\(\int x^2 \text {arcsinh}(\sqrt {x}) \, dx\) [292]

Optimal result

Integrand size = 10, antiderivative size = 72 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=-\frac {5}{48} \sqrt {x} \sqrt {1+x}+\frac {5}{72} x^{3/2} \sqrt {1+x}-\frac {1}{18} x^{5/2} \sqrt {1+x}+\frac {5 \text {arcsinh}\left (\sqrt {x}\right )}{48}+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right ) \]

[Out]

5/48*arcsinh(x^(1/2))+1/3*x^3*arcsinh(x^(1/2))+5/72*x^(3/2)*(1+x)^(1/2)-1/18*x^(5/2)*(1+x)^(1/2)-5/48*x^(1/2)*
(1+x)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5875, 12, 52, 56, 221} \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )+\frac {5 \text {arcsinh}\left (\sqrt {x}\right )}{48}-\frac {1}{18} \sqrt {x+1} x^{5/2}+\frac {5}{72} \sqrt {x+1} x^{3/2}-\frac {5}{48} \sqrt {x+1} \sqrt {x} \]

[In]

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

[Out]

(-5*Sqrt[x]*Sqrt[1 + x])/48 + (5*x^(3/2)*Sqrt[1 + x])/72 - (x^(5/2)*Sqrt[1 + x])/18 + (5*ArcSinh[Sqrt[x]])/48
+ (x^3*ArcSinh[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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 5875

Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m + 1)*((a + b*ArcSin
h[u])/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/Sqrt[1 + u^2]),
x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[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 {arcsinh}\left (\sqrt {x}\right )-\frac {1}{3} \int \frac {x^{5/2}}{2 \sqrt {1+x}} \, dx \\ & = \frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )-\frac {1}{6} \int \frac {x^{5/2}}{\sqrt {1+x}} \, dx \\ & = -\frac {1}{18} x^{5/2} \sqrt {1+x}+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )+\frac {5}{36} \int \frac {x^{3/2}}{\sqrt {1+x}} \, dx \\ & = \frac {5}{72} x^{3/2} \sqrt {1+x}-\frac {1}{18} x^{5/2} \sqrt {1+x}+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )-\frac {5}{48} \int \frac {\sqrt {x}}{\sqrt {1+x}} \, dx \\ & = -\frac {5}{48} \sqrt {x} \sqrt {1+x}+\frac {5}{72} x^{3/2} \sqrt {1+x}-\frac {1}{18} x^{5/2} \sqrt {1+x}+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )+\frac {5}{96} \int \frac {1}{\sqrt {x} \sqrt {1+x}} \, dx \\ & = -\frac {5}{48} \sqrt {x} \sqrt {1+x}+\frac {5}{72} x^{3/2} \sqrt {1+x}-\frac {1}{18} x^{5/2} \sqrt {1+x}+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )+\frac {5}{48} \text {Subst}\left (\int \frac {1}{\sqrt {1+x^2}} \, dx,x,\sqrt {x}\right ) \\ & = -\frac {5}{48} \sqrt {x} \sqrt {1+x}+\frac {5}{72} x^{3/2} \sqrt {1+x}-\frac {1}{18} x^{5/2} \sqrt {1+x}+\frac {5 \text {arcsinh}\left (\sqrt {x}\right )}{48}+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.60 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\frac {1}{144} \left (\sqrt {x} \sqrt {1+x} \left (-15+10 x-8 x^2\right )+3 \left (5+16 x^3\right ) \text {arcsinh}\left (\sqrt {x}\right )\right ) \]

[In]

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

[Out]

