3.705 \(\int e^x \sin ^{-1}(\tanh (x)) \, dx\)

Optimal. Leaf size=28 \[ e^x \sin ^{-1}(\tanh (x))-\log \left (e^{2 x}+1\right ) \cosh (x) \sqrt{\text{sech}^2(x)} \]

[Out]

E^x*ArcSin[Tanh[x]] - Cosh[x]*Log[1 + E^(2*x)]*Sqrt[Sech[x]^2]

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Rubi [A]  time = 0.0769321, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {2194, 4844, 6720, 2282, 12, 260} \[ e^x \sin ^{-1}(\tanh (x))-\log \left (e^{2 x}+1\right ) \cosh (x) \sqrt{\text{sech}^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[E^x*ArcSin[Tanh[x]],x]

[Out]

E^x*ArcSin[Tanh[x]] - Cosh[x]*Log[1 + E^(2*x)]*Sqrt[Sech[x]^2]

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 4844

Int[((a_.) + ArcSin[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcSin[u], w, x] - Dist
[b, Int[SimplifyIntegrand[(w*D[u, x])/Sqrt[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]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

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^x \sin ^{-1}(\tanh (x)) \, dx &=e^x \sin ^{-1}(\tanh (x))-\int e^x \sqrt{\text{sech}^2(x)} \, dx\\ &=e^x \sin ^{-1}(\tanh (x))-\left (\cosh (x) \sqrt{\text{sech}^2(x)}\right ) \int e^x \text{sech}(x) \, dx\\ &=e^x \sin ^{-1}(\tanh (x))-\left (\cosh (x) \sqrt{\text{sech}^2(x)}\right ) \operatorname{Subst}\left (\int \frac{2 x}{1+x^2} \, dx,x,e^x\right )\\ &=e^x \sin ^{-1}(\tanh (x))-\left (2 \cosh (x) \sqrt{\text{sech}^2(x)}\right ) \operatorname{Subst}\left (\int \frac{x}{1+x^2} \, dx,x,e^x\right )\\ &=e^x \sin ^{-1}(\tanh (x))-\cosh (x) \log \left (1+e^{2 x}\right ) \sqrt{\text{sech}^2(x)}\\ \end{align*}

Mathematica [B]  time = 0.837499, size = 64, normalized size = 2.29 \[ e^x \sin ^{-1}\left (\frac{e^{2 x}-1}{e^{2 x}+1}\right )-e^{-x} \sqrt{\frac{e^{2 x}}{\left (e^{2 x}+1\right )^2}} \left (e^{2 x}+1\right ) \log \left (e^{2 x}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^x*ArcSin[Tanh[x]],x]

[Out]

E^x*ArcSin[(-1 + E^(2*x))/(1 + E^(2*x))] - (Sqrt[E^(2*x)/(1 + E^(2*x))^2]*(1 + E^(2*x))*Log[1 + E^(2*x)])/E^x

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{x}}\arcsin \left ( \tanh \left ( x \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*arcsin(tanh(x)),x)

[Out]

int(exp(x)*arcsin(tanh(x)),x)

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Maxima [A]  time = 1.64726, size = 22, normalized size = 0.79 \begin{align*} \arcsin \left (\tanh \left (x\right )\right ) e^{x} - \log \left (e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*arcsin(tanh(x)),x, algorithm="maxima")

[Out]

arcsin(tanh(x))*e^x - log(e^(2*x) + 1)

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Fricas [A]  time = 2.21292, size = 100, normalized size = 3.57 \begin{align*}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\sinh \left (x\right )\right ) - \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*arcsin(tanh(x)),x, algorithm="fricas")

[Out]

(cosh(x) + sinh(x))*arctan(sinh(x)) - log(2*cosh(x)/(cosh(x) - sinh(x)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \operatorname{asin}{\left (\tanh{\left (x \right )} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*asin(tanh(x)),x)

[Out]

Integral(exp(x)*asin(tanh(x)), x)

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Giac [A]  time = 1.07862, size = 51, normalized size = 1.82 \begin{align*} \arcsin \left (\frac{e^{\left (2 \, x\right )}}{e^{\left (2 \, x\right )} + 1} - \frac{1}{e^{\left (2 \, x\right )} + 1}\right ) e^{x} - \log \left (e^{\left (2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*arcsin(tanh(x)),x, algorithm="giac")

[Out]

arcsin(e^(2*x)/(e^(2*x) + 1) - 1/(e^(2*x) + 1))*e^x - log(e^(2*x) + 1)