Integrand size = 29, antiderivative size = 69 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5 x}{2 a^2}-\frac {2 \cos (c+d x)}{a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {2 \cos (c+d x)}{a^2 d (1+\sin (c+d x))} \] Output:
-5/2*x/a^2-2*cos(d*x+c)/a^2/d+1/2*cos(d*x+c)*sin(d*x+c)/a^2/d-2*cos(d*x+c) /a^2/d/(1+sin(d*x+c))
Time = 0.42 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-10 (c+d x)-8 \cos (c+d x)+\frac {16 \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+\sin (2 (c+d x))}{4 a^2 d} \] Input:
Integrate[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
(-10*(c + d*x) - 8*Cos[c + d*x] + (16*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + Sin[2*(c + d*x)])/(4*a^2*d)
Time = 0.56 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3353, 3042, 3429, 3042, 3188, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^2}{(a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3353 |
| \(\displaystyle \frac {\int \frac {\sin ^2(c+d x) (a-a \sin (c+d x))}{\sin (c+d x) a+a}dx}{a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^2 (a-a \sin (c+d x))}{\sin (c+d x) a+a}dx}{a^2}\) |
| \(\Big \downarrow \) 3429 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x))^2 \tan ^2(c+d x)dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x))^2 \tan (c+d x)^2dx}{a^4}\) |
| \(\Big \downarrow \) 3188 |
| \(\displaystyle \frac {\int \left (-\sin ^2(c+d x)+2 \sin (c+d x)+\frac {2}{\sin (c+d x)+1}-2\right )dx}{a^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {2 \cos (c+d x)}{d}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}-\frac {2 \cos (c+d x)}{d (\sin (c+d x)+1)}-\frac {5 x}{2}}{a^2}\) |
Input:
Int[(Cos[c + d*x]^2*Sin[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
((-5*x)/2 - (2*Cos[c + d*x])/d + (Cos[c + d*x]*Sin[c + d*x])/(2*d) - (2*Co s[c + d*x])/(d*(1 + Sin[c + d*x])))/a^2
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_ ), x_Symbol] :> Simp[a^p Int[ExpandIntegrand[Sin[e + f*x]^p*((a + b*Sin[e + f*x])^(m - p/2)/(a - b*Sin[e + f*x])^(p/2)), x], x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/b^2 Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /; Fre eQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] || !IGtQ[ n, 0])
Int[sin[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^n*c^n Int[Tan[e + f*x]^p*(a + b*Sin[e + f*x])^(m - n), x], x] /; FreeQ[{a, b, c , d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[p + 2*n, 0] && IntegerQ[n]
Time = 0.46 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94
| method | result | size |
| parallelrisch | \(\frac {-24+17 \sin \left (d x +c \right )+\sin \left (3 d x +3 c \right )-8 \cos \left (2 d x +2 c \right )-32 \cos \left (d x +c \right )-20 d x \cos \left (d x +c \right )}{8 d \,a^{2} \cos \left (d x +c \right )}\) | \(65\) |
| risch | \(-\frac {5 x}{2 a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}-\frac {4}{d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {\sin \left (2 d x +2 c \right )}{4 d \,a^{2}}\) | \(81\) |
| derivativedivides | \(\frac {-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {4 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+1\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(91\) |
| default | \(\frac {-\frac {4}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}-\frac {4 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4}+1\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-5 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(91\) |
| norman | \(\frac {-\frac {8}{a d}-\frac {19 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d a}-\frac {41 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d a}-\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d a}-\frac {68 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}-\frac {36 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d a}-\frac {5 x}{2 a}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {35 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {65 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 a}-\frac {45 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {55 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a}-\frac {55 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {45 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a}-\frac {65 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}-\frac {35 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 a}-\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}-\frac {5 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 a}-\frac {84 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}-\frac {60 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d a}-\frac {94 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d a}-\frac {82 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(421\) |
Input:
int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/8*(-24+17*sin(d*x+c)+sin(3*d*x+3*c)-8*cos(2*d*x+2*c)-32*cos(d*x+c)-20*d* x*cos(d*x+c))/d/a^2/cos(d*x+c)
Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.45 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\cos \left (d x + c\right )^{3} + 5 \, d x + {\left (5 \, d x + 7\right )} \cos \left (d x + c\right ) + 4 \, \cos \left (d x + c\right )^{2} + {\left (5 \, d x - \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) + 4}{2 \, {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d \sin \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/2*(cos(d*x + c)^3 + 5*d*x + (5*d*x + 7)*cos(d*x + c) + 4*cos(d*x + c)^2 + (5*d*x - cos(d*x + c)^2 + 3*cos(d*x + c) - 4)*sin(d*x + c) + 4)/(a^2*d* cos(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. 1248 vs. \(2 (63) = 126\).
