Integrand size = 29, antiderivative size = 106 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {x}{a^2}-\frac {2 \sec (c+d x)}{a^2 d}+\frac {4 \sec ^3(c+d x)}{3 a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {\tan (c+d x)}{a^2 d}-\frac {\tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d} \] Output:
-x/a^2-2*sec(d*x+c)/a^2/d+4/3*sec(d*x+c)^3/a^2/d-2/5*sec(d*x+c)^5/a^2/d+ta n(d*x+c)/a^2/d-1/3*tan(d*x+c)^3/a^2/d+2/5*tan(d*x+c)^5/a^2/d
Time = 1.83 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.35 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\sec (c+d x) \left (20+\frac {5}{4} (-89+60 c+60 d x) \cos (c+d x)+44 \cos (2 (c+d x))+\frac {89}{4} \cos (3 (c+d x))-15 c \cos (3 (c+d x))-15 d x \cos (3 (c+d x))-10 \sin (c+d x)-89 \sin (2 (c+d x))+60 c \sin (2 (c+d x))+60 d x \sin (2 (c+d x))+26 \sin (3 (c+d x))\right )}{60 a^2 d (1+\sin (c+d x))^2} \] Input:
Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
-1/60*(Sec[c + d*x]*(20 + (5*(-89 + 60*c + 60*d*x)*Cos[c + d*x])/4 + 44*Co s[2*(c + d*x)] + (89*Cos[3*(c + d*x)])/4 - 15*c*Cos[3*(c + d*x)] - 15*d*x* Cos[3*(c + d*x)] - 10*Sin[c + d*x] - 89*Sin[2*(c + d*x)] + 60*c*Sin[2*(c + d*x)] + 60*d*x*Sin[2*(c + d*x)] + 26*Sin[3*(c + d*x)]))/(a^2*d*(1 + Sin[c + d*x])^2)
Time = 0.57 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3042, 3354, 3042, 3352, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^4}{\cos (c+d x)^2 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) (a-a \sin (c+d x))^2 \tan ^4(c+d x)dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^4 (a-a \sin (c+d x))^2}{\cos (c+d x)^6}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (a^2 \tan ^6(c+d x)-2 a^2 \sec (c+d x) \tan ^5(c+d x)+a^2 \sec ^2(c+d x) \tan ^4(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a^2 \tan ^5(c+d x)}{5 d}-\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \tan (c+d x)}{d}-\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {4 a^2 \sec ^3(c+d x)}{3 d}-\frac {2 a^2 \sec (c+d x)}{d}-a^2 x}{a^4}\) |
Input:
Int[(Sin[c + d*x]^2*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
(-(a^2*x) - (2*a^2*Sec[c + d*x])/d + (4*a^2*Sec[c + d*x]^3)/(3*d) - (2*a^2 *Sec[c + d*x]^5)/(5*d) + (a^2*Tan[c + d*x])/d - (a^2*Tan[c + d*x]^3)/(3*d) + (2*a^2*Tan[c + d*x]^5)/(5*d))/a^4
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig [(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x] /; F reeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(a/g)^(2* m) Int[(g*Cos[e + f*x])^(2*m + p)*((d*Sin[e + f*x])^n/(a - b*Sin[e + f*x] )^m), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]
Result contains complex when optimal does not.
