\(\int \frac {\sin (c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\) [52]

Optimal result

Integrand size = 22, antiderivative size = 18 \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\sec ^3(c+d x)}{3 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3254, 2686, 30} \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\sec ^3(c+d x)}{3 a^2 d} \]

[In]

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

[Out]

Sec[c + d*x]^3/(3*a^2*d)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2686

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])

Rule 3254

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^3(c+d x) \tan (c+d x) \, dx}{a^2} \\ & = \frac {\text {Subst}\left (\int x^2 \, dx,x,\sec (c+d x)\right )}{a^2 d} \\ & = \frac {\sec ^3(c+d x)}{3 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\sec ^3(c+d x)}{3 a^2 d} \]

[In]

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

[Out]

Sec[c + d*x]^3/(3*a^2*d)

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {1}{3 d \,a^{2} \cos \left (d x +c \right )^{3}}\)	\(17\)
default	\(\frac {1}{3 d \,a^{2} \cos \left (d x +c \right )^{3}}\)	\(17\)
risch	\(\frac {8 \,{\mathrm e}^{3 i \left (d x +c \right )}}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(31\)
parallelrisch	\(\frac {4+3 \cos \left (d x +c \right )+\cos \left (3 d x +3 c \right )}{3 a^{2} d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(48\)
norman	\(\frac {-\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2}{3 a d}-\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(101\)

[In]

int(sin(d*x+c)/(a-a*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {1}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/3/(a^2*d*cos(d*x + c)^3)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (14) = 28\).

Time = 1.92 (sec) , antiderivative size = 156, normalized size of antiderivative = 8.67 \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\begin {cases} - \frac {6 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} - \frac {2}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )}}{\left (- a \sin ^{2}{\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-6*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) - 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), Ne(d, 0)), (x*sin(c)/(-a*sin(c)**2 + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {1}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/3/(a^2*d*cos(d*x + c)^3)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {1}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

1/3/(a^2*d*cos(d*x + c)^3)

Mupad [B] (verification not implemented)

Time = 12.98 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {1}{3\,a^2\,d\,{\cos \left (c+d\,x\right )}^3} \]

[In]

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

[Out]

1/(3*a^2*d*cos(c + d*x)^3)