\(\int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [253]

Optimal result

Integrand size = 25, antiderivative size = 46 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {1}{3 a d (a+a \sin (c+d x))^3}-\frac {1}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2912, 12, 45} \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {1}{3 a d (a \sin (c+d x)+a)^3}-\frac {1}{2 d \left (a^2 \sin (c+d x)+a^2\right )^2} \]

[In]

Int[(Cos[c + d*x]*Sin[c + d*x])/(a + a*Sin[c + d*x])^4,x]

[Out]

1/(3*a*d*(a + a*Sin[c + d*x])^3) - 1/(2*d*(a^2 + a^2*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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)^4} \, dx,x,a \sin (c+d x)\right )}{a d} \\ & = \frac {\text {Subst}\left (\int \frac {x}{(a+x)^4} \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a}{(a+x)^4}+\frac {1}{(a+x)^3}\right ) \, dx,x,a \sin (c+d x)\right )}{a^2 d} \\ & = \frac {1}{3 a d (a+a \sin (c+d x))^3}-\frac {1}{2 d \left (a^2+a^2 \sin (c+d x)\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.65 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {1+3 \sin (c+d x)}{6 a^4 d (1+\sin (c+d x))^3} \]

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x])/(a + a*Sin[c + d*x])^4,x]

[Out]

-1/6*(1 + 3*Sin[c + d*x])/(a^4*d*(1 + Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\)	\(33\)
default	\(\frac {\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\)	\(33\)
risch	\(\frac {\frac {4 i {\mathrm e}^{3 i \left (d x +c \right )}}{3}+2 \,{\mathrm e}^{4 i \left (d x +c \right )}-2 \,{\mathrm e}^{2 i \left (d x +c \right )}}{d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}\)	\(58\)
parallelrisch	\(\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+3\right )}{3 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}\)	\(59\)
norman	\(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {22 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {22 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {10 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {26 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {26 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(186\)

[In]

int(cos(d*x+c)*sin(d*x+c)/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.35 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {3 \, \sin \left (d x + c\right ) + 1}{6 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

1/6*(3*sin(d*x + c) + 1)/(3*a^4*d*cos(d*x + c)^2 - 4*a^4*d + (a^4*d*cos(d*x + c)^2 - 4*a^4*d)*sin(d*x + c))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (37) = 74\).

Time = 0.84 (sec) , antiderivative size = 129, normalized size of antiderivative = 2.80 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\begin {cases} - \frac {3 \sin {\left (c + d x \right )}}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} - \frac {1}{6 a^{4} d \sin ^{3}{\left (c + d x \right )} + 18 a^{4} d \sin ^{2}{\left (c + d x \right )} + 18 a^{4} d \sin {\left (c + d x \right )} + 6 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)/(a+a*sin(d*x+c))**4,x)

[Out]

Piecewise((-3*sin(c + d*x)/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*
a**4*d) - 1/(6*a**4*d*sin(c + d*x)**3 + 18*a**4*d*sin(c + d*x)**2 + 18*a**4*d*sin(c + d*x) + 6*a**4*d), Ne(d,
0)), (x*sin(c)*cos(c)/(a*sin(c) + a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.24 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, \sin \left (d x + c\right ) + 1}{6 \, {\left (a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}\right )} d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/6*(3*sin(d*x + c) + 1)/((a^4*sin(d*x + c)^3 + 3*a^4*sin(d*x + c)^2 + 3*a^4*sin(d*x + c) + a^4)*d)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.61 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, \sin \left (d x + c\right ) + 1}{6 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

-1/6*(3*sin(d*x + c) + 1)/(a^4*d*(sin(d*x + c) + 1)^3)

Mupad [B] (verification not implemented)

Time = 9.97 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \frac {\cos (c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {1}{3\,a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^3}-\frac {1}{2\,a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2} \]

[In]

int((cos(c + d*x)*sin(c + d*x))/(a + a*sin(c + d*x))^4,x)

[Out]

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