\(\int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx\) [703]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [C] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 8, antiderivative size = 49 \[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=-\frac {2}{3} \arcsin \left (\frac {\cosh (x)}{\sqrt {2}}\right )+\frac {1}{6} \text {sech}(x) \sqrt {1-\sinh ^2(x)}+\arcsin (\sinh (x)) \tanh (x)-\frac {1}{3} \arcsin (\sinh (x)) \tanh ^3(x) \] Output:

-2/3*arcsin(1/2*cosh(x)*2^(1/2))+1/6*sech(x)*(1-sinh(x)^2)^(1/2)+arcsin(si 
nh(x))*tanh(x)-1/3*arcsin(sinh(x))*tanh(x)^3
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.35 \[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=\frac {1}{12} \left (8 i \log \left (i \sqrt {2} \cosh (x)+\sqrt {3-\cosh (2 x)}\right )+\sqrt {6-2 \cosh (2 x)} \text {sech}(x)+4 \arcsin (\sinh (x)) (2+\cosh (2 x)) \text {sech}^2(x) \tanh (x)\right ) \] Input:

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

Output:

((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 + Cosh[2*x])*Sech[x]^2*Tanh[x])/12
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5343, 27, 3042, 26, 4857, 358, 223}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \text {sech}^4(x) \arcsin (\sinh (x)) \, dx\)

\(\Big \downarrow \) 5343

\(\displaystyle -\int \frac {(\cosh (2 x)+2) \text {sech}(x) \tanh (x)}{3 \sqrt {1-\sinh ^2(x)}}dx-\frac {1}{3} \tanh ^3(x) \arcsin (\sinh (x))+\tanh (x) \arcsin (\sinh (x))\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{3} \int \frac {(\cosh (2 x)+2) \text {sech}(x) \tanh (x)}{\sqrt {1-\sinh ^2(x)}}dx-\frac {1}{3} \tanh ^3(x) \arcsin (\sinh (x))+\tanh (x) \arcsin (\sinh (x))\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{3} \int -\frac {i (\cos (2 i x)+2) \sin (i x)}{\cos (i x)^2 \sqrt {\sin (i x)^2+1}}dx-\frac {1}{3} \tanh ^3(x) \arcsin (\sinh (x))+\tanh (x) \arcsin (\sinh (x))\)

\(\Big \downarrow \) 26

\(\displaystyle \frac {1}{3} i \int \frac {(\cos (2 i x)+2) \sin (i x)}{\cos (i x)^2 \sqrt {\sin (i x)^2+1}}dx-\frac {1}{3} \tanh ^3(x) \arcsin (\sinh (x))+\tanh (x) \arcsin (\sinh (x))\)

\(\Big \downarrow \) 4857

\(\displaystyle -\frac {1}{3} \int \frac {\left (2 \cosh ^2(x)+1\right ) \text {sech}^2(x)}{\sqrt {2-\cosh ^2(x)}}d\cosh (x)-\frac {1}{3} \tanh ^3(x) \arcsin (\sinh (x))+\tanh (x) \arcsin (\sinh (x))\)

\(\Big \downarrow \) 358

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \sqrt {2-\cosh ^2(x)} \text {sech}(x)-2 \int \frac {1}{\sqrt {2-\cosh ^2(x)}}d\cosh (x)\right )-\frac {1}{3} \tanh ^3(x) \arcsin (\sinh (x))+\tanh (x) \arcsin (\sinh (x))\)

\(\Big \downarrow \) 223

\(\displaystyle -\frac {1}{3} \tanh ^3(x) \arcsin (\sinh (x))+\tanh (x) \arcsin (\sinh (x))+\frac {1}{3} \left (\frac {1}{2} \sqrt {2-\cosh ^2(x)} \text {sech}(x)-2 \arcsin \left (\frac {\cosh (x)}{\sqrt {2}}\right )\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 223
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]
 

rule 358
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x_ 
Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + S 
imp[d/e^2   Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, 
 -1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4857
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto 
rs[Cos[c*(a + b*x)], x]}, Simp[-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 5343
Int[((a_.) + ArcSin[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, 
 Simp[(a + b*ArcSin[u])   w, x] - Simp[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]]
 
Maple [F]

\[\int \arcsin \left (\sinh \left (x \right )\right ) \operatorname {sech}\left (x \right )^{4}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 519 vs. \(2 (40) = 80\).

