Integrand size = 29, antiderivative size = 151 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3 x}{a^3}+\frac {\cos (c+d x)}{a^3 d}+\frac {7 \sec (c+d x)}{a^3 d}-\frac {5 \sec ^3(c+d x)}{a^3 d}+\frac {13 \sec ^5(c+d x)}{5 a^3 d}-\frac {4 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \tan (c+d x)}{a^3 d}+\frac {\tan ^3(c+d x)}{a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}+\frac {4 \tan ^7(c+d x)}{7 a^3 d} \] Output:
3*x/a^3+cos(d*x+c)/a^3/d+7*sec(d*x+c)/a^3/d-5*sec(d*x+c)^3/a^3/d+13/5*sec( d*x+c)^5/a^3/d-4/7*sec(d*x+c)^7/a^3/d-3*tan(d*x+c)/a^3/d+tan(d*x+c)^3/a^3/ d-3/5*tan(d*x+c)^5/a^3/d+4/7*tan(d*x+c)^7/a^3/d
Time = 2.18 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.48 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {8400+14 (-1483+840 c+840 d x) \cos (c+d x)+5152 \cos (2 (c+d x))+8898 \cos (3 (c+d x))-5040 c \cos (3 (c+d x))-5040 d x \cos (3 (c+d x))-2288 \cos (4 (c+d x))+8008 \sin (c+d x)-20762 \sin (2 (c+d x))+11760 c \sin (2 (c+d x))+11760 d x \sin (2 (c+d x))+6588 \sin (3 (c+d x))+1483 \sin (4 (c+d x))-840 c \sin (4 (c+d x))-840 d x \sin (4 (c+d x))-140 \sin (5 (c+d x))}{2240 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^7} \] Input:
Integrate[(Sin[c + d*x]^4*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(8400 + 14*(-1483 + 840*c + 840*d*x)*Cos[c + d*x] + 5152*Cos[2*(c + d*x)] + 8898*Cos[3*(c + d*x)] - 5040*c*Cos[3*(c + d*x)] - 5040*d*x*Cos[3*(c + d* x)] - 2288*Cos[4*(c + d*x)] + 8008*Sin[c + d*x] - 20762*Sin[2*(c + d*x)] + 11760*c*Sin[2*(c + d*x)] + 11760*d*x*Sin[2*(c + d*x)] + 6588*Sin[3*(c + d *x)] + 1483*Sin[4*(c + d*x)] - 840*c*Sin[4*(c + d*x)] - 840*d*x*Sin[4*(c + d*x)] - 140*Sin[5*(c + d*x)])/(2240*a^3*d*(Cos[(c + d*x)/2] - Sin[(c + d* x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^7)
Time = 0.66 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.03, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^2 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) (a-a \sin (c+d x))^3 \tan ^6(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^6 (a-a \sin (c+d x))^3}{\cos (c+d x)^8}dx}{a^6}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (3 a^3 \tan ^8(c+d x)-a^3 \sin (c+d x) \tan ^8(c+d x)-3 a^3 \sec (c+d x) \tan ^7(c+d x)+a^3 \sec ^2(c+d x) \tan ^6(c+d x)\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^3 \cos (c+d x)}{d}+\frac {4 a^3 \tan ^7(c+d x)}{7 d}-\frac {3 a^3 \tan ^5(c+d x)}{5 d}+\frac {a^3 \tan ^3(c+d x)}{d}-\frac {3 a^3 \tan (c+d x)}{d}-\frac {4 a^3 \sec ^7(c+d x)}{7 d}+\frac {13 a^3 \sec ^5(c+d x)}{5 d}-\frac {5 a^3 \sec ^3(c+d x)}{d}+\frac {7 a^3 \sec (c+d x)}{d}+3 a^3 x}{a^6}\) |
Input:
Int[(Sin[c + d*x]^4*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(3*a^3*x + (a^3*Cos[c + d*x])/d + (7*a^3*Sec[c + d*x])/d - (5*a^3*Sec[c + d*x]^3)/d + (13*a^3*Sec[c + d*x]^5)/(5*d) - (4*a^3*Sec[c + d*x]^7)/(7*d) - (3*a^3*Tan[c + d*x])/d + (a^3*Tan[c + d*x]^3)/d - (3*a^3*Tan[c + d*x]^5)/ (5*d) + (4*a^3*Tan[c + d*x]^7)/(7*d))/a^6
