Integrand size = 29, antiderivative size = 141 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {9 x}{2 a^2}-\frac {2 \cos (c+d x)}{a^2 d}-\frac {6 \sec (c+d x)}{a^2 d}+\frac {2 \sec ^3(c+d x)}{a^2 d}-\frac {2 \sec ^5(c+d x)}{5 a^2 d}+\frac {\cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac {4 \tan (c+d x)}{a^2 d}-\frac {\tan ^3(c+d x)}{a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d} \] Output:
-9/2*x/a^2-2*cos(d*x+c)/a^2/d-6*sec(d*x+c)/a^2/d+2*sec(d*x+c)^3/a^2/d-2/5* sec(d*x+c)^5/a^2/d+1/2*cos(d*x+c)*sin(d*x+c)/a^2/d+4*tan(d*x+c)/a^2/d-tan( d*x+c)^3/a^2/d+2/5*tan(d*x+c)^5/a^2/d
Time = 2.02 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.35 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {500+10 (-103+90 c+90 d x) \cos (c+d x)+544 \cos (2 (c+d x))+206 \cos (3 (c+d x))-180 c \cos (3 (c+d x))-180 d x \cos (3 (c+d x))-20 \cos (4 (c+d x))+250 \sin (c+d x)-824 \sin (2 (c+d x))+720 c \sin (2 (c+d x))+720 d x \sin (2 (c+d x))+351 \sin (3 (c+d x))+5 \sin (5 (c+d x))}{160 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \] Input:
Integrate[(Sin[c + d*x]^4*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
-1/160*(500 + 10*(-103 + 90*c + 90*d*x)*Cos[c + d*x] + 544*Cos[2*(c + d*x) ] + 206*Cos[3*(c + d*x)] - 180*c*Cos[3*(c + d*x)] - 180*d*x*Cos[3*(c + d*x )] - 20*Cos[4*(c + d*x)] + 250*Sin[c + d*x] - 824*Sin[2*(c + d*x)] + 720*c *Sin[2*(c + d*x)] + 720*d*x*Sin[2*(c + d*x)] + 351*Sin[3*(c + d*x)] + 5*Si n[5*(c + d*x)])/(a^2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x )/2] + Sin[(c + d*x)/2])^5)
Time = 0.56 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.09, 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, 3189, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^2 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x))^2 \tan ^6(c+d x)dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x))^2 \tan (c+d x)^6dx}{a^4}\) |
| \(\Big \downarrow \) 3189 |
| \(\displaystyle \frac {\int \left (a^2 \tan ^6(c+d x)+a^2 \sin ^2(c+d x) \tan ^6(c+d x)-2 a^2 \sin (c+d x) \tan ^6(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {2 a^2 \cos (c+d x)}{d}+\frac {9 a^2 \tan ^5(c+d x)}{10 d}-\frac {3 a^2 \tan ^3(c+d x)}{2 d}+\frac {9 a^2 \tan (c+d x)}{2 d}-\frac {2 a^2 \sec ^5(c+d x)}{5 d}+\frac {2 a^2 \sec ^3(c+d x)}{d}-\frac {6 a^2 \sec (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x) \tan ^5(c+d x)}{2 d}-\frac {9 a^2 x}{2}}{a^4}\) |
Input:
Int[(Sin[c + d*x]^4*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^2,x]
Output:
((-9*a^2*x)/2 - (2*a^2*Cos[c + d*x])/d - (6*a^2*Sec[c + d*x])/d + (2*a^2*S ec[c + d*x]^3)/d - (2*a^2*Sec[c + d*x]^5)/(5*d) + (9*a^2*Tan[c + d*x])/(2* d) - (3*a^2*Tan[c + d*x]^3)/(2*d) + (9*a^2*Tan[c + d*x]^5)/(10*d) - (a^2*S in[c + d*x]^2*Tan[c + d*x]^5)/(2*d))/a^4
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*( x_)])^(p_.), x_Symbol] :> Int[ExpandIntegrand[(g*Tan[e + f*x])^p, (a + b*Si n[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, 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]
Time = 6.06 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.19
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {1}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{2}}\) | \(168\) |
| default | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}+\frac {1}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-9 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {31}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d \,a^{2}}\) | \(168\) |
| risch | \(-\frac {9 x}{2 a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{d \,a^{2}}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d \,a^{2}}-\frac {2 \left (-40 \,{\mathrm e}^{3 i \left (d x +c \right )}+75 i {\mathrm e}^{4 i \left (d x +c \right )}+30 \,{\mathrm e}^{5 i \left (d x +c \right )}-78 \,{\mathrm e}^{i \left (d x +c \right )}+60 i {\mathrm e}^{2 i \left (d x +c \right )}-27 i\right )}{5 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} d \,a^{2}}\) | \(174\) |
Input:
int(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
128/d/a^2*(-1/512/(tan(1/2*d*x+1/2*c)-1)-1/16*(1/8*tan(1/2*d*x+1/2*c)^3+1/ 2*tan(1/2*d*x+1/2*c)^2-1/8*tan(1/2*d*x+1/2*c)+1/2)/(1+tan(1/2*d*x+1/2*c)^2 )^2-9/128*arctan(tan(1/2*d*x+1/2*c))-1/160/(tan(1/2*d*x+1/2*c)+1)^5+1/64/( tan(1/2*d*x+1/2*c)+1)^4+1/128/(tan(1/2*d*x+1/2*c)+1)^3-7/256/(tan(1/2*d*x+ 1/2*c)+1)^2-31/512/(tan(1/2*d*x+1/2*c)+1))
Time = 0.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.94 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {45 \, d x \cos \left (d x + c\right )^{3} + 10 \, \cos \left (d x + c\right )^{4} - 90 \, d x \cos \left (d x + c\right ) - 78 \, \cos \left (d x + c\right )^{2} - {\left (5 \, \cos \left (d x + c\right )^{4} + 90 \, d x \cos \left (d x + c\right ) + 84 \, \cos \left (d x + c\right )^{2} - 6\right )} \sin \left (d x + c\right ) + 4}{10 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/10*(45*d*x*cos(d*x + c)^3 + 10*cos(d*x + c)^4 - 90*d*x*cos(d*x + c) - 7 8*cos(d*x + c)^2 - (5*cos(d*x + c)^4 + 90*d*x*cos(d*x + c) + 84*cos(d*x + c)^2 - 6)*sin(d*x + c) + 4)/(a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c)*s in(d*x + c) - 2*a^2*d*cos(d*x + c))
\[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sin(d*x+c)**4*tan(d*x+c)**2/(a+a*sin(d*x+c))**2,x)
Output:
Integral(sin(c + d*x)**4*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. 421 vs. \(2 (133) = 266\).
