/sinh(a*c + b*c*x))*exp(c*(a + b*x)),x)
Output:
(exp(b*c*x)*exp(a*c)*acot(1/((exp(b*c*x)*exp(a*c))/2 - (exp(-b*c*x)*exp(-a *c))/2)))/(b*c) - log(exp(2*b*c*x)*exp(2*a*c) + 1)/(b*c)
Time = 0.32 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.35 \[ \int e^{c (a+b x)} \cot ^{-1}(\text {csch}(a c+b c x)) \, dx=\frac {e^{b c x +a c} \mathit {atan} \left (\frac {e^{2 b c x +2 a c}-1}{2 e^{b c x +a c}}\right )-\mathrm {log}\left (e^{2 b c x +2 a c}+1\right )}{b c} \] Input:
int(exp(c*(b*x+a))*acot(csch(b*c*x+a*c)),x)
Output:
(e**(a*c + b*c*x)*atan((e**(2*a*c + 2*b*c*x) - 1)/(2*e**(a*c + b*c*x))) - log(e**(2*a*c + 2*b*c*x) + 1))/(b*c)