3.703 \(\int \sin ^{-1}(\sinh (x)) \text{sech}^4(x) \, dx\)

Optimal. Leaf size=49 \[ -\frac{2}{3} \sin ^{-1}\left (\frac{\cosh (x)}{\sqrt{2}}\right )+\frac{1}{6} \sqrt{1-\sinh ^2(x)} \text{sech}(x)-\frac{1}{3} \tanh ^3(x) \sin ^{-1}(\sinh (x))+\tanh (x) \sin ^{-1}(\sinh (x)) \]

[Out]

(-2*ArcSin[Cosh[x]/Sqrt[2]])/3 + (Sech[x]*Sqrt[1 - Sinh[x]^2])/6 + ArcSin[Sinh[x]]*Tanh[x] - (ArcSin[Sinh[x]]*
Tanh[x]^3)/3

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Rubi [A]  time = 0.140496, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3767, 4844, 12, 4357, 451, 216} \[ \frac{1}{6} \sqrt{2-\cosh ^2(x)} \text{sech}(x)-\frac{2}{3} \sin ^{-1}\left (\frac{\cosh (x)}{\sqrt{2}}\right )-\frac{1}{3} \tanh ^3(x) \sin ^{-1}(\sinh (x))+\tanh (x) \sin ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[Sinh[x]]*Sech[x]^4,x]

[Out]

(-2*ArcSin[Cosh[x]/Sqrt[2]])/3 + (Sqrt[2 - Cosh[x]^2]*Sech[x])/6 + ArcSin[Sinh[x]]*Tanh[x] - (ArcSin[Sinh[x]]*
Tanh[x]^3)/3

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4357

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rule 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \sin ^{-1}(\sinh (x)) \text{sech}^4(x) \, dx &=\sin ^{-1}(\sinh (x)) \tanh (x)-\frac{1}{3} \sin ^{-1}(\sinh (x)) \tanh ^3(x)-\int \frac{(2+\cosh (2 x)) \text{sech}(x) \tanh (x)}{3 \sqrt{1-\sinh ^2(x)}} \, dx\\ &=\sin ^{-1}(\sinh (x)) \tanh (x)-\frac{1}{3} \sin ^{-1}(\sinh (x)) \tanh ^3(x)-\frac{1}{3} \int \frac{(2+\cosh (2 x)) \text{sech}(x) \tanh (x)}{\sqrt{1-\sinh ^2(x)}} \, dx\\ &=\sin ^{-1}(\sinh (x)) \tanh (x)-\frac{1}{3} \sin ^{-1}(\sinh (x)) \tanh ^3(x)-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1+2 x^2}{x^2 \sqrt{2-x^2}} \, dx,x,\cosh (x)\right )\\ &=\frac{1}{6} \sqrt{2-\cosh ^2(x)} \text{sech}(x)+\sin ^{-1}(\sinh (x)) \tanh (x)-\frac{1}{3} \sin ^{-1}(\sinh (x)) \tanh ^3(x)-\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-x^2}} \, dx,x,\cosh (x)\right )\\ &=-\frac{2}{3} \sin ^{-1}\left (\frac{\cosh (x)}{\sqrt{2}}\right )+\frac{1}{6} \sqrt{2-\cosh ^2(x)} \text{sech}(x)+\sin ^{-1}(\sinh (x)) \tanh (x)-\frac{1}{3} \sin ^{-1}(\sinh (x)) \tanh ^3(x)\\ \end{align*}

Mathematica [C]  time = 0.228016, size = 66, normalized size = 1.35 \[ \frac{1}{12} \left (8 i \log \left (\sqrt{3-\cosh (2 x)}+i \sqrt{2} \cosh (x)\right )+\sqrt{6-2 \cosh (2 x)} \text{sech}(x)+4 (\cosh (2 x)+2) \tanh (x) \text{sech}^2(x) \sin ^{-1}(\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[Sinh[x]]*Sech[x]^4,x]

[Out]

((8*I)*Log[I*Sqrt[2]*Cosh[x] + Sqrt[3 - Cosh[2*x]]] + Sqrt[6 - 2*Cosh[2*x]]*Sech[x] + 4*ArcSin[Sinh[x]]*(2 + C
osh[2*x])*Sech[x]^2*Tanh[x])/12

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int \arcsin \left ( \sinh \left ( x \right ) \right ) \left ({\rm sech} \left (x\right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(sinh(x))*sech(x)^4,x)

