3.20 \(\int x^3 \text {sech}^{-1}(\sqrt {x}) \, dx\)

Optimal. Leaf size=164 \[ \frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {(1-x)^4}{28 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {3 (1-x)^3}{20 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}+\frac {(1-x)^2}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {1-x}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

[Out]

1/4*x^4*arcsech(x^(1/2))+1/4*(-1+x)/x^(1/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+1/4*(1-x)^2/x^(1/2)/(-1+1
/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)-3/20*(1-x)^3/x^(1/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)+1/28*(1-x)^4
/x^(1/2)/(-1+1/x^(1/2))^(1/2)/(1+1/x^(1/2))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 164, 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 = {6345, 12, 43} \[ \frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {(1-x)^4}{28 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {3 (1-x)^3}{20 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}+\frac {(1-x)^2}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {1-x}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - x)/(4*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (1 - x)^2/(4*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/S
qrt[x]]*Sqrt[x]) - (3*(1 - x)^3)/(20*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (1 - x)^4/(28*Sqrt[-1
+ 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (x^4*ArcSech[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 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^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx &=\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {1-x} \int \frac {x^3}{2 \sqrt {1-x}} \, dx}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {1-x} \int \frac {x^3}{\sqrt {1-x}} \, dx}{8 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {\sqrt {1-x} \int \left (\frac {1}{\sqrt {1-x}}-3 \sqrt {1-x}+3 (1-x)^{3/2}-(1-x)^{5/2}\right ) \, dx}{8 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}\\ &=-\frac {1-x}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {(1-x)^2}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}-\frac {3 (1-x)^3}{20 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {(1-x)^4}{28 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 84, normalized size = 0.51 \[ \frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )-\frac {1}{140} \sqrt {\frac {1-\sqrt {x}}{\sqrt {x}+1}} \left (5 x^{7/2}+6 x^{5/2}+8 x^{3/2}+5 x^3+6 x^2+8 x+16 \sqrt {x}+16\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/140*(Sqrt[(1 - Sqrt[x])/(1 + Sqrt[x])]*(16 + 16*Sqrt[x] + 8*x + 8*x^(3/2) + 6*x^2 + 6*x^(5/2) + 5*x^3 + 5*x
^(7/2))) + (x^4*ArcSech[Sqrt[x]])/4

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 57, normalized size = 0.35 \[ \frac {1}{4} \, x^{4} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right ) - \frac {1}{140} \, {\left (5 \, x^{3} + 6 \, x^{2} + 8 \, x + 16\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arsech}\left (\sqrt {x}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arcsech(sqrt(x)), x)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 54, normalized size = 0.33 \[ \frac {x^{4} \mathrm {arcsech}\left (\sqrt {x}\right )}{4}-\frac {\sqrt {-\frac {-1+\sqrt {x}}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {1+\sqrt {x}}{\sqrt {x}}}\, \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.32, size = 58, normalized size = 0.35 \[ \frac {1}{28} \, x^{\frac {7}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {7}{2}} - \frac {3}{20} \, x^{\frac {5}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {5}{2}} + \frac {1}{4} \, x^{4} \operatorname {arsech}\left (\sqrt {x}\right ) + \frac {1}{4} \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} - \frac {1}{4} \, \sqrt {x} \sqrt {\frac {1}{x} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(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*arcsech(sqrt(x)) + 1/4*x^(3/2)*(1/x - 1)
^(3/2) - 1/4*sqrt(x)*sqrt(1/x - 1)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*acosh(1/x^(1/2)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {asech}{\left (\sqrt {x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*asech(sqrt(x)), x)

________________________________________________________________________________________