\(\int x^3 \text {sech}^{-1}(\sqrt {x}) \, dx\) [20]

Optimal result

Integrand size = 10, antiderivative size = 164 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=-\frac {1-x}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {(1-x)^2}{4 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}-\frac {3 (1-x)^3}{20 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {(1-x)^4}{28 \sqrt {-1+\frac {1}{\sqrt {x}}} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right ) \]

[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] (verified)

Time = 0.02 (sec) , antiderivative size = 164, 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} \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right )+\frac {(1-x)^4}{28 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {3 (1-x)^3}{20 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}+\frac {(1-x)^2}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}}-\frac {1-x}{4 \sqrt {\frac {1}{\sqrt {x}}-1} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x}} \]

[In]

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

[Out]

-1/4*(1 - x)/(Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x]) + (1 - x)^2/(4*Sqrt[-1 + 1/Sqrt[x]]*Sqrt[1 + 1
/Sqrt[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 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}{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] (verified)

Time = 0.05 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.51 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=-\frac {1}{140} \sqrt {\frac {1-\sqrt {x}}{1+\sqrt {x}}} \left (16+16 \sqrt {x}+8 x+8 x^{3/2}+6 x^2+6 x^{5/2}+5 x^3+5 x^{7/2}\right )+\frac {1}{4} x^4 \text {sech}^{-1}\left (\sqrt {x}\right ) \]

[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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.33

method	result	size
derivativedivides	\(\frac {x^{4} \operatorname {arcsech}\left (\sqrt {x}\right )}{4}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140}\)	\(54\)
default	\(\frac {x^{4} \operatorname {arcsech}\left (\sqrt {x}\right )}{4}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140}\)	\(54\)
parts	\(\frac {x^{4} \operatorname {arcsech}\left (\sqrt {x}\right )}{4}-\frac {\sqrt {-\frac {\sqrt {x}-1}{\sqrt {x}}}\, \sqrt {x}\, \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \left (5 x^{3}+6 x^{2}+8 x +16\right )}{140}\)	\(54\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.35 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (\frac {x \sqrt {-\frac {x - 1}{x}} + \sqrt {x}}{x}\right ) - \frac {1}{140} \, {\left (5 \, x^{3} + 6 \, x^{2} + 8 \, x + 16\right )} \sqrt {x} \sqrt {-\frac {x - 1}{x}} \]

[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)

Sympy [F]

\[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int x^{3} \operatorname {asech}{\left (\sqrt {x} \right )}\, dx \]

[In]

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

[Out]

Integral(x**3*asech(sqrt(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.35 \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\frac {1}{28} \, x^{\frac {7}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {7}{2}} - \frac {3}{20} \, x^{\frac {5}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {5}{2}} + \frac {1}{4} \, x^{4} \operatorname {arsech}\left (\sqrt {x}\right ) + \frac {1}{4} \, x^{\frac {3}{2}} {\left (\frac {1}{x} - 1\right )}^{\frac {3}{2}} - \frac {1}{4} \, \sqrt {x} \sqrt {\frac {1}{x} - 1} \]

[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)

Giac [F]

\[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int { x^{3} \operatorname {arsech}\left (\sqrt {x}\right ) \,d x } \]

[In]

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

[Out]

integrate(x^3*arcsech(sqrt(x)), x)

Mupad [F(-1)]

Timed out. \[ \int x^3 \text {sech}^{-1}\left (\sqrt {x}\right ) \, dx=\int x^3\,\mathrm {acosh}\left (\frac {1}{\sqrt {x}}\right ) \,d x \]

[In]

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

[Out]

int(x^3*acosh(1/x^(1/2)), x)