3.2 \(\int x^3 \sec ^{-1}(\sqrt {x}) \, dx\)

Optimal. Leaf size=58 \[ \frac {1}{4} x^4 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{28} (x-1)^{7/2}-\frac {3}{20} (x-1)^{5/2}-\frac {1}{4} (x-1)^{3/2}-\frac {\sqrt {x-1}}{4} \]

[Out]

-1/4*(-1+x)^(3/2)-3/20*(-1+x)^(5/2)-1/28*(-1+x)^(7/2)+1/4*x^4*arcsec(x^(1/2))-1/4*(-1+x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 58, 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} \[ \frac {1}{4} x^4 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{28} (x-1)^{7/2}-\frac {3}{20} (x-1)^{5/2}-\frac {1}{4} (x-1)^{3/2}-\frac {\sqrt {x-1}}{4} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSec[Sqrt[x]],x]

[Out]

-Sqrt[-1 + x]/4 - (-1 + x)^(3/2)/4 - (3*(-1 + x)^(5/2))/20 - (-1 + x)^(7/2)/28 + (x^4*ArcSec[Sqrt[x]])/4

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^3 \sec ^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{4} x^4 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{4} \int \frac {x^3}{2 \sqrt {-1+x}} \, dx\\ &=\frac {1}{4} x^4 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{8} \int \frac {x^3}{\sqrt {-1+x}} \, dx\\ &=\frac {1}{4} x^4 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{8} \int \left (\frac {1}{\sqrt {-1+x}}+3 \sqrt {-1+x}+3 (-1+x)^{3/2}+(-1+x)^{5/2}\right ) \, dx\\ &=-\frac {1}{4} \sqrt {-1+x}-\frac {1}{4} (-1+x)^{3/2}-\frac {3}{20} (-1+x)^{5/2}-\frac {1}{28} (-1+x)^{7/2}+\frac {1}{4} x^4 \sec ^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 40, normalized size = 0.69 \[ \frac {1}{4} x^4 \sec ^{-1}\left (\sqrt {x}\right )-\frac {1}{140} \sqrt {x-1} \left (5 x^3+6 x^2+8 x+16\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSec[Sqrt[x]],x]

[Out]

-1/140*(Sqrt[-1 + x]*(16 + 8*x + 6*x^2 + 5*x^3)) + (x^4*ArcSec[Sqrt[x]])/4

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fricas [A]  time = 0.49, size = 32, normalized size = 0.55 \[ \frac {1}{4} \, x^{4} \operatorname {arcsec}\left (\sqrt {x}\right ) - \frac {1}{140} \, {\left (5 \, x^{3} + 6 \, x^{2} + 8 \, x + 16\right )} \sqrt {x - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arcsec(sqrt(x)) - 1/140*(5*x^3 + 6*x^2 + 8*x + 16)*sqrt(x - 1)

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giac [B]  time = 0.13, size = 152, normalized size = 2.62 \[ -\frac {1}{3584} \, x^{\frac {7}{2}} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{7} - \frac {7}{2560} \, x^{\frac {5}{2}} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{5} + \frac {1}{4} \, x^{4} \arccos \left (\frac {1}{\sqrt {x}}\right ) - \frac {7}{512} \, x^{\frac {3}{2}} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{3} - \frac {35}{512} \, \sqrt {x} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )} + \frac {1225 \, x^{3} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{6} + 245 \, x^{2} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{4} + 49 \, x {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{2} + 5}{17920 \, x^{\frac {7}{2}} {\left (\sqrt {-\frac {1}{x} + 1} - 1\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3584*x^(7/2)*(sqrt(-1/x + 1) - 1)^7 - 7/2560*x^(5/2)*(sqrt(-1/x + 1) - 1)^5 + 1/4*x^4*arccos(1/sqrt(x)) - 7
/512*x^(3/2)*(sqrt(-1/x + 1) - 1)^3 - 35/512*sqrt(x)*(sqrt(-1/x + 1) - 1) + 1/17920*(1225*x^3*(sqrt(-1/x + 1)
- 1)^6 + 245*x^2*(sqrt(-1/x + 1) - 1)^4 + 49*x*(sqrt(-1/x + 1) - 1)^2 + 5)/(x^(7/2)*(sqrt(-1/x + 1) - 1)^7)

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maple [A]  time = 0.05, size = 43, normalized size = 0.74 \[ \frac {x^{4} \mathrm {arcsec}\left (\sqrt {x}\right )}{4}-\frac {\left (x -1\right ) \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140 \sqrt {\frac {x -1}{x}}\, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsec(x^(1/2)),x)

[Out]

1/4*x^4*arcsec(x^(1/2))-1/140*(x-1)*(5*x^3+6*x^2+8*x+16)/((x-1)/x)^(1/2)/x^(1/2)

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maxima [A]  time = 0.33, size = 66, normalized size = 1.14 \[ -\frac {1}{28} \, x^{\frac {7}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {7}{2}} - \frac {3}{20} \, x^{\frac {5}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {5}{2}} + \frac {1}{4} \, x^{4} \operatorname {arcsec}\left (\sqrt {x}\right ) - \frac {1}{4} \, x^{\frac {3}{2}} {\left (-\frac {1}{x} + 1\right )}^{\frac {3}{2}} - \frac {1}{4} \, \sqrt {x} \sqrt {-\frac {1}{x} + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/28*x^(7/2)*(-1/x + 1)^(7/2) - 3/20*x^(5/2)*(-1/x + 1)^(5/2) + 1/4*x^4*arcsec(sqrt(x)) - 1/4*x^(3/2)*(-1/x +
1)^(3/2) - 1/4*sqrt(x)*sqrt(-1/x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^3\,\mathrm {acos}\left (\frac {1}{\sqrt {x}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acos(1/x^(1/2)),x)

[Out]

int(x^3*acos(1/x^(1/2)), x)

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sympy [C]  time = 73.95, size = 119, normalized size = 2.05 \[ \frac {x^{4} \operatorname {asec}{\left (\sqrt {x} \right )}}{4} - \frac {\begin {cases} \frac {2 x^{3} \sqrt {x - 1}}{7} + \frac {12 x^{2} \sqrt {x - 1}}{35} + \frac {16 x \sqrt {x - 1}}{35} + \frac {32 \sqrt {x - 1}}{35} & \text {for}\: \left |{x}\right | > 1 \\\frac {2 i x^{3} \sqrt {1 - x}}{7} + \frac {12 i x^{2} \sqrt {1 - x}}{35} + \frac {16 i x \sqrt {1 - x}}{35} + \frac {32 i \sqrt {1 - x}}{35} & \text {otherwise} \end {cases}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**4*asec(sqrt(x))/4 - Piecewise((2*x**3*sqrt(x - 1)/7 + 12*x**2*sqrt(x - 1)/35 + 16*x*sqrt(x - 1)/35 + 32*sqr
t(x - 1)/35, Abs(x) > 1), (2*I*x**3*sqrt(1 - x)/7 + 12*I*x**2*sqrt(1 - x)/35 + 16*I*x*sqrt(1 - x)/35 + 32*I*sq
rt(1 - x)/35, True))/8

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