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

Optimal. Leaf size=126 \[ \frac{1}{3} x^3 \text{sech}^{-1}\left (\sqrt{x}\right )-\frac{(1-x)^3}{15 \sqrt{\frac{1}{\sqrt{x}}-1} \sqrt{\frac{1}{\sqrt{x}}+1} \sqrt{x}}+\frac{2 (1-x)^2}{9 \sqrt{\frac{1}{\sqrt{x}}-1} \sqrt{\frac{1}{\sqrt{x}}+1} \sqrt{x}}-\frac{1-x}{3 \sqrt{\frac{1}{\sqrt{x}}-1} \sqrt{\frac{1}{\sqrt{x}}+1} \sqrt{x}} \]

[Out]

-(1 - x)/(3*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

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Rubi [A]  time = 0.0270286, antiderivative size = 126, 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 = {6345, 12, 43} \[ \frac{1}{3} x^3 \text{sech}^{-1}\left (\sqrt{x}\right )-\frac{(1-x)^3}{15 \sqrt{\frac{1}{\sqrt{x}}-1} \sqrt{\frac{1}{\sqrt{x}}+1} \sqrt{x}}+\frac{2 (1-x)^2}{9 \sqrt{\frac{1}{\sqrt{x}}-1} \sqrt{\frac{1}{\sqrt{x}}+1} \sqrt{x}}-\frac{1-x}{3 \sqrt{\frac{1}{\sqrt{x}}-1} \sqrt{\frac{1}{\sqrt{x}}+1} \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - x)/(3*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 6345

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]

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^2 \text{sech}^{-1}\left (\sqrt{x}\right ) \, dx &=\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]  time = 0.0395421, size = 72, normalized size = 0.57 \[ \frac{1}{3} x^3 \text{sech}^{-1}\left (\sqrt{x}\right )-\frac{1}{45} \sqrt{\frac{1-\sqrt{x}}{\sqrt{x}+1}} \left (3 x^{5/2}+3 x^2+4 x^{3/2}+4 x+8 \sqrt{x}+8\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.126, size = 49, normalized size = 0.4 \begin{align*}{\frac{{x}^{3}}{3}{\rm arcsech} \left (\sqrt{x}\right )}-{\frac{3\,{x}^{2}+4\,x+8}{45}\sqrt{-{ \left ( -1+\sqrt{x} \right ){\frac{1}{\sqrt{x}}}}}\sqrt{x}\sqrt{{ \left ( 1+\sqrt{x} \right ){\frac{1}{\sqrt{x}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.982506, size = 62, normalized size = 0.49 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.87177, size = 131, normalized size = 1.04 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{asech}{\left (\sqrt{x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*asech(sqrt(x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arsech}\left (\sqrt{x}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arcsech(sqrt(x)), x)