Integrand size = 21, antiderivative size = 104 \[ \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {3 x}{16 a^2}-\frac {3 \cos (c+d x) \sin (c+d x)}{16 a^2 d}-\frac {\cos (c+d x) \sin ^3(c+d x)}{8 a^2 d}-\frac {(a-a \cos (c+d x))^3 \sin ^3(c+d x)}{6 a^5 d}-\frac {\sin ^5(c+d x)}{10 a^2 d} \]
3/16*x/a^2-3/16*cos(d*x+c)*sin(d*x+c)/a^2/d-1/8*cos(d*x+c)*sin(d*x+c)^3/a^ 2/d-1/6*(a-a*cos(d*x+c))^3*sin(d*x+c)^3/a^5/d-1/10*sin(d*x+c)^5/a^2/d
Time = 0.62 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07 \[ \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (360 d x-480 \sin (c+d x)+30 \sin (2 (c+d x))+80 \sin (3 (c+d x))-90 \sin (4 (c+d x))+48 \sin (5 (c+d x))-10 \sin (6 (c+d x))+25 \tan \left (\frac {c}{2}\right )\right )}{480 a^2 d (1+\sec (c+d x))^2} \]
(Cos[(c + d*x)/2]^4*Sec[c + d*x]^2*(360*d*x - 480*Sin[c + d*x] + 30*Sin[2* (c + d*x)] + 80*Sin[3*(c + d*x)] - 90*Sin[4*(c + d*x)] + 48*Sin[5*(c + d*x )] - 10*Sin[6*(c + d*x)] + 25*Tan[c/2]))/(480*a^2*d*(1 + Sec[c + d*x])^2)
Time = 0.79 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {3042, 4360, 3042, 3354, 3042, 3349, 3042, 3148, 3042, 3115, 3042, 3115, 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.
| \(\displaystyle \int \frac {\sin ^6(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^6}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \frac {\sin ^6(c+d x) \cos ^2(c+d x)}{(a (-\cos (c+d x))-a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \cos \left (c+d x+\frac {\pi }{2}\right )^6}{\left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) (a-a \cos (c+d x))^2 \sin ^2(c+d x)dx}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \cos \left (c+d x+\frac {\pi }{2}\right )^2 \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2dx}{a^4}\) |
| \(\Big \downarrow \) 3349 |
| \(\displaystyle \frac {\frac {1}{2} a \int (a-a \cos (c+d x)) \sin ^4(c+d x)dx-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} a \int \cos \left (c+d x-\frac {\pi }{2}\right )^4 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )dx-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle \frac {\frac {1}{2} a \left (a \int \sin ^4(c+d x)dx-\frac {a \sin ^5(c+d x)}{5 d}\right )-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} a \left (a \int \sin (c+d x)^4dx-\frac {a \sin ^5(c+d x)}{5 d}\right )-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {1}{2} a \left (a \left (\frac {3}{4} \int \sin ^2(c+d x)dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \sin ^5(c+d x)}{5 d}\right )-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} a \left (a \left (\frac {3}{4} \int \sin (c+d x)^2dx-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \sin ^5(c+d x)}{5 d}\right )-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {1}{2} a \left (a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \sin ^5(c+d x)}{5 d}\right )-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{2} a \left (a \left (\frac {3}{4} \left (\frac {x}{2}-\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d}\right )-\frac {a \sin ^5(c+d x)}{5 d}\right )-\frac {\sin ^3(c+d x) (a-a \cos (c+d x))^3}{6 a d}}{a^4}\) |
(-1/6*((a - a*Cos[c + d*x])^3*Sin[c + d*x]^3)/(a*d) + (a*(-1/5*(a*Sin[c + d*x]^5)/d + a*(-1/4*(Cos[c + d*x]*Sin[c + d*x]^3)/d + (3*(x/2 - (Cos[c + d *x]*Sin[c + d*x])/(2*d)))/4)))/2)/a^4
3.1.84.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Simp[a Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && (IntegerQ[2*p] || NeQ[a^2 - b^2, 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[(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]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.93 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.74
| method | result | size |
| parallelrisch | \(\frac {180 d x -240 \sin \left (d x +c \right )-5 \sin \left (6 d x +6 c \right )-45 \sin \left (4 d x +4 c \right )+15 \sin \left (2 d x +2 c \right )+24 \sin \left (5 d x +5 c \right )+40 \sin \left (3 d x +3 c \right )}{960 a^{2} d}\) | \(77\) |
| risch | \(\frac {3 x}{16 a^{2}}-\frac {\sin \left (d x +c \right )}{4 a^{2} d}-\frac {\sin \left (6 d x +6 c \right )}{192 a^{2} d}+\frac {\sin \left (5 d x +5 c \right )}{40 a^{2} d}-\frac {3 \sin \left (4 d x +4 c \right )}{64 a^{2} d}+\frac {\sin \left (3 d x +3 c \right )}{24 a^{2} d}+\frac {\sin \left (2 d x +2 c \right )}{64 a^{2} d}\) | \(107\) |
| derivativedivides | \(\frac {\frac {32 \left (\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{256}-\frac {205 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{768}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{640}-\frac {99 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{640}-\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{256}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{256}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{a^{2} d}\) | \(115\) |
