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

Optimal. Leaf size=55 \[ (a+x) \sin ^{-1}\left (\sqrt {\frac {x-a}{a+x}}\right )-\frac {\sqrt {2} a \sqrt {\frac {x-a}{a+x}}}{\sqrt {\frac {a}{a+x}}} \]

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Rubi [B]  time = 0.84, antiderivative size = 118, normalized size of antiderivative = 2.15, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4840, 12, 6677, 6720, 385, 217, 206} \[ -\sqrt {2} \sqrt {\frac {a}{a+x}} \sqrt {-\frac {a-x}{a+x}} (a+x)+x \sin ^{-1}\left (\sqrt {-\frac {a-x}{a+x}}\right )-\frac {a \sqrt {\frac {a}{a+x}} \tanh ^{-1}\left (\frac {\sqrt {-\frac {a-x}{a+x}}}{\sqrt {2} \sqrt {-\frac {a}{a+x}}}\right )}{\sqrt {-\frac {a}{a+x}}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[Sqrt[(-a + x)/(a + x)]],x]

[Out]

-(Sqrt[2]*Sqrt[a/(a + x)]*Sqrt[-((a - x)/(a + x))]*(a + x)) + x*ArcSin[Sqrt[-((a - x)/(a + x))]] - (a*Sqrt[a/(
a + x)]*ArcTanh[Sqrt[-((a - x)/(a + x))]/(Sqrt[2]*Sqrt[-(a/(a + x))])])/Sqrt[-(a/(a + x))]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 4840

Int[ArcSin[u_], x_Symbol] :> Simp[x*ArcSin[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 - u^2], x], x] /;
InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 6677

Int[(u_)*((c_.)*((a_.) + (b_.)*(x_))^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c*(a + b*x)^n)^FracPart[p])/
(a + b*x)^(n*FracPart[p]), Int[u*(a + b*x)^(n*p), x], x] /; FreeQ[{a, b, c, n, p}, x] &&  !IntegerQ[p] &&  !Ma
tchQ[u, x^(n1_.)*(v_.) /; EqQ[n, n1 + 1]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 99, normalized size = 1.80 \[ x \sin ^{-1}\left (\sqrt {\frac {x-a}{a+x}}\right )+\frac {\sqrt {\frac {a}{a+x}} \left (\sqrt {2} \sqrt {a} \sqrt {x-a} \tan ^{-1}\left (\frac {\sqrt {x-a}}{\sqrt {2} \sqrt {a}}\right )+2 a-2 x\right )}{\sqrt {2} \sqrt {\frac {x-a}{a+x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[Sqrt[(-a + x)/(a + x)]],x]

[Out]

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

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

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[ArcSin[Sqrt[(-a + x)/(a + x)]],x]

[Out]

Could not integrate

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fricas [A]  time = 1.14, size = 51, normalized size = 0.93 \[ -\sqrt {2} {\left (a + x\right )} \sqrt {-\frac {a - x}{a + x}} \sqrt {\frac {a}{a + x}} + {\left (a + x\right )} \arcsin \left (\sqrt {-\frac {a - x}{a + x}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*(a + x)*sqrt(-(a - x)/(a + x))*sqrt(a/(a + x)) + (a + x)*arcsin(sqrt(-(a - x)/(a + x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(sqrt(-(a - x)/(a + x))), x)

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maple [A]  time = 0.06, size = 87, normalized size = 1.58

method	result	size
default	\(x \arcsin \left (\sqrt {\frac {-a +x}{a +x}}\right )-\frac {\sqrt {-a +x}\, \sqrt {2}\, \sqrt {\frac {a}{a +x}}\, \left (-\sqrt {a}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {-a +x}\, \sqrt {2}}{2 \sqrt {a}}\right )+2 \sqrt {-a +x}\right )}{2 \sqrt {-\frac {a -x}{a +x}}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.01, size = 103, normalized size = 1.87 \[ a {\left (\frac {2 \, \arcsin \left (\sqrt {-\frac {a - x}{a + x}}\right )}{\frac {a - x}{a + x} + 1} + \frac {\sqrt {\frac {a - x}{a + x} + 1}}{\sqrt {-\frac {a - x}{a + x}} + 1} + \frac {\sqrt {\frac {a - x}{a + x} + 1}}{\sqrt {-\frac {a - x}{a + x}} - 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(2*arcsin(sqrt(-(a - x)/(a + x)))/((a - x)/(a + x) + 1) + sqrt((a - x)/(a + x) + 1)/(sqrt(-(a - x)/(a + x))
+ 1) + sqrt((a - x)/(a + x) + 1)/(sqrt(-(a - x)/(a + x)) - 1))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {asin}\left (\sqrt {-\frac {a-x}{a+x}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin((-(a - x)/(a + x))^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {asin}{\left (\sqrt {\frac {- a + x}{a + x}} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(sqrt((-a + x)/(a + x))), x)

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