3.58 \(\int \frac{1}{(a-a \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=32 \[ \frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{\tan (c+d x)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0245773, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {3175, 3767} \[ \frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{\tan (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a - a*Sin[c + d*x]^2)^(-2),x]

[Out]

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

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

Mathematica [A]  time = 0.0382637, size = 26, normalized size = 0.81 \[ \frac{\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - a*Sin[c + d*x]^2)^(-2),x]

[Out]

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

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Maple [A]  time = 0.04, size = 25, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{2}d} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}+\tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-sin(d*x+c)^2*a)^2,x)

[Out]

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

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Maxima [A]  time = 0.956355, size = 34, normalized size = 1.06 \begin{align*} \frac{\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )}{3 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.57837, size = 86, normalized size = 2.69 \begin{align*} \frac{{\left (2 \, \cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right )}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*cos(d*x + c)^2 + 1)*sin(d*x + c)/(a^2*d*cos(d*x + c)^3)

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Sympy [A]  time = 13.4509, size = 238, normalized size = 7.44 \begin{align*} \begin{cases} - \frac{6 \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} + \frac{4 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} - \frac{6 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 3 a^{2} d} & \text{for}\: d \neq 0 \\\frac{x}{\left (- a \sin ^{2}{\left (c \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Piecewise((-6*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)**4 + 9*a**2*d*tan(
c/2 + d*x/2)**2 - 3*a**2*d) + 4*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*tan(c/2 + d*x/2)*
*4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d) - 6*tan(c/2 + d*x/2)/(3*a**2*d*tan(c/2 + d*x/2)**6 - 9*a**2*d*ta
n(c/2 + d*x/2)**4 + 9*a**2*d*tan(c/2 + d*x/2)**2 - 3*a**2*d), Ne(d, 0)), (x/(-a*sin(c)**2 + a)**2, True))

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Giac [A]  time = 1.129, size = 34, normalized size = 1.06 \begin{align*} \frac{\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )}{3 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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