3.118 \(\int \sqrt {a-a \sin ^2(x)} \, dx\)

Optimal. Leaf size=13 \[ \tan (x) \sqrt {a \cos ^2(x)} \]

[Out]

(a*cos(x)^2)^(1/2)*tan(x)

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Rubi [A]  time = 0.03, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3176, 3207, 2637} \[ \tan (x) \sqrt {a \cos ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*Sin[x]^2],x]

[Out]

Sqrt[a*Cos[x]^2]*Tan[x]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rubi steps

\begin {align*} \int \sqrt {a-a \sin ^2(x)} \, dx &=\int \sqrt {a \cos ^2(x)} \, dx\\ &=\left (\sqrt {a \cos ^2(x)} \sec (x)\right ) \int \cos (x) \, dx\\ &=\sqrt {a \cos ^2(x)} \tan (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \[ \tan (x) \sqrt {a \cos ^2(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - a*Sin[x]^2],x]

[Out]

Sqrt[a*Cos[x]^2]*Tan[x]

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fricas [A]  time = 0.42, size = 15, normalized size = 1.15 \[ \frac {\sqrt {a \cos \relax (x)^{2}} \sin \relax (x)}{\cos \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a*cos(x)^2)*sin(x)/cos(x)

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giac [B]  time = 0.15, size = 27, normalized size = 2.08 \[ -\frac {2 \, \sqrt {a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right )}{\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(a)*sgn(tan(1/2*x)^4 - 1)/(1/tan(1/2*x) + tan(1/2*x))

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maple [A]  time = 0.52, size = 15, normalized size = 1.15 \[ \frac {a \cos \relax (x ) \sin \relax (x )}{\sqrt {a \left (\cos ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*cos(x)*sin(x)/(a*cos(x)^2)^(1/2)

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maxima [A]  time = 0.48, size = 6, normalized size = 0.46 \[ \sqrt {a} \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*sin(x)

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mupad [B]  time = 0.22, size = 46, normalized size = 3.54 \[ \frac {\sqrt {2}\,\sqrt {a}\,\sqrt {\cos \left (2\,x\right )+1}\,\left (\cos \left (2\,x\right )-1+\sin \left (2\,x\right )\,1{}\mathrm {i}\right )}{2\,\left (\cos \left (2\,x\right )\,1{}\mathrm {i}-\sin \left (2\,x\right )+1{}\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - a*sin(x)^2)^(1/2),x)

[Out]

(2^(1/2)*a^(1/2)*(cos(2*x) + 1)^(1/2)*(cos(2*x) + sin(2*x)*1i - 1))/(2*(cos(2*x)*1i - sin(2*x) + 1i))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- a \sin ^{2}{\relax (x )} + a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-a*sin(x)**2 + a), x)

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