Integrand size = 29, antiderivative size = 142 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {x}{a^3}-\frac {3 \sec (c+d x)}{a^3 d}+\frac {10 \sec ^3(c+d x)}{3 a^3 d}-\frac {11 \sec ^5(c+d x)}{5 a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}+\frac {\tan (c+d x)}{a^3 d}-\frac {\tan ^3(c+d x)}{3 a^3 d}+\frac {\tan ^5(c+d x)}{5 a^3 d}-\frac {4 \tan ^7(c+d x)}{7 a^3 d} \] Output:
-x/a^3-3*sec(d*x+c)/a^3/d+10/3*sec(d*x+c)^3/a^3/d-11/5*sec(d*x+c)^5/a^3/d+ 4/7*sec(d*x+c)^7/a^3/d+tan(d*x+c)/a^3/d-1/3*tan(d*x+c)^3/a^3/d+1/5*tan(d*x +c)^5/a^3/d-4/7*tan(d*x+c)^7/a^3/d
Time = 1.95 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.51 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {4200+14 (-1663+840 c+840 d x) \cos (c+d x)+6272 \cos (2 (c+d x))+9978 \cos (3 (c+d x))-5040 c \cos (3 (c+d x))-5040 d x \cos (3 (c+d x))-1768 \cos (4 (c+d x))+2688 \sin (c+d x)-23282 \sin (2 (c+d x))+11760 c \sin (2 (c+d x))+11760 d x \sin (2 (c+d x))+5568 \sin (3 (c+d x))+1663 \sin (4 (c+d x))-840 c \sin (4 (c+d x))-840 d x \sin (4 (c+d x))}{6720 a^3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^7} \] Input:
Integrate[(Sin[c + d*x]^3*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
-1/6720*(4200 + 14*(-1663 + 840*c + 840*d*x)*Cos[c + d*x] + 6272*Cos[2*(c + d*x)] + 9978*Cos[3*(c + d*x)] - 5040*c*Cos[3*(c + d*x)] - 5040*d*x*Cos[3 *(c + d*x)] - 1768*Cos[4*(c + d*x)] + 2688*Sin[c + d*x] - 23282*Sin[2*(c + d*x)] + 11760*c*Sin[2*(c + d*x)] + 11760*d*x*Sin[2*(c + d*x)] + 5568*Sin[ 3*(c + d*x)] + 1663*Sin[4*(c + d*x)] - 840*c*Sin[4*(c + d*x)] - 840*d*x*Si n[4*(c + d*x)])/(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 = 0.65 (sec) , antiderivative size = 146, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^5}{\cos (c+d x)^2 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec ^3(c+d x) (a-a \sin (c+d x))^3 \tan ^5(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^5 (a-a \sin (c+d x))^3}{\cos (c+d x)^8}dx}{a^6}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (-a^3 \tan ^8(c+d x)+3 a^3 \sec (c+d x) \tan ^7(c+d x)-3 a^3 \sec ^2(c+d x) \tan ^6(c+d x)+a^3 \sec ^3(c+d x) \tan ^5(c+d x)\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {4 a^3 \tan ^7(c+d x)}{7 d}+\frac {a^3 \tan ^5(c+d x)}{5 d}-\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {a^3 \tan (c+d x)}{d}+\frac {4 a^3 \sec ^7(c+d x)}{7 d}-\frac {11 a^3 \sec ^5(c+d x)}{5 d}+\frac {10 a^3 \sec ^3(c+d x)}{3 d}-\frac {3 a^3 \sec (c+d x)}{d}-a^3 x}{a^6}\) |
Input:
Int[(Sin[c + d*x]^3*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(-(a^3*x) - (3*a^3*Sec[c + d*x])/d + (10*a^3*Sec[c + d*x]^3)/(3*d) - (11*a ^3*Sec[c + d*x]^5)/(5*d) + (4*a^3*Sec[c + d*x]^7)/(7*d) + (a^3*Tan[c + d*x ])/d - (a^3*Tan[c + d*x]^3)/(3*d) + (a^3*Tan[c + d*x]^5)/(5*d) - (4*a^3*Ta n[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]
Result contains complex when optimal does not.
