Integrand size = 29, antiderivative size = 121 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\sec (c+d x)}{a^3 d}+\frac {7 \sec ^3(c+d x)}{3 a^3 d}-\frac {3 \sec ^5(c+d x)}{a^3 d}+\frac {13 \sec ^7(c+d x)}{7 a^3 d}-\frac {4 \sec ^9(c+d x)}{9 a^3 d}+\frac {\tan ^7(c+d x)}{7 a^3 d}+\frac {4 \tan ^9(c+d x)}{9 a^3 d} \] Output:
-sec(d*x+c)/a^3/d+7/3*sec(d*x+c)^3/a^3/d-3*sec(d*x+c)^5/a^3/d+13/7*sec(d*x +c)^7/a^3/d-4/9*sec(d*x+c)^9/a^3/d+1/7*tan(d*x+c)^7/a^3/d+4/9*tan(d*x+c)^9 /a^3/d
Time = 1.75 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-9408+36252 \cos (c+d x)-12384 \cos (2 (c+d x))+2014 \cos (3 (c+d x))+4800 \cos (4 (c+d x))-6042 \cos (5 (c+d x))+608 \cos (6 (c+d x))-2304 \sin (c+d x)+27189 \sin (2 (c+d x))-16256 \sin (3 (c+d x))+12084 \sin (4 (c+d x))+384 \sin (5 (c+d x))-1007 \sin (6 (c+d x))}{64512 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 (a+a \sin (c+d x))^3} \] Input:
Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]
Output:
(-9408 + 36252*Cos[c + d*x] - 12384*Cos[2*(c + d*x)] + 2014*Cos[3*(c + d*x )] + 4800*Cos[4*(c + d*x)] - 6042*Cos[5*(c + d*x)] + 608*Cos[6*(c + d*x)] - 2304*Sin[c + d*x] + 27189*Sin[2*(c + d*x)] - 16256*Sin[3*(c + d*x)] + 12 084*Sin[4*(c + d*x)] + 384*Sin[5*(c + d*x)] - 1007*Sin[6*(c + d*x)])/(6451 2*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d *x)/2])^3*(a + a*Sin[c + d*x])^3)
Time = 0.62 (sec) , antiderivative size = 125, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^4 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec ^4(c+d x) (a-a \sin (c+d x))^3 \tan ^6(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^6 (a-a \sin (c+d x))^3}{\cos (c+d x)^{10}}dx}{a^6}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (-a^3 \sec (c+d x) \tan ^9(c+d x)+3 a^3 \sec ^2(c+d x) \tan ^8(c+d x)-3 a^3 \sec ^3(c+d x) \tan ^7(c+d x)+a^3 \sec ^4(c+d x) \tan ^6(c+d x)\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {4 a^3 \tan ^9(c+d x)}{9 d}+\frac {a^3 \tan ^7(c+d x)}{7 d}-\frac {4 a^3 \sec ^9(c+d x)}{9 d}+\frac {13 a^3 \sec ^7(c+d x)}{7 d}-\frac {3 a^3 \sec ^5(c+d x)}{d}+\frac {7 a^3 \sec ^3(c+d x)}{3 d}-\frac {a^3 \sec (c+d x)}{d}}{a^6}\) |
Input:
Int[(Sin[c + d*x]^2*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^3,x]
Output:
(-((a^3*Sec[c + d*x])/d) + (7*a^3*Sec[c + d*x]^3)/(3*d) - (3*a^3*Sec[c + d *x]^5)/d + (13*a^3*Sec[c + d*x]^7)/(7*d) - (4*a^3*Sec[c + d*x]^9)/(9*d) + (a^3*Tan[c + d*x]^7)/(7*d) + (4*a^3*Tan[c + d*x]^9)/(9*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]
Result contains complex when optimal does not.
Time = 1.29 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.37
| method | result | size |
| risch | \(-\frac {2 \left (63 \,{\mathrm e}^{11 i \left (d x +c \right )}-273 \,{\mathrm e}^{9 i \left (d x +c \right )}+51 \,{\mathrm e}^{i \left (d x +c \right )}+19 i+235 \,{\mathrm e}^{3 i \left (d x +c \right )}-39 i {\mathrm e}^{2 i \left (d x +c \right )}-294 i {\mathrm e}^{6 i \left (d x +c \right )}-450 i {\mathrm e}^{4 i \left (d x +c \right )}+189 i {\mathrm e}^{10 i \left (d x +c \right )}-306 \,{\mathrm e}^{5 i \left (d x +c \right )}-378 \,{\mathrm e}^{7 i \left (d x +c \right )}+63 i {\mathrm e}^{8 i \left (d x +c \right )}\right )}{63 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{9} d \,a^{3}}\) | \(166\) |
| derivativedivides | \(\frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {44}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {10}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a^{3} d}\) | \(190\) |
| default | \(\frac {-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {8}{9 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {44}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {10}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a^{3} d}\) | \(190\) |
Input:
int(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-2/63*(63*exp(11*I*(d*x+c))-273*exp(9*I*(d*x+c))+51*exp(I*(d*x+c))+19*I+23 5*exp(3*I*(d*x+c))-39*I*exp(2*I*(d*x+c))-294*I*exp(6*I*(d*x+c))-450*I*exp( 4*I*(d*x+c))+189*I*exp(10*I*(d*x+c))-306*exp(5*I*(d*x+c))-378*exp(7*I*(d*x +c))+63*I*exp(8*I*(d*x+c)))/(exp(I*(d*x+c))-I)^3/(exp(I*(d*x+c))+I)^9/d/a^ 3
Time = 0.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {19 \, \cos \left (d x + c\right )^{6} + 9 \, \cos \left (d x + c\right )^{4} - 51 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 34 \, \cos \left (d x + c\right )^{2} + 7\right )} \sin \left (d x + c\right ) + 7}{63 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/63*(19*cos(d*x + c)^6 + 9*cos(d*x + c)^4 - 51*cos(d*x + c)^2 + 2*(3*cos (d*x + c)^4 - 34*cos(d*x + c)^2 + 7)*sin(d*x + c) + 7)/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3)*sin(d*x + c))
\[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sin ^{2}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sin(d*x+c)**2*tan(d*x+c)**4/(a+a*sin(d*x+c))**3,x)
Output:
Integral(sin(c + d*x)**2*tan(c + d*x)**4/(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. 362 vs. \(2 (111) = 222\).
