Integrand size = 10, antiderivative size = 68 \[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {-1+x}}{18 x^3}+\frac {5 \sqrt {-1+x}}{72 x^2}+\frac {5 \sqrt {-1+x}}{48 x}-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}+\frac {5}{48} \arctan \left (\sqrt {-1+x}\right ) \] Output:
1/18*(-1+x)^(1/2)/x^3+5/72*(-1+x)^(1/2)/x^2+5/48*(-1+x)^(1/2)/x-1/3*arcsec (x^(1/2))/x^3+5/48*arctan((-1+x)^(1/2))
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.66 \[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {\sqrt {-1+x} \left (8+10 x+15 x^2\right )-48 \sec ^{-1}\left (\sqrt {x}\right )-15 x^3 \arcsin \left (\frac {1}{\sqrt {x}}\right )}{144 x^3} \] Input:
Integrate[ArcSec[Sqrt[x]]/x^4,x]
Output:
(Sqrt[-1 + x]*(8 + 10*x + 15*x^2) - 48*ArcSec[Sqrt[x]] - 15*x^3*ArcSin[1/S qrt[x]])/(144*x^3)
Time = 0.22 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5793, 27, 52, 52, 52, 73, 216}
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.
\(\displaystyle \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 5793 |
\(\displaystyle \frac {1}{3} \int \frac {1}{2 \sqrt {x-1} x^4}dx-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \int \frac {1}{\sqrt {x-1} x^4}dx-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \int \frac {1}{\sqrt {x-1} x^3}dx+\frac {\sqrt {x-1}}{3 x^3}\right )-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {3}{4} \int \frac {1}{\sqrt {x-1} x^2}dx+\frac {\sqrt {x-1}}{2 x^2}\right )+\frac {\sqrt {x-1}}{3 x^3}\right )-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {x-1} x}dx+\frac {\sqrt {x-1}}{x}\right )+\frac {\sqrt {x-1}}{2 x^2}\right )+\frac {\sqrt {x-1}}{3 x^3}\right )-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {3}{4} \left (\int \frac {1}{x}d\sqrt {x-1}+\frac {\sqrt {x-1}}{x}\right )+\frac {\sqrt {x-1}}{2 x^2}\right )+\frac {\sqrt {x-1}}{3 x^3}\right )-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {3}{4} \left (\arctan \left (\sqrt {x-1}\right )+\frac {\sqrt {x-1}}{x}\right )+\frac {\sqrt {x-1}}{2 x^2}\right )+\frac {\sqrt {x-1}}{3 x^3}\right )-\frac {\sec ^{-1}\left (\sqrt {x}\right )}{3 x^3}\) |
Input:
Int[ArcSec[Sqrt[x]]/x^4,x]
Output:
-1/3*ArcSec[Sqrt[x]]/x^3 + (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
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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_.) + ArcSec[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(c + d*x)^(m + 1)*((a + b*ArcSec[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*Sqrt[ u^2 - 1])), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && Invers eFunctionFreeQ[u, x] && !FunctionOfQ[(c + d*x)^(m + 1), u, x] && !Functio nOfExponentialQ[u, x]
Time = 0.12 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(-\frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{3 x^{3}}-\frac {\sqrt {-1+x}\, \left (15 \arctan \left (\frac {1}{\sqrt {-1+x}}\right ) x^{3}-15 \sqrt {-1+x}\, x^{2}-10 \sqrt {-1+x}\, x -8 \sqrt {-1+x}\right )}{144 \sqrt {\frac {-1+x}{x}}\, x^{\frac {7}{2}}}\) | \(67\) |
default | \(-\frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{3 x^{3}}-\frac {\sqrt {-1+x}\, \left (15 \arctan \left (\frac {1}{\sqrt {-1+x}}\right ) x^{3}-15 \sqrt {-1+x}\, x^{2}-10 \sqrt {-1+x}\, x -8 \sqrt {-1+x}\right )}{144 \sqrt {\frac {-1+x}{x}}\, x^{\frac {7}{2}}}\) | \(67\) |
parts | \(-\frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{3 x^{3}}+\frac {\sqrt {\frac {-1+x}{x}}\, \left (15 \arctan \left (\sqrt {-1+x}\right ) x^{3}+15 \sqrt {-1+x}\, x^{2}+10 \sqrt {-1+x}\, x +8 \sqrt {-1+x}\right )}{144 x^{\frac {5}{2}} \sqrt {-1+x}}\) | \(67\) |
Input:
int(arcsec(x^(1/2))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*arcsec(x^(1/2))/x^3-1/144*(-1+x)^(1/2)*(15*arctan(1/(-1+x)^(1/2))*x^3 -15*(-1+x)^(1/2)*x^2-10*(-1+x)^(1/2)*x-8*(-1+x)^(1/2))/((-1+x)/x)^(1/2)/x^ (7/2)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.51 \[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {3 \, {\left (5 \, x^{3} - 16\right )} \operatorname {arcsec}\left (\sqrt {x}\right ) + {\left (15 \, x^{2} + 10 \, x + 8\right )} \sqrt {x - 1}}{144 \, x^{3}} \] Input:
integrate(arcsec(x^(1/2))/x^4,x, algorithm="fricas")
Output:
1/144*(3*(5*x^3 - 16)*arcsec(sqrt(x)) + (15*x^2 + 10*x + 8)*sqrt(x - 1))/x ^3
Result contains complex when optimal does not.