Time = 1.34 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.89
| method | result | size |
| risch | \(-\frac {x}{a^{3}}-\frac {2 \left (1155 i {\mathrm e}^{6 i \left (d x +c \right )}+315 \,{\mathrm e}^{7 i \left (d x +c \right )}-525 i {\mathrm e}^{4 i \left (d x +c \right )}-1715 \,{\mathrm e}^{5 i \left (d x +c \right )}-1939 i {\mathrm e}^{2 i \left (d x +c \right )}-1379 \,{\mathrm e}^{3 i \left (d x +c \right )}+221 i+1011 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{105 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d \,a^{3}}\) | \(127\) |
| derivativedivides | \(\frac {-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {5}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a^{3} d}\) | \(142\) |
| default | \(\frac {-2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {18}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {5}{6 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{a^{3} d}\) | \(142\) |
Input:
int(sin(d*x+c)^3*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-x/a^3-2/105*(1155*I*exp(6*I*(d*x+c))+315*exp(7*I*(d*x+c))-525*I*exp(4*I*( d*x+c))-1715*exp(5*I*(d*x+c))-1939*I*exp(2*I*(d*x+c))-1379*exp(3*I*(d*x+c) )+221*I+1011*exp(I*(d*x+c)))/(exp(I*(d*x+c))-I)/(exp(I*(d*x+c))+I)^7/d/a^3
Time = 0.10 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.06 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {315 \, d x \cos \left (d x + c\right )^{3} + 221 \, \cos \left (d x + c\right )^{4} - 420 \, d x \cos \left (d x + c\right ) - 417 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (35 \, d x \cos \left (d x + c\right )^{3} - 140 \, d x \cos \left (d x + c\right ) - 116 \, \cos \left (d x + c\right )^{2} + 15\right )} \sin \left (d x + c\right ) + 60}{105 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
-1/105*(315*d*x*cos(d*x + c)^3 + 221*cos(d*x + c)^4 - 420*d*x*cos(d*x + c) - 417*cos(d*x + c)^2 + 3*(35*d*x*cos(d*x + c)^3 - 140*d*x*cos(d*x + c) - 116*cos(d*x + c)^2 + 15)*sin(d*x + c) + 60)/(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))
Timed out. \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate(sin(d*x+c)**3*tan(d*x+c)**2/(a+a*sin(d*x+c))**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (130) = 260\).
Time = 0.11 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.36 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {\frac {711 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {1274 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {469 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {1260 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {1435 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {630 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {105 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + 136}{a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} + \frac {105 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{105 \, d} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
-2/105*((711*sin(d*x + c)/(cos(d*x + c) + 1) + 1274*sin(d*x + c)^2/(cos(d* x + c) + 1)^2 + 469*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 1260*sin(d*x + c )^4/(cos(d*x + c) + 1)^4 - 1435*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 630* sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 105*sin(d*x + c)^7/(cos(d*x + c) + 1 )^7 + 136)/(a^3 + 6*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 14*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 14*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*a^3*sin(d*x + c)^6/(cos(d* x + c) + 1)^6 - 6*a^3*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) + 105*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a ^3)/d
Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {840 \, {\left (d x + c\right )}}{a^{3}} + \frac {105}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} + \frac {1575 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 10920 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31675 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 48160 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36981 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 14392 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2281}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{840 \, d} \] Input:
integrate(sin(d*x+c)^3*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
-1/840*(840*(d*x + c)/a^3 + 105/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) + (1575*t an(1/2*d*x + 1/2*c)^6 + 10920*tan(1/2*d*x + 1/2*c)^5 + 31675*tan(1/2*d*x + 1/2*c)^4 + 48160*tan(1/2*d*x + 1/2*c)^3 + 36981*tan(1/2*d*x + 1/2*c)^2 + 14392*tan(1/2*d*x + 1/2*c) + 2281)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d
Time = 34.97 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-12\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {82\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{3}-24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {134\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{15}+\frac {364\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{15}+\frac {474\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}+\frac {272}{105}}{a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7}-\frac {x}{a^3} \] Input:
int((sin(c + d*x)^3*tan(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)
Output:
((474*tan(c/2 + (d*x)/2))/35 + (364*tan(c/2 + (d*x)/2)^2)/15 + (134*tan(c/ 2 + (d*x)/2)^3)/15 - 24*tan(c/2 + (d*x)/2)^4 - (82*tan(c/2 + (d*x)/2)^5)/3 - 12*tan(c/2 + (d*x)/2)^6 - 2*tan(c/2 + (d*x)/2)^7 + 272/105)/(a^3*d*(tan (c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2) + 1)^7) - x/a^3
Time = 0.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.42 \[ \int \frac {\sin ^3(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {-105 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} d x -101 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-315 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x -303 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-315 \cos \left (d x +c \right ) \sin \left (d x +c \right ) d x -303 \cos \left (d x +c \right ) \sin \left (d x +c \right )-105 \cos \left (d x +c \right ) d x -101 \cos \left (d x +c \right )+221 \sin \left (d x +c \right )^{4}+348 \sin \left (d x +c \right )^{3}-25 \sin \left (d x +c \right )^{2}-303 \sin \left (d x +c \right )-136}{105 \cos \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1\right )} \] Input:
int(sin(d*x+c)^3*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x)
Output:
( - 105*cos(c + d*x)*sin(c + d*x)**3*d*x - 101*cos(c + d*x)*sin(c + d*x)** 3 - 315*cos(c + d*x)*sin(c + d*x)**2*d*x - 303*cos(c + d*x)*sin(c + d*x)** 2 - 315*cos(c + d*x)*sin(c + d*x)*d*x - 303*cos(c + d*x)*sin(c + d*x) - 10 5*cos(c + d*x)*d*x - 101*cos(c + d*x) + 221*sin(c + d*x)**4 + 348*sin(c + d*x)**3 - 25*sin(c + d*x)**2 - 303*sin(c + d*x) - 136)/(105*cos(c + d*x)*a **3*d*(sin(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1))