Integrand size = 27, antiderivative size = 60 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\log (1+\sin (c+d x))}{a^3 d}-\frac {1}{2 a d (a+a \sin (c+d x))^2}+\frac {2}{d \left (a^3+a^3 \sin (c+d x)\right )} \]
Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3+4 \sin (c+d x)+2 \log (1+\sin (c+d x)) (1+\sin (c+d x))^2}{2 a^3 d (1+\sin (c+d x))^2} \]
(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)
Time = 0.26 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 49, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos (c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)}{(a \sin (c+d x)+a)^3}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^2(c+d x)}{(\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^2 \sin ^2(c+d x)}{(\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\frac {a^2}{(\sin (c+d x) a+a)^3}-\frac {2 a}{(\sin (c+d x) a+a)^2}+\frac {1}{\sin (c+d x) a+a}\right )d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2}{2 (a \sin (c+d x)+a)^2}+\frac {2 a}{a \sin (c+d x)+a}+\log (a \sin (c+d x)+a)}{a^3 d}\) |
(Log[a + a*Sin[c + d*x]] - a^2/(2*(a + a*Sin[c + d*x])^2) + (2*a)/(a + a*S in[c + d*x]))/(a^3*d)
3.3.43.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[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]
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 | \(\frac {-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {2 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {8 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {26 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {26 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {30 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {30 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{3}}-\frac {\ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}}\) | \(265\) |
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, \sin \left (d x + c\right ) - 3}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \sin \left (d x + c\right ) - 2 \, a^{3} d\right )}} \]
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)
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 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\begin {cases} \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin ^{2}{\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {4 \sin {\left (c + d x \right )}}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} + \frac {3}{2 a^{3} d \sin ^{2}{\left (c + d x \right )} + 4 a^{3} d \sin {\left (c + d x \right )} + 2 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
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*log( 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*sin(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))
Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {4 \, \sin \left (d x + c\right ) + 3}{a^{3} \sin \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) + a^{3}} + \frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \]
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
Time = 0.36 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} + \frac {4 \, \sin \left (d x + c\right ) + 3}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{2 \, d} \]
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^3\,d}+\frac {2\,\sin \left (c+d\,x\right )+\frac {3}{2}}{a^3\,d\,{\left (\sin \left (c+d\,x\right )+1\right )}^2} \]