3.51 \(\int \frac{\sin ^3(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\)

Optimal. Leaf size=33 \[ \frac{\sec ^3(c+d x)}{3 a^2 d}-\frac{\sec (c+d x)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0615918, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {3175, 2606} \[ \frac{\sec ^3(c+d x)}{3 a^2 d}-\frac{\sec (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^3/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

-(Sec[c + d*x]/(a^2*d)) + Sec[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 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rubi steps

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

Mathematica [A]  time = 0.0296034, size = 31, normalized size = 0.94 \[ \frac{\frac{\sec ^3(c+d x)}{3 d}-\frac{\sec (c+d x)}{d}}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^3/(a - a*Sin[c + d*x]^2)^2,x]

[Out]

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

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Maple [A]  time = 0.04, size = 29, normalized size = 0.9 \begin{align*}{\frac{1}{{a}^{2}d} \left ( - \left ( \cos \left ( dx+c \right ) \right ) ^{-1}+{\frac{1}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^3/(a-sin(d*x+c)^2*a)^2,x)

[Out]

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

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Maxima [A]  time = 0.972657, size = 38, normalized size = 1.15 \begin{align*} -\frac{3 \, \cos \left (d x + c\right )^{2} - 1}{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(sin(d*x+c)^3/(a-a*sin(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.53299, size = 70, normalized size = 2.12 \begin{align*} -\frac{3 \, \cos \left (d x + c\right )^{2} - 1}{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(sin(d*x+c)^3/(a-a*sin(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 25.345, size = 309, normalized size = 9.36 \begin{align*} \begin{cases} - \frac{4 \tan ^{6}{\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{12 \tan ^{4}{\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{24 \tan ^{2}{\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{8}{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 \sin ^{3}{\left (c \right )}}{\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(sin(d*x+c)**3/(a-a*sin(d*x+c)**2)**2,x)

[Out]

Piecewise((-4*tan(c/2 + d*x/2)**6/(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) + 12*tan(c/2 + d*x/2)**4/(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) - 24*tan(c/2 + d*x/2)**2/(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) + 8/(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), Ne(d, 0)), (x*sin(c)**3/(-a*sin(c)**2 + a)**2
, True))

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Giac [A]  time = 1.15713, size = 38, normalized size = 1.15 \begin{align*} -\frac{3 \, \cos \left (d x + c\right )^{2} - 1}{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(sin(d*x+c)^3/(a-a*sin(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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