Integrand size = 27, antiderivative size = 66 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec (c+d x)}{a^2 d}-\frac {\sec ^3(c+d x)}{a^2 d}+\frac {2 \sec ^5(c+d x)}{5 a^2 d}-\frac {2 \tan ^5(c+d x)}{5 a^2 d} \]
Time = 0.64 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec (c+d x) (40-65 \cos (c+d x)-8 \cos (2 (c+d x))+13 \cos (3 (c+d x))+40 \sin (c+d x)-52 \sin (2 (c+d x))+8 \sin (3 (c+d x)))}{80 a^2 d (1+\sin (c+d x))^2} \]
(Sec[c + d*x]*(40 - 65*Cos[c + d*x] - 8*Cos[2*(c + d*x)] + 13*Cos[3*(c + d *x)] + 40*Sin[c + d*x] - 52*Sin[2*(c + d*x)] + 8*Sin[3*(c + d*x)]))/(80*a^ 2*d*(1 + Sin[c + d*x])^2)
Time = 0.49 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06, 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.
\(\displaystyle \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (c+d x)^3}{\cos (c+d x)^2 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3354 |
\(\displaystyle \frac {\int \sec ^3(c+d x) (a-a \sin (c+d x))^2 \tan ^3(c+d x)dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sin (c+d x)^3 (a-a \sin (c+d x))^2}{\cos (c+d x)^6}dx}{a^4}\) |
\(\Big \downarrow \) 3352 |
\(\displaystyle \frac {\int \left (a^2 \sec (c+d x) \tan ^5(c+d x)-2 a^2 \sec ^2(c+d x) \tan ^4(c+d x)+a^2 \sec ^3(c+d x) \tan ^3(c+d x)\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {2 a^2 \tan ^5(c+d x)}{5 d}+\frac {2 a^2 \sec ^5(c+d x)}{5 d}-\frac {a^2 \sec ^3(c+d x)}{d}+\frac {a^2 \sec (c+d x)}{d}}{a^4}\) |
((a^2*Sec[c + d*x])/d - (a^2*Sec[c + d*x]^3)/d + (2*a^2*Sec[c + d*x]^5)/(5 *d) - (2*a^2*Tan[c + d*x]^5)/(5*d))/a^4
3.8.81.3.1 Defintions of rubi rules used
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]
Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {-\frac {4}{5}-4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}}{d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\) | \(61\) |
risch | \(\frac {4 i {\mathrm e}^{4 i \left (d x +c \right )}+2 \,{\mathrm e}^{5 i \left (d x +c \right )}-4 \,{\mathrm e}^{3 i \left (d x +c \right )}-\frac {4 i}{5}-\frac {6 \,{\mathrm e}^{i \left (d x +c \right )}}{5}}{\left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) d \,a^{2}}\) | \(85\) |
derivativedivides | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{d \,a^{2}}\) | \(100\) |
default | \(\frac {-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {2}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {16}{64 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+64}}{d \,a^{2}}\) | \(100\) |
norman | \(\frac {-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {4}{5 a d}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5 d a}-\frac {24 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {16 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(129\) |
4/5*(-1-5*tan(1/2*d*x+1/2*c)^2-4*tan(1/2*d*x+1/2*c))/d/a^2/(tan(1/2*d*x+1/ 2*c)-1)/(tan(1/2*d*x+1/2*c)+1)^5
Time = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\cos \left (d x + c\right )^{2} - 2 \, {\left (\cos \left (d x + c\right )^{2} + 1\right )} \sin \left (d x + c\right ) - 3}{5 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \]
1/5*(cos(d*x + c)^2 - 2*(cos(d*x + c)^2 + 1)*sin(d*x + c) - 3)/(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))
Timed out. \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (62) = 124\).
Time = 0.22 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.48 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4 \, {\left (\frac {4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{5 \, {\left (a^{2} + \frac {4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \]
4/5*(4*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 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/(cos(d*x + c) + 1)^6)*d)
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.42 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {5}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 50 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{20 \, d} \]
-1/20*(5/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) - (5*tan(1/2*d*x + 1/2*c)^4 + 30 *tan(1/2*d*x + 1/2*c)^3 + 80*tan(1/2*d*x + 1/2*c)^2 + 50*tan(1/2*d*x + 1/2 *c) + 11)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d
Time = 11.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.68 \[ \int \frac {\sin (c+d x) \tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{5}+\frac {16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}+4\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a^2\,d\,\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5} \]