Integrand size = 10, antiderivative size = 90 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {\sqrt {-1-x}}{8 \sqrt {-x} x^{3/2}}-\frac {3 \sqrt {-1-x}}{16 \sqrt {-x} \sqrt {x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {3 \sqrt {x} \arctan \left (\sqrt {-1-x}\right )}{16 \sqrt {-x}} \]
-1/2*arccsch(x^(1/2))/x^2+1/8*(-1-x)^(1/2)/x^(3/2)/(-x)^(1/2)-3/16*(-1-x)^ (1/2)/(-x)^(1/2)/x^(1/2)+3/16*arctan((-1-x)^(1/2))*x^(1/2)/(-x)^(1/2)
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.52 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\frac {\sqrt {1+\frac {1}{x}} (2-3 x) \sqrt {x}-8 \text {csch}^{-1}\left (\sqrt {x}\right )+3 x^2 \text {arcsinh}\left (\frac {1}{\sqrt {x}}\right )}{16 x^2} \]
(Sqrt[1 + x^(-1)]*(2 - 3*x)*Sqrt[x] - 8*ArcCsch[Sqrt[x]] + 3*x^2*ArcSinh[1 /Sqrt[x]])/(16*x^2)
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6900, 27, 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.
\(\displaystyle \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 6900 |
\(\displaystyle \frac {\sqrt {x} \int \frac {1}{2 \sqrt {-x-1} x^3}dx}{2 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {x} \int \frac {1}{\sqrt {-x-1} x^3}dx}{4 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \int \frac {1}{\sqrt {-x-1} x^2}dx\right )}{4 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \left (\frac {\sqrt {-x-1}}{x}-\frac {1}{2} \int \frac {1}{\sqrt {-x-1} x}dx\right )\right )}{4 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \left (\int \frac {1}{x}d\sqrt {-x-1}+\frac {\sqrt {-x-1}}{x}\right )\right )}{4 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{2 x^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\sqrt {x} \left (\frac {\sqrt {-x-1}}{2 x^2}-\frac {3}{4} \left (\frac {\sqrt {-x-1}}{x}-\arctan \left (\sqrt {-x-1}\right )\right )\right )}{4 \sqrt {-x}}-\frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{2 x^2}\) |
-1/2*ArcCsch[Sqrt[x]]/x^2 + (Sqrt[x]*(Sqrt[-1 - x]/(2*x^2) - (3*(Sqrt[-1 - x]/x - ArcTan[Sqrt[-1 - x]]))/4))/(4*Sqrt[-x])
3.1.20.3.1 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[(-(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])
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]
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.63
method | result | size |
derivativedivides | \(-\frac {\operatorname {arccsch}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {1+x}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1+x}}\right ) x^{2}-3 \sqrt {1+x}\, x +2 \sqrt {1+x}\right )}{16 \sqrt {\frac {1+x}{x}}\, x^{\frac {5}{2}}}\) | \(57\) |
default | \(-\frac {\operatorname {arccsch}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {1+x}\, \left (3 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {1+x}}\right ) x^{2}-3 \sqrt {1+x}\, x +2 \sqrt {1+x}\right )}{16 \sqrt {\frac {1+x}{x}}\, x^{\frac {5}{2}}}\) | \(57\) |
parts | \(-\frac {\operatorname {arccsch}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {\frac {1+x}{x}}\, \sqrt {x}\, \left (3 \ln \left (\sqrt {1+x}+1\right ) x^{2}-3 \ln \left (\sqrt {1+x}-1\right ) x^{2}-6 \sqrt {1+x}\, x +4 \sqrt {1+x}\right )}{32 \sqrt {1+x}\, \left (\sqrt {1+x}+1\right )^{2} \left (\sqrt {1+x}-1\right )^{2}}\) | \(90\) |
-1/2*arccsch(x^(1/2))/x^2+1/16*(1+x)^(1/2)*(3*arctanh(1/(1+x)^(1/2))*x^2-3 *(1+x)^(1/2)*x+2*(1+x)^(1/2))/((1+x)/x)^(1/2)/x^(5/2)
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.59 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=-\frac {{\left (3 \, x - 2\right )} \sqrt {x} \sqrt {\frac {x + 1}{x}} - {\left (3 \, x^{2} - 8\right )} \log \left (\frac {x \sqrt {\frac {x + 1}{x}} + \sqrt {x}}{x}\right )}{16 \, x^{2}} \]
-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
\[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\int \frac {\operatorname {acsch}{\left (\sqrt {x} \right )}}{x^{3}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.02 \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=-\frac {3 \, x^{\frac {3}{2}} {\left (\frac {1}{x} + 1\right )}^{\frac {3}{2}} - 5 \, \sqrt {x} \sqrt {\frac {1}{x} + 1}}{16 \, {\left (x^{2} {\left (\frac {1}{x} + 1\right )}^{2} - 2 \, x {\left (\frac {1}{x} + 1\right )} + 1\right )}} - \frac {\operatorname {arcsch}\left (\sqrt {x}\right )}{2 \, x^{2}} + \frac {3}{32} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} + 1\right ) - \frac {3}{32} \, \log \left (\sqrt {x} \sqrt {\frac {1}{x} + 1} - 1\right ) \]
-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*arccsch(sqrt(x))/x^2 + 3/32*log(sqrt(x)*sqrt (1/x + 1) + 1) - 3/32*log(sqrt(x)*sqrt(1/x + 1) - 1)
\[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\int { \frac {\operatorname {arcsch}\left (\sqrt {x}\right )}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\text {csch}^{-1}\left (\sqrt {x}\right )}{x^3} \, dx=\int \frac {\mathrm {asinh}\left (\frac {1}{\sqrt {x}}\right )}{x^3} \,d x \]