3.295 \(\int \frac {\sqrt {\frac {x^2}{-1+a+(1+a) x^2}}}{1+x^2} \, dx\)

Optimal. Leaf size=68 \[ \frac {\sqrt {-\frac {x^2}{-\left ((a+1) x^2\right )-a+1}} \sqrt {(a+1) x^2+a-1} \tan ^{-1}\left (\frac {\sqrt {(a+1) x^2+a-1}}{\sqrt {2}}\right )}{\sqrt {2} x} \]

[Out]

1/2*arctan(1/2*(-1+a+(1+a)*x^2)^(1/2)*2^(1/2))*(-x^2/(1-a-(1+a)*x^2))^(1/2)*(-1+a+(1+a)*x^2)^(1/2)/x*2^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6719, 444, 63, 205} \[ \frac {\sqrt {-\frac {x^2}{-(a+1) x^2-a+1}} \sqrt {(a+1) x^2+a-1} \tan ^{-1}\left (\frac {\sqrt {(a+1) x^2+a-1}}{\sqrt {2}}\right )}{\sqrt {2} x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x^2/(-1 + a + (1 + a)*x^2)]/(1 + x^2),x]

[Out]

(Sqrt[-(x^2/(1 - a - (1 + a)*x^2))]*Sqrt[-1 + a + (1 + a)*x^2]*ArcTan[Sqrt[-1 + a + (1 + a)*x^2]/Sqrt[2]])/(Sq
rt[2]*x)

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 205

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

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]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 65, normalized size = 0.96 \[ \frac {\sqrt {a x^2+a+x^2-1} \sqrt {\frac {x^2}{(a+1) x^2+a-1}} \tan ^{-1}\left (\frac {\sqrt {(a+1) x^2+a-1}}{\sqrt {2}}\right )}{\sqrt {2} x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x^2/(-1 + a + (1 + a)*x^2)]/(1 + x^2),x]

[Out]

(Sqrt[-1 + a + x^2 + a*x^2]*Sqrt[x^2/(-1 + a + (1 + a)*x^2)]*ArcTan[Sqrt[-1 + a + (1 + a)*x^2]/Sqrt[2]])/(Sqrt
[2]*x)

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fricas [A]  time = 0.43, size = 42, normalized size = 0.62 \[ \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left ({\left (a + 1\right )} x^{2} + a - 3\right )} \sqrt {\frac {x^{2}}{{\left (a + 1\right )} x^{2} + a - 1}}}{4 \, x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.45, size = 61, normalized size = 0.90 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {a x^{2} + x^{2} + a - 1}\right ) \mathrm {sgn}\left (a x^{2} + x^{2} + a - 1\right ) \mathrm {sgn}\relax (x) - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} \sqrt {a - 1}\right ) \mathrm {sgn}\left (a - 1\right ) \mathrm {sgn}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 60, normalized size = 0.88 \[ \frac {\sqrt {\frac {x^{2}}{a \,x^{2}+x^{2}+a -1}}\, \sqrt {a \,x^{2}+x^{2}+a -1}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {a \,x^{2}+x^{2}+a -1}\, \sqrt {2}}{2}\right )}{2 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2/(-1+a+(1+a)*x^2))^(1/2)/(x^2+1),x)

[Out]

1/2*(x^2/(a*x^2+x^2+a-1))^(1/2)/x*(a*x^2+x^2+a-1)^(1/2)*2^(1/2)*arctan(1/2*(a*x^2+x^2+a-1)^(1/2)*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {x^{2}}{{\left (a + 1\right )} x^{2} + a - 1}}}{x^{2} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^2/((a + 1)*x^2 + a - 1))/(x^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {x^2}{\left (a+1\right )\,x^2+a-1}}}{x^2+1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2/(a + x^2*(a + 1) - 1))^(1/2)/(x^2 + 1),x)

[Out]

int((x^2/(a + x^2*(a + 1) - 1))^(1/2)/(x^2 + 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {x^{2}}{a x^{2} + a + x^{2} - 1}}}{x^{2} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(x**2/(a*x**2 + a + x**2 - 1))/(x**2 + 1), x)

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