\(\int \frac {\sin (x)}{(3+\cos (x))^2} \, dx\) [10]

Optimal result

Integrand size = 9, antiderivative size = 6 \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{3+\cos (x)} \]

[Out]

1/(3+cos(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2747, 32} \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{\cos (x)+3} \]

[In]

Int[Sin[x]/(3 + Cos[x])^2,x]

[Out]

(3 + Cos[x])^(-1)

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}& = -\text {Subst}\left (\int \frac {1}{(3+x)^2} \, dx,x,\cos (x)\right ) \\ & = \frac {1}{3+\cos (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{3+\cos (x)} \]

[In]

Integrate[Sin[x]/(3 + Cos[x])^2,x]

[Out]

(3 + Cos[x])^(-1)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {1}{3+\cos \left (x \right )}\)	\(7\)
default	\(\frac {1}{3+\cos \left (x \right )}\)	\(7\)
parallelrisch	\(-\frac {1}{2 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+4}\)	\(15\)
risch	\(\frac {2 \,{\mathrm e}^{i x}}{{\mathrm e}^{2 i x}+6 \,{\mathrm e}^{i x}+1}\)	\(24\)
norman	\(\frac {-\frac {\left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}-\frac {1}{2}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )+2\right )}\)	\(32\)

[In]

int(sin(x)/(3+cos(x))^2,x,method=_RETURNVERBOSE)

[Out]

1/(3+cos(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{\cos \left (x\right ) + 3} \]

[In]

integrate(sin(x)/(3+cos(x))^2,x, algorithm="fricas")

[Out]

1/(cos(x) + 3)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.83 \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{\cos {\left (x \right )} + 3} \]

[In]

integrate(sin(x)/(3+cos(x))**2,x)

[Out]

1/(cos(x) + 3)

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{\cos \left (x\right ) + 3} \]

[In]

integrate(sin(x)/(3+cos(x))^2,x, algorithm="maxima")

[Out]

1/(cos(x) + 3)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{\cos \left (x\right ) + 3} \]

[In]

integrate(sin(x)/(3+cos(x))^2,x, algorithm="giac")

[Out]

1/(cos(x) + 3)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.00 \[ \int \frac {\sin (x)}{(3+\cos (x))^2} \, dx=\frac {1}{\cos \left (x\right )+3} \]

[In]

int(sin(x)/(cos(x) + 3)^2,x)

[Out]

1/(cos(x) + 3)