Time = 0.10 (sec) , antiderivative size = 519, normalized size of antiderivative = 10.59 \[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=\text {Too large to display} \] Input:

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

Output:

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*(cosh(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)*sin 
h(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*cosh(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 + sin 
h(x)^2 - 3)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(cosh(x)^4 + 4*co 
sh(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)
 

Sympy [F]

\[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=\int \operatorname {asin}{\left (\sinh {\left (x \right )} \right )} \operatorname {sech}^{4}{\left (x \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=\int { \arcsin \left (\sinh \left (x\right )\right ) \operatorname {sech}\left (x\right )^{4} \,d x } \] Input:

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

Output:

-1/3*(4*(3*e^(2*x) + 1)*arctan2(e^(2*x) - 1, sqrt(e^(2*x) + 2*e^x - 1)*sqr 
t(-e^(2*x) + 2*e^x + 1)) + 3*(e^(6*x) + 3*e^(4*x) + 3*e^(2*x) + 1)*integra 
te(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)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 218, normalized size of antiderivative = 4.45 \[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=-\frac {16 \, {\left (-8 i \, \sqrt {2} \arctan \left (-i\right ) - 3 \, \sqrt {2} + 32 \, \arctan \left (-i\right ) - 3 i\right )}}{96 i \, \sqrt {2} - 384} + \frac {\sqrt {2} + \frac {2 \, \sqrt {2} - \sqrt {-e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} - 1}}{e^{\left (2 \, x\right )} - 3}}{6 \, {\left (\frac {\sqrt {2} {\left (2 \, \sqrt {2} - \sqrt {-e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} - 1}\right )}}{e^{\left (2 \, x\right )} - 3} + \frac {{\left (2 \, \sqrt {2} - \sqrt {-e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} - 1}\right )}^{2}}{{\left (e^{\left (2 \, x\right )} - 3\right )}^{2}} + 1\right )}} - \frac {4 \, {\left (3 \, e^{\left (2 \, x\right )} + 1\right )} \arcsin \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )}{3 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} - \frac {4}{3} \, \arctan \left (-2 \, \sqrt {2} - \frac {3 \, {\left (2 \, \sqrt {2} - \sqrt {-e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} - 1}\right )}}{e^{\left (2 \, x\right )} - 3}\right ) \] Input:

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

Output:

-16*(-8*I*sqrt(2)*arctan(-I) - 3*sqrt(2) + 32*arctan(-I) - 3*I)/(96*I*sqrt 
(2) - 384) + 1/6*(sqrt(2) + (2*sqrt(2) - sqrt(-e^(4*x) + 6*e^(2*x) - 1))/( 
e^(2*x) - 3))/(sqrt(2)*(2*sqrt(2) - sqrt(-e^(4*x) + 6*e^(2*x) - 1))/(e^(2* 
x) - 3) + (2*sqrt(2) - sqrt(-e^(4*x) + 6*e^(2*x) - 1))^2/(e^(2*x) - 3)^2 + 
 1) - 4/3*(3*e^(2*x) + 1)*arcsin(1/2*(e^(2*x) - 1)*e^(-x))/(e^(2*x) + 1)^3 
 - 4/3*arctan(-2*sqrt(2) - 3*(2*sqrt(2) - sqrt(-e^(4*x) + 6*e^(2*x) - 1))/ 
(e^(2*x) - 3))
 

Mupad [F(-1)]

Timed out. \[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=\int \frac {\mathrm {asin}\left (\mathrm {sinh}\left (x\right )\right )}{{\mathrm {cosh}\left (x\right )}^4} \,d x \] Input:

int(asin(sinh(x))/cosh(x)^4,x)
 

Output:

int(asin(sinh(x))/cosh(x)^4, x)
 

Reduce [F]

\[ \int \arcsin (\sinh (x)) \text {sech}^4(x) \, dx=\int \mathit {asin} \left (\sinh \left (x \right )\right ) \mathrm {sech}\left (x \right )^{4}d x \] Input:

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

Output:

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