Time = 0.13 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.99 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {\frac {211 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {268 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {212 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {84 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {174 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {300 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {300 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {180 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {45 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 64}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {8 \, a^{2} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {6 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {8 \, a^{2} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {7 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{5 \, d} \] Input:
integrate(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/5*((211*sin(d*x + c)/(cos(d*x + c) + 1) + 268*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 212*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 84*sin(d*x + c)^4/( cos(d*x + c) + 1)^4 - 174*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 300*sin(d* x + c)^6/(cos(d*x + c) + 1)^6 - 300*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 180*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 45*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 64)/(a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 7*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 8*a^2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 6*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 8*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 7*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 4*a^2*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a^ 2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) + 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.13 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {90 \, {\left (d x + c\right )}}{a^{2}} + \frac {20 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}} + \frac {5}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {155 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 690 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 750 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 181}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{20 \, d} \] Input:
integrate(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/20*(90*(d*x + c)/a^2 + 20*(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) + 5 /(a^2*(tan(1/2*d*x + 1/2*c) - 1)) + (155*tan(1/2*d*x + 1/2*c)^4 + 690*tan( 1/2*d*x + 1/2*c)^3 + 1120*tan(1/2*d*x + 1/2*c)^2 + 750*tan(1/2*d*x + 1/2*c ) + 181)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 36.52 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.22 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-60\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-60\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {174\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {84\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {212\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{5}+\frac {268\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{5}+\frac {211\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+\frac {64}{5}}{a^2\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {9\,x}{2\,a^2} \] Input:
int((sin(c + d*x)^4*tan(c + d*x)^2)/(a + a*sin(c + d*x))^2,x)
Output:
((211*tan(c/2 + (d*x)/2))/5 + (268*tan(c/2 + (d*x)/2)^2)/5 + (212*tan(c/2 + (d*x)/2)^3)/5 + (84*tan(c/2 + (d*x)/2)^4)/5 - (174*tan(c/2 + (d*x)/2)^5) /5 - 60*tan(c/2 + (d*x)/2)^6 - 60*tan(c/2 + (d*x)/2)^7 - 36*tan(c/2 + (d*x )/2)^8 - 9*tan(c/2 + (d*x)/2)^9 + 64/5)/(a^2*d*(tan(c/2 + (d*x)/2) - 1)*(t an(c/2 + (d*x)/2) + 1)^5*(tan(c/2 + (d*x)/2)^2 + 1)^2) - (9*x)/(2*a^2)
Time = 0.16 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.48 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-90 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c -90 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x -83 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-180 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -180 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -166 \cos \left (d x +c \right ) \sin \left (d x +c \right )-90 \cos \left (d x +c \right ) c -90 \cos \left (d x +c \right ) d x -83 \cos \left (d x +c \right )-10 \sin \left (d x +c \right )^{5}+20 \sin \left (d x +c \right )^{4}+188 \sin \left (d x +c \right )^{3}+116 \sin \left (d x +c \right )^{2}-166 \sin \left (d x +c \right )-128}{20 \cos \left (d x +c \right ) a^{2} d \left (\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1\right )} \] Input:
int(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^2,x)
Output:
( - 90*cos(c + d*x)*sin(c + d*x)**2*c - 90*cos(c + d*x)*sin(c + d*x)**2*d* x - 83*cos(c + d*x)*sin(c + d*x)**2 - 180*cos(c + d*x)*sin(c + d*x)*c - 18 0*cos(c + d*x)*sin(c + d*x)*d*x - 166*cos(c + d*x)*sin(c + d*x) - 90*cos(c + d*x)*c - 90*cos(c + d*x)*d*x - 83*cos(c + d*x) - 10*sin(c + d*x)**5 + 2 0*sin(c + d*x)**4 + 188*sin(c + d*x)**3 + 116*sin(c + d*x)**2 - 166*sin(c + d*x) - 128)/(20*cos(c + d*x)*a**2*d*(sin(c + d*x)**2 + 2*sin(c + d*x) + 1))