Integrand size = 10, antiderivative size = 72 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=-\frac {5}{48} \sqrt {x} \sqrt {1+x}+\frac {5}{72} x^{3/2} \sqrt {1+x}-\frac {1}{18} x^{5/2} \sqrt {1+x}+\frac {5 \text {arcsinh}\left (\sqrt {x}\right )}{48}+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right ) \]
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)
Time = 0.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.60 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\frac {1}{144} \left (\sqrt {x} \sqrt {1+x} \left (-15+10 x-8 x^2\right )+3 \left (5+16 x^3\right ) \text {arcsinh}\left (\sqrt {x}\right )\right ) \]
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6290, 27, 60, 60, 60, 63, 222}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx\) |
\(\Big \downarrow \) 6290 |
\(\displaystyle \frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )-\frac {1}{3} \int \frac {x^{5/2}}{2 \sqrt {x+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )-\frac {1}{6} \int \frac {x^{5/2}}{\sqrt {x+1}}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \int \frac {x^{3/2}}{\sqrt {x+1}}dx-\frac {1}{3} x^{5/2} \sqrt {x+1}\right )+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {1}{2} x^{3/2} \sqrt {x+1}-\frac {3}{4} \int \frac {\sqrt {x}}{\sqrt {x+1}}dx\right )-\frac {1}{3} x^{5/2} \sqrt {x+1}\right )+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {1}{2} x^{3/2} \sqrt {x+1}-\frac {3}{4} \left (\sqrt {x} \sqrt {x+1}-\frac {1}{2} \int \frac {1}{\sqrt {x} \sqrt {x+1}}dx\right )\right )-\frac {1}{3} x^{5/2} \sqrt {x+1}\right )+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 63 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {1}{2} x^{3/2} \sqrt {x+1}-\frac {3}{4} \left (\sqrt {x} \sqrt {x+1}-\int \frac {1}{\sqrt {x+1}}d\sqrt {x}\right )\right )-\frac {1}{3} x^{5/2} \sqrt {x+1}\right )+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 222 |
\(\displaystyle \frac {1}{6} \left (\frac {5}{6} \left (\frac {1}{2} x^{3/2} \sqrt {x+1}-\frac {3}{4} \left (\sqrt {x} \sqrt {x+1}-\text {arcsinh}\left (\sqrt {x}\right )\right )\right )-\frac {1}{3} x^{5/2} \sqrt {x+1}\right )+\frac {1}{3} x^3 \text {arcsinh}\left (\sqrt {x}\right )\) |
(-1/3*(x^(5/2)*Sqrt[1 + x]) + (5*((x^(3/2)*Sqrt[1 + x])/2 - (3*(Sqrt[x]*Sq rt[1 + x] - ArcSinh[Sqrt[x]]))/4))/6)/6 + (x^3*ArcSinh[Sqrt[x]])/3
3.3.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2/b S ubst[Int[1/Sqrt[c + d*(x^2/b)], x], x, Sqrt[b*x]], x] /; FreeQ[{b, c, d}, x ] && GtQ[c, 0]
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]
Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Si mp[(c + d*x)^(m + 1)*((a + b*ArcSinh[u])/(d*(m + 1))), x] - Simp[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]
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 | \(\frac {x^{3} \operatorname {arcsinh}\left (\sqrt {x}\right )}{3}-\frac {x^{\frac {5}{2}} \sqrt {1+x}}{18}+\frac {5 x^{\frac {3}{2}} \sqrt {1+x}}{72}-\frac {5 \sqrt {x}\, \sqrt {1+x}}{48}+\frac {5 \sqrt {x \left (1+x \right )}\, \ln \left (\frac {1}{2}+x +\sqrt {x^{2}+x}\right )}{96 \sqrt {x}\, \sqrt {1+x}}\) | \(69\) |
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)
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.56 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=-\frac {1}{144} \, {\left (8 \, x^{2} - 10 \, x + 15\right )} \sqrt {x + 1} \sqrt {x} + \frac {1}{48} \, {\left (16 \, x^{3} + 5\right )} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) \]
-1/144*(8*x^2 - 10*x + 15)*sqrt(x + 1)*sqrt(x) + 1/48*(16*x^3 + 5)*log(sqr t(x + 1) + sqrt(x))
\[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\int x^{2} \operatorname {asinh}{\left (\sqrt {x} \right )}\, dx \]
Time = 0.37 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.64 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \operatorname {arsinh}\left (\sqrt {x}\right ) - \frac {1}{18} \, \sqrt {x + 1} x^{\frac {5}{2}} + \frac {5}{72} \, \sqrt {x + 1} x^{\frac {3}{2}} - \frac {5}{48} \, \sqrt {x + 1} \sqrt {x} + \frac {5}{48} \, \operatorname {arsinh}\left (\sqrt {x}\right ) \]
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))
Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.69 \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) - \frac {1}{144} \, {\left (2 \, {\left (4 \, x - 5\right )} x + 15\right )} \sqrt {x + 1} \sqrt {x} - \frac {5}{48} \, \log \left (\sqrt {x + 1} - \sqrt {x}\right ) \]
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) - sqrt(x))
Timed out. \[ \int x^2 \text {arcsinh}\left (\sqrt {x}\right ) \, dx=\int x^2\,\mathrm {asinh}\left (\sqrt {x}\right ) \,d x \]