\(\int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx\) [546]

Optimal result

Integrand size = 27, antiderivative size = 55 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^2(c+d x)}{2 a^2 d}-\frac {2 \sin ^3(c+d x)}{3 a^2 d}+\frac {\sin ^4(c+d x)}{4 a^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2915, 12, 45} \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^4(c+d x)}{4 a^2 d}-\frac {2 \sin ^3(c+d x)}{3 a^2 d}+\frac {\sin ^2(c+d x)}{2 a^2 d} \]

[In]

Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {(a-x)^2 x}{a} \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int (a-x)^2 x \, dx,x,a \sin (c+d x)\right )}{a^6 d} \\ & = \frac {\text {Subst}\left (\int \left (a^2 x-2 a x^2+x^3\right ) \, dx,x,a \sin (c+d x)\right )}{a^6 d} \\ & = \frac {\sin ^2(c+d x)}{2 a^2 d}-\frac {2 \sin ^3(c+d x)}{3 a^2 d}+\frac {\sin ^4(c+d x)}{4 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sin ^2(c+d x) \left (6-8 \sin (c+d x)+3 \sin ^2(c+d x)\right )}{12 a^2 d} \]

[In]

Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

(Sin[c + d*x]^2*(6 - 8*Sin[c + d*x] + 3*Sin[c + d*x]^2))/(12*a^2*d)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71

method	result	size
derivativedivides	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d \,a^{2}}\)	\(39\)
default	\(\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}}{d \,a^{2}}\)	\(39\)
parallelrisch	\(\frac {16 \sin \left (3 d x +3 c \right )-48 \sin \left (d x +c \right )+3 \cos \left (4 d x +4 c \right )+33-36 \cos \left (2 d x +2 c \right )}{96 d \,a^{2}}\)	\(52\)
risch	\(-\frac {\sin \left (d x +c \right )}{2 a^{2} d}+\frac {\cos \left (4 d x +4 c \right )}{32 d \,a^{2}}+\frac {\sin \left (3 d x +3 c \right )}{6 d \,a^{2}}-\frac {3 \cos \left (2 d x +2 c \right )}{8 d \,a^{2}}\)	\(67\)
norman	\(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {6 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {6 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {2 \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\)	\(262\)

[In]

int(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, \cos \left (d x + c\right )^{4} - 12 \, \cos \left (d x + c\right )^{2} + 8 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{12 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(3*cos(d*x + c)^4 - 12*cos(d*x + c)^2 + 8*(cos(d*x + c)^2 - 1)*sin(d*x + c))/(a^2*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 493 vs. \(2 (46) = 92\).

Time = 20.27 (sec) , antiderivative size = 493, normalized size of antiderivative = 8.96 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} \frac {6 \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} - \frac {16 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} + \frac {24 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} - \frac {16 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} + \frac {6 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{2} d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a^{2} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin {\left (c \right )} \cos ^{5}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5*sin(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((6*tan(c/2 + d*x/2)**6/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan
(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 16*tan(c/2 + d*x/2)**5/(3*a**2*d*tan(c/2 + d*x/
2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2
*d) + 24*tan(c/2 + d*x/2)**4/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2
+ d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d) - 16*tan(c/2 + d*x/2)**3/(3*a**2*d*tan(c/2 + d*x/2)**
8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d)
+ 6*tan(c/2 + d*x/2)**2/(3*a**2*d*tan(c/2 + d*x/2)**8 + 12*a**2*d*tan(c/2 + d*x/2)**6 + 18*a**2*d*tan(c/2 + d*
x/2)**4 + 12*a**2*d*tan(c/2 + d*x/2)**2 + 3*a**2*d), Ne(d, 0)), (x*sin(c)*cos(c)**5/(a*sin(c) + a)**2, True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, \sin \left (d x + c\right )^{4} - 8 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2}}{12 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(3*sin(d*x + c)^4 - 8*sin(d*x + c)^3 + 6*sin(d*x + c)^2)/(a^2*d)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.71 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {3 \, \sin \left (d x + c\right )^{4} - 8 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2}}{12 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^5*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(3*sin(d*x + c)^4 - 8*sin(d*x + c)^3 + 6*sin(d*x + c)^2)/(a^2*d)

Mupad [B] (verification not implemented)

Time = 9.86 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.65 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (3\,{\sin \left (c+d\,x\right )}^2-8\,\sin \left (c+d\,x\right )+6\right )}{12\,a^2\,d} \]

[In]

int((cos(c + d*x)^5*sin(c + d*x))/(a + a*sin(c + d*x))^2,x)

[Out]

(sin(c + d*x)^2*(3*sin(c + d*x)^2 - 8*sin(c + d*x) + 6))/(12*a^2*d)