3.1.100 \(\int \frac {x^2}{\sqrt {1-(1+x)^2}} \, dx\) [100]

Optimal. Leaf size=44 \[ \frac {3}{2} \sqrt {1-(1+x)^2}-\frac {1}{2} x \sqrt {1-(1+x)^2}+\frac {3}{2} \sin ^{-1}(1+x) \]

[Out]

3/2*arcsin(1+x)+3/2*(1-(1+x)^2)^(1/2)-1/2*x*(1-(1+x)^2)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {378, 685, 655, 222} \begin {gather*} \frac {3}{2} \text {ArcSin}(x+1)-\frac {1}{2} \sqrt {1-(x+1)^2} x+\frac {3}{2} \sqrt {1-(x+1)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[1 - (1 + x)^2],x]

[Out]

(3*Sqrt[1 - (1 + x)^2])/2 - (x*Sqrt[1 - (1 + x)^2])/2 + (3*ArcSin[1 + x])/2

Rule 222

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]

Rule 378

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 655

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[e*((a + c*x^2)^(p + 1)/(2*c*(p + 1))),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 685

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*((a + c*x^2)^(p
+ 1)/(c*(m + 2*p + 1))), x] + Dist[2*c*d*((m + p)/(c*(m + 2*p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p, x]
, x] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p
]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {1-(1+x)^2}} \, dx &=\text {Subst}\left (\int \frac {(-1+x)^2}{\sqrt {1-x^2}} \, dx,x,1+x\right )\\ &=-\frac {1}{2} x \sqrt {1-(1+x)^2}-\frac {3}{2} \text {Subst}\left (\int \frac {-1+x}{\sqrt {1-x^2}} \, dx,x,1+x\right )\\ &=\frac {3}{2} \sqrt {1-(1+x)^2}-\frac {1}{2} x \sqrt {1-(1+x)^2}+\frac {3}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,1+x\right )\\ &=\frac {3}{2} \sqrt {1-(1+x)^2}-\frac {1}{2} x \sqrt {1-(1+x)^2}+\frac {3}{2} \sin ^{-1}(1+x)\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 51, normalized size = 1.16 \begin {gather*} \frac {x \left (-6-x+x^2\right )+6 \sqrt {x} \sqrt {2+x} \tanh ^{-1}\left (\sqrt {\frac {x}{2+x}}\right )}{2 \sqrt {-x (2+x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[1 - (1 + x)^2],x]

[Out]

(x*(-6 - x + x^2) + 6*Sqrt[x]*Sqrt[2 + x]*ArcTanh[Sqrt[x/(2 + x)]])/(2*Sqrt[-(x*(2 + x))])

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Maple [A]
time = 0.18, size = 35, normalized size = 0.80

method	result	size
risch	\(\frac {\left (x -3\right ) x \left (x +2\right )}{2 \sqrt {-x \left (x +2\right )}}+\frac {3 \arcsin \left (1+x \right )}{2}\)	\(25\)
default	\(-\frac {x \sqrt {-x^{2}-2 x}}{2}+\frac {3 \sqrt {-x^{2}-2 x}}{2}+\frac {3 \arcsin \left (1+x \right )}{2}\)	\(35\)
meijerg	\(-\frac {4 i \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-5 x +15\right ) \sqrt {1+\frac {x}{2}}}{40}+\frac {3 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {2}\, \sqrt {x}}{2}\right )}{4}\right )}{\sqrt {\pi }}\)	\(45\)
trager	\(\left (-\frac {x}{2}+\frac {3}{2}\right ) \sqrt {-x^{2}-2 x}-\frac {3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (x \RootOf \left (\textit {\_Z}^{2}+1\right )+\sqrt {-x^{2}-2 x}+\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{2}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(1-(1+x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*x*(-x^2-2*x)^(1/2)+3/2*(-x^2-2*x)^(1/2)+3/2*arcsin(1+x)

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Maxima [A]
time = 0.48, size = 36, normalized size = 0.82 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} - 2 \, x} x + \frac {3}{2} \, \sqrt {-x^{2} - 2 \, x} - \frac {3}{2} \, \arcsin \left (-x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(1-(1+x)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(-x^2 - 2*x)*x + 3/2*sqrt(-x^2 - 2*x) - 3/2*arcsin(-x - 1)

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Fricas [A]
time = 0.37, size = 35, normalized size = 0.80 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} - 2 \, x} {\left (x - 3\right )} - 3 \, \arctan \left (\frac {\sqrt {-x^{2} - 2 \, x}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(1-(1+x)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(-x^2 - 2*x)*(x - 3) - 3*arctan(sqrt(-x^2 - 2*x)/x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\sqrt {- x \left (x + 2\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(1-(1+x)**2)**(1/2),x)

[Out]

Integral(x**2/sqrt(-x*(x + 2)), x)

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Giac [A]
time = 5.58, size = 23, normalized size = 0.52 \begin {gather*} -\frac {1}{2} \, \sqrt {-x^{2} - 2 \, x} {\left (x - 3\right )} + \frac {3}{2} \, \arcsin \left (x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(1-(1+x)^2)^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(-x^2 - 2*x)*(x - 3) + 3/2*arcsin(x + 1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{\sqrt {1-{\left (x+1\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(1 - (x + 1)^2)^(1/2),x)

[Out]

int(x^2/(1 - (x + 1)^2)^(1/2), x)

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