\(\int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [452]

Optimal result

Integrand size = 19, antiderivative size = 22 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {1}{2 b d (a+b \sin (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, 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} \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {1}{2 b d (a+b \sin (c+d x))^2} \]

[In]

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

[Out]

-1/2*1/(b*d*(a + b*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 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)^3} \, dx,x,b \sin (c+d x)\right )}{b d} \\ & = -\frac {1}{2 b d (a+b \sin (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {1}{2 b d (a+b \sin (c+d x))^2} \]

[In]

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

[Out]

-1/2*1/(b*d*(a + b*Sin[c + d*x])^2)

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(-\frac {1}{2 b d \left (a +b \sin \left (d x +c \right )\right )^{2}}\)	\(21\)
default	\(-\frac {1}{2 b d \left (a +b \sin \left (d x +c \right )\right )^{2}}\)	\(21\)
risch	\(\frac {2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d b}\)	\(48\)
parallelrisch	\(\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{2} d {\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}\)	\(73\)
norman	\(\frac {\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}+\frac {2 b \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2} d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) {\left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}^{2}}\)	\(142\)

[In]

int(cos(d*x+c)/(a+b*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (20) = 40\).

Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.95 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {1}{2 \, {\left (b^{3} d \cos \left (d x + c\right )^{2} - 2 \, a b^{2} d \sin \left (d x + c\right ) - {\left (a^{2} b + b^{3}\right )} d\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (19) = 38\).

Time = 0.71 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.32 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\begin {cases} \frac {x \cos {\left (c \right )}}{a^{3}} & \text {for}\: b = 0 \wedge d = 0 \\\frac {\sin {\left (c + d x \right )}}{a^{3} d} & \text {for}\: b = 0 \\\frac {x \cos {\left (c \right )}}{\left (a + b \sin {\left (c \right )}\right )^{3}} & \text {for}\: d = 0 \\- \frac {1}{2 a^{2} b d + 4 a b^{2} d \sin {\left (c + d x \right )} + 2 b^{3} d \sin ^{2}{\left (c + d x \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {1}{2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} b d} \]

[In]

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

[Out]

-1/2/((b*sin(d*x + c) + a)^2*b*d)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {1}{2 \, {\left (b \sin \left (d x + c\right ) + a\right )}^{2} b d} \]

[In]

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

[Out]

-1/2/((b*sin(d*x + c) + a)^2*b*d)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {\cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {1}{d\,\left (2\,a^2\,b+4\,a\,b^2\,\sin \left (c+d\,x\right )+2\,b^3\,{\sin \left (c+d\,x\right )}^2\right )} \]

[In]

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

[Out]

-1/(d*(2*a^2*b + 2*b^3*sin(c + d*x)^2 + 4*a*b^2*sin(c + d*x)))