Time = 7.89 (sec) , antiderivative size = 1248, normalized size of antiderivative = 18.09 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**2*sin(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Piecewise((-5*d*x*tan(c/2 + d*x/2)**5/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a* *2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 5*d*x*tan(c/2 + d*x /2)**4/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a* *2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 10*d*x*tan(c/2 + d*x/2)**3/(2*a**2*d*tan(c/2 + d*x /2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a **2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 10*d*x *tan(c/2 + d*x/2)**2/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d* x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2* a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 5*d*x*tan(c/2 + d*x/2)/(2*a**2*d*tan (c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2 )**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d ) - 5*d*x/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4 *a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan( c/2 + d*x/2) + 2*a**2*d) - 10*tan(c/2 + d*x/2)**4/(2*a**2*d*tan(c/2 + d*x/ 2)**5 + 2*a**2*d*tan(c/2 + d*x/2)**4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a* *2*d*tan(c/2 + d*x/2)**2 + 2*a**2*d*tan(c/2 + d*x/2) + 2*a**2*d) - 10*tan( c/2 + d*x/2)**3/(2*a**2*d*tan(c/2 + d*x/2)**5 + 2*a**2*d*tan(c/2 + d*x/2)* *4 + 4*a**2*d*tan(c/2 + d*x/2)**3 + 4*a**2*d*tan(c/2 + d*x/2)**2 + 2*a*...
Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (65) = 130\).
Time = 0.11 (sec) , antiderivative size = 226, normalized size of antiderivative = 3.28 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {11 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 8}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {5 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-((3*sin(d*x + c)/(cos(d*x + c) + 1) + 11*sin(d*x + c)^2/(cos(d*x + c) + 1 )^2 + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 8)/(a^2 + a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 5*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {5 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {8}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/2*(5*(d*x + c)/a^2 + 2*(tan(1/2*d*x + 1/2*c)^3 + 4*tan(1/2*d*x + 1/2*c) ^2 - tan(1/2*d*x + 1/2*c) + 4)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2) + 8/(a ^2*(tan(1/2*d*x + 1/2*c) + 1)))/d
Time = 18.99 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {5\,x}{2\,a^2}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \] Input:
int((cos(c + d*x)^2*sin(c + d*x)^2)/(a + a*sin(c + d*x))^2,x)
Output:
- (5*x)/(2*a^2) - (3*tan(c/2 + (d*x)/2) + 11*tan(c/2 + (d*x)/2)^2 + 5*tan( c/2 + (d*x)/2)^3 + 5*tan(c/2 + (d*x)/2)^4 + 8)/(a^2*d*(tan(c/2 + (d*x)/2) + 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^2(c+d x) \sin ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+3 \cos \left (d x +c \right ) \sin \left (d x +c \right )-5 \cos \left (d x +c \right ) d x +10 \cos \left (d x +c \right )-\sin \left (d x +c \right )^{3}+4 \sin \left (d x +c \right )^{2}+5 \sin \left (d x +c \right ) d x +3 \sin \left (d x +c \right )+5 d x -10}{2 a^{2} d \left (\cos \left (d x +c \right )-\sin \left (d x +c \right )-1\right )} \] Input:
int(cos(d*x+c)^2*sin(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
( - cos(c + d*x)*sin(c + d*x)**2 + 3*cos(c + d*x)*sin(c + d*x) - 5*cos(c + d*x)*d*x + 10*cos(c + d*x) - sin(c + d*x)**3 + 4*sin(c + d*x)**2 + 5*sin( c + d*x)*d*x + 3*sin(c + d*x) + 5*d*x - 10)/(2*a**2*d*(cos(c + d*x) - sin( c + d*x) - 1))