(Sqrt[x]*Sqrt[1 + x]*(-15 + 10*x - 8*x^2) + 3*(5 + 16*x^3)*ArcSinh[Sqrt[x]])/144

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {5 \,\operatorname {arcsinh}\left (\sqrt {x}\right )}{48}+\frac {x^{3} \operatorname {arcsinh}\left (\sqrt {x}\right )}{3}+\frac {5 x^{\frac {3}{2}} \sqrt {1+x}}{72}-\frac {x^{\frac {5}{2}} \sqrt {1+x}}{18}-\frac {5 \sqrt {x}\, \sqrt {1+x}}{48}\)	\(47\)
default	\(\frac {5 \,\operatorname {arcsinh}\left (\sqrt {x}\right )}{48}+\frac {x^{3} \operatorname {arcsinh}\left (\sqrt {x}\right )}{3}+\frac {5 x^{\frac {3}{2}} \sqrt {1+x}}{72}-\frac {x^{\frac {5}{2}} \sqrt {1+x}}{18}-\frac {5 \sqrt {x}\, \sqrt {1+x}}{48}\)	\(47\)
parts	\(\frac {x^{3} \operatorname {arcsinh}\left (\sqrt {x}\right )}{3}-\frac {x^{\frac {5}{2}} \sqrt {1+x}}{18}+\frac {5 x^{\frac {3}{2}} \sqrt {1+x}}{72}-\frac {5 \sqrt {x}\, \sqrt {1+x}}{48}+\frac {5 \sqrt {x \left (1+x \right )}\, \ln \left (\frac {1}{2}+x +\sqrt {x^{2}+x}\right )}{96 \sqrt {x}\, \sqrt {1+x}}\)	\(69\)

[In]

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

[Out]

5/48*arcsinh(x^(1/2))+1/3*x^3*arcsinh(x^(1/2))+5/72*x^(3/2)*(1+x)^(1/2)-1/18*x^(5/2)*(1+x)^(1/2)-5/48*x^(1/2)*
(1+x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=-\frac {1}{144} \, {\left (8 \, x^{2} - 10 \, x + 15\right )} \sqrt {x + 1} \sqrt {x} + \frac {1}{48} \, {\left (16 \, x^{3} + 5\right )} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) \]

[In]

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

[Out]

-1/144*(8*x^2 - 10*x + 15)*sqrt(x + 1)*sqrt(x) + 1/48*(16*x^3 + 5)*log(sqrt(x + 1) + sqrt(x))

Sympy [F]

\[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\int x^{2} \operatorname {asinh}{\left (\sqrt {x} \right )}\, dx \]

[In]

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

[Out]

Integral(x**2*asinh(sqrt(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.64 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \operatorname {arsinh}\left (\sqrt {x}\right ) - \frac {1}{18} \, \sqrt {x + 1} x^{\frac {5}{2}} + \frac {5}{72} \, \sqrt {x + 1} x^{\frac {3}{2}} - \frac {5}{48} \, \sqrt {x + 1} \sqrt {x} + \frac {5}{48} \, \operatorname {arsinh}\left (\sqrt {x}\right ) \]

[In]

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

[Out]

1/3*x^3*arcsinh(sqrt(x)) - 1/18*sqrt(x + 1)*x^(5/2) + 5/72*sqrt(x + 1)*x^(3/2) - 5/48*sqrt(x + 1)*sqrt(x) + 5/
48*arcsinh(sqrt(x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.69 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) - \frac {1}{144} \, {\left (2 \, {\left (4 \, x - 5\right )} x + 15\right )} \sqrt {x + 1} \sqrt {x} - \frac {5}{48} \, \log \left (\sqrt {x + 1} - \sqrt {x}\right ) \]

[In]

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

[Out]

1/3*x^3*log(sqrt(x + 1) + sqrt(x)) - 1/144*(2*(4*x - 5)*x + 15)*sqrt(x + 1)*sqrt(x) - 5/48*log(sqrt(x + 1) - s
qrt(x))

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\int x^2\,\mathrm {asinh}\left (\sqrt {x}\right ) \,d x \]

[In]

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

[Out]

int(x^2*asinh(x^(1/2)), x)