Integrand size = 29, antiderivative size = 155 \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 x}{a^2}-\frac {\cos (c+d x)}{a^2 d}-\frac {5 \sec (c+d x)}{a^2 d}+\frac {3 \sec ^3(c+d x)}{a^2 d}-\frac {7 \sec ^5(c+d x)}{5 a^2 d}+\frac {2 \sec ^7(c+d x)}{7 a^2 d}+\frac {2 \tan (c+d x)}{a^2 d}-\frac {2 \tan ^3(c+d x)}{3 a^2 d}+\frac {2 \tan ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^7(c+d x)}{7 a^2 d} \] Output:
-2*x/a^2-cos(d*x+c)/a^2/d-5*sec(d*x+c)/a^2/d+3*sec(d*x+c)^3/a^2/d-7/5*sec( d*x+c)^5/a^2/d+2/7*sec(d*x+c)^7/a^2/d+2*tan(d*x+c)/a^2/d-2/3*tan(d*x+c)^3/ a^2/d+2/5*tan(d*x+c)^5/a^2/d-2/7*tan(d*x+c)^7/a^2/d
Time = 2.21 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.72 \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {11172+42 (-551+280 c+280 d x) \cos (c+d x)+14834 \cos (2 (c+d x))-4959 \cos (3 (c+d x))+2520 c \cos (3 (c+d x))+2520 d x \cos (3 (c+d x))+1852 \cos (4 (c+d x))+1653 \cos (5 (c+d x))-840 c \cos (5 (c+d x))-840 d x \cos (5 (c+d x))-210 \cos (6 (c+d x))+5488 \sin (c+d x)-13224 \sin (2 (c+d x))+6720 c \sin (2 (c+d x))+6720 d x \sin (2 (c+d x))+8376 \sin (3 (c+d x))-6612 \sin (4 (c+d x))+3360 c \sin (4 (c+d x))+3360 d x \sin (4 (c+d x))+2248 \sin (5 (c+d x))}{6720 a^2 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^7} \] Input:
Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
-1/6720*(11172 + 42*(-551 + 280*c + 280*d*x)*Cos[c + d*x] + 14834*Cos[2*(c + d*x)] - 4959*Cos[3*(c + d*x)] + 2520*c*Cos[3*(c + d*x)] + 2520*d*x*Cos[ 3*(c + d*x)] + 1852*Cos[4*(c + d*x)] + 1653*Cos[5*(c + d*x)] - 840*c*Cos[5 *(c + d*x)] - 840*d*x*Cos[5*(c + d*x)] - 210*Cos[6*(c + d*x)] + 5488*Sin[c + d*x] - 13224*Sin[2*(c + d*x)] + 6720*c*Sin[2*(c + d*x)] + 6720*d*x*Sin[ 2*(c + d*x)] + 8376*Sin[3*(c + d*x)] - 6612*Sin[4*(c + d*x)] + 3360*c*Sin[ 4*(c + d*x)] + 3360*d*x*Sin[4*(c + d*x)] + 2248*Sin[5*(c + d*x)])/(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 = 159, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^7}{\cos (c+d x)^4 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec (c+d x) (a-a \sin (c+d x))^2 \tan ^7(c+d x)dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^7 (a-a \sin (c+d x))^2}{\cos (c+d x)^8}dx}{a^4}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (-2 a^2 \tan ^8(c+d x)+a^2 \sin (c+d x) \tan ^8(c+d x)+a^2 \sec (c+d x) \tan ^7(c+d x)\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^2 \cos (c+d x)}{d}-\frac {2 a^2 \tan ^7(c+d x)}{7 d}+\frac {2 a^2 \tan ^5(c+d x)}{5 d}-\frac {2 a^2 \tan ^3(c+d x)}{3 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {2 a^2 \sec ^7(c+d x)}{7 d}-\frac {7 a^2 \sec ^5(c+d x)}{5 d}+\frac {3 a^2 \sec ^3(c+d x)}{d}-\frac {5 a^2 \sec (c+d x)}{d}-2 a^2 x}{a^4}\) |
Input:
Int[(Sin[c + d*x]^3*Tan[c + d*x]^4)/(a + a*Sin[c + d*x])^2,x]
Output:
(-2*a^2*x - (a^2*Cos[c + d*x])/d - (5*a^2*Sec[c + d*x])/d + (3*a^2*Sec[c + d*x]^3)/d - (7*a^2*Sec[c + d*x]^5)/(5*d) + (2*a^2*Sec[c + d*x]^7)/(7*d) + (2*a^2*Tan[c + d*x])/d - (2*a^2*Tan[c + d*x]^3)/(3*d) + (2*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 = 2.58 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12
