3.61 \(\int \frac {1}{(a-a \sin ^2(x))^3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, 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} \[ \frac {\tan ^5(x)}{5 a^3}+\frac {2 \tan ^3(x)}{3 a^3}+\frac {\tan (x)}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sin[x]^2)^(-3),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 31, normalized size = 1.07 \[ \frac {\frac {8 \tan (x)}{15}+\frac {1}{5} \tan (x) \sec ^4(x)+\frac {4}{15} \tan (x) \sec ^2(x)}{a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sin[x]^2)^(-3),x]

[Out]

((8*Tan[x])/15 + (4*Sec[x]^2*Tan[x])/15 + (Sec[x]^4*Tan[x])/5)/a^3

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fricas [A]  time = 0.40, size = 25, normalized size = 0.86 \[ \frac {{\left (8 \, \cos \relax (x)^{4} + 4 \, \cos \relax (x)^{2} + 3\right )} \sin \relax (x)}{15 \, a^{3} \cos \relax (x)^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 22, normalized size = 0.76 \[ \frac {3 \, \tan \relax (x)^{5} + 10 \, \tan \relax (x)^{3} + 15 \, \tan \relax (x)}{15 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.16, size = 20, normalized size = 0.69 \[ \frac {\frac {\left (\tan ^{5}\relax (x )\right )}{5}+\frac {2 \left (\tan ^{3}\relax (x )\right )}{3}+\tan \relax (x )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-a*sin(x)^2)^3,x)

[Out]

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

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maxima [A]  time = 0.37, size = 22, normalized size = 0.76 \[ \frac {3 \, \tan \relax (x)^{5} + 10 \, \tan \relax (x)^{3} + 15 \, \tan \relax (x)}{15 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)^2)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 13.41, size = 21, normalized size = 0.72 \[ \frac {\mathrm {tan}\relax (x)\,\left (3\,{\mathrm {tan}\relax (x)}^4+10\,{\mathrm {tan}\relax (x)}^2+15\right )}{15\,a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a - a*sin(x)^2)^3,x)

[Out]

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

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sympy [B]  time = 7.35, size = 362, normalized size = 12.48 \[ - \frac {30 \tan ^{9}{\left (\frac {x}{2} \right )}}{15 a^{3} \tan ^{10}{\left (\frac {x}{2} \right )} - 75 a^{3} \tan ^{8}{\left (\frac {x}{2} \right )} + 150 a^{3} \tan ^{6}{\left (\frac {x}{2} \right )} - 150 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 75 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} - 15 a^{3}} + \frac {40 \tan ^{7}{\left (\frac {x}{2} \right )}}{15 a^{3} \tan ^{10}{\left (\frac {x}{2} \right )} - 75 a^{3} \tan ^{8}{\left (\frac {x}{2} \right )} + 150 a^{3} \tan ^{6}{\left (\frac {x}{2} \right )} - 150 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 75 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} - 15 a^{3}} - \frac {116 \tan ^{5}{\left (\frac {x}{2} \right )}}{15 a^{3} \tan ^{10}{\left (\frac {x}{2} \right )} - 75 a^{3} \tan ^{8}{\left (\frac {x}{2} \right )} + 150 a^{3} \tan ^{6}{\left (\frac {x}{2} \right )} - 150 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 75 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} - 15 a^{3}} + \frac {40 \tan ^{3}{\left (\frac {x}{2} \right )}}{15 a^{3} \tan ^{10}{\left (\frac {x}{2} \right )} - 75 a^{3} \tan ^{8}{\left (\frac {x}{2} \right )} + 150 a^{3} \tan ^{6}{\left (\frac {x}{2} \right )} - 150 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 75 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} - 15 a^{3}} - \frac {30 \tan {\left (\frac {x}{2} \right )}}{15 a^{3} \tan ^{10}{\left (\frac {x}{2} \right )} - 75 a^{3} \tan ^{8}{\left (\frac {x}{2} \right )} + 150 a^{3} \tan ^{6}{\left (\frac {x}{2} \right )} - 150 a^{3} \tan ^{4}{\left (\frac {x}{2} \right )} + 75 a^{3} \tan ^{2}{\left (\frac {x}{2} \right )} - 15 a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(x)**2)**3,x)

[Out]

-30*tan(x/2)**9/(15*a**3*tan(x/2)**10 - 75*a**3*tan(x/2)**8 + 150*a**3*tan(x/2)**6 - 150*a**3*tan(x/2)**4 + 75
*a**3*tan(x/2)**2 - 15*a**3) + 40*tan(x/2)**7/(15*a**3*tan(x/2)**10 - 75*a**3*tan(x/2)**8 + 150*a**3*tan(x/2)*
*6 - 150*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**2 - 15*a**3) - 116*tan(x/2)**5/(15*a**3*tan(x/2)**10 - 75*a**3*t
an(x/2)**8 + 150*a**3*tan(x/2)**6 - 150*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**2 - 15*a**3) + 40*tan(x/2)**3/(15
*a**3*tan(x/2)**10 - 75*a**3*tan(x/2)**8 + 150*a**3*tan(x/2)**6 - 150*a**3*tan(x/2)**4 + 75*a**3*tan(x/2)**2 -
15*a**3) - 30*tan(x/2)/(15*a**3*tan(x/2)**10 - 75*a**3*tan(x/2)**8 + 150*a**3*tan(x/2)**6 - 150*a**3*tan(x/2)
**4 + 75*a**3*tan(x/2)**2 - 15*a**3)

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