Integrand size = 27, antiderivative size = 52 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \log (1+\sin (c+d x))}{a^2 d}+\frac {\sin (c+d x)}{a^2 d}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \] Output:
-2*ln(1+sin(d*x+c))/a^2/d+sin(d*x+c)/a^2/d-1/d/(a^2+a^2*sin(d*x+c))
Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-1-2 \log (1+\sin (c+d x))+(1-2 \log (1+\sin (c+d x))) \sin (c+d x)+\sin ^2(c+d x)}{a^2 d (1+\sin (c+d x))} \] Input:
Integrate[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
(-1 - 2*Log[1 + Sin[c + d*x]] + (1 - 2*Log[1 + Sin[c + d*x]])*Sin[c + d*x] + Sin[c + d*x]^2)/(a^2*d*(1 + Sin[c + d*x]))
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90, 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.
\(\displaystyle \int \frac {\sin ^2(c+d x) \cos (c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)}{(a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\sin ^2(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a^2 \sin ^2(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\frac {a^2}{(\sin (c+d x) a+a)^2}-\frac {2 a}{\sin (c+d x) a+a}+1\right )d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {a^2}{a \sin (c+d x)+a}+a \sin (c+d x)-2 a \log (a \sin (c+d x)+a)}{a^3 d}\) |
Input:
Int[(Cos[c + d*x]*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
(-2*a*Log[a + a*Sin[c + d*x]] + a*Sin[c + d*x] - a^2/(a + a*Sin[c + d*x])) /(a^3*d)
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 = 0.38 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(\frac {\sin \left (d x +c \right )-2 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\) | \(38\) |
default | \(\frac {\sin \left (d x +c \right )-2 \ln \left (1+\sin \left (d x +c \right )\right )-\frac {1}{1+\sin \left (d x +c \right )}}{d \,a^{2}}\) | \(38\) |
parallelrisch | \(\frac {\left (4 \sin \left (d x +c \right )+4\right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-8 \sin \left (d x +c \right )-8\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )+1}{2 \left (1+\sin \left (d x +c \right )\right ) d \,a^{2}}\) | \(86\) |
risch | \(\frac {2 i x}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {4 i c}{d \,a^{2}}-\frac {2 i {\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2}}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}\) | \(108\) |
norman | \(\frac {\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}+\frac {20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}+\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}+\frac {2 \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,a^{2}}\) | \(227\) |
Input:
int(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d/a^2*(sin(d*x+c)-2*ln(1+sin(d*x+c))-1/(1+sin(d*x+c)))
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.10 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{2} + 2 \, {\left (\sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - \sin \left (d x + c\right )}{a^{2} d \sin \left (d x + c\right ) + a^{2} d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
-(cos(d*x + c)^2 + 2*(sin(d*x + c) + 1)*log(sin(d*x + c) + 1) - sin(d*x + c))/(a^2*d*sin(d*x + c) + a^2*d)
Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (44) = 88\).
Time = 0.53 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.42 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\begin {cases} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )} \sin {\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} - \frac {2 \log {\left (\sin {\left (c + d x \right )} + 1 \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} + \frac {\sin ^{2}{\left (c + d x \right )}}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} - \frac {2}{a^{2} d \sin {\left (c + d x \right )} + a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{2}{\left (c \right )} \cos {\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((-2*log(sin(c + d*x) + 1)*sin(c + d*x)/(a**2*d*sin(c + d*x) + a* *2*d) - 2*log(sin(c + d*x) + 1)/(a**2*d*sin(c + d*x) + a**2*d) + sin(c + d *x)**2/(a**2*d*sin(c + d*x) + a**2*d) - 2/(a**2*d*sin(c + d*x) + a**2*d), Ne(d, 0)), (x*sin(c)**2*cos(c)/(a*sin(c) + a)**2, True))
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {1}{a^{2} \sin \left (d x + c\right ) + a^{2}} + \frac {2 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} - \frac {\sin \left (d x + c\right )}{a^{2}}}{d} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
-(1/(a^2*sin(d*x + c) + a^2) + 2*log(sin(d*x + c) + 1)/a^2 - sin(d*x + c)/ a^2)/d
Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} d} + \frac {\sin \left (d x + c\right )}{a^{2} d} - \frac {1}{a^{2} d {\left (\sin \left (d x + c\right ) + 1\right )}} \] Input:
integrate(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac")
Output:
-2*log(abs(sin(d*x + c) + 1))/(a^2*d) + sin(d*x + c)/(a^2*d) - 1/(a^2*d*(s in(d*x + c) + 1))
Time = 18.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^2-2}{a^2\,d\,\left (\sin \left (c+d\,x\right )+1\right )}-\frac {2\,\ln \left (\sin \left (c+d\,x\right )+1\right )}{a^2\,d} \] Input:
int((cos(c + d*x)*sin(c + d*x)^2)/(a + a*sin(c + d*x))^2,x)
Output:
(sin(c + d*x)^2 - 2)/(a^2*d*(sin(c + d*x) + 1)) - (2*log(sin(c + d*x) + 1) )/(a^2*d)
Time = 0.16 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.19 \[ \int \frac {\cos (c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-2 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right ) \sin \left (d x +c \right )-2 \,\mathrm {log}\left (\sin \left (d x +c \right )+1\right )+\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )}{a^{2} d \left (\sin \left (d x +c \right )+1\right )} \] Input:
int(cos(d*x+c)*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
( - 2*log(sin(c + d*x) + 1)*sin(c + d*x) - 2*log(sin(c + d*x) + 1) + sin(c + d*x)**2 + 2*sin(c + d*x))/(a**2*d*(sin(c + d*x) + 1))