\(\int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx\) [272]

Optimal result

Integrand size = 16, antiderivative size = 29 \[ \int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx=\frac {\tan (x)}{a}+\frac {2 \tan ^3(x)}{3 a}+\frac {\tan ^5(x)}{5 a} \]

[Out]

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

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} \[ \int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx=\frac {\tan ^5(x)}{5 a}+\frac {2 \tan ^3(x)}{3 a}+\frac {\tan (x)}{a} \]

[In]

Int[Sec[x]^4/(a - a*Sin[x]^2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx=\frac {\tan (x)+\frac {2 \tan ^3(x)}{3}+\frac {\tan ^5(x)}{5}}{a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.65 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69

method	result	size
default	\(\frac {\frac {\left (\tan ^{5}\left (x \right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (x \right )\right )}{3}+\tan \left (x \right )}{a}\)	\(20\)
parallelrisch	\(\frac {\tan \left (x \right ) \left (3 \left (\sec ^{4}\left (x \right )\right )+4 \left (\sec ^{2}\left (x \right )\right )+8\right )}{15 a}\)	\(22\)
risch	\(\frac {16 i \left (10 \,{\mathrm e}^{4 i x}+5 \,{\mathrm e}^{2 i x}+1\right )}{15 \left ({\mathrm e}^{2 i x}+1\right )^{5} a}\)	\(32\)
norman	\(\frac {-\frac {2 \tan \left (\frac {x}{2}\right )}{a}+\frac {8 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3 a}-\frac {116 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{15 a}+\frac {8 \left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{3 a}-\frac {2 \left (\tan ^{9}\left (\frac {x}{2}\right )\right )}{a}}{\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right )^{5}}\)	\(66\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx=\frac {{\left (8 \, \cos \left (x\right )^{4} + 4 \, \cos \left (x\right )^{2} + 3\right )} \sin \left (x\right )}{15 \, a \cos \left (x\right )^{5}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx=\frac {3 \, \tan \left (x\right )^{5} + 10 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )}{15 \, a} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx=\frac {3 \, \tan \left (x\right )^{5} + 10 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )}{15 \, a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 13.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^4(x)}{a-a \sin ^2(x)} \, dx=\frac {\mathrm {tan}\left (x\right )\,\left (3\,{\mathrm {tan}\left (x\right )}^4+10\,{\mathrm {tan}\left (x\right )}^2+15\right )}{15\,a} \]

[In]

int(1/(cos(x)^4*(a - a*sin(x)^2)),x)

[Out]

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