Integrand size = 27, antiderivative size = 88 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec ^3(c+d x)}{a^3 d}-\frac {7 \sec ^5(c+d x)}{5 a^3 d}+\frac {4 \sec ^7(c+d x)}{7 a^3 d}-\frac {3 \tan ^5(c+d x)}{5 a^3 d}-\frac {4 \tan ^7(c+d x)}{7 a^3 d} \] Output:
sec(d*x+c)^3/a^3/d-7/5*sec(d*x+c)^5/a^3/d+4/7*sec(d*x+c)^7/a^3/d-3/5*tan(d *x+c)^5/a^3/d-4/7*tan(d*x+c)^7/a^3/d
Time = 1.25 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.18 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\sec (c+d x) (840-602 \cos (c+d x)-448 \cos (2 (c+d x))+258 \cos (3 (c+d x))-8 \cos (4 (c+d x))+1008 \sin (c+d x)-602 \sin (2 (c+d x))+48 \sin (3 (c+d x))+43 \sin (4 (c+d x)))}{2240 a^3 d (1+\sin (c+d x))^3} \] Input:
Integrate[(Sin[c + d*x]*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
(Sec[c + d*x]*(840 - 602*Cos[c + d*x] - 448*Cos[2*(c + d*x)] + 258*Cos[3*( c + d*x)] - 8*Cos[4*(c + d*x)] + 1008*Sin[c + d*x] - 602*Sin[2*(c + d*x)] + 48*Sin[3*(c + d*x)] + 43*Sin[4*(c + d*x)]))/(2240*a^3*d*(1 + Sin[c + d*x ])^3)
Time = 0.59 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, 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.
| \(\displaystyle \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x)^2 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \sec ^5(c+d x) (a-a \sin (c+d x))^3 \tan ^3(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x)^3 (a-a \sin (c+d x))^3}{\cos (c+d x)^8}dx}{a^6}\) |
| \(\Big \downarrow \) 3352 |
| \(\displaystyle \frac {\int \left (-a^3 \sec ^2(c+d x) \tan ^6(c+d x)+3 a^3 \sec ^3(c+d x) \tan ^5(c+d x)-3 a^3 \sec ^4(c+d x) \tan ^4(c+d x)+a^3 \sec ^5(c+d x) \tan ^3(c+d x)\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {4 a^3 \tan ^7(c+d x)}{7 d}-\frac {3 a^3 \tan ^5(c+d x)}{5 d}+\frac {4 a^3 \sec ^7(c+d x)}{7 d}-\frac {7 a^3 \sec ^5(c+d x)}{5 d}+\frac {a^3 \sec ^3(c+d x)}{d}}{a^6}\) |
Input:
Int[(Sin[c + d*x]*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^3,x]
Output:
((a^3*Sec[c + d*x]^3)/d - (7*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]^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]
Result contains complex when optimal does not.
Time = 1.26 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.24
| method | result | size |
| risch | \(\frac {2 i \left (-105 \,{\mathrm e}^{4 i \left (d x +c \right )}-56 i {\mathrm e}^{3 i \left (d x +c \right )}+21 \,{\mathrm e}^{2 i \left (d x +c \right )}-6 i {\mathrm e}^{i \left (d x +c \right )}+1+70 i {\mathrm e}^{5 i \left (d x +c \right )}+35 \,{\mathrm e}^{6 i \left (d x +c \right )}\right )}{35 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{3} d}\) | \(109\) |
| derivativedivides | \(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {26}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+128}}{a^{3} d}\) | \(130\) |
| default | \(\frac {-\frac {1}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}+\frac {26}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+128}}{a^{3} d}\) | \(130\) |
Input:
int(sin(d*x+c)*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
2/35*I*(-105*exp(4*I*(d*x+c))-56*I*exp(3*I*(d*x+c))+21*exp(2*I*(d*x+c))-6* I*exp(I*(d*x+c))+1+70*I*exp(5*I*(d*x+c))+35*exp(6*I*(d*x+c)))/(exp(I*(d*x+ c))+I)^7/(exp(I*(d*x+c))-I)/a^3/d
Time = 0.08 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.16 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\cos \left (d x + c\right )^{4} + 13 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{2} + 5\right )} \sin \left (d x + c\right ) - 20}{35 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right ) + {\left (a^{3} d \cos \left (d x + c\right )^{3} - 4 \, a^{3} d \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="fricas" )
Output:
1/35*(cos(d*x + c)^4 + 13*cos(d*x + c)^2 - 3*(cos(d*x + c)^2 + 5)*sin(d*x + c) - 20)/(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))
\[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )} \tan ^{2}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)**2/(a+a*sin(d*x+c))**3,x)
Output:
Integral(sin(c + d*x)*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. 250 vs. \(2 (80) = 160\).
