Integrand size = 8, antiderivative size = 60 \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=-\frac {3}{16} \sqrt {1-x} \sqrt {x}-\frac {1}{8} \sqrt {1-x} x^{3/2}+\frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )-\frac {3}{32} \arcsin (1-2 x) \] Output:
-3/16*(1-x)^(1/2)*x^(1/2)-1/8*(1-x)^(1/2)*x^(3/2)+1/2*x^2*arccos(x^(1/2))+ 3/32*arcsin(-1+2*x)
Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68 \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=\frac {1}{16} \left (-\sqrt {-((-1+x) x)} (3+2 x)+8 x^2 \arccos \left (\sqrt {x}\right )+3 \arcsin \left (\sqrt {x}\right )\right ) \] Input:
Integrate[x*ArcCos[Sqrt[x]],x]
Output:
(-(Sqrt[-((-1 + x)*x)]*(3 + 2*x)) + 8*x^2*ArcCos[Sqrt[x]] + 3*ArcSin[Sqrt[ x]])/16
Time = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5342, 27, 60, 60, 62, 1090, 223}
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.
\(\displaystyle \int x \arccos \left (\sqrt {x}\right ) \, dx\) |
\(\Big \downarrow \) 5342 |
\(\displaystyle \frac {1}{2} \int \frac {x^{3/2}}{2 \sqrt {1-x}}dx+\frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {x^{3/2}}{\sqrt {1-x}}dx+\frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{4} \int \frac {\sqrt {x}}{\sqrt {1-x}}dx-\frac {1}{2} \sqrt {1-x} x^{3/2}\right )+\frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-x} \sqrt {x}}dx-\sqrt {1-x} \sqrt {x}\right )-\frac {1}{2} \sqrt {1-x} x^{3/2}\right )+\frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 62 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {x-x^2}}dx-\sqrt {1-x} \sqrt {x}\right )-\frac {1}{2} \sqrt {1-x} x^{3/2}\right )+\frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{4} \left (\frac {3}{4} \left (-\frac {1}{2} \int \frac {1}{\sqrt {1-(1-2 x)^2}}d(1-2 x)-\sqrt {1-x} \sqrt {x}\right )-\frac {1}{2} \sqrt {1-x} x^{3/2}\right )+\frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{2} x^2 \arccos \left (\sqrt {x}\right )+\frac {1}{4} \left (\frac {3}{4} \left (-\frac {1}{2} \arcsin (1-2 x)-\sqrt {1-x} \sqrt {x}\right )-\frac {1}{2} \sqrt {1-x} x^{3/2}\right )\) |
Input:
Int[x*ArcCos[Sqrt[x]],x]
Output:
(x^2*ArcCos[Sqrt[x]])/2 + (-1/2*(Sqrt[1 - x]*x^(3/2)) + (3*(-(Sqrt[1 - x]* Sqrt[x]) - ArcSin[1 - 2*x]/2))/4)/4
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[(a_.) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Int[ 1/Sqrt[a*c - b*(a - c)*x - b^2*x^2], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/(2*c*(-4* (c/(b^2 - 4*a*c)))^p) Subst[Int[Simp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]
Int[((a_.) + ArcCos[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Sim p[(c + d*x)^(m + 1)*((a + b*ArcCos[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 = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {x^{2} \arccos \left (\sqrt {x}\right )}{2}-\frac {\sqrt {1-x}\, x^{\frac {3}{2}}}{8}-\frac {3 \sqrt {1-x}\, \sqrt {x}}{16}+\frac {3 \arcsin \left (\sqrt {x}\right )}{16}\) | \(41\) |