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]
Time = 8.94 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05
| method | result | size |
| derivativedivides | \(\frac {-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {14}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {17}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {49}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {128}{64+64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\) | \(159\) |
| default | \(\frac {-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {14}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {17}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {49}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {128}{64+64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\) | \(159\) |
| risch | \(\frac {3 x}{a^{3}}+\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{3}}+\frac {-46 i {\mathrm e}^{4 i \left (d x +c \right )}-94 \,{\mathrm e}^{5 i \left (d x +c \right )}-\frac {254 \,{\mathrm e}^{3 i \left (d x +c \right )}}{5}+58 i {\mathrm e}^{6 i \left (d x +c \right )}+14 \,{\mathrm e}^{7 i \left (d x +c \right )}-\frac {434 i {\mathrm e}^{2 i \left (d x +c \right )}}{5}+\frac {1682 \,{\mathrm e}^{i \left (d x +c \right )}}{35}+\frac {362 i}{35}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{3}}\) | \(161\) |
Input:
int(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
128/d/a^3*(-1/112/(tan(1/2*d*x+1/2*c)+1)^7+1/32/(tan(1/2*d*x+1/2*c)+1)^6-7 /320/(tan(1/2*d*x+1/2*c)+1)^5-3/128/(tan(1/2*d*x+1/2*c)+1)^4+1/256/(tan(1/ 2*d*x+1/2*c)+1)^3+17/512/(tan(1/2*d*x+1/2*c)+1)^2+49/1024/(tan(1/2*d*x+1/2 *c)+1)-1/1024/(tan(1/2*d*x+1/2*c)-1)+1/64/(1+tan(1/2*d*x+1/2*c)^2)+3/64*ar ctan(tan(1/2*d*x+1/2*c)))
Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {315 \, d x \cos \left (d x + c\right )^{3} + 286 \, \cos \left (d x + c\right )^{4} - 420 \, d x \cos \left (d x + c\right ) - 447 \, \cos \left (d x + c\right )^{2} + {\left (105 \, d x \cos \left (d x + c\right )^{3} + 35 \, \cos \left (d x + c\right )^{4} - 420 \, d x \cos \left (d x + c\right ) - 438 \, \cos \left (d x + c\right )^{2} - 20\right )} \sin \left (d x + c\right ) - 15}{35 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
1/35*(315*d*x*cos(d*x + c)^3 + 286*cos(d*x + c)^4 - 420*d*x*cos(d*x + c) - 447*cos(d*x + c)^2 + (105*d*x*cos(d*x + c)^3 + 35*cos(d*x + c)^4 - 420*d* x*cos(d*x + c) - 438*cos(d*x + c)^2 - 20)*sin(d*x + c) - 15)/(3*a^3*d*cos( d*x + c)^3 - 4*a^3*d*cos(d*x + c) + (a^3*d*cos(d*x + c)^3 - 4*a^3*d*cos(d* x + c))*sin(d*x + c))
\[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sin(d*x+c)**4*tan(d*x+c)**2/(a+a*sin(d*x+c))**3,x)
Output:
Integral(sin(c + d*x)**4*tan(c + d*x)**2/(sin(c + d*x)**3 + 3*sin(c + d*x) **2 + 3*sin(c + d*x) + 1), x)/a**3
Leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (143) = 286\).
