\(\int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx\) [243]

Optimal result

Integrand size = 27, antiderivative size = 60 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\log (1+\sin (c+d x))}{a^3 d}-\frac {1}{2 a d (a+a \sin (c+d x))^2}+\frac {2}{d \left (a^3+a^3 \sin (c+d x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 60, 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 = {2912, 12, 45} \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2}{d \left (a^3 \sin (c+d x)+a^3\right )}+\frac {\log (\sin (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \sin (c+d x)+a)^2} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3+4 \sin (c+d x)+2 \log (1+\sin (c+d x)) (1+\sin (c+d x))^2}{2 a^3 d (1+\sin (c+d x))^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.70

method	result	size
derivativedivides	\(\frac {\frac {2}{1+\sin \left (d x +c \right )}+\ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{3}}\)	\(42\)
default	\(\frac {\frac {2}{1+\sin \left (d x +c \right )}+\ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{2 \left (1+\sin \left (d x +c \right )\right )^{2}}}{d \,a^{3}}\)	\(42\)
risch	\(-\frac {i x}{a^{3}}-\frac {2 i c}{d \,a^{3}}+\frac {2 i \left (3 i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}-2 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{3}}\)	\(98\)
parallelrisch	\(\frac {\left (6-2 \cos \left (2 d x +2 c \right )+8 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 \cos \left (2 d x +2 c \right )-16 \sin \left (d x +c \right )-12\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )+3}{2 d \,a^{3} \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right )}\)	\(119\)
norman	\(\frac {-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {2 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {8 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {26 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {26 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {30 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {30 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\)	\(265\)

[In]

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

[Out]

1/d/a^3*(2/(1+sin(d*x+c))+ln(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 = 75, normalized size of antiderivative = 1.25 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, \sin \left (d x + c\right ) - 3}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]

[In]

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

[Out]

1/2*(2*(cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d*x + c) + 1) - 4*sin(d*x + c) - 3)/(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. 257 vs. \(2 (49) = 98\).

Time = 0.58 (sec) , antiderivative size = 257, normalized size of antiderivative = 4.28 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {3}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*log(sin(c + d*x) + 1)*sin(c + d*x)**2/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*
d) + 4*log(sin(c + d*x) + 1)*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) + 2*lo
g(sin(c + d*x) + 1)/(2*a**3*d*sin(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 4*sin(c + d*x)/(2*a**3*d*s
in(c + d*x)**2 + 4*a**3*d*sin(c + d*x) + 2*a**3*d) + 3/(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)**2*cos(c)/(a*sin(c) + a)**3, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {4 \, \sin \left (d x + c\right ) + 3}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac {4 \, \sin \left (d x + c\right ) + 3}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d}+\frac {2\,\sin \left (c+d\,x\right )+\frac {3}{2}}{a^3\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2} \]

[In]

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

[Out]

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