default | \(\frac {x^{2} \arccos \left (\sqrt {x}\right )}{2}-\frac {\sqrt {1-x}\, x^{\frac {3}{2}}}{8}-\frac {3 \sqrt {1-x}\, \sqrt {x}}{16}+\frac {3 \arcsin \left (\sqrt {x}\right )}{16}\) | \(41\) |
parts | \(\frac {x^{2} \arccos \left (\sqrt {x}\right )}{2}-\frac {\sqrt {1-x}\, x^{\frac {3}{2}}}{8}-\frac {3 \sqrt {1-x}\, \sqrt {x}}{16}+\frac {3 \sqrt {x \left (1-x \right )}\, \arcsin \left (-1+2 x \right )}{32 \sqrt {x}\, \sqrt {1-x}}\) | \(62\) |
Input:
int(x*arccos(x^(1/2)),x,method=_RETURNVERBOSE)
Output:
1/2*x^2*arccos(x^(1/2))-1/8*(1-x)^(1/2)*x^(3/2)-3/16*(1-x)^(1/2)*x^(1/2)+3 /16*arcsin(x^(1/2))
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.52 \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=-\frac {1}{16} \, {\left (2 \, x + 3\right )} \sqrt {x} \sqrt {-x + 1} + \frac {1}{16} \, {\left (8 \, x^{2} - 3\right )} \arccos \left (\sqrt {x}\right ) \] Input:
integrate(x*arccos(x^(1/2)),x, algorithm="fricas")
Output:
-1/16*(2*x + 3)*sqrt(x)*sqrt(-x + 1) + 1/16*(8*x^2 - 3)*arccos(sqrt(x))
Time = 0.42 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.77 \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=\frac {x^{2} \operatorname {acos}{\left (\sqrt {x} \right )}}{2} + \frac {\sqrt {1 - x} \left (- \frac {x^{\frac {3}{2}}}{4} - \frac {3 \sqrt {x}}{8}\right )}{2} + \frac {3 \operatorname {asin}{\left (\sqrt {x} \right )}}{16} \] Input:
integrate(x*acos(x**(1/2)),x)
Output:
x**2*acos(sqrt(x))/2 + sqrt(1 - x)*(-x**(3/2)/4 - 3*sqrt(x)/8)/2 + 3*asin( sqrt(x))/16
Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.67 \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=\frac {1}{2} \, x^{2} \arccos \left (\sqrt {x}\right ) - \frac {1}{8} \, x^{\frac {3}{2}} \sqrt {-x + 1} - \frac {3}{16} \, \sqrt {x} \sqrt {-x + 1} + \frac {3}{16} \, \arcsin \left (\sqrt {x}\right ) \] Input:
integrate(x*arccos(x^(1/2)),x, algorithm="maxima")
Output:
1/2*x^2*arccos(sqrt(x)) - 1/8*x^(3/2)*sqrt(-x + 1) - 3/16*sqrt(x)*sqrt(-x + 1) + 3/16*arcsin(sqrt(x))
Time = 0.11 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.67 \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=\frac {1}{2} \, x^{2} \arccos \left (\sqrt {x}\right ) - \frac {1}{8} \, x^{\frac {3}{2}} \sqrt {-x + 1} - \frac {3}{16} \, \sqrt {x} \sqrt {-x + 1} - \frac {3}{16} \, \arccos \left (\sqrt {x}\right ) \] Input:
integrate(x*arccos(x^(1/2)),x, algorithm="giac")
Output:
1/2*x^2*arccos(sqrt(x)) - 1/8*x^(3/2)*sqrt(-x + 1) - 3/16*sqrt(x)*sqrt(-x + 1) - 3/16*arccos(sqrt(x))
Timed out. \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=\int x\,\mathrm {acos}\left (\sqrt {x}\right ) \,d x \] Input:
int(x*acos(x^(1/2)),x)
Output:
int(x*acos(x^(1/2)), x)
Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.58 \[ \int x \arccos \left (\sqrt {x}\right ) \, dx=\frac {\mathit {acos} \left (\sqrt {x}\right ) x^{2}}{2}+\frac {3 \mathit {asin} \left (\sqrt {x}\right )}{16}-\frac {\sqrt {x}\, \sqrt {1-x}\, x}{8}-\frac {3 \sqrt {x}\, \sqrt {1-x}}{16} \] Input:
int(x*acos(x^(1/2)),x)
Output:
(8*acos(sqrt(x))*x**2 + 3*asin(sqrt(x)) - 2*sqrt(x)*sqrt( - x + 1)*x - 3*s qrt(x)*sqrt( - x + 1))/16