Integrand size = 29, antiderivative size = 143 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {4 \sec ^7(c+d x)}{7 a^4 d}+\frac {4 \sec ^9(c+d x)}{3 a^4 d}-\frac {8 \sec ^{11}(c+d x)}{11 a^4 d}+\frac {\tan ^3(c+d x)}{3 a^4 d}+\frac {2 \tan ^5(c+d x)}{a^4 d}+\frac {25 \tan ^7(c+d x)}{7 a^4 d}+\frac {8 \tan ^9(c+d x)}{3 a^4 d}+\frac {8 \tan ^{11}(c+d x)}{11 a^4 d} \] Output:
-4/7*sec(d*x+c)^7/a^4/d+4/3*sec(d*x+c)^9/a^4/d-8/11*sec(d*x+c)^11/a^4/d+1/ 3*tan(d*x+c)^3/a^4/d+2*tan(d*x+c)^5/a^4/d+25/7*tan(d*x+c)^7/a^4/d+8/3*tan( d*x+c)^9/a^4/d+8/11*tan(d*x+c)^11/a^4/d
Time = 1.96 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\sec ^3(c+d x) (11264-1287 \cos (c+d x)-5632 \cos (2 (c+d x))+143 \cos (3 (c+d x))-2048 \cos (4 (c+d x))+325 \cos (5 (c+d x))+512 \cos (6 (c+d x))-13 \cos (7 (c+d x))+26048 \sin (c+d x)-1144 \sin (2 (c+d x))-704 \sin (3 (c+d x))-416 \sin (4 (c+d x))-1600 \sin (5 (c+d x))+104 \sin (6 (c+d x))+64 \sin (7 (c+d x)))}{118272 a^4 d (1+\sin (c+d x))^4} \] Input:
Integrate[(Sec[c + d*x]^2*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]
Output:
(Sec[c + d*x]^3*(11264 - 1287*Cos[c + d*x] - 5632*Cos[2*(c + d*x)] + 143*C os[3*(c + d*x)] - 2048*Cos[4*(c + d*x)] + 325*Cos[5*(c + d*x)] + 512*Cos[6 *(c + d*x)] - 13*Cos[7*(c + d*x)] + 26048*Sin[c + d*x] - 1144*Sin[2*(c + d *x)] - 704*Sin[3*(c + d*x)] - 416*Sin[4*(c + d*x)] - 1600*Sin[5*(c + d*x)] + 104*Sin[6*(c + d*x)] + 64*Sin[7*(c + d*x)]))/(118272*a^4*d*(1 + Sin[c + d*x])^4)
Time = 1.15 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.47, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.517, Rules used = {3042, 3349, 3042, 3151, 3042, 3151, 3042, 3151, 3042, 3151, 3042, 3151, 3042, 4254, 24}
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.
