\(\int \frac {\sec ^4(x)}{(a-a \sin ^2(x))^2} \, dx\) [284]

Optimal result

Integrand size = 16, antiderivative size = 37 \[ \int \frac {\sec ^4(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan (x)}{a^2}+\frac {\tan ^3(x)}{a^2}+\frac {3 \tan ^5(x)}{5 a^2}+\frac {\tan ^7(x)}{7 a^2} \]

[Out]

tan(x)/a^2+tan(x)^3/a^2+3/5*tan(x)^5/a^2+1/7*tan(x)^7/a^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 37, 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} \[ \int \frac {\sec ^4(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan ^7(x)}{7 a^2}+\frac {3 \tan ^5(x)}{5 a^2}+\frac {\tan ^3(x)}{a^2}+\frac {\tan (x)}{a^2} \]

[In]

Int[Sec[x]^4/(a - a*Sin[x]^2)^2,x]

[Out]

Tan[x]/a^2 + Tan[x]^3/a^2 + (3*Tan[x]^5)/(5*a^2) + Tan[x]^7/(7*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 ^8(x) \, dx}{a^2} \\ & = -\frac {\text {Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (x)\right )}{a^2} \\ & = \frac {\tan (x)}{a^2}+\frac {\tan ^3(x)}{a^2}+\frac {3 \tan ^5(x)}{5 a^2}+\frac {\tan ^7(x)}{7 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.73 \[ \int \frac {\sec ^4(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\tan (x)+\tan ^3(x)+\frac {3 \tan ^5(x)}{5}+\frac {\tan ^7(x)}{7}}{a^2} \]

[In]

Integrate[Sec[x]^4/(a - a*Sin[x]^2)^2,x]

[Out]

(Tan[x] + Tan[x]^3 + (3*Tan[x]^5)/5 + Tan[x]^7/7)/a^2

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.65

method	result	size
default	\(\frac {\frac {\left (\tan ^{7}\left (x \right )\right )}{7}+\frac {3 \left (\tan ^{5}\left (x \right )\right )}{5}+\tan ^{3}\left (x \right )+\tan \left (x \right )}{a^{2}}\)	\(24\)
parallelrisch	\(\frac {\tan \left (x \right ) \left (\sec ^{6}\left (x \right )\right ) \left (32+\cos \left (6 x \right )+8 \cos \left (4 x \right )+29 \cos \left (2 x \right )\right )}{70 a^{2}}\)	\(30\)
risch	\(\frac {32 i \left (35 \,{\mathrm e}^{6 i x}+21 \,{\mathrm e}^{4 i x}+7 \,{\mathrm e}^{2 i x}+1\right )}{35 \left ({\mathrm e}^{2 i x}+1\right )^{7} a^{2}}\)	\(39\)
norman	\(\frac {-\frac {2 \tan \left (\frac {x}{2}\right )}{a}+\frac {4 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{a}-\frac {86 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{5 a}+\frac {424 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{35 a}-\frac {86 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{5 a}+\frac {4 \left (\tan ^{11}\left (\frac {x}{2}\right )\right )}{a}-\frac {2 \left (\tan ^{13}\left (\frac {x}{2}\right )\right )}{a}}{\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right )^{7} a}\)	\(91\)

[In]

int(sec(x)^4/(a-a*sin(x)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^2*(1/7*tan(x)^7+3/5*tan(x)^5+tan(x)^3+tan(x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.84 \[ \int \frac {\sec ^4(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {{\left (16 \, \cos \left (x\right )^{6} + 8 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 5\right )} \sin \left (x\right )}{35 \, a^{2} \cos \left (x\right )^{7}} \]

[In]

integrate(sec(x)^4/(a-a*sin(x)^2)^2,x, algorithm="fricas")

[Out]

1/35*(16*cos(x)^6 + 8*cos(x)^4 + 6*cos(x)^2 + 5)*sin(x)/(a^2*cos(x)^7)

Sympy [F]

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

[In]

integrate(sec(x)**4/(a-a*sin(x)**2)**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^4(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {5 \, \tan \left (x\right )^{7} + 21 \, \tan \left (x\right )^{5} + 35 \, \tan \left (x\right )^{3} + 35 \, \tan \left (x\right )}{35 \, a^{2}} \]

[In]

integrate(sec(x)^4/(a-a*sin(x)^2)^2,x, algorithm="maxima")

[Out]

1/35*(5*tan(x)^7 + 21*tan(x)^5 + 35*tan(x)^3 + 35*tan(x))/a^2

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^4(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {5 \, \tan \left (x\right )^{7} + 21 \, \tan \left (x\right )^{5} + 35 \, \tan \left (x\right )^{3} + 35 \, \tan \left (x\right )}{35 \, a^{2}} \]

[In]

integrate(sec(x)^4/(a-a*sin(x)^2)^2,x, algorithm="giac")

[Out]

1/35*(5*tan(x)^7 + 21*tan(x)^5 + 35*tan(x)^3 + 35*tan(x))/a^2

Mupad [B] (verification not implemented)

Time = 13.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^4(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\mathrm {tan}\left (x\right )}{a^2}+\frac {{\mathrm {tan}\left (x\right )}^3}{a^2}+\frac {3\,{\mathrm {tan}\left (x\right )}^5}{5\,a^2}+\frac {{\mathrm {tan}\left (x\right )}^7}{7\,a^2} \]

[In]

int(1/(cos(x)^4*(a - a*sin(x)^2)^2),x)

[Out]

tan(x)/a^2 + tan(x)^3/a^2 + (3*tan(x)^5)/(5*a^2) + tan(x)^7/(7*a^2)