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

Optimal. Leaf size=88 \[ \frac {1}{2} x^2 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {(1-x)^2}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {1-x}{2 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

[Out]

1/2*x^2*arcsech(x^(1/2))+1/2*(-1+x)/x^(1/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+1/6*(1-x)^2/x^(1/2)/(-1+1
/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6345, 12, 43} \[ \frac {1}{2} x^2 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {(1-x)^2}{6 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {1-x}{2 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - x)/(2*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (1 - x)^2/(6*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/S
qrt[x]]*Sqrt[x]) + (x^2*ArcSech[Sqrt[x]])/2

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 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]

Rubi steps

\begin {align*} \int x \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{2} x^2 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {1-x} \int \frac {x}{2 \sqrt {1-x}} \, dx}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1}{2} x^2 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {1-x} \int \frac {x}{\sqrt {1-x}} \, dx}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1}{2} x^2 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {1-x} \int \left (\frac {1}{\sqrt {1-x}}-\sqrt {1-x}\right ) \, dx}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=-\frac {1-x}{2 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {(1-x)^2}{6 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {1}{2} x^2 \text {sech}^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 56, normalized size = 0.64 \[ \frac {1}{2} x^2 \text {sech}^{-1}\left (\sqrt {x}\right )-\frac {1}{6} \sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}} \left (x^{3/2}+x+2 \sqrt {x}+2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(2 + 2*Sqrt[x] + x + x^(3/2))) + (x^2*ArcSech[Sqrt[x]])/2

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fricas [A]  time = 0.63, size = 45, normalized size = 0.51 \[ \frac {1}{2} \, x^{2} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right ) - \frac {1}{6} \, {\left (x + 2\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*log((x*sqrt(-(x - 1)/x) + sqrt(x))/x) - 1/6*(x + 2)*sqrt(x)*sqrt(-(x - 1)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {arsech}\left (\sqrt {x}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arcsech(sqrt(x)), x)

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maple [A]  time = 0.06, size = 42, normalized size = 0.48 \[ \frac {x^{2} \mathrm {arcsech}\left (\sqrt {x}\right )}{2}-\frac {\sqrt {-\frac {-1+\sqrt {x}}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {1+\sqrt {x}}{\sqrt {x}}}\, \left (x +2\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*arcsech(x^(1/2))-1/6*(-(-1+x^(1/2))/x^(1/2))^(1/2)*x^(1/2)*((1+x^(1/2))/x^(1/2))^(1/2)*(x+2)

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maxima [A]  time = 0.38, size = 34, normalized size = 0.39 \[ \frac {1}{6} \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} + \frac {1}{2} \, x^{2} \operatorname {arsech}\left (\sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x} \sqrt {\frac {1}{x} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^(3/2)*(1/x - 1)^(3/2) + 1/2*x^2*arcsech(sqrt(x)) - 1/2*sqrt(x)*sqrt(1/x - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*acosh(1/x^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {asech}{\left (\sqrt {x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*asech(sqrt(x)), x)

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