3.2.19 \(\int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx\) [119]

Optimal. Leaf size=16 \[ \frac {\tanh ^{-1}(\sin (x)) \cos (x)}{\sqrt {a \cos ^2(x)}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 16, 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 = {3255, 3286, 3855} \begin {gather*} \frac {\cos (x) \tanh ^{-1}(\sin (x))}{\sqrt {a \cos ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a - a*Sin[x]^2],x]

[Out]

(ArcTanh[Sin[x]]*Cos[x])/Sqrt[a*Cos[x]^2]

Rule 3255

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 3286

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
]])

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx &=\int \frac {1}{\sqrt {a \cos ^2(x)}} \, dx\\ &=\frac {\cos (x) \int \sec (x) \, dx}{\sqrt {a \cos ^2(x)}}\\ &=\frac {\tanh ^{-1}(\sin (x)) \cos (x)}{\sqrt {a \cos ^2(x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(46\) vs. \(2(16)=32\).
time = 0.02, size = 46, normalized size = 2.88 \begin {gather*} \frac {\cos (x) \left (-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )\right )}{\sqrt {a \cos ^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a - a*Sin[x]^2],x]

[Out]

(Cos[x]*(-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]]))/Sqrt[a*Cos[x]^2]

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.14, size = 20, normalized size = 1.25

method	result	size
default	\(\frac {\cos \left (x \right ) \mathrm {am}^{-1}\left (x | 1\right )}{\sqrt {a \left (\cos ^{2}\left (x \right )\right )}\, \mathrm {csgn}\left (\cos \left (x \right )\right )}\)	\(20\)
risch	\(-\frac {2 \ln \left ({\mathrm e}^{i x}-i\right ) \cos \left (x \right )}{\sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {2 \ln \left ({\mathrm e}^{i x}+i\right ) \cos \left (x \right )}{\sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*cos(x)^2)^(1/2)/csgn(cos(x))*cos(x)*InverseJacobiAM(x,1)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (14) = 28\).
time = 0.61, size = 38, normalized size = 2.38 \begin {gather*} \frac {\log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) - \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right )}{2 \, \sqrt {a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).
time = 0.42, size = 65, normalized size = 4.06 \begin {gather*} \left [-\frac {\sqrt {a \cos \left (x\right )^{2}} \log \left (-\frac {\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right )}{2 \, a \cos \left (x\right )}, -\frac {\sqrt {-a} \arctan \left (\frac {\sqrt {a \cos \left (x\right )^{2}} \sqrt {-a} \sin \left (x\right )}{a \cos \left (x\right )}\right )}{a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*sqrt(a*cos(x)^2)*log(-(sin(x) - 1)/(sin(x) + 1))/(a*cos(x)), -sqrt(-a)*arctan(sqrt(a*cos(x)^2)*sqrt(-a)*
sin(x)/(a*cos(x)))/a]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {- a \sin ^{2}{\left (x \right )} + a}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(-a*sin(x)**2 + a), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(-a*sin(x)^2 + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} \int \frac {1}{\sqrt {a-a\,{\sin \left (x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - a*sin(x)^2)^(1/2),x)

[Out]

int(1/(a - a*sin(x)^2)^(1/2), x)

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