Integrand size = 29, antiderivative size = 140 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {x}{a^2}+\frac {2 \sec (c+d x)}{a^2 d}-\frac {2 \sec ^3(c+d x)}{a^2 d}+\frac {6 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \sec ^7(c+d x)}{7 a^2 d}-\frac {\tan (c+d x)}{a^2 d}+\frac {\tan ^3(c+d x)}{3 a^2 d}-\frac {\tan ^5(c+d x)}{5 a^2 d}+\frac {2 \tan ^7(c+d x)}{7 a^2 d} \] Output:
x/a^2+2*sec(d*x+c)/a^2/d-2*sec(d*x+c)^3/a^2/d+6/5*sec(d*x+c)^5/a^2/d-2/7*s ec(d*x+c)^7/a^2/d-tan(d*x+c)/a^2/d+1/3*tan(d*x+c)^3/a^2/d-1/5*tan(d*x+c)^5 /a^2/d+2/7*tan(d*x+c)^7/a^2/d
Time = 1.95 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.84 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4032+42 (-381+280 c+280 d x) \cos (c+d x)+5504 \cos (2 (c+d x))-3429 \cos (3 (c+d x))+2520 c \cos (3 (c+d x))+2520 d x \cos (3 (c+d x))+2752 \cos (4 (c+d x))+1143 \cos (5 (c+d x))-840 c \cos (5 (c+d x))-840 d x \cos (5 (c+d x))+2128 \sin (c+d x)-9144 \sin (2 (c+d x))+6720 c \sin (2 (c+d x))+6720 d x \sin (2 (c+d x))+456 \sin (3 (c+d x))-4572 \sin (4 (c+d x))+3360 c \sin (4 (c+d x))+3360 d x \sin (4 (c+d x))+1528 \sin (5 (c+d x))}{13440 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^7} \] Input:
Integrate[(Sin[c + d*x]^2*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
(4032 + 42*(-381 + 280*c + 280*d*x)*Cos[c + d*x] + 5504*Cos[2*(c + d*x)] - 3429*Cos[3*(c + d*x)] + 2520*c*Cos[3*(c + d*x)] + 2520*d*x*Cos[3*(c + d*x )] + 2752*Cos[4*(c + d*x)] + 1143*Cos[5*(c + d*x)] - 840*c*Cos[5*(c + d*x) ] - 840*d*x*Cos[5*(c + d*x)] + 2128*Sin[c + d*x] - 9144*Sin[2*(c + d*x)] + 6720*c*Sin[2*(c + d*x)] + 6720*d*x*Sin[2*(c + d*x)] + 456*Sin[3*(c + d*x) ] - 4572*Sin[4*(c + d*x)] + 3360*c*Sin[4*(c + d*x)] + 3360*d*x*Sin[4*(c + d*x)] + 1528*Sin[5*(c + d*x)])/(13440*a^2*d*(Cos[(c + d*x)/2] - Sin[(c + d *x)/2])^3*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^7)
Time = 0.58 (sec) , antiderivative size = 144, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^6}{\cos (c+d x)^4 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec ^2(c+d x) (a-a \sin (c+d x))^2 \tan ^6(c+d x)dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^6 (a-a \sin (c+d x))^2}{\cos (c+d x)^8}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (a^2 \tan ^8(c+d x)-2 a^2 \sec (c+d x) \tan ^7(c+d x)+a^2 \sec ^2(c+d x) \tan ^6(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a^2 \tan ^7(c+d x)}{7 d}-\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}-\frac {2 a^2 \sec ^7(c+d x)}{7 d}+\frac {6 a^2 \sec ^5(c+d x)}{5 d}-\frac {2 a^2 \sec ^3(c+d x)}{d}+\frac {2 a^2 \sec (c+d x)}{d}+a^2 x}{a^4}\) |
Input:
Int[(Sin[c + d*x]^2*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