Time = 108.73 (sec) , antiderivative size = 180, normalized size of antiderivative = 2.65 \[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {\begin {cases} \frac {5 i \operatorname {acosh}{\left (\frac {1}{\sqrt {x}} \right )}}{8} - \frac {5 i}{8 \sqrt {x} \sqrt {-1 + \frac {1}{x}}} + \frac {5 i}{24 x^{\frac {3}{2}} \sqrt {-1 + \frac {1}{x}}} + \frac {i}{12 x^{\frac {5}{2}} \sqrt {-1 + \frac {1}{x}}} + \frac {i}{3 x^{\frac {7}{2}} \sqrt {-1 + \frac {1}{x}}} & \text {for}\: \frac {1}{\left |{x}\right |} > 1 \\- \frac {5 \operatorname {asin}{\left (\frac {1}{\sqrt {x}} \right )}}{8} + \frac {5}{8 \sqrt {x} \sqrt {1 - \frac {1}{x}}} - \frac {5}{24 x^{\frac {3}{2}} \sqrt {1 - \frac {1}{x}}} - \frac {1}{12 x^{\frac {5}{2}} \sqrt {1 - \frac {1}{x}}} - \frac {1}{3 x^{\frac {7}{2}} \sqrt {1 - \frac {1}{x}}} & \text {otherwise} \end {cases}}{6} - \frac {\operatorname {asec}{\left (\sqrt {x} \right )}}{3 x^{3}} \] Input:
integrate(asec(x**(1/2))/x**4,x)
Output:
Piecewise((5*I*acosh(1/sqrt(x))/8 - 5*I/(8*sqrt(x)*sqrt(-1 + 1/x)) + 5*I/( 24*x**(3/2)*sqrt(-1 + 1/x)) + I/(12*x**(5/2)*sqrt(-1 + 1/x)) + I/(3*x**(7/ 2)*sqrt(-1 + 1/x)), 1/Abs(x) > 1), (-5*asin(1/sqrt(x))/8 + 5/(8*sqrt(x)*sq rt(1 - 1/x)) - 5/(24*x**(3/2)*sqrt(1 - 1/x)) - 1/(12*x**(5/2)*sqrt(1 - 1/x )) - 1/(3*x**(7/2)*sqrt(1 - 1/x)), True))/6 - asec(sqrt(x))/(3*x**3)
Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (48) = 96\).
Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.56 \[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=-\frac {15 \, x^{\frac {5}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{2}} + 40 \, x^{\frac {3}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{2}} + 33 \, \sqrt {x} \sqrt {-\frac {1}{x} + 1}}{144 \, {\left (x^{3} {\left (\frac {1}{x} - 1\right )}^{3} - 3 \, x^{2} {\left (\frac {1}{x} - 1\right )}^{2} + 3 \, x {\left (\frac {1}{x} - 1\right )} - 1\right )}} - \frac {\operatorname {arcsec}\left (\sqrt {x}\right )}{3 \, x^{3}} + \frac {5}{48} \, \arctan \left (\sqrt {x} \sqrt {-\frac {1}{x} + 1}\right ) \] Input:
integrate(arcsec(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*sqr t(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*arcsec(sqrt(x))/x^3 + 5/48*arctan(sqrt(x)*sqrt(-1/x + 1))
Time = 0.39 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\frac {5 \, \sqrt {-\frac {1}{x} + 1}}{48 \, \sqrt {x}} + \frac {5 \, \sqrt {-\frac {1}{x} + 1}}{72 \, x^{\frac {3}{2}}} + \frac {\sqrt {-\frac {1}{x} + 1}}{18 \, x^{\frac {5}{2}}} - \frac {\arccos \left (\frac {1}{\sqrt {x}}\right )}{3 \, x^{3}} + \frac {5}{48} \, \arccos \left (\frac {1}{\sqrt {x}}\right ) \] Input:
integrate(arcsec(x^(1/2))/x^4,x, algorithm="giac")
Output:
5/48*sqrt(-1/x + 1)/sqrt(x) + 5/72*sqrt(-1/x + 1)/x^(3/2) + 1/18*sqrt(-1/x + 1)/x^(5/2) - 1/3*arccos(1/sqrt(x))/x^3 + 5/48*arccos(1/sqrt(x))
Timed out. \[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right )}{x^4} \,d x \] Input:
int(acos(1/x^(1/2))/x^4,x)
Output:
int(acos(1/x^(1/2))/x^4, x)
\[ \int \frac {\sec ^{-1}\left (\sqrt {x}\right )}{x^4} \, dx=\int \frac {\mathit {asec} \left (\sqrt {x}\right )}{x^{4}}d x \] Input:
int(asec(x^(1/2))/x^4,x)
Output:
int(asec(sqrt(x))/x**4,x)