Integrand size = 10, antiderivative size = 117 \[ \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx=-\frac {5}{48} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} \sqrt {x}-\frac {5}{72} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} x^{3/2}-\frac {1}{18} \sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} x^{5/2}-\frac {5 \text {arccosh}\left (\sqrt {x}\right )}{48}+\frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right ) \]
-5/48*arccosh(x^(1/2))+1/3*x^3*arccosh(x^(1/2))-5/72*x^(3/2)*(-1+x^(1/2))^ (1/2)*(1+x^(1/2))^(1/2)-1/18*x^(5/2)*(-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)- 5/48*x^(1/2)*(-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.68 \[ \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx=\frac {1}{144} \left (-\sqrt {-1+\sqrt {x}} \sqrt {1+\sqrt {x}} \sqrt {x} \left (15+10 x+8 x^2\right )+48 x^3 \text {arccosh}\left (\sqrt {x}\right )-30 \text {arctanh}\left (\sqrt {\frac {-1+\sqrt {x}}{1+\sqrt {x}}}\right )\right ) \]
(-(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x]*(15 + 10*x + 8*x^2)) + 48* x^3*ArcCosh[Sqrt[x]] - 30*ArcTanh[Sqrt[(-1 + Sqrt[x])/(1 + Sqrt[x])]])/144
Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {6432, 27, 845, 845, 845, 852, 43}
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.
\(\displaystyle \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx\) |
\(\Big \downarrow \) 6432 |
\(\displaystyle \frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right )-\frac {1}{3} \int \frac {x^{5/2}}{2 \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right )-\frac {1}{6} \int \frac {x^{5/2}}{\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}dx\) |
\(\Big \downarrow \) 845 |
\(\displaystyle \frac {1}{6} \left (-\frac {5}{6} \int \frac {x^{3/2}}{\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}dx-\frac {1}{3} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{5/2}\right )+\frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 845 |
\(\displaystyle \frac {1}{6} \left (-\frac {5}{6} \left (\frac {3}{4} \int \frac {\sqrt {x}}{\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}dx+\frac {1}{2} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}\right )-\frac {1}{3} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{5/2}\right )+\frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 845 |
\(\displaystyle \frac {1}{6} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}}dx+\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}\right )+\frac {1}{2} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}\right )-\frac {1}{3} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{5/2}\right )+\frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 852 |
\(\displaystyle \frac {1}{6} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\int \frac {1}{\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}d\sqrt {x}+\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}\right )+\frac {1}{2} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}\right )-\frac {1}{3} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{5/2}\right )+\frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 43 |
\(\displaystyle \frac {1}{6} \left (-\frac {5}{6} \left (\frac {3}{4} \left (\text {arccosh}\left (\sqrt {x}\right )+\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x}\right )+\frac {1}{2} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{3/2}\right )-\frac {1}{3} \sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} x^{5/2}\right )+\frac {1}{3} x^3 \text {arccosh}\left (\sqrt {x}\right )\) |
(x^3*ArcCosh[Sqrt[x]])/3 + (-1/3*(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*x^( 5/2)) - (5*((Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*x^(3/2))/2 + (3*(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]]*Sqrt[x] + ArcCosh[Sqrt[x]]))/4))/6)/6
3.3.32.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ ArcCosh[b*(x/a)]/(b*Sqrt[d/b]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a *d, 0] && GtQ[a, 0] && GtQ[d/b, 0]
Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^ (n_))^(p_), x_Symbol] :> Simp[c^(2*n - 1)*(c*x)^(m - 2*n + 1)*(a1 + b1*x^n) ^(p + 1)*((a2 + b2*x^n)^(p + 1)/(b1*b2*(m + 2*n*p + 1))), x] - Simp[a1*a2*c ^(2*n)*((m - 2*n + 1)/(b1*b2*(m + 2*n*p + 1))) Int[(c*x)^(m - 2*n)*(a1 + b1*x^n)^p*(a2 + b2*x^n)^p, x], x] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && Eq Q[a2*b1 + a1*b2, 0] && IGtQ[2*n, 0] && GtQ[m, 2*n - 1] && NeQ[m + 2*n*p + 1 , 0] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]
Int[((c_.)*(x_))^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^ (n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/c Subst[Int[x^ (k*(m + 1) - 1)*(a1 + b1*(x^(k*n)/c^n))^p*(a2 + b2*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a1, b1, a2, b2, c, p}, x] && EqQ[a2*b1 + a1*b2 , 0] && IGtQ[2*n, 0] && FractionQ[m] && IntBinomialQ[a1*a2, b1*b2, c, 2*n, m, p, x]
Int[((a_.) + ArcCosh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Si mp[(c + d*x)^(m + 1)*((a + b*ArcCosh[u])/(d*(m + 1))), x] - Simp[b/(d*(m + 1)) Int[SimplifyIntegrand[(c + d*x)^(m + 1)*(D[u, x]/(Sqrt[-1 + u]*Sqrt[1 + u])), x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFu nctionFreeQ[u, x] && !FunctionOfQ[(c + d*x)^(m + 1), u, x] && !FunctionOf ExponentialQ[u, x]
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.64
method | result | size |
derivativedivides | \(\frac {x^{3} \operatorname {arccosh}\left (\sqrt {x}\right )}{3}-\frac {\sqrt {-1+\sqrt {x}}\, \sqrt {1+\sqrt {x}}\, \left (8 \sqrt {x -1}\, x^{\frac {5}{2}}+10 x^{\frac {3}{2}} \sqrt {x -1}+15 \sqrt {x}\, \sqrt {x -1}+15 \ln \left (\sqrt {x}+\sqrt {x -1}\right )\right )}{144 \sqrt {x -1}}\) | \(75\) |
default | \(\frac {x^{3} \operatorname {arccosh}\left (\sqrt {x}\right )}{3}-\frac {\sqrt {-1+\sqrt {x}}\, \sqrt {1+\sqrt {x}}\, \left (8 \sqrt {x -1}\, x^{\frac {5}{2}}+10 x^{\frac {3}{2}} \sqrt {x -1}+15 \sqrt {x}\, \sqrt {x -1}+15 \ln \left (\sqrt {x}+\sqrt {x -1}\right )\right )}{144 \sqrt {x -1}}\) | \(75\) |
parts | \(\frac {x^{3} \operatorname {arccosh}\left (\sqrt {x}\right )}{3}-\frac {\sqrt {-1+\sqrt {x}}\, \sqrt {1+\sqrt {x}}\, \left (8 \sqrt {x -1}\, x^{\frac {5}{2}}+10 x^{\frac {3}{2}} \sqrt {x -1}+15 \sqrt {x}\, \sqrt {x -1}+15 \ln \left (\sqrt {x}+\sqrt {x -1}\right )\right )}{144 \sqrt {x -1}}\) | \(75\) |
1/3*x^3*arccosh(x^(1/2))-1/144*(-1+x^(1/2))^(1/2)*(1+x^(1/2))^(1/2)*(8*(x- 1)^(1/2)*x^(5/2)+10*x^(3/2)*(x-1)^(1/2)+15*x^(1/2)*(x-1)^(1/2)+15*ln(x^(1/ 2)+(x-1)^(1/2)))/(x-1)^(1/2)
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.34 \[ \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx=-\frac {1}{144} \, {\left (8 \, x^{2} + 10 \, x + 15\right )} \sqrt {x - 1} \sqrt {x} + \frac {1}{48} \, {\left (16 \, x^{3} - 5\right )} \log \left (\sqrt {x - 1} + \sqrt {x}\right ) \]
-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))
\[ \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx=\int x^{2} \operatorname {acosh}{\left (\sqrt {x} \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.48 \[ \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \operatorname {arcosh}\left (\sqrt {x}\right ) - \frac {1}{18} \, \sqrt {x - 1} x^{\frac {5}{2}} - \frac {5}{72} \, \sqrt {x - 1} x^{\frac {3}{2}} - \frac {5}{48} \, \sqrt {x - 1} \sqrt {x} - \frac {5}{48} \, \log \left (2 \, \sqrt {x - 1} + 2 \, \sqrt {x}\right ) \]
1/3*x^3*arccosh(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*log(2*sqrt(x - 1) + 2*sqrt(x))
Time = 0.49 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.51 \[ \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (\sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} + \sqrt {x}\right ) - \frac {1}{144} \, {\left (2 \, {\left (4 \, x + 5\right )} x + 15\right )} \sqrt {x - 1} \sqrt {x} + \frac {5}{48} \, \log \left (-\sqrt {x - 1} + \sqrt {x}\right ) \]
1/3*x^3*log(sqrt(sqrt(x) + 1)*sqrt(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. \[ \int x^2 \text {arccosh}\left (\sqrt {x}\right ) \, dx=\int x^2\,\mathrm {acosh}\left (\sqrt {x}\right ) \,d x \]