3.69 \(\int \frac{\cos (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=21 \[ -\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )} \]

[Out]

-(1/(d*(a^2 + a^2*Sin[c + d*x])))

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Rubi [A]  time = 0.0261668, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2667, 32} \[ -\frac{1}{d \left (a^2 \sin (c+d x)+a^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(d*(a^2 + a^2*Sin[c + d*x])))

Rule 2667

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 + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

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]

Rubi steps

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

Mathematica [A]  time = 0.0704079, size = 31, normalized size = 1.48 \[ -\frac{1}{a^2 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(a^2*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2))

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Maple [A]  time = 0.015, size = 21, normalized size = 1. \begin{align*} -{\frac{1}{d \left ( a+a\sin \left ( dx+c \right ) \right ) a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.960756, size = 27, normalized size = 1.29 \begin{align*} -\frac{1}{{\left (a \sin \left (d x + c\right ) + a\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((a*sin(d*x + c) + a)*a*d)

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Fricas [A]  time = 1.85078, size = 45, normalized size = 2.14 \begin{align*} -\frac{1}{a^{2} d \sin \left (d x + c\right ) + a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(a^2*d*sin(d*x + c) + a^2*d)

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Sympy [A]  time = 0.990904, size = 32, normalized size = 1.52 \begin{align*} \begin{cases} - \frac{1}{a^{2} d \sin{\left (c + d x \right )} + a^{2} d} & \text{for}\: d \neq 0 \\\frac{x \cos{\left (c \right )}}{\left (a \sin{\left (c \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15113, size = 27, normalized size = 1.29 \begin{align*} -\frac{1}{{\left (a \sin \left (d x + c\right ) + a\right )} a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((a*sin(d*x + c) + a)*a*d)