\(\int \frac {\sec ^2(x)}{(a-a \sin ^2(x))^2} \, dx\) [283]

Optimal result

Integrand size = 16, antiderivative size = 29 \[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan (x)}{a^2}+\frac {2 \tan ^3(x)}{3 a^2}+\frac {\tan ^5(x)}{5 a^2} \]

[Out]

tan(x)/a^2+2/3*tan(x)^3/a^2+1/5*tan(x)^5/a^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3254, 3852} \[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan ^5(x)}{5 a^2}+\frac {2 \tan ^3(x)}{3 a^2}+\frac {\tan (x)}{a^2} \]

[In]

Int[Sec[x]^2/(a - a*Sin[x]^2)^2,x]

[Out]

Tan[x]/a^2 + (2*Tan[x]^3)/(3*a^2) + Tan[x]^5/(5*a^2)

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 ^6(x) \, dx}{a^2} \\ & = -\frac {\text {Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (x)\right )}{a^2} \\ & = \frac {\tan (x)}{a^2}+\frac {2 \tan ^3(x)}{3 a^2}+\frac {\tan ^5(x)}{5 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan (x)+\frac {2 \tan ^3(x)}{3}+\frac {\tan ^5(x)}{5}}{a^2} \]

[In]

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

[Out]

(Tan[x] + (2*Tan[x]^3)/3 + Tan[x]^5/5)/a^2

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {\frac {\left (\tan ^{5}\left (x \right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (x \right )\right )}{3}+\tan \left (x \right )}{a^{2}}\)	\(20\)
parallelrisch	\(\frac {\tan \left (x \right ) \left (3 \left (\sec ^{4}\left (x \right )\right )+4 \left (\sec ^{2}\left (x \right )\right )+8\right )}{15 a^{2}}\)	\(22\)
risch	\(\frac {16 i \left (10 \,{\mathrm e}^{4 i x}+5 \,{\mathrm e}^{2 i x}+1\right )}{15 \left ({\mathrm e}^{2 i x}+1\right )^{5} a^{2}}\)	\(32\)
norman	\(\frac {-\frac {2 \tan \left (\frac {x}{2}\right )}{a}+\frac {8 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3 a}-\frac {116 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{15 a}+\frac {8 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{3 a}-\frac {2 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{a}}{\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right )^{5} a}\)	\(69\)

[In]

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

[Out]

1/a^2*(1/5*tan(x)^5+2/3*tan(x)^3+tan(x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {{\left (8 \, \cos \left (x\right )^{4} + 4 \, \cos \left (x\right )^{2} + 3\right )} \sin \left (x\right )}{15 \, a^{2} \cos \left (x\right )^{5}} \]

[In]

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

[Out]

1/15*(8*cos(x)^4 + 4*cos(x)^2 + 3)*sin(x)/(a^2*cos(x)^5)

Sympy [F]

\[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\int \frac {\sec ^{2}{\left (x \right )}}{\sin ^{4}{\left (x \right )} - 2 \sin ^{2}{\left (x \right )} + 1}\, dx}{a^{2}} \]

[In]

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

[Out]

Integral(sec(x)**2/(sin(x)**4 - 2*sin(x)**2 + 1), x)/a**2

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {3 \, \tan \left (x\right )^{5} + 10 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )}{15 \, a^{2}} \]

[In]

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

[Out]

1/15*(3*tan(x)^5 + 10*tan(x)^3 + 15*tan(x))/a^2

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {3 \, \tan \left (x\right )^{5} + 10 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )}{15 \, a^{2}} \]

[In]

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

[Out]

1/15*(3*tan(x)^5 + 10*tan(x)^3 + 15*tan(x))/a^2

Mupad [B] (verification not implemented)

Time = 13.66 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\mathrm {tan}\left (x\right )\,\left (3\,{\mathrm {tan}\left (x\right )}^4+10\,{\mathrm {tan}\left (x\right )}^2+15\right )}{15\,a^2} \]

[In]

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

[Out]

(tan(x)*(10*tan(x)^2 + 3*tan(x)^4 + 15))/(15*a^2)