| \(\displaystyle \int \frac {\tan ^2(c+d x) \sec ^2(c+d x)}{(a \sin (c+d x)+a)^4} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{\cos (c+d x)^4 (a \sin (c+d x)+a)^4}dx\) |
| \(\Big \downarrow \) 3349 |
| \(\displaystyle \frac {1}{2} a \int \frac {\sec ^2(c+d x)}{(\sin (c+d x) a+a)^5}dx+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)^5}dx+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \int \frac {\sec ^2(c+d x)}{(\sin (c+d x) a+a)^4}dx}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)^4}dx}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \int \frac {\sec ^2(c+d x)}{(\sin (c+d x) a+a)^3}dx}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)^3}dx}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \int \frac {\sec ^2(c+d x)}{(\sin (c+d x) a+a)^2}dx}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)^2}dx}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \left (\frac {3 \int \frac {\sec ^2(c+d x)}{\sin (c+d x) a+a}dx}{5 a}-\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\right )}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \left (\frac {3 \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)}dx}{5 a}-\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\right )}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \left (\frac {3 \left (\frac {2 \int \sec ^2(c+d x)dx}{3 a}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}-\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\right )}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \left (\frac {3 \left (\frac {2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{3 a}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}-\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\right )}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \left (\frac {3 \left (-\frac {2 \int 1d(-\tan (c+d x))}{3 a d}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}-\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\right )}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )+\frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\sec ^3(c+d x)}{6 a d (a \sin (c+d x)+a)^3}+\frac {1}{2} a \left (\frac {6 \left (\frac {5 \left (\frac {4 \left (\frac {3 \left (\frac {2 \tan (c+d x)}{3 a d}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}-\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\right )}{7 a}-\frac {\sec (c+d x)}{7 d (a \sin (c+d x)+a)^3}\right )}{9 a}-\frac {\sec (c+d x)}{9 d (a \sin (c+d x)+a)^4}\right )}{11 a}-\frac {\sec (c+d x)}{11 d (a \sin (c+d x)+a)^5}\right )\) |
Input:
Int[(Sec[c + d*x]^2*Tan[c + d*x]^2)/(a + a*Sin[c + d*x])^4,x]
Output:
Sec[c + d*x]^3/(6*a*d*(a + a*Sin[c + d*x])^3) + (a*(-1/11*Sec[c + d*x]/(d* (a + a*Sin[c + d*x])^5) + (6*(-1/9*Sec[c + d*x]/(d*(a + a*Sin[c + d*x])^4) + (5*(-1/7*Sec[c + d*x]/(d*(a + a*Sin[c + d*x])^3) + (4*(-1/5*Sec[c + d*x ]/(d*(a + a*Sin[c + d*x])^2) + (3*(-1/3*Sec[c + d*x]/(d*(a + a*Sin[c + d*x ])) + (2*Tan[c + d*x])/(3*a*d)))/(5*a)))/(7*a)))/(9*a)))/(11*a)))/2
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^m/(a*f*g*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x] , x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-(g*Cos[e + f*x])^( p + 1))*((a + b*Sin[e + f*x])^(m + 1)/(2*b*f*g*(m + 1))), x] + Simp[a/(2*g^ 2) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; F reeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[m - p, 0]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Result contains complex when optimal does not.
Time = 28.54 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.92
| method | result | size |
| risch | \(-\frac {16 i \left (176 i {\mathrm e}^{7 i \left (d x +c \right )}+154 \,{\mathrm e}^{8 i \left (d x +c \right )}-88 i {\mathrm e}^{5 i \left (d x +c \right )}-253 \,{\mathrm e}^{6 i \left (d x +c \right )}-32 i {\mathrm e}^{3 i \left (d x +c \right )}+11 \,{\mathrm e}^{4 i \left (d x +c \right )}+8 i {\mathrm e}^{i \left (d x +c \right )}+25 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{231 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{11} d \,a^{4}}\) | \(132\) |
| derivativedivides | \(\frac {-\frac {16}{11 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{11}}+\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{10}}-\frac {64}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {36}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {295}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {71}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {43}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {109}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+128}-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d \,a^{4}}\) | \(218\) |