| default | \(\frac {\frac {32 \left (\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{256}-\frac {205 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{768}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{640}-\frac {99 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{640}-\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{256}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{256}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}}{a^{2} d}\) | \(115\) |
| norman | \(\frac {\frac {3 x}{16 a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 a d}-\frac {99 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 a d}-\frac {29 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}-\frac {205 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 a d}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 a d}+\frac {9 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 a}+\frac {45 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16 a}+\frac {15 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{4 a}+\frac {45 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16 a}+\frac {9 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8 a}+\frac {3 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} a}\) | \(241\) |
1/960*(180*d*x-240*sin(d*x+c)-5*sin(6*d*x+6*c)-45*sin(4*d*x+4*c)+15*sin(2* d*x+2*c)+24*sin(5*d*x+5*c)+40*sin(3*d*x+3*c))/a^2/d
Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.68 \[ \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {45 \, d x - {\left (40 \, \cos \left (d x + c\right )^{5} - 96 \, \cos \left (d x + c\right )^{4} + 50 \, \cos \left (d x + c\right )^{3} + 32 \, \cos \left (d x + c\right )^{2} - 45 \, \cos \left (d x + c\right ) + 64\right )} \sin \left (d x + c\right )}{240 \, a^{2} d} \]
1/240*(45*d*x - (40*cos(d*x + c)^5 - 96*cos(d*x + c)^4 + 50*cos(d*x + c)^3 + 32*cos(d*x + c)^2 - 45*cos(d*x + c) + 64)*sin(d*x + c))/(a^2*d)
\[ \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sin ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (95) = 190\).
Time = 0.29 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.81 \[ \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {\frac {45 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {255 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {594 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {174 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {1025 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {45 \, \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}}}{a^{2} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {15 \, a^{2} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {6 \, a^{2} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} + \frac {a^{2} \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}} - \frac {45 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{120 \, d} \]
-1/120*((45*sin(d*x + c)/(cos(d*x + c) + 1) + 255*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 594*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 174*sin(d*x + c)^7 /(cos(d*x + c) + 1)^7 + 1025*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - 45*sin( d*x + c)^11/(cos(d*x + c) + 1)^11)/(a^2 + 6*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 15*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 20*a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 15*a^2*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 6 *a^2*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + a^2*sin(d*x + c)^12/(cos(d*x + c) + 1)^12) - 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.34 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.09 \[ \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {45 \, {\left (d x + c\right )}}{a^{2}} + \frac {2 \, {\left (45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 1025 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 174 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 594 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 255 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 45 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6} a^{2}}}{240 \, d} \]
1/240*(45*(d*x + c)/a^2 + 2*(45*tan(1/2*d*x + 1/2*c)^11 - 1025*tan(1/2*d*x + 1/2*c)^9 - 174*tan(1/2*d*x + 1/2*c)^7 - 594*tan(1/2*d*x + 1/2*c)^5 - 25 5*tan(1/2*d*x + 1/2*c)^3 - 45*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c) ^2 + 1)^6*a^2))/d
Time = 16.58 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int \frac {\sin ^6(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {3\,x}{16\,a^2}-\frac {-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{8}+\frac {205\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{24}+\frac {29\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{20}+\frac {99\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{20}+\frac {17\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{a^2\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^6} \]