\(\int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [357]

Optimal result

Integrand size = 29, antiderivative size = 37 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^4(c+d x)}{4 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2915, 12, 45} \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^4(c+d x)}{4 a d} \]

[In]

Int[(Cos[c + d*x]^3*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

Sin[c + d*x]^3/(3*a*d) - Sin[c + d*x]^4/(4*a*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) x^2}{a^2} \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int (a-x) x^2 \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\text {Subst}\left (\int \left (a x^2-x^3\right ) \, dx,x,a \sin (c+d x)\right )}{a^5 d} \\ & = \frac {\sin ^3(c+d x)}{3 a d}-\frac {\sin ^4(c+d x)}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {(4-3 \sin (c+d x)) \sin ^3(c+d x)}{12 a d} \]

[In]

Integrate[(Cos[c + d*x]^3*Sin[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

((4 - 3*Sin[c + d*x])*Sin[c + d*x]^3)/(12*a*d)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(-\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d a}\)	\(30\)
default	\(-\frac {\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\sin ^{3}\left (d x +c \right )\right )}{3}}{d a}\)	\(30\)
parallelrisch	\(\frac {-9+12 \cos \left (2 d x +2 c \right )-3 \cos \left (4 d x +4 c \right )+24 \sin \left (d x +c \right )-8 \sin \left (3 d x +3 c \right )}{96 d a}\)	\(52\)
risch	\(\frac {\sin \left (d x +c \right )}{4 a d}-\frac {\cos \left (4 d x +4 c \right )}{32 a d}-\frac {\sin \left (3 d x +3 c \right )}{12 d a}+\frac {\cos \left (2 d x +2 c \right )}{8 a d}\)	\(67\)
norman	\(\frac {-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {4 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {4 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(145\)

[In]

int(cos(d*x+c)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right )}{12 \, a d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (26) = 52\).

Time = 6.52 (sec) , antiderivative size = 277, normalized size of antiderivative = 7.49 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {8 \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {12 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {8 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 18 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 12 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )} \cos ^{3}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*sin(d*x+c)**2/(a+a*sin(d*x+c)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3}}{12 \, a d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {3 \, \sin \left (d x + c\right )^{4} - 4 \, \sin \left (d x + c\right )^{3}}{12 \, a d} \]

[In]

integrate(cos(d*x+c)^3*sin(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^3(c+d x) \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {{\sin \left (c+d\,x\right )}^3\,\left (3\,\sin \left (c+d\,x\right )-4\right )}{12\,a\,d} \]

[In]

int((cos(c + d*x)^3*sin(c + d*x)^2)/(a + a*sin(c + d*x)),x)

[Out]

-(sin(c + d*x)^3*(3*sin(c + d*x) - 4))/(12*a*d)