| default | \(\frac {-\frac {16}{11 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{11}}+\frac {8}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{10}}-\frac {64}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{9}}+\frac {36}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {295}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {71}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {43}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {9}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {109}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {5}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{128 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+128}-\frac {1}{48 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {1}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {1}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}}{d \,a^{4}}\) | \(218\) |
Input:
int(sec(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
-16/231*I*(176*I*exp(7*I*(d*x+c))+154*exp(8*I*(d*x+c))-88*I*exp(5*I*(d*x+c ))-253*exp(6*I*(d*x+c))-32*I*exp(3*I*(d*x+c))+11*exp(4*I*(d*x+c))+8*I*exp( I*(d*x+c))+25*exp(2*I*(d*x+c))-1)/(exp(I*(d*x+c))-I)^3/(exp(I*(d*x+c))+I)^ 11/d/a^4
Time = 0.09 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {32 \, \cos \left (d x + c\right )^{6} - 80 \, \cos \left (d x + c\right )^{4} + 28 \, \cos \left (d x + c\right )^{2} + {\left (8 \, \cos \left (d x + c\right )^{6} - 60 \, \cos \left (d x + c\right )^{4} + 35 \, \cos \left (d x + c\right )^{2} + 49\right )} \sin \left (d x + c\right ) + 28}{231 \, {\left (a^{4} d \cos \left (d x + c\right )^{7} - 8 \, a^{4} d \cos \left (d x + c\right )^{5} + 8 \, a^{4} d \cos \left (d x + c\right )^{3} - 4 \, {\left (a^{4} d \cos \left (d x + c\right )^{5} - 2 \, a^{4} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(sec(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="frica s")
Output:
1/231*(32*cos(d*x + c)^6 - 80*cos(d*x + c)^4 + 28*cos(d*x + c)^2 + (8*cos( d*x + c)^6 - 60*cos(d*x + c)^4 + 35*cos(d*x + c)^2 + 49)*sin(d*x + c) + 28 )/(a^4*d*cos(d*x + c)^7 - 8*a^4*d*cos(d*x + c)^5 + 8*a^4*d*cos(d*x + c)^3 - 4*(a^4*d*cos(d*x + c)^5 - 2*a^4*d*cos(d*x + c)^3)*sin(d*x + c))
\[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\int \frac {\tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} + 4 \sin ^{3}{\left (c + d x \right )} + 6 \sin ^{2}{\left (c + d x \right )} + 4 \sin {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:
integrate(sec(d*x+c)**2*tan(d*x+c)**2/(a+a*sin(d*x+c))**4,x)
Output:
Integral(tan(c + d*x)**2*sec(c + d*x)**2/(sin(c + d*x)**4 + 4*sin(c + d*x) **3 + 6*sin(c + d*x)**2 + 4*sin(c + d*x) + 1), x)/a**4
Leaf count of result is larger than twice the leaf count of optimal. 528 vs. \(2 (129) = 258\).
Time = 0.05 (sec) , antiderivative size = 528, normalized size of antiderivative = 3.69 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="maxim a")
Output:
8/231*(16*sin(d*x + c)/(cos(d*x + c) + 1) + 50*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + 141*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 132*sin(d*x + c)^4/(c os(d*x + c) + 1)^4 + 132*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 44*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 110*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 15 4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 308*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 154*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 77*sin(d*x + c)^11/(co s(d*x + c) + 1)^11 + 2)/((a^4 + 8*a^4*sin(d*x + c)/(cos(d*x + c) + 1) + 25 *a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 32*a^4*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 11*a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 88*a^4*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 99*a^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 99*a^4*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 88*a^4*sin(d*x + c)^9/(cos(d *x + c) + 1)^9 + 11*a^4*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 32*a^4*sin (d*x + c)^11/(cos(d*x + c) + 1)^11 - 25*a^4*sin(d*x + c)^12/(cos(d*x + c) + 1)^12 - 8*a^4*sin(d*x + c)^13/(cos(d*x + c) + 1)^13 - a^4*sin(d*x + c)^1 4/(cos(d*x + c) + 1)^14)*d)
