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]

-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

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Rubi [A]  time = 0.0197305, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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]

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])

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*}

Mathematica [A]  time = 0.0261725, 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]

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

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Maple [A]  time = 0.114, size = 43, normalized size = 0.7 \begin{align*}{\frac{{x}^{4}}{4}{\rm arcsec} \left (\sqrt{x}\right )}-{\frac{ \left ( x-1 \right ) \left ( 5\,{x}^{3}+6\,{x}^{2}+8\,x+16 \right ) }{140}{\frac{1}{\sqrt{{\frac{x-1}{x}}}}}{\frac{1}{\sqrt{x}}}} \end{align*}

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 = 1.00133, size = 89, normalized size = 1.53 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 2.28065, size = 97, normalized size = 1.67 \begin{align*} \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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.1011, size = 55, normalized size = 0.95 \begin{align*} \frac{1}{4} \, x^{4} \arccos \left (\frac{1}{\sqrt{x}}\right ) - \frac{1}{28} \,{\left (x - 1\right )}^{\frac{7}{2}} - \frac{3}{20} \,{\left (x - 1\right )}^{\frac{5}{2}} - \frac{1}{4} \,{\left (x - 1\right )}^{\frac{3}{2}} + \frac{4}{35} \, i - \frac{1}{4} \, \sqrt{x - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*arccos(1/sqrt(x)) - 1/28*(x - 1)^(7/2) - 3/20*(x - 1)^(5/2) - 1/4*(x - 1)^(3/2) + 4/35*i - 1/4*sqrt(x
- 1)