\(\int \frac {\cos ^6(x)}{(a-a \sin ^2(x))^2} \, dx\) [280]

Optimal result

Integrand size = 16, antiderivative size = 20 \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {x}{2 a^2}+\frac {\cos (x) \sin (x)}{2 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, 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, 2715, 8} \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {x}{2 a^2}+\frac {\sin (x) \cos (x)}{2 a^2} \]

[In]

Int[Cos[x]^6/(a - a*Sin[x]^2)^2,x]

[Out]

x/(2*a^2) + (Cos[x]*Sin[x])/(2*a^2)

Rule 8

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

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \cos ^2(x) \, dx}{a^2} \\ & = \frac {\cos (x) \sin (x)}{2 a^2}+\frac {\int 1 \, dx}{2 a^2} \\ & = \frac {x}{2 a^2}+\frac {\cos (x) \sin (x)}{2 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\frac {x}{2}+\frac {1}{4} \sin (2 x)}{a^2} \]

[In]

Integrate[Cos[x]^6/(a - a*Sin[x]^2)^2,x]

[Out]

(x/2 + Sin[2*x]/4)/a^2

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70

method	result	size
parallelrisch	\(\frac {2 x +\sin \left (2 x \right )}{4 a^{2}}\)	\(14\)
risch	\(\frac {x}{2 a^{2}}+\frac {\sin \left (2 x \right )}{4 a^{2}}\)	\(17\)
default	\(\frac {\frac {\tan \left (x \right )}{2+2 \left (\tan ^{2}\left (x \right )\right )}+\frac {\arctan \left (\tan \left (x \right )\right )}{2}}{a^{2}}\)	\(23\)

[In]

int(cos(x)^6/(a-a*sin(x)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/4*(2*x+sin(2*x))/a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\cos \left (x\right ) \sin \left (x\right ) + x}{2 \, a^{2}} \]

[In]

integrate(cos(x)^6/(a-a*sin(x)^2)^2,x, algorithm="fricas")

[Out]

1/2*(cos(x)*sin(x) + x)/a^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (17) = 34\).

Time = 9.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 8.90 \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {x \tan ^{4}{\left (\frac {x}{2} \right )}}{2 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 4 a^{2} \tan ^{2}{\left (\frac {x}{2} \right )} + 2 a^{2}} + \frac {2 x \tan ^{2}{\left (\frac {x}{2} \right )}}{2 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 4 a^{2} \tan ^{2}{\left (\frac {x}{2} \right )} + 2 a^{2}} + \frac {x}{2 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 4 a^{2} \tan ^{2}{\left (\frac {x}{2} \right )} + 2 a^{2}} - \frac {2 \tan ^{3}{\left (\frac {x}{2} \right )}}{2 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 4 a^{2} \tan ^{2}{\left (\frac {x}{2} \right )} + 2 a^{2}} + \frac {2 \tan {\left (\frac {x}{2} \right )}}{2 a^{2} \tan ^{4}{\left (\frac {x}{2} \right )} + 4 a^{2} \tan ^{2}{\left (\frac {x}{2} \right )} + 2 a^{2}} \]

[In]

integrate(cos(x)**6/(a-a*sin(x)**2)**2,x)

[Out]

x*tan(x/2)**4/(2*a**2*tan(x/2)**4 + 4*a**2*tan(x/2)**2 + 2*a**2) + 2*x*tan(x/2)**2/(2*a**2*tan(x/2)**4 + 4*a**
2*tan(x/2)**2 + 2*a**2) + x/(2*a**2*tan(x/2)**4 + 4*a**2*tan(x/2)**2 + 2*a**2) - 2*tan(x/2)**3/(2*a**2*tan(x/2
)**4 + 4*a**2*tan(x/2)**2 + 2*a**2) + 2*tan(x/2)/(2*a**2*tan(x/2)**4 + 4*a**2*tan(x/2)**2 + 2*a**2)

Maxima [A] (verification not implemented)

none

Time = 0.53 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan \left (x\right )}{2 \, {\left (a^{2} \tan \left (x\right )^{2} + a^{2}\right )}} + \frac {x}{2 \, a^{2}} \]

[In]

integrate(cos(x)^6/(a-a*sin(x)^2)^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {x}{2 \, a^{2}} + \frac {\tan \left (x\right )}{2 \, {\left (\tan \left (x\right )^{2} + 1\right )} a^{2}} \]

[In]

integrate(cos(x)^6/(a-a*sin(x)^2)^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.59 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^6(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {2\,x+\sin \left (2\,x\right )}{4\,a^2} \]

[In]

int(cos(x)^6/(a - a*sin(x)^2)^2,x)

[Out]

(2*x + sin(2*x))/(4*a^2)