Time = 0.12 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.79 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (\frac {\frac {951 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {2010 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {1980 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {574 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {966 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {1890 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {1540 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {630 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {105 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 176}{a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {15 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {20 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {20 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {15 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{35 \, d} \] Input:
integrate(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
2/35*((951*sin(d*x + c)/(cos(d*x + c) + 1) + 2010*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1980*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 574*sin(d*x + c)^ 4/(cos(d*x + c) + 1)^4 - 966*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1890*si n(d*x + c)^6/(cos(d*x + c) + 1)^6 - 1540*sin(d*x + c)^7/(cos(d*x + c) + 1) ^7 - 630*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 105*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 176)/(a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 15*a^3* sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 20*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 14*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 14*a^3*sin(d*x + c)^ 6/(cos(d*x + c) + 1)^6 - 20*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 15*a ^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*a^3*sin(d*x + c)^9/(cos(d*x + c ) + 1)^9 - a^3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) + 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d
Time = 0.30 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.17 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {840 \, {\left (d x + c\right )}}{a^{3}} - \frac {35 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 16 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )} a^{3}} + \frac {1715 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 11480 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31815 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 45920 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35161 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 13832 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2221}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \] Input:
integrate(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
1/280*(840*(d*x + c)/a^3 - 35*(tan(1/2*d*x + 1/2*c)^2 - 16*tan(1/2*d*x + 1 /2*c) + 17)/((tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c)^2 + tan(1/2*d* x + 1/2*c) - 1)*a^3) + (1715*tan(1/2*d*x + 1/2*c)^6 + 11480*tan(1/2*d*x + 1/2*c)^5 + 31815*tan(1/2*d*x + 1/2*c)^4 + 45920*tan(1/2*d*x + 1/2*c)^3 + 3 5161*tan(1/2*d*x + 1/2*c)^2 + 13832*tan(1/2*d*x + 1/2*c) + 2221)/(a^3*(tan (1/2*d*x + 1/2*c) + 1)^7))/d
Time = 34.87 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.21 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {3\,x}{a^3}-\frac {-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-36\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-88\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-108\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {276\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {164\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{5}+\frac {792\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{7}+\frac {804\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{7}+\frac {1902\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+\frac {352}{35}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \] Input:
int((sin(c + d*x)^4*tan(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)
Output:
(3*x)/a^3 - ((1902*tan(c/2 + (d*x)/2))/35 + (804*tan(c/2 + (d*x)/2)^2)/7 + (792*tan(c/2 + (d*x)/2)^3)/7 + (164*tan(c/2 + (d*x)/2)^4)/5 - (276*tan(c/ 2 + (d*x)/2)^5)/5 - 108*tan(c/2 + (d*x)/2)^6 - 88*tan(c/2 + (d*x)/2)^7 - 3 6*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 + (d*x)/2)^9 + 352/35)/(a^3*d*(tan(c/2 + (d*x)/2) + 1)^7*(tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^2 + tan(c/2 + ( d*x)/2)^3 - 1))
Time = 0.17 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.79 \[ \int \frac {\sin ^4(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} d x +141 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+315 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c +315 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x +423 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+315 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +315 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x +423 \cos \left (d x +c \right ) \sin \left (d x +c \right )+105 \cos \left (d x +c \right ) c +105 \cos \left (d x +c \right ) d x +141 \cos \left (d x +c \right )-35 \sin \left (d x +c \right )^{5}-286 \sin \left (d x +c \right )^{4}-368 \sin \left (d x +c \right )^{3}+125 \sin \left (d x +c \right )^{2}+423 \sin \left (d x +c \right )+176}{35 \cos \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1\right )} \] Input:
int(sin(d*x+c)^4*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x)
Output:
(105*cos(c + d*x)*sin(c + d*x)**3*c + 105*cos(c + d*x)*sin(c + d*x)**3*d*x + 141*cos(c + d*x)*sin(c + d*x)**3 + 315*cos(c + d*x)*sin(c + d*x)**2*c + 315*cos(c + d*x)*sin(c + d*x)**2*d*x + 423*cos(c + d*x)*sin(c + d*x)**2 + 315*cos(c + d*x)*sin(c + d*x)*c + 315*cos(c + d*x)*sin(c + d*x)*d*x + 423 *cos(c + d*x)*sin(c + d*x) + 105*cos(c + d*x)*c + 105*cos(c + d*x)*d*x + 1 41*cos(c + d*x) - 35*sin(c + d*x)**5 - 286*sin(c + d*x)**4 - 368*sin(c + d *x)**3 + 125*sin(c + d*x)**2 + 423*sin(c + d*x) + 176)/(35*cos(c + d*x)*a* *3*d*(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1))