\(\int \frac {\text {csch}^{-1}(\sqrt {x})}{x^4} \, dx\) [21]

Optimal result

Integrand size = 10, antiderivative size = 115 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {-1-x}}{18 \sqrt {-x} x^{5/2}}-\frac {5 \sqrt {-1-x}}{72 \sqrt {-x} x^{3/2}}+\frac {5 \sqrt {-1-x}}{48 \sqrt {-x} \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {5 \sqrt {x} \arctan \left (\sqrt {-1-x}\right )}{48 \sqrt {-x}} \] Output:

1/18*(-1-x)^(1/2)/(-x)^(1/2)/x^(5/2)-5/72*(-1-x)^(1/2)/(-x)^(1/2)/x^(3/2)+ 
5/48*(-1-x)^(1/2)/(-x)^(1/2)/x^(1/2)-1/3*arccsch(x^(1/2))/x^3-5/48*x^(1/2) 
*arctan((-1-x)^(1/2))/(-x)^(1/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.45 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {1+\frac {1}{x}} \sqrt {x} \left (8-10 x+15 x^2\right )-48 \text {csch}^{-1}\left (\sqrt {x}\right )-15 x^3 \text {arcsinh}\left (\frac {1}{\sqrt {x}}\right )}{144 x^3} \] Input:

Integrate[ArcCsch[Sqrt[x]]/x^4,x]
 

Output:

(Sqrt[1 + x^(-1)]*Sqrt[x]*(8 - 10*x + 15*x^2) - 48*ArcCsch[Sqrt[x]] - 15*x 
^3*ArcSinh[1/Sqrt[x]])/(144*x^3)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.85, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6900, 27, 52, 52, 52, 73, 217}

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.

\(\displaystyle \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx\)

\(\Big \downarrow \) 6900

\(\displaystyle \frac {\sqrt {x} \int \frac {1}{2 \sqrt {-x-1} x^4}dx}{3 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {x} \int \frac {1}{\sqrt {-x-1} x^4}dx}{6 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{3 x^3}-\frac {5}{6} \int \frac {1}{\sqrt {-x-1} x^3}dx\right )}{6 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{3 x^3}-\frac {5}{6} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \int \frac {1}{\sqrt {-x-1} x^2}dx\right )\right )}{6 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{3 x^3}-\frac {5}{6} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \left (\frac {\sqrt {-x-1}}{x}-\frac {1}{2} \int \frac {1}{\sqrt {-x-1} x}dx\right )\right )\right )}{6 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{3 x^3}-\frac {5}{6} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \left (\int \frac {1}{x}d\sqrt {-x-1}+\frac {\sqrt {-x-1}}{x}\right )\right )\right )}{6 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{3 x^3}-\frac {5}{6} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \left (\frac {\sqrt {-x-1}}{x}-\arctan \left (\sqrt {-x-1}\right )\right )\right )\right )}{6 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{3 x^3}\)

Input:

Int[ArcCsch[Sqrt[x]]/x^4,x]
 

Output:

-1/3*ArcCsch[Sqrt[x]]/x^3 + (Sqrt[x]*(Sqrt[-1 - x]/(3*x^3) - (5*(Sqrt[-1 - 
 x]/(2*x^2) - (3*(Sqrt[-1 - x]/x - ArcTan[Sqrt[-1 - x]]))/4))/6))/(6*Sqrt[ 
-x])
 

Defintions of rubi rules used

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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 6900

Int[((a_.) + ArcCsch[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*((a + b*ArcCsch[u])/(d*(m + 1))), x] - Simp[b*(u/(d*(m 
 + 1)*Sqrt[-u^2]))   Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/(u*Sq 
rt[-1 - u^2])), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && In 
verseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)^(m + 1), u, x] &&  !Fun 
ctionOfExponentialQ[u, x]
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.58

