3.152 \(\int e^{c (a+b x)} \tan ^{-1}(\text{csch}(a c+b c x)) \, dx\)

Optimal. Leaf size=47 \[ \frac{\log \left (e^{2 c (a+b x)}+1\right )}{b c}+\frac{e^{a c+b c x} \tan ^{-1}(\text{csch}(c (a+b x)))}{b c} \]

[Out]

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

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Rubi [A]  time = 0.0793913, antiderivative size = 47, 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, 5207, 2282, 12, 260} \[ \frac{\log \left (e^{2 c (a+b x)}+1\right )}{b c}+\frac{e^{a c+b c x} \tan ^{-1}(\text{csch}(c (a+b x)))}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*ArcTan[Csch[a*c + b*c*x]],x]

[Out]

(E^(a*c + b*c*x)*ArcTan[Csch[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 5207

Int[((a_.) + ArcTan[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTan[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*ArcTan[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)} \tan ^{-1}(\text{csch}(a c+b c x)) \, dx &=\frac{\operatorname{Subst}\left (\int e^x \tan ^{-1}(\text{csch}(x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\text{csch}(c (a+b x)))}{b c}+\frac{\operatorname{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} \tan ^{-1}(\text{csch}(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} \tan ^{-1}(\text{csch}(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} \tan ^{-1}(\text{csch}(c (a+b x)))}{b c}+\frac{\log \left (1+e^{2 c (a+b x)}\right )}{b c}\\ \end{align*}

Mathematica [A]  time = 0.106656, size = 57, normalized size = 1.21 \[ \frac{\log \left (e^{2 c (a+b x)}+1\right )+e^{c (a+b x)} \tan ^{-1}\left (\frac{2 e^{c (a+b x)}}{e^{2 c (a+b x)}-1}\right )}{b c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(c*(a + b*x))*ArcTan[Csch[a*c + b*c*x]],x]

[Out]

(E^(c*(a + b*x))*ArcTan[(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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Maple [C]  time = 0.513, size = 885, normalized size = 18.8 \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))*arctan(csch(b*c*x+a*c)),x)

[Out]

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

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Maxima [A]  time = 1.51905, size = 63, normalized size = 1.34 \begin{align*} \frac{\arctan \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(csch(b*c*x+a*c)),x, algorithm="maxima")

[Out]

arctan(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)

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Fricas [B]  time = 1.90537, size = 343, normalized size = 7.3 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(csch(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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a c} \int e^{b c x} \operatorname{atan}{\left (\operatorname{csch}{\left (a c + b c x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*atan(csch(b*c*x+a*c)),x)

[Out]

exp(a*c)*Integral(exp(b*c*x)*atan(csch(a*c + b*c*x)), x)

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Giac [A]  time = 1.15124, size = 89, normalized size = 1.89 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctan(csch(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)