Integrand size = 25, antiderivative size = 71 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}-\frac {2 \sec (c+d x)}{15 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac {4 \tan (c+d x)}{15 a^2 d} \]
1/5*sec(d*x+c)/d/(a+a*sin(d*x+c))^2-2/15*sec(d*x+c)/d/(a^2+a^2*sin(d*x+c)) +4/15*tan(d*x+c)/a^2/d
Time = 0.49 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\sec (c+d x) (-80-5 \cos (c+d x)+64 \cos (2 (c+d x))+\cos (3 (c+d x))-80 \sin (c+d x)-4 \sin (2 (c+d x))+16 \sin (3 (c+d x)))}{240 a^2 d (1+\sin (c+d x))^2} \]
-1/240*(Sec[c + d*x]*(-80 - 5*Cos[c + d*x] + 64*Cos[2*(c + d*x)] + Cos[3*( c + d*x)] - 80*Sin[c + d*x] - 4*Sin[2*(c + d*x)] + 16*Sin[3*(c + d*x)]))/( a^2*d*(1 + Sin[c + d*x])^2)
Time = 0.42 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 3338, 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.
\(\displaystyle \int \frac {\tan (c+d x) \sec (c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)}{\cos (c+d x)^2 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3338 |
\(\displaystyle \frac {2 \int \frac {\sec ^2(c+d x)}{\sin (c+d x) a+a}dx}{5 a}+\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {1}{\cos (c+d x)^2 (\sin (c+d x) a+a)}dx}{5 a}+\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3151 |
\(\displaystyle \frac {2 \left (\frac {2 \int \sec ^2(c+d x)dx}{3 a}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}+\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{3 a}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}+\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {2 \left (-\frac {2 \int 1d(-\tan (c+d x))}{3 a d}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}+\frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2}+\frac {2 \left (\frac {2 \tan (c+d x)}{3 a d}-\frac {\sec (c+d x)}{3 d (a \sin (c+d x)+a)}\right )}{5 a}\) |
Sec[c + d*x]/(5*d*(a + a*Sin[c + d*x])^2) + (2*(-1/3*Sec[c + d*x]/(d*(a + a*Sin[c + d*x])) + (2*Tan[c + d*x])/(3*a*d)))/(5*a)
3.8.83.3.1 Defintions of rubi rules used
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_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) )), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 ]) && NeQ[2*m + p + 1, 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 = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {8 \left (5 i {\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{3 i \left (d x +c \right )}-i-4 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} d \,a^{2}}\) | \(74\) |
parallelrisch | \(\frac {-2-30 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-40 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(87\) |
derivativedivides | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {7}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{d \,a^{2}}\) | \(100\) |
default | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {7}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {3}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16}}{d \,a^{2}}\) | \(100\) |
norman | \(\frac {-\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}-\frac {2}{15 a d}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 d a}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} a}\) | \(114\) |
-8/15*(5*I*exp(2*I*(d*x+c))+5*exp(3*I*(d*x+c))-I-4*exp(I*(d*x+c)))/(exp(I* (d*x+c))-I)/(exp(I*(d*x+c))+I)^5/d/a^2
Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {8 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) - 9}{15 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \]
1/15*(8*cos(d*x + c)^2 + 2*(2*cos(d*x + c)^2 - 3)*sin(d*x + c) - 9)/(a^2*d *cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c)*sin(d*x + c) - 2*a^2*d*cos(d*x + c) )
\[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (65) = 130\).
Time = 0.23 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.87 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}{15 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \]
2/15*(4*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d *x + c) + 1)^4 + 1)/((a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^2* sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 5*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^2*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - a^2*sin(d*x + c)^6/(co s(d*x + c) + 1)^6)*d)
Time = 0.40 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.32 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {15}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 50 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \]
-1/60*(15/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) - (15*tan(1/2*d*x + 1/2*c)^4 - 30*tan(1/2*d*x + 1/2*c)^3 - 40*tan(1/2*d*x + 1/2*c)^2 - 50*tan(1/2*d*x + 1 /2*c) - 7)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 11.71 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.24 \[ \int \frac {\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{15}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{15}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}+\frac {8\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^2\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5} \]
((2*cos(c/2 + (d*x)/2)^6)/15 + (8*cos(c/2 + (d*x)/2)^5*sin(c/2 + (d*x)/2)) /15 + 2*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4 + (8*cos(c/2 + (d*x)/2)^ 3*sin(c/2 + (d*x)/2)^3)/3 + (8*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^2)/ 3)/(a^2*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))^5)