3.62 \(\int \frac{1}{(a-a \sin ^2(x))^4} \, dx\)

Optimal. Leaf size=37 \[ \frac{\tan ^7(x)}{7 a^4}+\frac{3 \tan ^5(x)}{5 a^4}+\frac{\tan ^3(x)}{a^4}+\frac{\tan (x)}{a^4} \]

[Out]

Tan[x]/a^4 + Tan[x]^3/a^4 + (3*Tan[x]^5)/(5*a^4) + Tan[x]^7/(7*a^4)

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Rubi [A]  time = 0.02165, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3175, 3767} \[ \frac{\tan ^7(x)}{7 a^4}+\frac{3 \tan ^5(x)}{5 a^4}+\frac{\tan ^3(x)}{a^4}+\frac{\tan (x)}{a^4} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sin[x]^2)^(-4),x]

[Out]

Tan[x]/a^4 + Tan[x]^3/a^4 + (3*Tan[x]^5)/(5*a^4) + Tan[x]^7/(7*a^4)

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{1}{\left (a-a \sin ^2(x)\right )^4} \, dx &=\frac{\int \sec ^8(x) \, dx}{a^4}\\ &=-\frac{\operatorname{Subst}\left (\int \left (1+3 x^2+3 x^4+x^6\right ) \, dx,x,-\tan (x)\right )}{a^4}\\ &=\frac{\tan (x)}{a^4}+\frac{\tan ^3(x)}{a^4}+\frac{3 \tan ^5(x)}{5 a^4}+\frac{\tan ^7(x)}{7 a^4}\\ \end{align*}

Mathematica [A]  time = 0.0051274, size = 41, normalized size = 1.11 \[ \frac{\frac{16 \tan (x)}{35}+\frac{1}{7} \tan (x) \sec ^6(x)+\frac{6}{35} \tan (x) \sec ^4(x)+\frac{8}{35} \tan (x) \sec ^2(x)}{a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sin[x]^2)^(-4),x]

[Out]

((16*Tan[x])/35 + (8*Sec[x]^2*Tan[x])/35 + (6*Sec[x]^4*Tan[x])/35 + (Sec[x]^6*Tan[x])/7)/a^4

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Maple [A]  time = 0.032, size = 24, normalized size = 0.7 \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{ \left ( \tan \left ( x \right ) \right ) ^{7}}{7}}+{\frac{3\, \left ( \tan \left ( x \right ) \right ) ^{5}}{5}}+ \left ( \tan \left ( x \right ) \right ) ^{3}+\tan \left ( x \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/7*tan(x)^7+3/5*tan(x)^5+tan(x)^3+tan(x))

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Maxima [A]  time = 0.941432, size = 38, normalized size = 1.03 \begin{align*} \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^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^4,x, algorithm="maxima")

[Out]

1/35*(5*tan(x)^7 + 21*tan(x)^5 + 35*tan(x)^3 + 35*tan(x))/a^4

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Fricas [A]  time = 1.89384, size = 97, normalized size = 2.62 \begin{align*} \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^{4} \cos \left (x\right )^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^4,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 108.262, size = 675, normalized size = 18.24 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-70*tan(x/2)**13/(35*a**4*tan(x/2)**14 - 245*a**4*tan(x/2)**12 + 735*a**4*tan(x/2)**10 - 1225*a**4*tan(x/2)**8
+ 1225*a**4*tan(x/2)**6 - 735*a**4*tan(x/2)**4 + 245*a**4*tan(x/2)**2 - 35*a**4) + 140*tan(x/2)**11/(35*a**4*
tan(x/2)**14 - 245*a**4*tan(x/2)**12 + 735*a**4*tan(x/2)**10 - 1225*a**4*tan(x/2)**8 + 1225*a**4*tan(x/2)**6 -
735*a**4*tan(x/2)**4 + 245*a**4*tan(x/2)**2 - 35*a**4) - 602*tan(x/2)**9/(35*a**4*tan(x/2)**14 - 245*a**4*tan
(x/2)**12 + 735*a**4*tan(x/2)**10 - 1225*a**4*tan(x/2)**8 + 1225*a**4*tan(x/2)**6 - 735*a**4*tan(x/2)**4 + 245
*a**4*tan(x/2)**2 - 35*a**4) + 424*tan(x/2)**7/(35*a**4*tan(x/2)**14 - 245*a**4*tan(x/2)**12 + 735*a**4*tan(x/
2)**10 - 1225*a**4*tan(x/2)**8 + 1225*a**4*tan(x/2)**6 - 735*a**4*tan(x/2)**4 + 245*a**4*tan(x/2)**2 - 35*a**4
) - 602*tan(x/2)**5/(35*a**4*tan(x/2)**14 - 245*a**4*tan(x/2)**12 + 735*a**4*tan(x/2)**10 - 1225*a**4*tan(x/2)
**8 + 1225*a**4*tan(x/2)**6 - 735*a**4*tan(x/2)**4 + 245*a**4*tan(x/2)**2 - 35*a**4) + 140*tan(x/2)**3/(35*a**
4*tan(x/2)**14 - 245*a**4*tan(x/2)**12 + 735*a**4*tan(x/2)**10 - 1225*a**4*tan(x/2)**8 + 1225*a**4*tan(x/2)**6
- 735*a**4*tan(x/2)**4 + 245*a**4*tan(x/2)**2 - 35*a**4) - 70*tan(x/2)/(35*a**4*tan(x/2)**14 - 245*a**4*tan(x
/2)**12 + 735*a**4*tan(x/2)**10 - 1225*a**4*tan(x/2)**8 + 1225*a**4*tan(x/2)**6 - 735*a**4*tan(x/2)**4 + 245*a
**4*tan(x/2)**2 - 35*a**4)

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Giac [A]  time = 1.11095, size = 38, normalized size = 1.03 \begin{align*} \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^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^4,x, algorithm="giac")

[Out]

1/35*(5*tan(x)^7 + 21*tan(x)^5 + 35*tan(x)^3 + 35*tan(x))/a^4