| method | result | size |
| risch | \(-\frac {2 x}{a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {2 \left (1260 i {\mathrm e}^{8 i \left (d x +c \right )}+525 \,{\mathrm e}^{9 i \left (d x +c \right )}+2940 i {\mathrm e}^{6 i \left (d x +c \right )}+1988 i {\mathrm e}^{4 i \left (d x +c \right )}-2058 \,{\mathrm e}^{5 i \left (d x +c \right )}-204 i {\mathrm e}^{2 i \left (d x +c \right )}-2816 \,{\mathrm e}^{3 i \left (d x +c \right )}-352 i-883 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{105 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} d \,a^{2}}\) | \(173\) |
| derivativedivides | \(\frac {-\frac {2}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {6}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {23}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {9}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {256}{512 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-512}}{d \,a^{2}}\) | \(189\) |
| default | \(\frac {-\frac {2}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {6}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {23}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {9}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {1}{12 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {256}{512 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-512}}{d \,a^{2}}\) | \(189\) |
Input:
int(sin(d*x+c)^3*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-2*x/a^2-1/2/d/a^2*exp(I*(d*x+c))-1/2/d/a^2*exp(-I*(d*x+c))-2/105*(1260*I* exp(8*I*(d*x+c))+525*exp(9*I*(d*x+c))+2940*I*exp(6*I*(d*x+c))+1988*I*exp(4 *I*(d*x+c))-2058*exp(5*I*(d*x+c))-204*I*exp(2*I*(d*x+c))-2816*exp(3*I*(d*x +c))-352*I-883*exp(I*(d*x+c)))/(exp(I*(d*x+c))+I)^7/(exp(I*(d*x+c))-I)^3/d /a^2
Time = 0.09 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.97 \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {210 \, d x \cos \left (d x + c\right )^{5} + 105 \, \cos \left (d x + c\right )^{6} - 420 \, d x \cos \left (d x + c\right )^{3} - 389 \, \cos \left (d x + c\right )^{4} - 173 \, \cos \left (d x + c\right )^{2} - 2 \, {\left (210 \, d x \cos \left (d x + c\right )^{3} + 281 \, \cos \left (d x + c\right )^{4} + 51 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) + 25}{105 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - 2 \, a^{2} d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/105*(210*d*x*cos(d*x + c)^5 + 105*cos(d*x + c)^6 - 420*d*x*cos(d*x + c) ^3 - 389*cos(d*x + c)^4 - 173*cos(d*x + c)^2 - 2*(210*d*x*cos(d*x + c)^3 + 281*cos(d*x + c)^4 + 51*cos(d*x + c)^2 - 5)*sin(d*x + c) + 25)/(a^2*d*cos (d*x + c)^5 - 2*a^2*d*cos(d*x + c)^3*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^3 )
Timed out. \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**3*tan(d*x+c)**4/(a+a*sin(d*x+c))**2,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 507 vs. \(2 (145) = 290\).