Time = 0.27 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.38 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=-\frac {\frac {77 \, {\left (6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5\right )}}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}} - \frac {462 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{10} + 5775 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 14399 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 29260 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30800 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 27874 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12650 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6556 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 935 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 127}{a^{4} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{11}}}{7392 \, d} \] Input:
integrate(sec(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x, algorithm="giac" )
Output:
-1/7392*(77*(6*tan(1/2*d*x + 1/2*c)^2 - 9*tan(1/2*d*x + 1/2*c) + 5)/(a^4*( tan(1/2*d*x + 1/2*c) - 1)^3) - (462*tan(1/2*d*x + 1/2*c)^10 + 5775*tan(1/2 *d*x + 1/2*c)^9 + 14399*tan(1/2*d*x + 1/2*c)^8 + 29260*tan(1/2*d*x + 1/2*c )^7 + 30800*tan(1/2*d*x + 1/2*c)^6 + 27874*tan(1/2*d*x + 1/2*c)^5 + 12650* tan(1/2*d*x + 1/2*c)^4 + 6556*tan(1/2*d*x + 1/2*c)^3 + 1210*tan(1/2*d*x + 1/2*c)^2 + 935*tan(1/2*d*x + 1/2*c) + 127)/(a^4*(tan(1/2*d*x + 1/2*c) + 1) ^11))/d
Time = 36.03 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.29 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}}{231}+\frac {128\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{231}+\frac {400\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{231}+\frac {376\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{77}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{7}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{7}-\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{21}+\frac {80\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{21}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}+\frac {32\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{3}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{3}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{3}}{a^4\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^3\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^{11}} \] Input:
int(tan(c + d*x)^2/(cos(c + d*x)^2*(a + a*sin(c + d*x))^4),x)
Output:
((16*cos(c/2 + (d*x)/2)^14)/231 + (128*cos(c/2 + (d*x)/2)^13*sin(c/2 + (d* x)/2))/231 + (8*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^11)/3 + (16*cos(c/ 2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^10)/3 + (32*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)^9)/3 + (16*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2)^8)/3 + (80* cos(c/2 + (d*x)/2)^7*sin(c/2 + (d*x)/2)^7)/21 - (32*cos(c/2 + (d*x)/2)^8*s in(c/2 + (d*x)/2)^6)/21 + (32*cos(c/2 + (d*x)/2)^9*sin(c/2 + (d*x)/2)^5)/7 + (32*cos(c/2 + (d*x)/2)^10*sin(c/2 + (d*x)/2)^4)/7 + (376*cos(c/2 + (d*x )/2)^11*sin(c/2 + (d*x)/2)^3)/77 + (400*cos(c/2 + (d*x)/2)^12*sin(c/2 + (d *x)/2)^2)/231)/(a^4*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))^11)
Time = 0.20 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.55 \[ \int \frac {\sec ^2(c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^4} \, dx=\frac {8 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6}+32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5}+40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4}-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-32 \cos \left (d x +c \right ) \sin \left (d x +c \right )-8 \cos \left (d x +c \right )+8 \sin \left (d x +c \right )^{7}+32 \sin \left (d x +c \right )^{6}+36 \sin \left (d x +c \right )^{5}-16 \sin \left (d x +c \right )^{4}-61 \sin \left (d x +c \right )^{3}-36 \sin \left (d x +c \right )^{2}-32 \sin \left (d x +c \right )-8}{231 \cos \left (d x +c \right ) a^{4} d \left (\sin \left (d x +c \right )^{6}+4 \sin \left (d x +c \right )^{5}+5 \sin \left (d x +c \right )^{4}-5 \sin \left (d x +c \right )^{2}-4 \sin \left (d x +c \right )-1\right )} \] Input:
int(sec(d*x+c)^2*tan(d*x+c)^2/(a+a*sin(d*x+c))^4,x)
Output:
(8*cos(c + d*x)*sin(c + d*x)**6 + 32*cos(c + d*x)*sin(c + d*x)**5 + 40*cos (c + d*x)*sin(c + d*x)**4 - 40*cos(c + d*x)*sin(c + d*x)**2 - 32*cos(c + d *x)*sin(c + d*x) - 8*cos(c + d*x) + 8*sin(c + d*x)**7 + 32*sin(c + d*x)**6 + 36*sin(c + d*x)**5 - 16*sin(c + d*x)**4 - 61*sin(c + d*x)**3 - 36*sin(c + d*x)**2 - 32*sin(c + d*x) - 8)/(231*cos(c + d*x)*a**4*d*(sin(c + d*x)** 6 + 4*sin(c + d*x)**5 + 5*sin(c + d*x)**4 - 5*sin(c + d*x)**2 - 4*sin(c + d*x) - 1))