(a^2*x + (2*a^2*Sec[c + d*x])/d - (2*a^2*Sec[c + d*x]^3)/d + (6*a^2*Sec[c + d*x]^5)/(5*d) - (2*a^2*Sec[c + d*x]^7)/(7*d) - (a^2*Tan[c + d*x])/d + (a ^2*Tan[c + d*x]^3)/(3*d) - (a^2*Tan[c + d*x]^5)/(5*d) + (2*a^2*Tan[c + d*x ]^7)/(7*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 = 0.56 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.06
| method | result | size |
| risch | \(\frac {x}{a^{2}}+\frac {-\frac {48 \,{\mathrm e}^{5 i \left (d x +c \right )}}{5}+28 i {\mathrm e}^{6 i \left (d x +c \right )}+\frac {344 i {\mathrm e}^{4 i \left (d x +c \right )}}{15}+8 \,{\mathrm e}^{7 i \left (d x +c \right )}-\frac {2216 \,{\mathrm e}^{3 i \left (d x +c \right )}}{105}+6 i {\mathrm e}^{8 i \left (d x +c \right )}+4 \,{\mathrm e}^{9 i \left (d x +c \right )}+\frac {172 i {\mathrm e}^{2 i \left (d x +c \right )}}{35}-\frac {1108 \,{\mathrm e}^{i \left (d x +c \right )}}{105}-\frac {382 i}{105}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{2}}\) | \(149\) |
| derivativedivides | \(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {11}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {13}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d \,a^{2}}\) | \(172\) |
| default | \(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {5}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {11}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {13}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d \,a^{2}}\) | \(172\) |
Input:
int(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
x/a^2+2/105*(-504*exp(5*I*(d*x+c))+1470*I*exp(6*I*(d*x+c))+1204*I*exp(4*I* (d*x+c))+420*exp(7*I*(d*x+c))-1108*exp(3*I*(d*x+c))+315*I*exp(8*I*(d*x+c)) +210*exp(9*I*(d*x+c))+258*I*exp(2*I*(d*x+c))-554*exp(I*(d*x+c))-191*I)/(ex p(I*(d*x+c))-I)^3/(exp(I*(d*x+c))+I)^7/d/a^2
Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {105 \, d x \cos \left (d x + c\right )^{5} - 210 \, d x \cos \left (d x + c\right )^{3} - 172 \, \cos \left (d x + c\right )^{4} + 86 \, \cos \left (d x + c\right )^{2} - {\left (210 \, d x \cos \left (d x + c\right )^{3} + 191 \, \cos \left (d x + c\right )^{4} - 129 \, \cos \left (d x + c\right )^{2} + 25\right )} \sin \left (d x + c\right ) - 10}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/105*(105*d*x*cos(d*x + c)^5 - 210*d*x*cos(d*x + c)^3 - 172*cos(d*x + c)^ 4 + 86*cos(d*x + c)^2 - (210*d*x*cos(d*x + c)^3 + 191*cos(d*x + c)^4 - 129 *cos(d*x + c)^2 + 25)*sin(d*x + c) - 10)/(a^2*d*cos(d*x + c)^5 - 2*a^2*d*c os(d*x + c)^3*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^3)
\[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sin ^{2}{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sin(d*x+c)**2*tan(d*x+c)**4/(a+a*sin(d*x+c))**2,x)
Output:
Integral(sin(c + d*x)**2*tan(c + d*x)**4/(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. \(2 (130) = 260\).
