\(\int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx\) [252]

Optimal result

Integrand size = 27, antiderivative size = 30 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^3(c+d x)}{3 a d (a+a \sin (c+d x))^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2912, 12, 37} \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sin ^3(c+d x)}{3 a d (a \sin (c+d x)+a)^3} \]

[In]

Int[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {1+3 \sin (c+d x)+3 \sin ^2(c+d x)}{3 a^4 d (1+\sin (c+d x))^3} \]

[In]

Integrate[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]

[Out]

-1/3*(1 + 3*Sin[c + d*x] + 3*Sin[c + d*x]^2)/(a^4*d*(1 + Sin[c + d*x])^3)

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10

method	result	size
parallelrisch	\(\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}\)	\(33\)
derivativedivides	\(\frac {-\frac {1}{1+\sin \left (d x +c \right )}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{\left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\)	\(43\)
default	\(\frac {-\frac {1}{1+\sin \left (d x +c \right )}-\frac {1}{3 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{\left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{4}}\)	\(43\)
risch	\(-\frac {2 i \left (3 \,{\mathrm e}^{5 i \left (d x +c \right )}-10 \,{\mathrm e}^{3 i \left (d x +c \right )}+6 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}-6 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6}}\)	\(82\)
norman	\(\frac {\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {8 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {8 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}\)	\(186\)

[In]

int(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (28) = 56\).

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.40 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, \cos \left (d x + c\right )^{2} - 3 \, \sin \left (d x + c\right ) - 4}{3 \, {\left (3 \, a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d + {\left (a^{4} d \cos \left (d x + c\right )^{2} - 4 \, a^{4} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/3*(3*cos(d*x + c)^2 - 3*sin(d*x + c) - 4)/(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. 192 vs. \(2 (24) = 48\).

Time = 0.84 (sec) , antiderivative size = 192, normalized size of antiderivative = 6.40 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\begin {cases} - \frac {3 \sin ^{2}{\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {3 \sin {\left (c + d x \right )}}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} - \frac {1}{3 a^{4} d \sin ^{3}{\left (c + d x \right )} + 9 a^{4} d \sin ^{2}{\left (c + d x \right )} + 9 a^{4} d \sin {\left (c + d x \right )} + 3 a^{4} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{4}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)**2/(a+a*sin(d*x+c))**4,x)

[Out]

Piecewise((-3*sin(c + d*x)**2/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3
*a**4*d) - 3*sin(c + d*x)/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**
4*d) - 1/(3*a**4*d*sin(c + d*x)**3 + 9*a**4*d*sin(c + d*x)**2 + 9*a**4*d*sin(c + d*x) + 3*a**4*d), Ne(d, 0)),
(x*sin(c)**2*cos(c)/(a*sin(c) + a)**4, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).

Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.23 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) + 1}{3 \, {\left (a^{4} \sin \left (d x + c\right )^{3} + 3 \, a^{4} \sin \left (d x + c\right )^{2} + 3 \, a^{4} \sin \left (d x + c\right ) + a^{4}\right )} d} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/3*(3*sin(d*x + c)^2 + 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 = 38, normalized size of antiderivative = 1.27 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) + 1}{3 \, a^{4} d {\left (\sin \left (d x + c\right ) + 1\right )}^{3}} \]

[In]

integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.72 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.80 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {1}{a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2}-\frac {1}{a^4\,d\,\left (\sin \left (c+d\,x\right )+1\right )}-\frac {1}{3\,a^4\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^3} \]

[In]

int((cos(c + d*x)*sin(c + d*x)^2)/(a + a*sin(c + d*x))^4,x)

[Out]

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