### 3.137 $$\int \frac{x^2}{\sqrt{5-x^2}} \, dx$$

Optimal. Leaf size=29 $\frac{5}{2} \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right )-\frac{1}{2} x \sqrt{5-x^2}$

[Out]

-(x*Sqrt[5 - x^2])/2 + (5*ArcSin[x/Sqrt[5]])/2

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Rubi [A]  time = 0.0061221, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.133, Rules used = {321, 216} $\frac{5}{2} \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right )-\frac{1}{2} x \sqrt{5-x^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2/Sqrt[5 - x^2],x]

[Out]

-(x*Sqrt[5 - x^2])/2 + (5*ArcSin[x/Sqrt[5]])/2

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

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]

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{5-x^2}} \, dx &=-\frac{1}{2} x \sqrt{5-x^2}+\frac{5}{2} \int \frac{1}{\sqrt{5-x^2}} \, dx\\ &=-\frac{1}{2} x \sqrt{5-x^2}+\frac{5}{2} \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0089389, size = 29, normalized size = 1. $\frac{5}{2} \sin ^{-1}\left (\frac{x}{\sqrt{5}}\right )-\frac{1}{2} x \sqrt{5-x^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2/Sqrt[5 - x^2],x]

[Out]

-(x*Sqrt[5 - x^2])/2 + (5*ArcSin[x/Sqrt[5]])/2

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Maple [A]  time = 0.005, size = 23, normalized size = 0.8 \begin{align*}{\frac{5}{2}\arcsin \left ({\frac{x\sqrt{5}}{5}} \right ) }-{\frac{x}{2}\sqrt{-{x}^{2}+5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-x^2+5)^(1/2),x)

[Out]

5/2*arcsin(1/5*x*5^(1/2))-1/2*x*(-x^2+5)^(1/2)

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Maxima [A]  time = 1.40043, size = 30, normalized size = 1.03 \begin{align*} -\frac{1}{2} \, \sqrt{-x^{2} + 5} x + \frac{5}{2} \, \arcsin \left (\frac{1}{5} \, \sqrt{5} x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^2+5)^(1/2),x, algorithm="maxima")

[Out]

-1/2*sqrt(-x^2 + 5)*x + 5/2*arcsin(1/5*sqrt(5)*x)

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Fricas [A]  time = 2.27144, size = 73, normalized size = 2.52 \begin{align*} -\frac{1}{2} \, \sqrt{-x^{2} + 5} x - \frac{5}{2} \, \arctan \left (\frac{\sqrt{-x^{2} + 5}}{x}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^2+5)^(1/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(-x^2 + 5)*x - 5/2*arctan(sqrt(-x^2 + 5)/x)

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Sympy [A]  time = 0.202197, size = 24, normalized size = 0.83 \begin{align*} - \frac{x \sqrt{5 - x^{2}}}{2} + \frac{5 \operatorname{asin}{\left (\frac{\sqrt{5} x}{5} \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(-x**2+5)**(1/2),x)

[Out]

-x*sqrt(5 - x**2)/2 + 5*asin(sqrt(5)*x/5)/2

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Giac [A]  time = 1.08117, size = 30, normalized size = 1.03 \begin{align*} -\frac{1}{2} \, \sqrt{-x^{2} + 5} x + \frac{5}{2} \, \arcsin \left (\frac{1}{5} \, \sqrt{5} x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-x^2+5)^(1/2),x, algorithm="giac")

[Out]

-1/2*sqrt(-x^2 + 5)*x + 5/2*arcsin(1/5*sqrt(5)*x)