Integrand size = 17, antiderivative size = 54 \[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=\sqrt {x-x^2}-\frac {3}{2} \arcsin (1-2 x)+\sqrt {2} \arctan \left (\frac {1-3 x}{2 \sqrt {2} \sqrt {x-x^2}}\right ) \] Output:
(-x^2+x)^(1/2)+3/2*arcsin(-1+2*x)+2^(1/2)*arctan(1/4*(1-3*x)*2^(1/2)/(-x^2 +x)^(1/2))
Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.52 \[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=\frac {\sqrt {-((-1+x) x)} \left (\sqrt {-1+x} \sqrt {x}-6 \text {arctanh}\left (\frac {\sqrt {-1+x}}{-1+\sqrt {x}}\right )+2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2}}{\sqrt {\frac {-1+x}{x}}}\right )\right )}{\sqrt {-1+x} \sqrt {x}} \] Input:
Integrate[Sqrt[x - x^2]/(1 + x),x]
Output:
(Sqrt[-((-1 + x)*x)]*(Sqrt[-1 + x]*Sqrt[x] - 6*ArcTanh[Sqrt[-1 + x]/(-1 + Sqrt[x])] + 2*Sqrt[2]*ArcTanh[Sqrt[2]/Sqrt[(-1 + x)/x]]))/(Sqrt[-1 + x]*Sq rt[x])
Time = 0.36 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1162, 1269, 1090, 223, 1154, 217}
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.
\(\displaystyle \int \frac {\sqrt {x-x^2}}{x+1} \, dx\) |
\(\Big \downarrow \) 1162 |
\(\displaystyle \sqrt {x-x^2}-\frac {1}{2} \int \frac {1-3 x}{(x+1) \sqrt {x-x^2}}dx\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle \frac {1}{2} \left (3 \int \frac {1}{\sqrt {x-x^2}}dx-4 \int \frac {1}{(x+1) \sqrt {x-x^2}}dx\right )+\sqrt {x-x^2}\) |
\(\Big \downarrow \) 1090 |
\(\displaystyle \frac {1}{2} \left (-4 \int \frac {1}{(x+1) \sqrt {x-x^2}}dx-3 \int \frac {1}{\sqrt {1-(1-2 x)^2}}d(1-2 x)\right )+\sqrt {x-x^2}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {1}{2} \left (-4 \int \frac {1}{(x+1) \sqrt {x-x^2}}dx-3 \arcsin (1-2 x)\right )+\sqrt {x-x^2}\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{-\frac {(1-3 x)^2}{x-x^2}-8}d\left (-\frac {1-3 x}{\sqrt {x-x^2}}\right )-3 \arcsin (1-2 x)\right )+\sqrt {x-x^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (2 \sqrt {2} \arctan \left (\frac {1-3 x}{2 \sqrt {2} \sqrt {x-x^2}}\right )-3 \arcsin (1-2 x)\right )+\sqrt {x-x^2}\) |
Input:
Int[Sqrt[x - x^2]/(1 + x),x]
Output:
Sqrt[x - x^2] + (-3*ArcSin[1 - 2*x] + 2*Sqrt[2]*ArcTan[(1 - 3*x)/(2*Sqrt[2 ]*Sqrt[x - x^2])])/2
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 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[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p/(e*(m + 2*p + 1)) Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x ] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || LtQ[m, 1]) && !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
Time = 0.52 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(2 \sqrt {2}\, \arctan \left (\frac {\sqrt {-\left (-1+x \right ) x}\, \sqrt {2}}{2 x}\right )+\sqrt {-\left (-1+x \right ) x}-3 \arctan \left (\frac {\sqrt {-\left (-1+x \right ) x}}{x}\right )\) | \(47\) |
default | \(\sqrt {-\left (1+x \right )^{2}+1+3 x}+\frac {3 \arcsin \left (2 x -1\right )}{2}-\sqrt {2}\, \arctan \left (\frac {\left (3 x -1\right ) \sqrt {2}}{4 \sqrt {-\left (1+x \right )^{2}+1+3 x}}\right )\) | \(54\) |
