\(\int x^2 \text {sech}^{-1}(\sqrt {x}) \, dx\) [21]

Optimal result

Integrand size = 10, antiderivative size = 126 \[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=-\frac {1-x}{3 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {2 (1-x)^2}{9 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}-\frac {(1-x)^3}{15 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {1}{3} x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6480, 12, 45} \[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} x^3 \text {sech}^{-1}\left (\sqrt {x}\right )-\frac {(1-x)^3}{15 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}+\frac {2 (1-x)^2}{9 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {1-x}{3 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

[In]

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

[Out]

-1/3*(1 - x)/(Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (2*(1 - x)^2)/(9*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1
+ 1/Sqrt[x]]*Sqrt[x]) - (1 - x)^3/(15*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (x^3*ArcSech[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 45

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 6480

Int[((a_.) + ArcSech[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((a + b*ArcSec
h[u])/(d*(m + 1))), x] + Dist[b*(Sqrt[1 - u^2]/(d*(m + 1)*u*Sqrt[-1 + 1/u]*Sqrt[1 + 1/u])), Int[SimplifyIntegr
and[(c + d*x)^(m + 1)*(D[u, x]/(u*Sqrt[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] &&  !FunctionOfExponentialQ[u, x]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.57 \[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=-\frac {1}{45} \sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}} \left (8+8 \sqrt {x}+4 x+4 x^{3/2}+3 x^2+3 x^{5/2}\right )+\frac {1}{3} x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \]

[In]

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

[Out]

-1/45*(Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(8 + 8*Sqrt[x] + 4*x + 4*x^(3/2) + 3*x^2 + 3*x^(5/2))) + (x^3*ArcSech
[Sqrt[x]])/3

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.39

method	result	size
derivativedivides	\(\frac {x^{3} \operatorname {arcsech}\left (\sqrt {x}\right )}{3}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (3 x^{2}+4 x +8\right )}{45}\)	\(49\)
default	\(\frac {x^{3} \operatorname {arcsech}\left (\sqrt {x}\right )}{3}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (3 x^{2}+4 x +8\right )}{45}\)	\(49\)
parts	\(\frac {x^{3} \operatorname {arcsech}\left (\sqrt {x}\right )}{3}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (3 x^{2}+4 x +8\right )}{45}\)	\(49\)

[In]

int(x^2*arcsech(x^(1/2)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.41 \[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right ) - \frac {1}{45} \, {\left (3 \, x^{2} + 4 \, x + 8\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int x^{2} \operatorname {asech}{\left (\sqrt {x} \right )}\, dx \]

[In]

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

[Out]

Integral(x**2*asech(sqrt(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.37 \[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=-\frac {1}{15} \, x^{\frac {5}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {5}{2}} + \frac {1}{3} \, x^{3} \operatorname {arsech}\left (\sqrt {x}\right ) + \frac {2}{9} \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} - \frac {1}{3} \, \sqrt {x} \sqrt {\frac {1}{x} - 1} \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int { x^{2} \operatorname {arsech}\left (\sqrt {x}\right ) \,d x } \]

[In]

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

[Out]

integrate(x^2*arcsech(sqrt(x)), x)

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int x^2\,\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right ) \,d x \]

[In]

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

[Out]

int(x^2*acosh(1/x^(1/2)), x)