\(\int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [38]

Optimal result

Integrand size = 22, antiderivative size = 13 \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\sec (c+d x)}{a d} \]

[Out]

sec(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 13, 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, 8} \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\sec (c+d x)}{a d} \]

[In]

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

[Out]

Sec[c + d*x]/(a*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 (c+d x) \tan (c+d x) \, dx}{a} \\ & = \frac {\text {Subst}(\int 1 \, dx,x,\sec (c+d x))}{a d} \\ & = \frac {\sec (c+d x)}{a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\sec (c+d x)}{a d} \]

[In]

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

[Out]

Sec[c + d*x]/(a*d)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {1}{d a \cos \left (d x +c \right )}\)	\(16\)
default	\(\frac {1}{d a \cos \left (d x +c \right )}\)	\(16\)
parallelrisch	\(-\frac {2}{d a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(24\)
risch	\(\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(31\)
norman	\(\frac {-\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(60\)

[In]

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

[Out]

1/d/a/cos(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {1}{a d \cos \left (d x + c\right )} \]

[In]

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

[Out]

1/(a*d*cos(d*x + c))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (8) = 16\).

Time = 0.59 (sec) , antiderivative size = 34, normalized size of antiderivative = 2.62 \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\begin {cases} - \frac {2}{a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - a d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )}}{- a \sin ^{2}{\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {1}{a d \cos \left (d x + c\right )} \]

[In]

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

[Out]

1/(a*d*cos(d*x + c))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {1}{a d \cos \left (d x + c\right )} \]

[In]

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

[Out]

1/(a*d*cos(d*x + c))

Mupad [B] (verification not implemented)

Time = 13.81 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\sin (c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {1}{a\,d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

1/(a*d*cos(c + d*x))