\(\int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [244]

Optimal result

Integrand size = 25, antiderivative size = 30 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^2(c+d x)}{2 a d (a+a \sin (c+d x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2912, 12, 37} \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^2(c+d x)}{2 a d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x}{a (a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \frac {x}{(a+x)^3} \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {\sin ^2(c+d x)}{2 a d (a+a \sin (c+d x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sin ^2(c+d x)}{2 a d (a+a \sin (c+d x))^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {1}{1+\sin \left (d x +c \right )}+\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{3}}\)	\(33\)
default	\(\frac {-\frac {1}{1+\sin \left (d x +c \right )}+\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{3}}\)	\(33\)
parallelrisch	\(\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}\)	\(33\)
risch	\(-\frac {2 i \left (i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4}}\)	\(57\)
norman	\(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(148\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.53 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (24) = 48\).

Time = 0.55 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.30 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} - \frac {2 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} - \frac {1}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, \sin \left (d x + c\right ) + 1}{2 \, {\left (a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}\right )} d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, \sin \left (d x + c\right ) + 1}{2 \, a^{3} d {\left (\sin \left (d x + c\right ) + 1\right )}^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {1}{2\,a^3\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2}-\frac {1}{a^3\,d\,\left (\sin \left (c+d\,x\right )+1\right )} \]

[In]

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

[Out]

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