3.241 \(\int \frac {x}{\sqrt {3-x^2} (5-x^2)} \, dx\)

Optimal. Leaf size=25 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3-x^2}}{\sqrt {2}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {444, 63, 203} \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3-x^2}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[x/(Sqrt[3 - x^2]*(5 - x^2)),x]

[Out]

-(ArcTan[Sqrt[3 - x^2]/Sqrt[2]]/Sqrt[2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {3-x^2} \left (5-x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-x} (5-x)} \, dx,x,x^2\right )\\ &=-\operatorname {Subst}\left (\int \frac {1}{2+x^2} \, dx,x,\sqrt {3-x^2}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {3-x^2}}{\sqrt {2}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3-x^2}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(Sqrt[3 - x^2]*(5 - x^2)),x]

[Out]

-(ArcTan[Sqrt[3 - x^2]/Sqrt[2]]/Sqrt[2])

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IntegrateAlgebraic [A]  time = 0.03, size = 25, normalized size = 1.00 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3-x^2}}{\sqrt {2}}\right )}{\sqrt {2}} \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/(Sqrt[3 - x^2]*(5 - x^2)),x]

[Out]

-(ArcTan[Sqrt[3 - x^2]/Sqrt[2]]/Sqrt[2])

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fricas [A]  time = 1.15, size = 32, normalized size = 1.28 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{2} - 1\right )} \sqrt {-x^{2} + 3}}{4 \, {\left (x^{2} - 3\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x^2+5)/(-x^2+3)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(2)*arctan(1/4*sqrt(2)*(x^2 - 1)*sqrt(-x^2 + 3)/(x^2 - 3))

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giac [A]  time = 0.59, size = 20, normalized size = 0.80 \[ -\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {-x^{2} + 3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x^2+5)/(-x^2+3)^(1/2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.30, size = 48, normalized size = 1.92




method result size



trager \(\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x^{2}-\RootOf \left (\textit {\_Z}^{2}+2\right )-4 \sqrt {-x^{2}+3}}{x^{2}-5}\right )}{4}\) \(48\)
default \(-\frac {\sqrt {2}\, \arctan \left (\frac {\left (-4+2 \sqrt {5}\, \left (x +\sqrt {5}\right )\right ) \sqrt {2}}{4 \sqrt {-\left (x +\sqrt {5}\right )^{2}+2 \sqrt {5}\, \left (x +\sqrt {5}\right )-2}}\right )}{4}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (-4-2 \sqrt {5}\, \left (x -\sqrt {5}\right )\right ) \sqrt {2}}{4 \sqrt {-\left (x -\sqrt {5}\right )^{2}-2 \sqrt {5}\, \left (x -\sqrt {5}\right )-2}}\right )}{4}\) \(100\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(-x^2+5)/(-x^2+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/4*RootOf(_Z^2+2)*ln((RootOf(_Z^2+2)*x^2-RootOf(_Z^2+2)-4*(-x^2+3)^(1/2))/(x^2-5))

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maxima [B]  time = 1.37, size = 101, normalized size = 4.04 \[ -\frac {1}{20} \, \sqrt {5} {\left (\sqrt {5} \sqrt {2} \arcsin \left (\frac {2 \, \sqrt {5} \sqrt {3} x}{3 \, {\left | 2 \, x + 2 \, \sqrt {5} \right |}} + \frac {2 \, \sqrt {3}}{{\left | 2 \, x + 2 \, \sqrt {5} \right |}}\right ) - \sqrt {5} \sqrt {2} \arcsin \left (\frac {2 \, \sqrt {5} \sqrt {3} x}{3 \, {\left | 2 \, x - 2 \, \sqrt {5} \right |}} - \frac {2 \, \sqrt {3}}{{\left | 2 \, x - 2 \, \sqrt {5} \right |}}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x^2+5)/(-x^2+3)^(1/2),x, algorithm="maxima")

[Out]

-1/20*sqrt(5)*(sqrt(5)*sqrt(2)*arcsin(2/3*sqrt(5)*sqrt(3)*x/abs(2*x + 2*sqrt(5)) + 2*sqrt(3)/abs(2*x + 2*sqrt(
5))) - sqrt(5)*sqrt(2)*arcsin(2/3*sqrt(5)*sqrt(3)*x/abs(2*x - 2*sqrt(5)) - 2*sqrt(3)/abs(2*x - 2*sqrt(5))))

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mupad [B]  time = 0.76, size = 83, normalized size = 3.32 \[ -\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (\sqrt {5}\,x+3\right )}{2}+\sqrt {3-x^2}\,1{}\mathrm {i}}{x+\sqrt {5}}\right )\,1{}\mathrm {i}}{4}-\frac {\sqrt {2}\,\ln \left (\frac {\frac {\sqrt {2}\,\left (\sqrt {5}\,x-3\right )}{2}-\sqrt {3-x^2}\,1{}\mathrm {i}}{x-\sqrt {5}}\right )\,1{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x/((3 - x^2)^(1/2)*(x^2 - 5)),x)

[Out]

- (2^(1/2)*log(((2^(1/2)*(5^(1/2)*x + 3))/2 + (3 - x^2)^(1/2)*1i)/(x + 5^(1/2)))*1i)/4 - (2^(1/2)*log(((2^(1/2
)*(5^(1/2)*x - 3))/2 - (3 - x^2)^(1/2)*1i)/(x - 5^(1/2)))*1i)/4

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sympy [A]  time = 5.54, size = 24, normalized size = 0.96 \[ - \frac {\sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} \sqrt {3 - x^{2}}}{2} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(-x**2+5)/(-x**2+3)**(1/2),x)

[Out]

-sqrt(2)*atan(sqrt(2)*sqrt(3 - x**2)/2)/2

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