[Out]

int(arcsin(sinh(x))*sech(x)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{4 \,{\left (3 \, e^{\left (2 \, x\right )} + 1\right )} \arctan \left (e^{\left (2 \, x\right )} - 1, \sqrt{e^{\left (2 \, x\right )} + 2 \, e^{x} - 1} \sqrt{-e^{\left (2 \, x\right )} + 2 \, e^{x} + 1}\right ) + 16 \,{\left (e^{\left (6 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} + 1\right )} \int -\frac{{\left (3 \, e^{\left (4 \, x\right )} + e^{\left (2 \, x\right )}\right )} e^{\left (\frac{1}{2} \, \log \left (e^{\left (2 \, x\right )} + 2 \, e^{x} - 1\right ) + \frac{1}{2} \, \log \left (-e^{\left (2 \, x\right )} + 2 \, e^{x} + 1\right )\right )}}{{\left (e^{\left (8 \, x\right )} - 4 \, e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} - 4 \, e^{\left (2 \, x\right )} + 1\right )}{\left (e^{\left (2 \, x\right )} + 2 \, e^{x} - 1\right )}{\left (e^{\left (2 \, x\right )} - 2 \, e^{x} - 1\right )} - e^{\left (12 \, x\right )} + 6 \, e^{\left (10 \, x\right )} + e^{\left (8 \, x\right )} - 12 \, e^{\left (6 \, x\right )} + e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} - 1}\,{d x}}{3 \,{\left (e^{\left (6 \, x\right )} + 3 \, e^{\left (4 \, x\right )} + 3 \, e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(sinh(x))*sech(x)^4,x, algorithm="maxima")

[Out]

-1/3*(4*(3*e^(2*x) + 1)*arctan2(e^(2*x) - 1, sqrt(e^(2*x) + 2*e^x - 1)*sqrt(-e^(2*x) + 2*e^x + 1)) + 3*(e^(6*x
) + 3*e^(4*x) + 3*e^(2*x) + 1)*integrate(16/3*(3*e^(4*x) + e^(2*x))*e^(1/2*log(e^(2*x) + 2*e^x - 1) + 1/2*log(
-e^(2*x) + 2*e^x + 1))/((e^(8*x) - 4*e^(6*x) - 10*e^(4*x) - 4*e^(2*x) + 1)*e^(log(e^(2*x) + 2*e^x - 1) + log(-
e^(2*x) + 2*e^x + 1)) + e^(12*x) - 6*e^(10*x) - e^(8*x) + 12*e^(6*x) - e^(4*x) - 6*e^(2*x) + 1), x))/(e^(6*x)
+ 3*e^(4*x) + 3*e^(2*x) + 1)

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Fricas [B]  time = 2.55435, size = 1785, normalized size = 36.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(sinh(x))*sech(x)^4,x, algorithm="fricas")

[Out]

1/6*(sqrt(2)*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(c
osh(x)^3 + cosh(x))*sinh(x) + 1)*sqrt(-(cosh(x)^2 + sinh(x)^2 - 3)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)
) - 4*(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x)^2 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(5*cosh(x
)^3 + 3*cosh(x))*sinh(x)^3 + 3*(5*cosh(x)^4 + 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 + 2*cosh
(x)^3 + cosh(x))*sinh(x) + 1)*arctan(sqrt(2)*(3*cosh(x)^2 + 6*cosh(x)*sinh(x) + 3*sinh(x)^2 - 1)*sqrt(-(cosh(x
)^2 + sinh(x)^2 - 3)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4
+ 6*(cosh(x)^2 - 1)*sinh(x)^2 - 6*cosh(x)^2 + 4*(cosh(x)^3 - 3*cosh(x))*sinh(x) + 1)) + 8*(3*cosh(x)^2 + 6*co
sh(x)*sinh(x) + 3*sinh(x)^2 + 1)*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-(cosh(x)
^2 + sinh(x)^2 - 3)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4
+ 6*(cosh(x)^2 - 1)*sinh(x)^2 - 6*cosh(x)^2 + 4*(cosh(x)^3 - 3*cosh(x))*sinh(x) + 1)))/(cosh(x)^6 + 6*cosh(x)*
sinh(x)^5 + sinh(x)^6 + 3*(5*cosh(x)^2 + 1)*sinh(x)^4 + 3*cosh(x)^4 + 4*(5*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 +
3*(5*cosh(x)^4 + 6*cosh(x)^2 + 1)*sinh(x)^2 + 3*cosh(x)^2 + 6*(cosh(x)^5 + 2*cosh(x)^3 + cosh(x))*sinh(x) + 1)

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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(asin(sinh(x))*sech(x)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \arcsin \left (\sinh \left (x\right )\right ) \operatorname{sech}\left (x\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(sinh(x))*sech(x)^4,x, algorithm="giac")

[Out]

integrate(arcsin(sinh(x))*sech(x)^4, x)