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

Optimal. Leaf size=20 \[ -\frac {1}{b d (a+b \sin (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 20, normalized size of antiderivative = 1.00, 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 = {2668, 32} \[ -\frac {1}{b d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[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 2668

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*} \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.00 \[ -\frac {1}{b d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(b*d*(a + b*Sin[c + d*x])))

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fricas [A]  time = 0.45, size = 20, normalized size = 1.00 \[ -\frac {1}{b^{2} d \sin \left (d x + c\right ) + a b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.36, size = 20, normalized size = 1.00 \[ -\frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.13, size = 21, normalized size = 1.05 \[ -\frac {1}{b d \left (a +b \sin \left (d x +c \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 20, normalized size = 1.00 \[ -\frac {1}{{\left (b \sin \left (d x + c\right ) + a\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 5.07, size = 20, normalized size = 1.00 \[ -\frac {1}{b\,d\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(b*d*(a + b*sin(c + d*x)))

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sympy [A]  time = 1.24, size = 51, normalized size = 2.55 \[ \begin {cases} \frac {x \cos {\relax (c )}}{a^{2}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin {\left (c + d x \right )}}{a^{2} d} & \text {for}\: b = 0 \\\frac {x \cos {\relax (c )}}{\left (a + b \sin {\relax (c )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {1}{a b d + b^{2} d \sin {\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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