Integrand size = 10, antiderivative size = 172 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {1-x}{18 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{7/2}}+\frac {5 (1-x)}{72 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{5/2}}+\frac {5 (1-x)}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5 \sqrt {1-x} \text {arctanh}\left (\sqrt {1-x}\right )}{48 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}} \]
-1/3*arcsech(x^(1/2))/x^3+1/18*(1-x)/x^(7/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^( 1/2))^(1/2)+5/72*(1-x)/x^(5/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+5/ 48*(1-x)/x^(3/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+5/48*arctanh((1- x)^(1/2))*(1-x)^(1/2)/x^(1/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)
Time = 0.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.81 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}} \left (8+8 \sqrt {x}+10 x+10 x^{3/2}+15 x^2+15 x^{5/2}\right )-48 \text {sech}^{-1}\left (\sqrt {x}\right )+15 x^3 \log \left (1+\sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}}+\sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}} \sqrt {x}\right )-\frac {15}{2} x^3 \log (x)}{144 x^3} \]
(Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(8 + 8*Sqrt[x] + 10*x + 10*x^(3/2) + 15 *x^2 + 15*x^(5/2)) - 48*ArcSech[Sqrt[x]] + 15*x^3*Log[1 + Sqrt[(1 - Sqrt[x ])/(1 + Sqrt[x])] + Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*Sqrt[x]] - (15*x^3*L og[x])/2)/(144*x^3)
Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.72, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6899, 27, 52, 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.
| \(\displaystyle \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx\) |
| \(\Big \downarrow \) 6899 |
| \(\displaystyle -\frac {\sqrt {1-x} \int \frac {1}{2 \sqrt {1-x} x^4}dx}{3 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\sqrt {1-x} \int \frac {1}{\sqrt {1-x} x^4}dx}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {\sqrt {1-x} \left (\frac {5}{6} \int \frac {1}{\sqrt {1-x} x^3}dx-\frac {\sqrt {1-x}}{3 x^3}\right )}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {\sqrt {1-x} \left (\frac {5}{6} \left (\frac {3}{4} \int \frac {1}{\sqrt {1-x} x^2}dx-\frac {\sqrt {1-x}}{2 x^2}\right )-\frac {\sqrt {1-x}}{3 x^3}\right )}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {\sqrt {1-x} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-x} x}dx-\frac {\sqrt {1-x}}{x}\right )-\frac {\sqrt {1-x}}{2 x^2}\right )-\frac {\sqrt {1-x}}{3 x^3}\right )}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {\sqrt {1-x} \left (\frac {5}{6} \left (\frac {3}{4} \left (-\int \frac {1}{x}d\sqrt {1-x}-\frac {\sqrt {1-x}}{x}\right )-\frac {\sqrt {1-x}}{2 x^2}\right )-\frac {\sqrt {1-x}}{3 x^3}\right )}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\sqrt {1-x} \left (\frac {5}{6} \left (\frac {3}{4} \left (-\text {arctanh}\left (\sqrt {1-x}\right )-\frac {\sqrt {1-x}}{x}\right )-\frac {\sqrt {1-x}}{2 x^2}\right )-\frac {\sqrt {1-x}}{3 x^3}\right )}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
-1/3*ArcSech[Sqrt[x]]/x^3 - (Sqrt[1 - x]*(-1/3*Sqrt[1 - x]/x^3 + (5*(-1/2* Sqrt[1 - x]/x^2 + (3*(-(Sqrt[1 - x]/x) - ArcTanh[Sqrt[1 - x]]))/4))/6))/(6 *Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x])
3.1.27.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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])
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]
Time = 0.23 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.53
| method | result | size |
| derivativedivides | \(-\frac {\operatorname {arcsech}\left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (15 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1-x}}\right ) x^{3}+15 \sqrt {1-x}\, x^{2}+10 \sqrt {1-x}\, x +8 \sqrt {1-x}\right )}{144 x^{\frac {5}{2}} \sqrt {1-x}}\) | \(91\) |
| default | \(-\frac {\operatorname {arcsech}\left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (15 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1-x}}\right ) x^{3}+15 \sqrt {1-x}\, x^{2}+10 \sqrt {1-x}\, x +8 \sqrt {1-x}\right )}{144 x^{\frac {5}{2}} \sqrt {1-x}}\) | \(91\) |
| parts | \(-\frac {\operatorname {arcsech}\left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (15 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1-x}}\right ) x^{3}+15 \sqrt {1-x}\, x^{2}+10 \sqrt {1-x}\, x +8 \sqrt {1-x}\right )}{144 x^{\frac {5}{2}} \sqrt {1-x}}\) | \(91\) |
-1/3*arcsech(x^(1/2))/x^3+1/144*(-(x^(1/2)-1)/x^(1/2))^(1/2)/x^(5/2)*((x^( 1/2)+1)/x^(1/2))^(1/2)*(15*arctanh(1/(1-x)^(1/2))*x^3+15*(1-x)^(1/2)*x^2+1 0*(1-x)^(1/2)*x+8*(1-x)^(1/2))/(1-x)^(1/2)
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.35 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {{\left (15 \, x^{2} + 10 \, x + 8\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} + 3 \, {\left (5 \, x^{3} - 16\right )} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right )}{144 \, x^{3}} \]
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
\[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\operatorname {asech}{\left (\sqrt {x} \right )}}{x^{4}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.67 \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=-\frac {15 \, x^{\frac {5}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {5}{2}} - 40 \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {x} \sqrt {\frac {1}{x} - 1}}{144 \, {\left (x^{3} {\left (\frac {1}{x} - 1\right )}^{3} - 3 \, x^{2} {\left (\frac {1}{x} - 1\right )}^{2} + 3 \, x {\left (\frac {1}{x} - 1\right )} - 1\right )}} - \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{3 \, x^{3}} + \frac {5}{96} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} + 1\right ) - \frac {5}{96} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} - 1} - 1\right ) \]
-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*arcsech(sqrt(x))/x^3 + 5/96*log(sqrt(x)*sqrt(1/x - 1) + 1) - 5/96* log(sqrt(x)*sqrt(1/x - 1) - 1)
\[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int { \frac {\operatorname {arsech}\left (\sqrt {x}\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\text {sech}^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right )}{x^4} \,d x \]