\(\int \frac {\text {sech}^{-1}(\sqrt {x})}{x^3} \, dx\) [26]

Optimal result

Integrand size = 10, antiderivative size = 126 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {\sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}}}{8 x^{3/2}}+\frac {3 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}}}{16 \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {3 \sqrt {1-x} \text {arctanh}\left (\sqrt {1-x}\right )}{16 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}} \] Output:

1/8*(-1+1/x^(1/2))^(1/2)*(1/x^(1/2)+1)^(1/2)/x^(3/2)+3/16*(-1+1/x^(1/2))^( 
1/2)*(1/x^(1/2)+1)^(1/2)/x^(1/2)-1/2*arcsech(x^(1/2))/x^2+3/16*(1-x)^(1/2) 
*arctanh((1-x)^(1/2))/(-1+1/x^(1/2))^(1/2)/(1/x^(1/2)+1)^(1/2)/x^(1/2)
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.99 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {1}{16} \left (\frac {\sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}} \left (2+2 \sqrt {x}+3 x+3 x^{3/2}\right )}{x^2}-\frac {8 \text {sech}^{-1}\left (\sqrt {x}\right )}{x^2}+3 \log \left (1+\sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}}+\sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}} \sqrt {x}\right )-\frac {3 \log (x)}{2}\right ) \] Input:

Integrate[ArcSech[Sqrt[x]]/x^3,x]
 

Output:

((Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(2 + 2*Sqrt[x] + 3*x + 3*x^(3/2)))/x^2 
 - (8*ArcSech[Sqrt[x]])/x^2 + 3*Log[1 + Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])] 
+ Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*Sqrt[x]] - (3*Log[x])/2)/16
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.81, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6899, 27, 52, 52, 73, 219}

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.

\(\displaystyle \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx\)

\(\Big \downarrow \) 6899

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\sqrt {1-x} \left (\frac {3}{4} \left (-\text {arctanh}\left (\sqrt {1-x}\right )-\frac {\sqrt {1-x}}{x}\right )-\frac {\sqrt {1-x}}{2 x^2}\right )}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{2 x^2}\)

Input:

Int[ArcSech[Sqrt[x]]/x^3,x]
 

Output:

-1/2*ArcSech[Sqrt[x]]/x^2 - (Sqrt[1 - x]*(-1/2*Sqrt[1 - x]/x^2 + (3*(-(Sqr 
t[1 - x]/x) - ArcTanh[Sqrt[1 - x]]))/4))/(4*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 
1/Sqrt[x]]*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 219

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

rule 6899

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(-\frac {\operatorname {arcsech}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1-x}}\right ) x^{2}+3 \sqrt {1-x}\, x +2 \sqrt {1-x}\right )}{16 x^{\frac {3}{2}} \sqrt {1-x}}\)	\(79\)
default	\(-\frac {\operatorname {arcsech}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1-x}}\right ) x^{2}+3 \sqrt {1-x}\, x +2 \sqrt {1-x}\right )}{16 x^{\frac {3}{2}} \sqrt {1-x}}\)	\(79\)
parts	\(-\frac {\operatorname {arcsech}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1-x}}\right ) x^{2}+3 \sqrt {1-x}\, x +2 \sqrt {1-x}\right )}{16 x^{\frac {3}{2}} \sqrt {1-x}}\)	\(79\)

Input:

int(arcsech(x^(1/2))/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*arcsech(x^(1/2))/x^2+1/16*(-(x^(1/2)-1)/x^(1/2))^(1/2)/x^(3/2)*((x^(1 
/2)+1)/x^(1/2))^(1/2)*(3*arctanh(1/(1-x)^(1/2))*x^2+3*(1-x)^(1/2)*x+2*(1-x 
)^(1/2))/(1-x)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.43 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {{\left (3 \, x + 2\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} + {\left (3 \, x^{2} - 8\right )} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right )}{16 \, x^{2}} \] Input:

integrate(arcsech(x^(1/2))/x^3,x, algorithm="fricas")
 

Output:

1/16*((3*x + 2)*sqrt(x)*sqrt(-(x - 1)/x) + (3*x^2 - 8)*log((x*sqrt(-(x - 1 
)/x) + sqrt(x))/x))/x^2
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\int \frac {\operatorname {asech}{\left (\sqrt {x} \right )}}{x^{3}}\, dx \] Input:

integrate(asech(x**(1/2))/x**3,x)
 

Output:

Integral(asech(sqrt(x))/x**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.73 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=-\frac {3 \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} - 5 \, \sqrt {x} \sqrt {\frac {1}{x} - 1}}{16 \, {\left (x^{2} {\left (\frac {1}{x} - 1\right )}^{2} - 2 \, x {\left (\frac {1}{x} - 1\right )} + 1\right )}} - \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{2 \, x^{2}} + \frac {3}{32} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} + 1\right ) - \frac {3}{32} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} - 1\right ) \] Input:

integrate(arcsech(x^(1/2))/x^3,x, algorithm="maxima")
 

Output:

-1/16*(3*x^(3/2)*(1/x - 1)^(3/2) - 5*sqrt(x)*sqrt(1/x - 1))/(x^2*(1/x - 1) 
^2 - 2*x*(1/x - 1) + 1) - 1/2*arcsech(sqrt(x))/x^2 + 3/32*log(sqrt(x)*sqrt 
(1/x - 1) + 1) - 3/32*log(sqrt(x)*sqrt(1/x - 1) - 1)
 

Giac [F]

\[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\int { \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x^{3}} \,d x } \] Input:

integrate(arcsech(x^(1/2))/x^3,x, algorithm="giac")
 

Output:

integrate(arcsech(sqrt(x))/x^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\int \frac {\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right )}{x^3} \,d x \] Input:

int(acosh(1/x^(1/2))/x^3,x)
 

Output:

int(acosh(1/x^(1/2))/x^3, x)
 

Reduce [F]

\[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\int \frac {\mathit {asech} \left (\sqrt {x}\right )}{x^{3}}d x \] Input:

int(asech(x^(1/2))/x^3,x)
 

Output:

int(asech(sqrt(x))/x**3,x)