method	result	size
derivativedivides	\(-\frac {\operatorname {arccsch}\left (\sqrt {x}\right )}{3 x^{3}}-\frac {\sqrt {x +1}\, \left (15 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {x +1}}\right ) x^{3}-15 \sqrt {x +1}\, x^{2}+10 \sqrt {x +1}\, x -8 \sqrt {x +1}\right )}{144 \sqrt {\frac {x +1}{x}}\, x^{\frac {7}{2}}}\)	\(67\)
default	\(-\frac {\operatorname {arccsch}\left (\sqrt {x}\right )}{3 x^{3}}-\frac {\sqrt {x +1}\, \left (15 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {x +1}}\right ) x^{3}-15 \sqrt {x +1}\, x^{2}+10 \sqrt {x +1}\, x -8 \sqrt {x +1}\right )}{144 \sqrt {\frac {x +1}{x}}\, x^{\frac {7}{2}}}\)	\(67\)
parts	\(-\frac {\operatorname {arccsch}\left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {\frac {x +1}{x}}\, \sqrt {x}\, \left (15 \ln \left (\sqrt {x +1}-1\right ) x^{3}-15 \ln \left (\sqrt {x +1}+1\right ) x^{3}+30 \sqrt {x +1}\, x^{2}-20 \sqrt {x +1}\, x +16 \sqrt {x +1}\right )}{288 \sqrt {x +1}\, \left (\sqrt {x +1}-1\right )^{3} \left (\sqrt {x +1}+1\right )^{3}}\)	\(100\)

Input:

int(arccsch(x^(1/2))/x^4,x,method=_RETURNVERBOSE)
 

Output:

-1/3*arccsch(x^(1/2))/x^3-1/144*(x+1)^(1/2)*(15*arctanh(1/(x+1)^(1/2))*x^3 
-15*(x+1)^(1/2)*x^2+10*(x+1)^(1/2)*x-8*(x+1)^(1/2))/((x+1)/x)^(1/2)/x^(7/2 
)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.50 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {{\left (15 \, x^{2} - 10 \, x + 8\right )} \sqrt {x} \sqrt {\frac {x + 1}{x}} - 3 \, {\left (5 \, x^{3} + 16\right )} \log \left (\frac {x \sqrt {\frac {x + 1}{x}} + \sqrt {x}}{x}\right )}{144 \, x^{3}} \] Input:

integrate(arccsch(x^(1/2))/x^4,x, algorithm="fricas")
 

Output:

1/144*((15*x^2 - 10*x + 8)*sqrt(x)*sqrt((x + 1)/x) - 3*(5*x^3 + 16)*log((x 
*sqrt((x + 1)/x) + sqrt(x))/x))/x^3
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\operatorname {acsch}{\left (\sqrt {x} \right )}}{x^{4}}\, dx \] Input:

integrate(acsch(x**(1/2))/x**4,x)
 

Output:

Integral(acsch(sqrt(x))/x**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.01 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {15 \, x^{\frac {5}{2}} {\left (\frac {1}{x} + 1\right )}^{\frac {5}{2}} - 40 \, x^{\frac {3}{2}} {\left (\frac {1}{x} + 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {x} \sqrt {\frac {1}{x} + 1}}{144 \, {\left (x^{3} {\left (\frac {1}{x} + 1\right )}^{3} - 3 \, x^{2} {\left (\frac {1}{x} + 1\right )}^{2} + 3 \, x {\left (\frac {1}{x} + 1\right )} - 1\right )}} - \frac {\operatorname {arcsch}\left (\sqrt {x}\right )}{3 \, x^{3}} - \frac {5}{96} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} + 1\right ) + \frac {5}{96} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} - 1\right ) \] Input:

integrate(arccsch(x^(1/2))/x^4,x, algorithm="maxima")
 

Output:

1/144*(15*x^(5/2)*(1/x + 1)^(5/2) - 40*x^(3/2)*(1/x + 1)^(3/2) + 33*sqrt(x 
)*sqrt(1/x + 1))/(x^3*(1/x + 1)^3 - 3*x^2*(1/x + 1)^2 + 3*x*(1/x + 1) - 1) 
 - 1/3*arccsch(sqrt(x))/x^3 - 5/96*log(sqrt(x)*sqrt(1/x + 1) + 1) + 5/96*l 
og(sqrt(x)*sqrt(1/x + 1) - 1)
 

Giac [F]

\[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int { \frac {\operatorname {arcsch}\left (\sqrt {x}\right )}{x^{4}} \,d x } \] Input:

integrate(arccsch(x^(1/2))/x^4,x, algorithm="giac")
 

Output:

integrate(arccsch(sqrt(x))/x^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\mathrm {asinh}\left (\frac {1}{\sqrt {x}}\right )}{x^4} \,d x \] Input:

int(asinh(1/x^(1/2))/x^4,x)
 

Output:

int(asinh(1/x^(1/2))/x^4, x)
 

Reduce [F]

\[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\mathit {acsch} \left (\sqrt {x}\right )}{x^{4}}d x \] Input:

int(acsch(x^(1/2))/x^4,x)
 

Output:

int(acsch(sqrt(x))/x**4,x)