Time = 0.12 (sec) , antiderivative size = 421, normalized size of antiderivative = 3.01 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {\frac {279 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {132 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1048 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {364 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {1554 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {980 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {280 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {420 \, \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}} + 96}{a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, 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 {14 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, 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 {3 \, 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 {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{105 \, d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
2/105*((279*sin(d*x + c)/(cos(d*x + c) + 1) - 132*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1048*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 364*sin(d*x + c)^ 4/(cos(d*x + c) + 1)^4 + 1554*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 980*si n(d*x + c)^6/(cos(d*x + c) + 1)^6 - 280*sin(d*x + c)^7/(cos(d*x + c) + 1)^ 7 - 420*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 105*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 96)/(a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 3*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 - 14*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 14*a^2*sin(d*x + c)^6/(c os(d*x + c) + 1)^6 + 8*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 3*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) + 105*arctan(sin(d*x + c)/ (cos(d*x + c) + 1))/a^2)/d
Time = 0.29 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.11 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {840 \, {\left (d x + c\right )}}{a^{2}} + \frac {35 \, {\left (9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {1365 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 9345 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 26600 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 39410 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30261 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 11837 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1886}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \] Input:
integrate(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/840*(840*(d*x + c)/a^2 + 35*(9*tan(1/2*d*x + 1/2*c)^2 - 21*tan(1/2*d*x + 1/2*c) + 10)/(a^2*(tan(1/2*d*x + 1/2*c) - 1)^3) + (1365*tan(1/2*d*x + 1/2 *c)^6 + 9345*tan(1/2*d*x + 1/2*c)^5 + 26600*tan(1/2*d*x + 1/2*c)^4 + 39410 *tan(1/2*d*x + 1/2*c)^3 + 30261*tan(1/2*d*x + 1/2*c)^2 + 11837*tan(1/2*d*x + 1/2*c) + 1886)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^7))/d
Time = 37.28 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.11 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {x}{a^2}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {56\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}-\frac {148\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{5}+\frac {104\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}+\frac {2096\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}+\frac {88\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}-\frac {186\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}-\frac {64}{35}}{a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7} \] Input:
int((sin(c + d*x)^2*tan(c + d*x)^4)/(a + a*sin(c + d*x))^2,x)
Output:
x/a^2 + ((88*tan(c/2 + (d*x)/2)^2)/35 - (186*tan(c/2 + (d*x)/2))/35 + (209 6*tan(c/2 + (d*x)/2)^3)/105 + (104*tan(c/2 + (d*x)/2)^4)/15 - (148*tan(c/2 + (d*x)/2)^5)/5 - (56*tan(c/2 + (d*x)/2)^6)/3 + (16*tan(c/2 + (d*x)/2)^7) /3 + 8*tan(c/2 + (d*x)/2)^8 + 2*tan(c/2 + (d*x)/2)^9 - 64/35)/(a^2*d*(tan( c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^7)
Time = 0.18 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.69 \[ \int \frac {\sin ^2(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {35 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} \tan \left (d x +c \right )^{3}-105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} \tan \left (d x +c \right )+105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} d x +96 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}+70 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} \tan \left (d x +c \right )^{3}-210 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} \tan \left (d x +c \right )+210 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} d x +192 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-70 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \tan \left (d x +c \right )^{3}+210 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \tan \left (d x +c \right )-210 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -192 \cos \left (d x +c \right ) \sin \left (d x +c \right )-35 \cos \left (d x +c \right ) \tan \left (d x +c \right )^{3}+105 \cos \left (d x +c \right ) \tan \left (d x +c \right )-105 \cos \left (d x +c \right ) d x -96 \cos \left (d x +c \right )-51 \sin \left (d x +c \right )^{5}+108 \sin \left (d x +c \right )^{4}+288 \sin \left (d x +c \right )^{3}+48 \sin \left (d x +c \right )^{2}-192 \sin \left (d x +c \right )-96}{105 \cos \left (d x +c \right ) a^{2} d \left (\sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )-1\right )} \] Input:
int(sin(d*x+c)^2*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x)
Output:
(35*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x)**3 - 105*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x) + 105*cos(c + d*x)*sin(c + d*x)**4*d*x + 96*cos(c + d*x)*sin(c + d*x)**4 + 70*cos(c + d*x)*sin(c + d*x)**3*tan(c + d*x)**3 - 210*cos(c + d*x)*sin(c + d*x)**3*tan(c + d*x) + 210*cos(c + d*x)*sin(c + d *x)**3*d*x + 192*cos(c + d*x)*sin(c + d*x)**3 - 70*cos(c + d*x)*sin(c + d* x)*tan(c + d*x)**3 + 210*cos(c + d*x)*sin(c + d*x)*tan(c + d*x) - 210*cos( c + d*x)*sin(c + d*x)*d*x - 192*cos(c + d*x)*sin(c + d*x) - 35*cos(c + d*x )*tan(c + d*x)**3 + 105*cos(c + d*x)*tan(c + d*x) - 105*cos(c + d*x)*d*x - 96*cos(c + d*x) - 51*sin(c + d*x)**5 + 108*sin(c + d*x)**4 + 288*sin(c + d*x)**3 + 48*sin(c + d*x)**2 - 192*sin(c + d*x) - 96)/(105*cos(c + d*x)*a* *2*d*(sin(c + d*x)**4 + 2*sin(c + d*x)**3 - 2*sin(c + d*x) - 1))