3.82 \(\int \frac {\cos (c+d x)}{(a+a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=22 \[ -\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[Out]

-1/2/a/d/(a+a*sin(d*x+c))^2

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Rubi [A]  time = 0.02, antiderivative size = 22, 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 = {2667, 32} \[ -\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/(2*a*d*(a + a*Sin[c + d*x])^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]

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])

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{(a+a \sin (c+d x))^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{(a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=-\frac {1}{2 a d (a+a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 33, normalized size = 1.50 \[ -\frac {1}{2 a^3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/(a + a*Sin[c + d*x])^3,x]

[Out]

-1/2*1/(a^3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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fricas [A]  time = 0.70, size = 36, normalized size = 1.64 \[ \frac {1}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2/(a^3*d*cos(d*x + c)^2 - 2*a^3*d*sin(d*x + c) - 2*a^3*d)

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giac [A]  time = 2.91, size = 20, normalized size = 0.91 \[ -\frac {1}{2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2/((a*sin(d*x + c) + a)^2*a*d)

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maple [A]  time = 0.07, size = 21, normalized size = 0.95 \[ -\frac {1}{2 a d \left (a +a \sin \left (d x +c \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(a+a*sin(d*x+c))^3,x)

[Out]

-1/2/a/d/(a+a*sin(d*x+c))^2

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maxima [A]  time = 0.38, size = 20, normalized size = 0.91 \[ -\frac {1}{2 \, {\left (a \sin \left (d x + c\right ) + a\right )}^{2} a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/2/((a*sin(d*x + c) + a)^2*a*d)

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mupad [B]  time = 4.46, size = 18, normalized size = 0.82 \[ -\frac {1}{2\,a^3\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)/(a + a*sin(c + d*x))^3,x)

[Out]

-1/(2*a^3*d*(sin(c + d*x) + 1)^2)

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sympy [A]  time = 1.89, size = 51, normalized size = 2.32 \[ \begin {cases} - \frac {1}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos {\relax (c )}}{\left (a \sin {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((-1/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d), Ne(d, 0)), (x*cos(c)/(a*sin(c) +
a)**3, True))

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