\(\int \frac {\sin ^2(c+d x)}{(a-a \sin ^2(c+d x))^2} \, dx\) [57]

Optimal result

Integrand size = 24, antiderivative size = 18 \[ \int \frac {\sin ^2(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\tan ^3(c+d x)}{3 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3254, 2687, 30} \[ \int \frac {\sin ^2(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\tan ^3(c+d x)}{3 a^2 d} \]

[In]

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

[Out]

Tan[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 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 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 ^2(c+d x) \tan ^2(c+d x) \, dx}{a^2} \\ & = \frac {\text {Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{a^2 d} \\ & = \frac {\tan ^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 \[ \int \frac {\sin ^2(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\tan ^3(c+d x)}{3 a^2 d} \]

[In]

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

[Out]

Tan[c + d*x]^3/(3*a^2*d)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {\tan ^{3}\left (d x +c \right )}{3 a^{2} d}\)	\(17\)
default	\(\frac {\tan ^{3}\left (d x +c \right )}{3 a^{2} d}\)	\(17\)
risch	\(-\frac {2 i \left (3 \,{\mathrm e}^{4 i \left (d x +c \right )}+1\right )}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}\)	\(36\)
parallelrisch	\(\frac {-\sin \left (3 d x +3 c \right )+3 \sin \left (d x +c \right )}{3 a^{2} d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(49\)
norman	\(\frac {-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {16 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}\)	\(93\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.78 \[ \int \frac {\sin ^2(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=-\frac {{\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{3 \, a^{2} d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

-1/3*(cos(d*x + c)^2 - 1)*sin(d*x + c)/(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. 94 vs. \(2 (14) = 28\).

Time = 2.79 (sec) , antiderivative size = 94, normalized size of antiderivative = 5.22 \[ \int \frac {\sin ^2(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\begin {cases} - \frac {8 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 9 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )}}{\left (- a \sin ^{2}{\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-8*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), Ne(d, 0)), (x*sin(c)**2/(-a*sin(c)**2 + a)**2, True))

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

1/3*tan(d*x + c)^3/(a^2*d)

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

1/3*tan(d*x + c)^3/(a^2*d)

Mupad [B] (verification not implemented)

Time = 12.60 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^2(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,a^2\,d} \]

[In]

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

[Out]

tan(c + d*x)^3/(3*a^2*d)