Time = 0.04 (sec) , antiderivative size = 250, normalized size of antiderivative = 2.84 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4 \, {\left (\frac {18 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {42 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {35 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 3\right )}}{35 \, {\left (a^{3} + \frac {6 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {14 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {14 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac {6 \, a^{3} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="maxima" )
Output:
4/35*(18*sin(d*x + c)/(cos(d*x + c) + 1) + 42*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 42*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 35*sin(d*x + c)^4/(cos( d*x + c) + 1)^4 + 3)/((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)*d)
Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {35}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 280 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1015 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 1120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1001 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 392 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 61}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{280 \, d} \] Input:
integrate(sin(d*x+c)*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) - 1)) - (35*tan(1/2*d*x + 1/2*c)^6 + 280*tan(1/2*d*x + 1/2*c)^5 + 1015*tan(1/2*d*x + 1/2*c)^4 + 1120*tan(1/2*d *x + 1/2*c)^3 + 1001*tan(1/2*d*x + 1/2*c)^2 + 392*tan(1/2*d*x + 1/2*c) + 6 1)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^7))/d
Time = 31.77 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.80 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+18\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+42\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+42\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+35\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{35\,a^3\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^7} \] Input:
int((sin(c + d*x)*tan(c + d*x)^2)/(a + a*sin(c + d*x))^3,x)
Output:
(4*cos(c/2 + (d*x)/2)^4*(3*cos(c/2 + (d*x)/2)^4 + 35*sin(c/2 + (d*x)/2)^4 + 42*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^3 + 18*cos(c/2 + (d*x)/2)^3*sin (c/2 + (d*x)/2) + 42*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2))/(35*a^3*d *(cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))*(cos(c/2 + (d*x)/2) + sin(c/2 + (d*x)/2))^7)
Time = 0.17 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.59 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}+18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}+18 \cos \left (d x +c \right ) \sin \left (d x +c \right )+6 \cos \left (d x +c \right )-\sin \left (d x +c \right )^{4}-3 \sin \left (d x +c \right )^{3}+15 \sin \left (d x +c \right )^{2}+18 \sin \left (d x +c \right )+6}{35 \cos \left (d x +c \right ) a^{3} d \left (\sin \left (d x +c \right )^{3}+3 \sin \left (d x +c \right )^{2}+3 \sin \left (d x +c \right )+1\right )} \] Input:
int(sin(d*x+c)*tan(d*x+c)^2/(a+a*sin(d*x+c))^3,x)
Output:
(6*cos(c + d*x)*sin(c + d*x)**3 + 18*cos(c + d*x)*sin(c + d*x)**2 + 18*cos (c + d*x)*sin(c + d*x) + 6*cos(c + d*x) - sin(c + d*x)**4 - 3*sin(c + d*x) **3 + 15*sin(c + d*x)**2 + 18*sin(c + d*x) + 6)/(35*cos(c + d*x)*a**3*d*(s in(c + d*x)**3 + 3*sin(c + d*x)**2 + 3*sin(c + d*x) + 1))