3.3 \(\int x^2 \sec ^{-1}(\sqrt {x}) \, dx\)

Optimal. Leaf size=47 \[ \frac {1}{3} x^3 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{15} (x-1)^{5/2}-\frac {2}{9} (x-1)^{3/2}-\frac {\sqrt {x-1}}{3} \]

[Out]

-2/9*(-1+x)^(3/2)-1/15*(-1+x)^(5/2)+1/3*x^3*arcsec(x^(1/2))-1/3*(-1+x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5270, 12, 43} \[ \frac {1}{3} x^3 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{15} (x-1)^{5/2}-\frac {2}{9} (x-1)^{3/2}-\frac {\sqrt {x-1}}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSec[Sqrt[x]],x]

[Out]

-Sqrt[-1 + x]/3 - (2*(-1 + x)^(3/2))/9 - (-1 + x)^(5/2)/15 + (x^3*ArcSec[Sqrt[x]])/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5270

Int[((a_.) + ArcSec[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSec[
u]))/(d*(m + 1)), x] - Dist[(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] && InverseFunctionFreeQ[u, x] &&  !Funct
ionOfQ[(c + d*x)^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

\begin {align*} \int x^2 \sec ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{3} x^3 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{3} \int \frac {x^2}{2 \sqrt {-1+x}} \, dx\\ &=\frac {1}{3} x^3 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{6} \int \frac {x^2}{\sqrt {-1+x}} \, dx\\ &=\frac {1}{3} x^3 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{6} \int \left (\frac {1}{\sqrt {-1+x}}+2 \sqrt {-1+x}+(-1+x)^{3/2}\right ) \, dx\\ &=-\frac {1}{3} \sqrt {-1+x}-\frac {2}{9} (-1+x)^{3/2}-\frac {1}{15} (-1+x)^{5/2}+\frac {1}{3} x^3 \sec ^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 35, normalized size = 0.74 \[ \frac {1}{3} x^3 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{45} \sqrt {x-1} \left (3 x^2+4 x+8\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSec[Sqrt[x]],x]

[Out]

-1/45*(Sqrt[-1 + x]*(8 + 4*x + 3*x^2)) + (x^3*ArcSec[Sqrt[x]])/3

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fricas [A]  time = 0.77, size = 27, normalized size = 0.57 \[ \frac {1}{3} \, x^{3} \operatorname {arcsec}\left (\sqrt {x}\right ) - \frac {1}{45} \, {\left (3 \, x^{2} + 4 \, x + 8\right )} \sqrt {x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsec(x^(1/2)),x, algorithm="fricas")

[Out]

1/3*x^3*arcsec(sqrt(x)) - 1/45*(3*x^2 + 4*x + 8)*sqrt(x - 1)

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giac [B]  time = 0.14, size = 116, normalized size = 2.47 \[ -\frac {1}{480} \, x^{\frac {5}{2}} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{5} - \frac {5}{288} \, x^{\frac {3}{2}} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{3} + \frac {1}{3} \, x^{3} \arccos \left (\frac {1}{\sqrt {x}}\right ) - \frac {5}{48} \, \sqrt {x} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )} + \frac {150 \, x^{2} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{4} + 25 \, x {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{2} + 3}{1440 \, x^{\frac {5}{2}} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsec(x^(1/2)),x, algorithm="giac")

[Out]

-1/480*x^(5/2)*(sqrt(-1/x + 1) - 1)^5 - 5/288*x^(3/2)*(sqrt(-1/x + 1) - 1)^3 + 1/3*x^3*arccos(1/sqrt(x)) - 5/4
8*sqrt(x)*(sqrt(-1/x + 1) - 1) + 1/1440*(150*x^2*(sqrt(-1/x + 1) - 1)^4 + 25*x*(sqrt(-1/x + 1) - 1)^2 + 3)/(x^
(5/2)*(sqrt(-1/x + 1) - 1)^5)

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maple [A]  time = 0.05, size = 38, normalized size = 0.81 \[ \frac {x^{3} \mathrm {arcsec}\left (\sqrt {x}\right )}{3}-\frac {\left (x -1\right ) \left (3 x^{2}+4 x +8\right )}{45 \sqrt {\frac {x -1}{x}}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsec(x^(1/2)),x)

[Out]

1/3*x^3*arcsec(x^(1/2))-1/45*(x-1)*(3*x^2+4*x+8)/((x-1)/x)^(1/2)/x^(1/2)

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maxima [A]  time = 0.33, size = 52, normalized size = 1.11 \[ -\frac {1}{15} \, x^{\frac {5}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{2}} + \frac {1}{3} \, x^{3} \operatorname {arcsec}\left (\sqrt {x}\right ) - \frac {2}{9} \, x^{\frac {3}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{2}} - \frac {1}{3} \, \sqrt {x} \sqrt {-\frac {1}{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsec(x^(1/2)),x, algorithm="maxima")

[Out]

-1/15*x^(5/2)*(-1/x + 1)^(5/2) + 1/3*x^3*arcsec(sqrt(x)) - 2/9*x^(3/2)*(-1/x + 1)^(3/2) - 1/3*sqrt(x)*sqrt(-1/
x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^2\,\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acos(1/x^(1/2)),x)

[Out]

int(x^2*acos(1/x^(1/2)), x)

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sympy [C]  time = 25.69, size = 90, normalized size = 1.91 \[ \frac {x^{3} \operatorname {asec}{\left (\sqrt {x} \right )}}{3} - \frac {\begin {cases} \frac {2 x^{2} \sqrt {x - 1}}{5} + \frac {8 x \sqrt {x - 1}}{15} + \frac {16 \sqrt {x - 1}}{15} & \text {for}\: \left |{x}\right | > 1 \\\frac {2 i x^{2} \sqrt {1 - x}}{5} + \frac {8 i x \sqrt {1 - x}}{15} + \frac {16 i \sqrt {1 - x}}{15} & \text {otherwise} \end {cases}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asec(x**(1/2)),x)

[Out]

x**3*asec(sqrt(x))/3 - Piecewise((2*x**2*sqrt(x - 1)/5 + 8*x*sqrt(x - 1)/15 + 16*sqrt(x - 1)/15, Abs(x) > 1),
(2*I*x**2*sqrt(1 - x)/5 + 8*I*x*sqrt(1 - x)/15 + 16*I*sqrt(1 - x)/15, True))/6

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