risch | \(-\frac {\left (-1+x \right ) x}{\sqrt {-\left (-1+x \right ) x}}+\frac {3 \arcsin \left (2 x -1\right )}{2}-\sqrt {2}\, \arctan \left (\frac {\left (3 x -1\right ) \sqrt {2}}{4 \sqrt {-\left (1+x \right )^{2}+1+3 x}}\right )\) | \(54\) |
trager | \(\sqrt {-x^{2}+x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +4 \sqrt {-x^{2}+x}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )}{1+x}\right )+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {-x^{2}+x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )\right )}{2}\) | \(90\) |
Input:
int((-x^2+x)^(1/2)/(1+x),x,method=_RETURNVERBOSE)
Output:
2*2^(1/2)*arctan(1/2*(-(-1+x)*x)^(1/2)/x*2^(1/2))+(-(-1+x)*x)^(1/2)-3*arct an((-(-1+x)*x)^(1/2)/x)
Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-x^{2} + x}}{x - 1}\right ) + \sqrt {-x^{2} + x} - 3 \, \arctan \left (\frac {\sqrt {-x^{2} + x}}{x - 1}\right ) \] Input:
integrate((-x^2+x)^(1/2)/(1+x),x, algorithm="fricas")
Output:
2*sqrt(2)*arctan(sqrt(2)*sqrt(-x^2 + x)/(x - 1)) + sqrt(-x^2 + x) - 3*arct an(sqrt(-x^2 + x)/(x - 1))
\[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=\int \frac {\sqrt {- x \left (x - 1\right )}}{x + 1}\, dx \] Input:
integrate((-x**2+x)**(1/2)/(1+x),x)
Output:
Integral(sqrt(-x*(x - 1))/(x + 1), x)
Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=-\sqrt {2} \arcsin \left (\frac {3 \, x}{{\left | x + 1 \right |}} - \frac {1}{{\left | x + 1 \right |}}\right ) + \sqrt {-x^{2} + x} + \frac {3}{2} \, \arcsin \left (2 \, x - 1\right ) \] Input:
integrate((-x^2+x)^(1/2)/(1+x),x, algorithm="maxima")
Output:
-sqrt(2)*arcsin(3*x/abs(x + 1) - 1/abs(x + 1)) + sqrt(-x^2 + x) + 3/2*arcs in(2*x - 1)
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=2 \, \sqrt {2} \arctan \left (\frac {1}{4} \, \sqrt {2} {\left (\frac {3 \, {\left (2 \, \sqrt {-x^{2} + x} - 1\right )}}{2 \, x - 1} - 1\right )}\right ) + \sqrt {-x^{2} + x} + \frac {3}{2} \, \arcsin \left (2 \, x - 1\right ) \] Input:
integrate((-x^2+x)^(1/2)/(1+x),x, algorithm="giac")
Output:
2*sqrt(2)*arctan(1/4*sqrt(2)*(3*(2*sqrt(-x^2 + x) - 1)/(2*x - 1) - 1)) + s qrt(-x^2 + x) + 3/2*arcsin(2*x - 1)
Timed out. \[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=\int \frac {\sqrt {x-x^2}}{x+1} \,d x \] Input:
int((x - x^2)^(1/2)/(x + 1),x)
Output:
int((x - x^2)^(1/2)/(x + 1), x)
Time = 0.18 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.98 \[ \int \frac {\sqrt {x-x^2}}{1+x} \, dx=\sqrt {x}\, \sqrt {1-x}-\sqrt {2}\, \mathrm {log}\left (\sqrt {1-x}+\sqrt {x}\, i -\sqrt {2}-1\right ) i +\sqrt {2}\, \mathrm {log}\left (\sqrt {1-x}+\sqrt {x}\, i -\sqrt {2}+1\right ) i +\sqrt {2}\, \mathrm {log}\left (\sqrt {1-x}+\sqrt {x}\, i +\sqrt {2}-1\right ) i -\sqrt {2}\, \mathrm {log}\left (\sqrt {1-x}+\sqrt {x}\, i +\sqrt {2}+1\right ) i -3 \,\mathrm {log}\left (\sqrt {1-x}+\sqrt {x}\, i \right ) i \] Input:
int((-x^2+x)^(1/2)/(1+x),x)
Output:
sqrt(x)*sqrt( - x + 1) - sqrt(2)*log(sqrt( - x + 1) + sqrt(x)*i - sqrt(2) - 1)*i + sqrt(2)*log(sqrt( - x + 1) + sqrt(x)*i - sqrt(2) + 1)*i + sqrt(2) *log(sqrt( - x + 1) + sqrt(x)*i + sqrt(2) - 1)*i - sqrt(2)*log(sqrt( - x + 1) + sqrt(x)*i + sqrt(2) + 1)*i - 3*log(sqrt( - x + 1) + sqrt(x)*i)*i