\(\int \frac {\cos ^2(x)}{(a-a \sin ^2(x))^2} \, dx\) [282]

Optimal result

Integrand size = 16, antiderivative size = 6 \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan (x)}{a^2} \]

[Out]

tan(x)/a^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3254, 3852, 8} \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan (x)}{a^2} \]

[In]

Int[Cos[x]^2/(a - a*Sin[x]^2)^2,x]

[Out]

Tan[x]/a^2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3254

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^2(x) \, dx}{a^2} \\ & = -\frac {\text {Subst}(\int 1 \, dx,x,-\tan (x))}{a^2} \\ & = \frac {\tan (x)}{a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan (x)}{a^2} \]

[In]

Integrate[Cos[x]^2/(a - a*Sin[x]^2)^2,x]

[Out]

Tan[x]/a^2

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.17

method	result	size
default	\(\frac {\tan \left (x \right )}{a^{2}}\)	\(7\)
parallelrisch	\(\frac {\tan \left (x \right )}{a^{2}}\)	\(7\)
risch	\(\frac {2 i}{a^{2} \left ({\mathrm e}^{2 i x}+1\right )}\)	\(16\)
norman	\(\frac {-\frac {2 \tan \left (\frac {x}{2}\right )}{a}+\frac {4 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{a}-\frac {2 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{a}}{a \left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )-1\right )^{3}}\)	\(57\)

[In]

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

[Out]

tan(x)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.67 \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\sin \left (x\right )}{a^{2} \cos \left (x\right )} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 20 vs. \(2 (5) = 10\).

Time = 1.62 (sec) , antiderivative size = 20, normalized size of antiderivative = 3.33 \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=- \frac {2 \tan {\left (\frac {x}{2} \right )}}{a^{2} \tan ^{2}{\left (\frac {x}{2} \right )} - a^{2}} \]

[In]

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

[Out]

-2*tan(x/2)/(a**2*tan(x/2)**2 - a**2)

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan \left (x\right )}{a^{2}} \]

[In]

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

[Out]

tan(x)/a^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan \left (x\right )}{a^{2}} \]

[In]

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

[Out]

tan(x)/a^2

Mupad [B] (verification not implemented)

Time = 13.89 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\mathrm {tan}\left (x\right )}{a^2} \]

[In]

int(cos(x)^2/(a - a*sin(x)^2)^2,x)

[Out]

tan(x)/a^2