Time = 0.04 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.99 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {32 \, {\left (\frac {6 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 1\right )}}{63 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {12 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {27 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {36 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {27 \, a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {12 \, a^{3} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {a^{3} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
-32/63*(6*sin(d*x + c)/(cos(d*x + c) + 1) + 12*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + 2*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27*sin(d*x + c)^4/(cos( d*x + c) + 1)^4 - 36*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1)/((a^3 + 6*a^ 3*sin(d*x + c)/(cos(d*x + c) + 1) + 12*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 2*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 27*a^3*sin(d*x + c)^4/( cos(d*x + c) + 1)^4 - 36*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 36*a^3* sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 27*a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 2*a^3*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 12*a^3*sin(d*x + c)^1 0/(cos(d*x + c) + 1)^10 - 6*a^3*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - a^ 3*sin(d*x + c)^12/(cos(d*x + c) + 1)^12)*d)
Time = 0.33 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.42 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {21 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 36 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 17\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {315 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 3024 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 13020 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 32760 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 51282 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 43008 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 20988 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 5688 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 667}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{9}}}{2016 \, d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
1/2016*(21*(15*tan(1/2*d*x + 1/2*c)^2 - 36*tan(1/2*d*x + 1/2*c) + 17)/(a^3 *(tan(1/2*d*x + 1/2*c) - 1)^3) - (315*tan(1/2*d*x + 1/2*c)^8 + 3024*tan(1/ 2*d*x + 1/2*c)^7 + 13020*tan(1/2*d*x + 1/2*c)^6 + 32760*tan(1/2*d*x + 1/2* c)^5 + 51282*tan(1/2*d*x + 1/2*c)^4 + 43008*tan(1/2*d*x + 1/2*c)^3 + 20988 *tan(1/2*d*x + 1/2*c)^2 + 5688*tan(1/2*d*x + 1/2*c) + 667)/(a^3*(tan(1/2*d *x + 1/2*c) + 1)^9))/d
Time = 35.22 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.52 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}}{63}+\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}+\frac {128\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{21}+\frac {64\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{63}-\frac {96\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}-\frac {128\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{7}}{a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^9} \] Input:
int((sin(c + d*x)^2*tan(c + d*x)^4)/(a + a*sin(c + d*x))^3,x)
Output:
-((32*cos(c/2 + (d*x)/2)^12)/63 + (64*cos(c/2 + (d*x)/2)^11*sin(c/2 + (d*x )/2))/21 - (128*cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^5)/7 - (96*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2)^4)/7 + (64*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^3)/63 + (128*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^2)/21)/(a^ 3*d*(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^3*(cos(c/2 + (d*x)/2) + sin( c/2 + (d*x)/2))^9)
Time = 0.18 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.75 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}-48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+48 \cos \left (d x +c \right ) \sin \left (d x +c \right )+16 \cos \left (d x +c \right )+19 \sin \left (d x +c \right )^{6}-6 \sin \left (d x +c \right )^{5}-66 \sin \left (d x +c \right )^{4}-56 \sin \left (d x +c \right )^{3}+24 \sin \left (d x +c \right )^{2}+48 \sin \left (d x +c \right )+16}{63 \cos \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right )^{5}+3 \sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )^{2}-3 \sin \left (d x +c \right )-1\right )} \] Input:
int(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^3,x)
Output:
( - 16*cos(c + d*x)*sin(c + d*x)**5 - 48*cos(c + d*x)*sin(c + d*x)**4 - 32 *cos(c + d*x)*sin(c + d*x)**3 + 32*cos(c + d*x)*sin(c + d*x)**2 + 48*cos(c + d*x)*sin(c + d*x) + 16*cos(c + d*x) + 19*sin(c + d*x)**6 - 6*sin(c + d* x)**5 - 66*sin(c + d*x)**4 - 56*sin(c + d*x)**3 + 24*sin(c + d*x)**2 + 48* sin(c + d*x) + 16)/(63*cos(c + d*x)*a**3*d*(sin(c + d*x)**5 + 3*sin(c + d* x)**4 + 2*sin(c + d*x)**3 - 2*sin(c + d*x)**2 - 3*sin(c + d*x) - 1))