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

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\frac {\text {arctanh}(\sin (x)) \cos (x)}{\sqrt {a \cos ^2(x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3255, 3286, 3855} \[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\frac {\cos (x) \text {arctanh}(\sin (x))}{\sqrt {a \cos ^2(x)}} \]

[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*} \text {integral}& = \int \frac {1}{\sqrt {a \cos ^2(x)}} \, dx \\ & = \frac {\cos (x) \int \sec (x) \, dx}{\sqrt {a \cos ^2(x)}} \\ & = \frac {\text {arctanh}(\sin (x)) \cos (x)}{\sqrt {a \cos ^2(x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\frac {\text {arctanh}(\sin (x)) \cos (x)}{\sqrt {a \cos ^2(x)}} \]

[In]

Integrate[1/Sqrt[a - a*Sin[x]^2],x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.38

method	result	size
default	\(\frac {\operatorname {am}^{-1}\left (x | 1\right )}{\sec \left (x \right ) \sqrt {a \left (\cos ^{2}\left (x \right )\right )}\, \operatorname {csgn}\left (\cos \left (x \right )\right )}\)	\(22\)
risch	\(-\frac {2 \ln \left ({\mathrm e}^{i x}-i\right ) \cos \left (x \right )}{\sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {2 \ln \left ({\mathrm e}^{i x}+i\right ) \cos \left (x \right )}{\sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(64\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (14) = 28\).

Time = 0.27 (sec) , antiderivative size = 65, normalized size of antiderivative = 4.06 \[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\left [-\frac {\sqrt {a \cos \left (x\right )^{2}} \log \left (-\frac {\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right )}{2 \, a \cos \left (x\right )}, -\frac {\sqrt {-a} \arctan \left (\frac {\sqrt {a \cos \left (x\right )^{2}} \sqrt {-a} \sin \left (x\right )}{a \cos \left (x\right )}\right )}{a}\right ] \]

[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]

Sympy [F]

\[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\int \frac {1}{\sqrt {- a \sin ^{2}{\left (x \right )} + a}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(-a*sin(x)**2 + a), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (14) = 28\).

Time = 0.40 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.38 \[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\frac {\log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} + 2 \, \sin \left (x\right ) + 1\right ) - \log \left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \sin \left (x\right ) + 1\right )}{2 \, \sqrt {a}} \]

[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)

Giac [F]

\[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\int { \frac {1}{\sqrt {-a \sin \left (x\right )^{2} + a}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(-a*sin(x)^2 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a-a \sin ^2(x)}} \, dx=\int \frac {1}{\sqrt {a-a\,{\sin \left (x\right )}^2}} \,d x \]

[In]

int(1/(a - a*sin(x)^2)^(1/2),x)

[Out]

int(1/(a - a*sin(x)^2)^(1/2), x)