3.147 \(\int e^{c (a+b x)} \tan ^{-1}(\sinh (a c+b c x)) \, dx\)
Optimal. Leaf size=48 \[ \frac{e^{a c+b c x} \tan ^{-1}(\sinh (c (a+b x)))}{b c}-\frac{\log \left (e^{2 c (a+b x)}+1\right )}{b c} \]
[Out]
(E^(a*c + b*c*x)*ArcTan[Sinh[c*(a + b*x)]])/(b*c) - Log[1 + E^(2*c*(a + b*x))]/(b*c)
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Rubi [A] time = 0.0753947, antiderivative size = 48, 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} \[ \frac{e^{a c+b c x} \tan ^{-1}(\sinh (c (a+b x)))}{b c}-\frac{\log \left (e^{2 c (a+b x)}+1\right )}{b c} \]
Antiderivative was successfully verified.
[In]
Int[E^(c*(a + b*x))*ArcTan[Sinh[a*c + b*c*x]],x]
[Out]
(E^(a*c + b*c*x)*ArcTan[Sinh[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}(\sinh (a c+b c x)) \, dx &=\frac{\operatorname{Subst}\left (\int e^x \tan ^{-1}(\sinh (x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \tan ^{-1}(\sinh (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}(\sinh (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}(\sinh (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}(\sinh (c (a+b x)))}{b c}-\frac{\log \left (1+e^{2 c (a+b x)}\right )}{b c}\\ \end{align*}
Mathematica [A] time = 0.10271, size = 61, normalized size = 1.27 \[ -\frac{\log \left (e^{2 c (a+b x)}+1\right )+e^{c (a+b x)} \tan ^{-1}\left (\frac{1}{2} e^{-c (a+b x)}-\frac{1}{2} e^{c (a+b x)}\right )}{b c} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[E^(c*(a + b*x))*ArcTan[Sinh[a*c + b*c*x]],x]
[Out]
-((E^(c*(a + b*x))*ArcTan[1/(2*E^(c*(a + b*x))) - E^(c*(a + b*x))/2] + Log[1 + E^(2*c*(a + b*x))])/(b*c))
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Maple [C] time = 0.671, size = 1299, normalized size = 27.1 \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(sinh(b*c*x+a*c)),x)
[Out]
2*a/b+1/2/c/b*exp(c*(b*x+a))*Pi-I/c/b*exp(c*(b*x+a))*ln(exp(c*(b*x+a))-I)+I/c/b*exp(c*(b*x+a))*ln(exp(c*(b*x+a
))+I)-ln(1+exp(2*c*(b*x+a)))/b/c+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))-1/4/c/b*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/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(-c*(b*x+a))*(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(c*(b*x+a))-I)^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)))*cs
gn(I*exp(-c*(b*x+a))*(exp(c*(b*x+a))-I)^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)))*csgn(I*exp(-c*(b*x+a))*(exp(c*(b*x+a))+I)^2)*exp(c*(b*x+a))+1/4/c/b*Pi*csgn(exp(-c*(b*x+a))*(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*ex
p(-c*(b*x+a))*(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))*(exp(c*(b*x+a))-I)^2)^3*exp(c*(b*x+a))+1/4/c/b*Pi*csgn(exp(-c*(b*x+a))*(exp(
c*(b*x+a))-I)^2)^3*exp(c*(b*x+a))-1/4/c/b*Pi*csgn(exp(-c*(b*x+a))*(exp(c*(b*x+a))+I)^2)^2*exp(c*(b*x+a))-1/4/c
/b*Pi*csgn(exp(-c*(b*x+a))*(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)))*csgn(I*ex
p(-c*(b*x+a))*(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))*(exp(c*(b*x+a))-I)^2)*c
sgn(exp(-c*(b*x+a))*(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))*(exp(c*(b*x+a))+I
)^2)*csgn(exp(-c*(b*x+a))*(exp(c*(b*x+a))+I)^2)*exp(c*(b*x+a))-1/4/c/b*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)*exp(c*(b*x+a))-1/4/c/b*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))-1/2/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(-c*(b*x+a))*
(exp(c*(b*x+a))+I)^2)^2*exp(c*(b*x+a))
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Maxima [A] time = 1.51609, size = 65, normalized size = 1.35 \begin{align*} \frac{\arctan \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(c*(b*x+a))*arctan(sinh(b*c*x+a*c)),x, algorithm="maxima")
[Out]
arctan(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)
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Fricas [A] time = 1.9274, size = 192, normalized size = 4. \begin{align*} \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} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(c*(b*x+a))*arctan(sinh(b*c*x+a*c)),x, algorithm="fricas")
[Out]
((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)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(exp(c*(b*x+a))*atan(sinh(b*c*x+a*c)),x)
[Out]
Timed out
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Giac [A] time = 1.11608, size = 88, normalized size = 1.83 \begin{align*} \frac{{\left (\arctan \left (\frac{1}{2} \, e^{\left (b c x + a c\right )} - \frac{1}{2} \, 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(sinh(b*c*x+a*c)),x, algorithm="giac")
[Out]
(arctan(1/2*e^(b*c*x + a*c) - 1/2*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)