Time = 0.13 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.27 \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-4/105*((759*sin(d*x + c)/(cos(d*x + c) + 1) + 444*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1249*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1816*sin(d*x + c )^4/(cos(d*x + c) + 1)^4 - 454*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 616*s in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1274*sin(d*x + c)^7/(cos(d*x + c) + 1 )^7 + 560*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 385*sin(d*x + c)^9/(cos(d* x + c) + 1)^9 - 420*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 105*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 + 216)/(a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 4*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 4*a^2*sin(d*x + c)^3/( cos(d*x + c) + 1)^3 - 11*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 8*a^2*s in(d*x + c)^5/(cos(d*x + c) + 1)^5 + 8*a^2*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 11*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 - 4*a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 4*a^2* sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) + 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.31 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.13 \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {1680 \, {\left (d x + c\right )}}{a^{2}} + \frac {1680}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} - \frac {35 \, {\left (12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 13\right )}}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} + \frac {3780 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 25095 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 68845 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 98350 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75222 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 29659 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4777}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-1/840*(1680*(d*x + c)/a^2 + 1680/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) - 35* (12*tan(1/2*d*x + 1/2*c)^2 - 27*tan(1/2*d*x + 1/2*c) + 13)/(a^2*(tan(1/2*d *x + 1/2*c) - 1)^3) + (3780*tan(1/2*d*x + 1/2*c)^6 + 25095*tan(1/2*d*x + 1 /2*c)^5 + 68845*tan(1/2*d*x + 1/2*c)^4 + 98350*tan(1/2*d*x + 1/2*c)^3 + 75 222*tan(1/2*d*x + 1/2*c)^2 + 29659*tan(1/2*d*x + 1/2*c) + 4777)/(a^2*(tan( 1/2*d*x + 1/2*c) + 1)^7))/d
Time = 38.14 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.28 \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-\frac {44\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}+\frac {64\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {728\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{15}+\frac {352\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}-\frac {1816\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{105}-\frac {7264\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{105}-\frac {4996\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{105}+\frac {592\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{35}+\frac {1012\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+\frac {288}{35}}{a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {2\,x}{a^2} \] Input:
int((sin(c + d*x)^3*tan(c + d*x)^4)/(a + a*sin(c + d*x))^2,x)
Output:
((1012*tan(c/2 + (d*x)/2))/35 + (592*tan(c/2 + (d*x)/2)^2)/35 - (4996*tan( c/2 + (d*x)/2)^3)/105 - (7264*tan(c/2 + (d*x)/2)^4)/105 - (1816*tan(c/2 + (d*x)/2)^5)/105 + (352*tan(c/2 + (d*x)/2)^6)/15 + (728*tan(c/2 + (d*x)/2)^ 7)/15 + (64*tan(c/2 + (d*x)/2)^8)/3 - (44*tan(c/2 + (d*x)/2)^9)/3 - 16*tan (c/2 + (d*x)/2)^10 - 4*tan(c/2 + (d*x)/2)^11 + 288/35)/(a^2*d*(tan(c/2 + ( d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2) + 1)^7*(tan(c/2 + (d*x)/2)^2 + 1)) - (2 *x)/a^2
Time = 0.20 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.81 \[ \int \frac {\sin ^3(c+d x) \tan ^4(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {-210 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} c -210 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} d x -327 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-420 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c -420 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} d x -654 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+420 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +420 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x +654 \cos \left (d x +c \right ) \sin \left (d x +c \right )+210 \cos \left (d x +c \right ) c +210 \cos \left (d x +c \right ) d x +327 \cos \left (d x +c \right )+105 \sin \left (d x +c \right )^{6}+562 \sin \left (d x +c \right )^{5}+74 \sin \left (d x +c \right )^{4}-1226 \sin \left (d x +c \right )^{3}-636 \sin \left (d x +c \right )^{2}+654 \sin \left (d x +c \right )+432}{105 \cos \left (d x +c \right ) a^{2} d \left (\sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{3}-2 \sin \left (d x +c \right )-1\right )} \] Input:
int(sin(d*x+c)^3*tan(d*x+c)^4/(a+a*sin(d*x+c))^2,x)
Output:
( - 210*cos(c + d*x)*sin(c + d*x)**4*c - 210*cos(c + d*x)*sin(c + d*x)**4* d*x - 327*cos(c + d*x)*sin(c + d*x)**4 - 420*cos(c + d*x)*sin(c + d*x)**3* c - 420*cos(c + d*x)*sin(c + d*x)**3*d*x - 654*cos(c + d*x)*sin(c + d*x)** 3 + 420*cos(c + d*x)*sin(c + d*x)*c + 420*cos(c + d*x)*sin(c + d*x)*d*x + 654*cos(c + d*x)*sin(c + d*x) + 210*cos(c + d*x)*c + 210*cos(c + d*x)*d*x + 327*cos(c + d*x) + 105*sin(c + d*x)**6 + 562*sin(c + d*x)**5 + 74*sin(c + d*x)**4 - 1226*sin(c + d*x)**3 - 636*sin(c + d*x)**2 + 654*sin(c + d*x) + 432)/(105*cos(c + d*x)*a**2*d*(sin(c + d*x)**4 + 2*sin(c + d*x)**3 - 2*s in(c + d*x) - 1))