\(\int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx\) [229]

Optimal result

Integrand size = 20, antiderivative size = 47 \[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=\frac {e^{a c+b c x} \cot ^{-1}(\sinh (c (a+b x)))}{b c}+\frac {\log \left (1+e^{2 c (a+b x)}\right )}{b c} \]

[Out]

exp(b*c*x+a*c)*arccot(sinh(c*(b*x+a)))/b/c+ln(1+exp(2*c*(b*x+a)))/b/c

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2225, 5316, 2320, 12, 266} \[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=\frac {\log \left (e^{2 c (a+b x)}+1\right )}{b c}+\frac {e^{a c+b c x} \cot ^{-1}(\sinh (c (a+b x)))}{b c} \]

[In]

Int[E^(c*(a + b*x))*ArcCot[Sinh[a*c + b*c*x]],x]

[Out]

(E^(a*c + b*c*x)*ArcCot[Sinh[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 266

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 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 2320

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 5316

Int[((a_.) + ArcCot[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcCot[u], w, x] + Dist
[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[Functi
onOfLinear[v*(a + b*ArcCot[u]), x]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int e^x \cot ^{-1}(\sinh (x)) \, dx,x,a c+b c x\right )}{b c} \\ & = \frac {e^{a c+b c x} \cot ^{-1}(\sinh (c (a+b x)))}{b c}+\frac {\text {Subst}\left (\int e^x \text {sech}(x) \, dx,x,a c+b c x\right )}{b c} \\ & = \frac {e^{a c+b c x} \cot ^{-1}(\sinh (c (a+b x)))}{b c}+\frac {\text {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} \cot ^{-1}(\sinh (c (a+b x)))}{b c}+\frac {2 \text {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} \cot ^{-1}(\sinh (c (a+b x)))}{b c}+\frac {\log \left (1+e^{2 c (a+b x)}\right )}{b c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.30 \[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=\frac {-e^{c (a+b x)} \cot ^{-1}\left (\frac {1}{2} e^{-c (a+b x)}-\frac {1}{2} e^{c (a+b x)}\right )+\log \left (1+e^{2 c (a+b x)}\right )}{b c} \]

[In]

Integrate[E^(c*(a + b*x))*ArcCot[Sinh[a*c + b*c*x]],x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.25 (sec) , antiderivative size = 1281, normalized size of antiderivative = 27.26

method	result	size
risch	\(\text {Expression too large to display}\)	\(1281\)

[In]

int(exp(c*(b*x+a))*arccot(sinh(b*c*x+a*c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (45) = 90\).

Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 2.79 \[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=\frac {{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \arctan \left (\frac {2 \, {\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )}}{\cosh \left (b c x + a c\right )^{2} + 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2} - 1}\right ) + \log \left (\frac {2 \, \cosh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{b c} \]

[In]

integrate(exp(c*(b*x+a))*arccot(sinh(b*c*x+a*c)),x, algorithm="fricas")

[Out]

((cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*arctan(2*(cosh(b*c*x + a*c) + sinh(b*c*x + a*c))/(cosh(b*c*x + a*c)^2
+ 2*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + sinh(b*c*x + a*c)^2 - 1)) + log(2*cosh(b*c*x + a*c)/(cosh(b*c*x + a
*c) - sinh(b*c*x + a*c))))/(b*c)

Sympy [F]

\[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=e^{a c} \int e^{b c x} \operatorname {acot}{\left (\sinh {\left (a c + b c x \right )} \right )}\, dx \]

[In]

integrate(exp(c*(b*x+a))*acot(sinh(b*c*x+a*c)),x)

[Out]

exp(a*c)*Integral(exp(b*c*x)*acot(sinh(a*c + b*c*x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=\frac {\operatorname {arccot}\left (\sinh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac {\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}{b c} \]

[In]

integrate(exp(c*(b*x+a))*arccot(sinh(b*c*x+a*c)),x, algorithm="maxima")

[Out]

arccot(sinh(b*c*x + a*c))*e^((b*x + a)*c)/(b*c) + log(e^(2*b*c*x + 2*a*c) + 1)/(b*c)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40 \[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=\frac {{\left (\arctan \left (\frac {2}{e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}}\right ) e^{\left (b c x\right )} + e^{\left (-a c\right )} \log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )\right )} e^{\left (a c\right )}}{b c} \]

[In]

integrate(exp(c*(b*x+a))*arccot(sinh(b*c*x+a*c)),x, algorithm="giac")

[Out]

(arctan(2/(e^(b*c*x + a*c) - 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)

Mupad [B] (verification not implemented)

Time = 0.88 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.38 \[ \int e^{c (a+b x)} \cot ^{-1}(\sinh (a c+b c x)) \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}+1\right )}{b\,c}+\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,\mathrm {acot}\left (\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}-\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}\right )}{b\,c} \]

[In]

int(exp(c*(a + b*x))*acot(sinh(a*c + b*c*x)),x)

[Out]

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