\(\int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [440]

Optimal result

Integrand size = 19, antiderivative size = 20 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {1}{b d (a+b \sin (c+d x))} \]

[Out]

-1/b/d/(a+b*sin(d*x+c))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2747, 32} \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {1}{b d (a+b \sin (c+d x))} \]

[In]

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

[Out]

-(1/(b*d*(a + b*Sin[c + d*x])))

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2747

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{(a+x)^2} \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = -\frac {1}{b d (a+b \sin (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {1}{b d (a+b \sin (c+d x))} \]

[In]

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

[Out]

-(1/(b*d*(a + b*Sin[c + d*x])))

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(-\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )}\)	\(21\)
default	\(-\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )}\)	\(21\)
parallelrisch	\(-\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )}\)	\(21\)
risch	\(-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{b d \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}\)	\(49\)
norman	\(\frac {\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}\)	\(83\)

[In]

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

[Out]

-1/b/d/(a+b*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {1}{b^{2} d \sin \left (d x + c\right ) + a b d} \]

[In]

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

[Out]

-1/(b^2*d*sin(d*x + c) + a*b*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (15) = 30\).

Time = 0.53 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.55 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\begin {cases} \frac {x \cos {\left (c \right )}}{a^{2}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin {\left (c + d x \right )}}{a^{2} d} & \text {for}\: b = 0 \\\frac {x \cos {\left (c \right )}}{\left (a + b \sin {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {1}{a b d + b^{2} d \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x*cos(c)/a**2, Eq(b, 0) & Eq(d, 0)), (sin(c + d*x)/(a**2*d), Eq(b, 0)), (x*cos(c)/(a + b*sin(c))**2
, Eq(d, 0)), (-1/(a*b*d + b**2*d*sin(c + d*x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )} b d} \]

[In]

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

[Out]

-1/((b*sin(d*x + c) + a)*b*d)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )} b d} \]

[In]

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

[Out]

-1/((b*sin(d*x + c) + a)*b*d)

Mupad [B] (verification not implemented)

Time = 4.51 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {1}{b\,d\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \]

[In]

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

[Out]

-1/(b*d*(a + b*sin(c + d*x)))