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

Optimal. Leaf size=29 \[ \frac{\tan ^5(x)}{5 a^2}+\frac{2 \tan ^3(x)}{3 a^2}+\frac{\tan (x)}{a^2} \]

[Out]

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

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Rubi [A]  time = 0.046838, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3175, 3767} \[ \frac{\tan ^5(x)}{5 a^2}+\frac{2 \tan ^3(x)}{3 a^2}+\frac{\tan (x)}{a^2} \]

Antiderivative was successfully verified.

[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 3175

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 3767

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*} \int \frac{\sec ^2(x)}{\left (a-a \sin ^2(x)\right )^2} \, dx &=\frac{\int \sec ^6(x) \, dx}{a^2}\\ &=-\frac{\operatorname{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]  time = 0.0034332, size = 31, normalized size = 1.07 \[ \frac{\frac{8 \tan (x)}{15}+\frac{1}{5} \tan (x) \sec ^4(x)+\frac{4}{15} \tan (x) \sec ^2(x)}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*Tan[x])/15 + (4*Sec[x]^2*Tan[x])/15 + (Sec[x]^4*Tan[x])/5)/a^2

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Maple [A]  time = 0.043, size = 20, normalized size = 0.7 \begin{align*}{\frac{1}{{a}^{2}} \left ({\frac{ \left ( \tan \left ( x \right ) \right ) ^{5}}{5}}+{\frac{2\, \left ( \tan \left ( x \right ) \right ) ^{3}}{3}}+\tan \left ( x \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.978235, size = 30, normalized size = 1.03 \begin{align*} \frac{3 \, \tan \left (x\right )^{5} + 10 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )}{15 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.78191, size = 78, normalized size = 2.69 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sec ^{2}{\left (x \right )}}{\sin ^{4}{\left (x \right )} - 2 \sin ^{2}{\left (x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.11368, size = 30, normalized size = 1.03 \begin{align*} \frac{3 \, \tan \left (x\right )^{5} + 10 \, \tan \left (x\right )^{3} + 15 \, \tan \left (x\right )}{15 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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