Time = 0.52 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.98
| method | result | size |
| risch | \(-\frac {x}{a^{2}}-\frac {4 \left (-10 \,{\mathrm e}^{3 i \left (d x +c \right )}+30 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{5 i \left (d x +c \right )}-37 \,{\mathrm e}^{i \left (d x +c \right )}+35 i {\mathrm e}^{2 i \left (d x +c \right )}-13 i\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} d \,a^{2}}\) | \(104\) |
| derivativedivides | \(\frac {-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d \,a^{2}}\) | \(112\) |
| default | \(\frac {-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d \,a^{2}}\) | \(112\) |
Input:
int(sin(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-x/a^2-4/15*(-10*exp(3*I*(d*x+c))+30*I*exp(4*I*(d*x+c))+15*exp(5*I*(d*x+c) )-37*exp(I*(d*x+c))+35*I*exp(2*I*(d*x+c))-13*I)/(exp(I*(d*x+c))-I)/(exp(I* (d*x+c))+I)^5/d/a^2
Time = 0.08 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {15 \, d x \cos \left (d x + c\right )^{3} - 30 \, d x \cos \left (d x + c\right ) - 22 \, \cos \left (d x + c\right )^{2} - {\left (30 \, d x \cos \left (d x + c\right ) + 26 \, \cos \left (d x + c\right )^{2} - 9\right )} \sin \left (d x + c\right ) + 6}{15 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/15*(15*d*x*cos(d*x + c)^3 - 30*d*x*cos(d*x + c) - 22*cos(d*x + c)^2 - ( 30*d*x*cos(d*x + c) + 26*cos(d*x + c)^2 - 9)*sin(d*x + c) + 6)/(a^2*d*cos( d*x + c)^3 - 2*a^2*d*cos(d*x + c)*sin(d*x + c) - 2*a^2*d*cos(d*x + c))
\[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sin ^{2}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sin(d*x+c)**2*tan(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Integral(sin(c + d*x)**2*tan(c + d*x)**2/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2
Leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (98) = 196\).
Time = 0.12 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.35 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 \, {\left (\frac {\frac {49 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {70 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {60 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 16}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{15 \, d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-2/15*((49*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 70*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 60*sin(d*x + c)^4/(co s(d*x + c) + 1)^4 - 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 16)/(a^2 + 4* a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 5*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^2*sin(d*x + c)^5/( cos(d*x + c) + 1)^5 - a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 15*arctan (sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.97 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {60 \, {\left (d x + c\right )}}{a^{2}} + \frac {15}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {105 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 510 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 920 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 610 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 143}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/60*(60*(d*x + c)/a^2 + 15/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) + (105*tan(1 /2*d*x + 1/2*c)^4 + 510*tan(1/2*d*x + 1/2*c)^3 + 920*tan(1/2*d*x + 1/2*c)^ 2 + 610*tan(1/2*d*x + 1/2*c) + 143)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 34.00 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.99 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {98\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {32}{15}}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5}-\frac {x}{a^2} \] Input:
int((sin(c + d*x)^2*tan(c + d*x)^2)/(a + a*sin(c + d*x))^2,x)
Output:
((98*tan(c/2 + (d*x)/2))/15 + (8*tan(c/2 + (d*x)/2)^2)/3 - (28*tan(c/2 + ( d*x)/2)^3)/3 - 8*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^5 + 32/15)/(a ^2*d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2) + 1)^5) - x/a^2
Time = 0.16 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.92 \[ \int \frac {\sin ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x -16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \tan \left (d x +c \right )-30 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -32 \cos \left (d x +c \right ) \sin \left (d x +c \right )+15 \cos \left (d x +c \right ) \tan \left (d x +c \right )-15 \cos \left (d x +c \right ) d x -16 \cos \left (d x +c \right )+11 \sin \left (d x +c \right )^{3}-8 \sin \left (d x +c \right )^{2}-32 \sin \left (d x +c \right )-16}{15 \cos \left (d x +c \right ) a^{2} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sin(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
(15*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x) - 15*cos(c + d*x)*sin(c + d* x)**2*d*x - 16*cos(c + d*x)*sin(c + d*x)**2 + 30*cos(c + d*x)*sin(c + d*x) *tan(c + d*x) - 30*cos(c + d*x)*sin(c + d*x)*d*x - 32*cos(c + d*x)*sin(c + d*x) + 15*cos(c + d*x)*tan(c + d*x) - 15*cos(c + d*x)*d*x - 16*cos(c + d* x) + 11*sin(c + d*x)**3 - 8*sin(c + d*x)**2 - 32*sin(c + d*x) - 16)/(15*co s(c + d*x)*a**2*d*(sin(c + d*x)**2 + 2*sin(c + d*x) + 1))