Integrand size = 21, antiderivative size = 157 \[ \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx=-\frac {25 a^2 x}{16}+\frac {2 a^2 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{16 d}+\frac {7 a^2 \cos ^3(c+d x) \sin (c+d x)}{24 d}-\frac {a^2 \cos ^5(c+d x) \sin (c+d x)}{6 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {2 a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \tan (c+d x)}{d} \]
-25/16*a^2*x+2*a^2*arctanh(sin(d*x+c))/d-2*a^2*sin(d*x+c)/d+7/16*a^2*cos(d *x+c)*sin(d*x+c)/d+7/24*a^2*cos(d*x+c)^3*sin(d*x+c)/d-1/6*a^2*cos(d*x+c)^5 *sin(d*x+c)/d-2/3*a^2*sin(d*x+c)^3/d-2/5*a^2*sin(d*x+c)^5/d+a^2*tan(d*x+c) /d
Time = 0.47 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79 \[ \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx=-\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (1080 c+1080 d x+420 \arctan (\tan (c+d x))-1920 \text {arctanh}(\sin (c+d x))+1920 \sin (c+d x)+640 \sin ^3(c+d x)+384 \sin ^5(c+d x)-255 \sin (2 (c+d x))-15 \sin (4 (c+d x))+5 \sin (6 (c+d x))-960 \tan (c+d x)\right )}{3840 d} \]
-1/3840*(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(1080*c + 1080*d*x + 420*ArcTan[Tan[c + d*x]] - 1920*ArcTanh[Sin[c + d*x]] + 1920*Sin[c + d*x] + 640*Sin[c + d*x]^3 + 384*Sin[c + d*x]^5 - 255*Sin[2*(c + d*x)] - 15*Sin[ 4*(c + d*x)] + 5*Sin[6*(c + d*x)] - 960*Tan[c + d*x]))/d
Time = 0.56 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4360, 3042, 3351, 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.
| \(\displaystyle \int \sin ^6(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^6 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \sin ^4(c+d x) \tan ^2(c+d x) (a (-\cos (c+d x))-a)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x+\frac {\pi }{2}\right )^6 \left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle \frac {\int \left (-\cos ^6(c+d x) a^8-2 \cos ^5(c+d x) a^8+2 \cos ^4(c+d x) a^8+6 \cos ^3(c+d x) a^8+\sec ^2(c+d x) a^8-6 \cos (c+d x) a^8+2 \sec (c+d x) a^8-2 a^8\right )dx}{a^6}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 a^8 \text {arctanh}(\sin (c+d x))}{d}-\frac {2 a^8 \sin ^5(c+d x)}{5 d}-\frac {2 a^8 \sin ^3(c+d x)}{3 d}-\frac {2 a^8 \sin (c+d x)}{d}+\frac {a^8 \tan (c+d x)}{d}-\frac {a^8 \sin (c+d x) \cos ^5(c+d x)}{6 d}+\frac {7 a^8 \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac {7 a^8 \sin (c+d x) \cos (c+d x)}{16 d}-\frac {25 a^8 x}{16}}{a^6}\) |
((-25*a^8*x)/16 + (2*a^8*ArcTanh[Sin[c + d*x]])/d - (2*a^8*Sin[c + d*x])/d + (7*a^8*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (7*a^8*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) - (a^8*Cos[c + d*x]^5*Sin[c + d*x])/(6*d) - (2*a^8*Sin[c + d*x]^3)/(3*d) - (2*a^8*Sin[c + d*x]^5)/(5*d) + (a^8*Tan[c + d*x])/d)/a^6
3.1.30.3.1 Defintions of rubi rules used
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/a^p Int[Expan dTrig[(d*sin[e + f*x])^n*(a - b*sin[e + f*x])^(p/2)*(a + b*sin[e + f*x])^(m + p/2), x], x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && In tegersQ[m, n, p/2] && ((GtQ[m, 0] && GtQ[p, 0] && LtQ[-m - p, n, -1]) || (G tQ[m, 2] && LtQ[p, 0] && GtQ[m + p/2, 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 = 2.13 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.90
| method | result | size |
| parallelrisch | \(\frac {a^{2} \left (-3000 d x \cos \left (d x +c \right )-3840 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+3840 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+2175 \sin \left (d x +c \right )-2360 \sin \left (2 d x +2 c \right )-24 \sin \left (6 d x +6 c \right )+256 \sin \left (4 d x +4 c \right )-5 \sin \left (7 d x +7 c \right )+10 \sin \left (5 d x +5 c \right )+270 \sin \left (3 d x +3 c \right )\right )}{1920 d \cos \left (d x +c \right )}\) | \(142\) |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(164\) |
| default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(164\) |
| parts | \(\frac {a^{2} \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}+\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{7}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )}{d}+\frac {2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{5}}{5}-\frac {\sin \left (d x +c \right )^{3}}{3}-\sin \left (d x +c \right )+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(169\) |
| risch | \(-\frac {25 a^{2} x}{16}+\frac {17 i a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{128 d}-\frac {17 i a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{128 d}-\frac {11 i a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {11 i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {2 i a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{2} \sin \left (6 d x +6 c \right )}{192 d}-\frac {a^{2} \sin \left (5 d x +5 c \right )}{40 d}+\frac {a^{2} \sin \left (4 d x +4 c \right )}{64 d}+\frac {7 a^{2} \sin \left (3 d x +3 c \right )}{24 d}\) | \(212\) |
| norman | \(\frac {\frac {25 a^{2} x}{16}+\frac {7 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}+\frac {27 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {797 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{40 d}-\frac {91 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{2 d}-\frac {8041 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{120 d}-\frac {431 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 d}+\frac {125 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{16}+\frac {225 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{16}+\frac {125 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{16}-\frac {125 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{16}-\frac {225 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{16}-\frac {125 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{16}-\frac {25 a^{2} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{16}-\frac {57 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{8 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(330\) |
1/1920*a^2*(-3000*d*x*cos(d*x+c)-3840*ln(tan(1/2*d*x+1/2*c)-1)*cos(d*x+c)+ 3840*ln(tan(1/2*d*x+1/2*c)+1)*cos(d*x+c)+2175*sin(d*x+c)-2360*sin(2*d*x+2* c)-24*sin(6*d*x+6*c)+256*sin(4*d*x+4*c)-5*sin(7*d*x+7*c)+10*sin(5*d*x+5*c) +270*sin(3*d*x+3*c))/d/cos(d*x+c)
Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.01 \[ \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx=-\frac {375 \, a^{2} d x \cos \left (d x + c\right ) - 240 \, a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + 240 \, a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (40 \, a^{2} \cos \left (d x + c\right )^{6} + 96 \, a^{2} \cos \left (d x + c\right )^{5} - 70 \, a^{2} \cos \left (d x + c\right )^{4} - 352 \, a^{2} \cos \left (d x + c\right )^{3} - 105 \, a^{2} \cos \left (d x + c\right )^{2} + 736 \, a^{2} \cos \left (d x + c\right ) - 240 \, a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )} \]
-1/240*(375*a^2*d*x*cos(d*x + c) - 240*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) + 240*a^2*cos(d*x + c)*log(-sin(d*x + c) + 1) + (40*a^2*cos(d*x + c)^6 + 96*a^2*cos(d*x + c)^5 - 70*a^2*cos(d*x + c)^4 - 352*a^2*cos(d*x + c)^3 - 105*a^2*cos(d*x + c)^2 + 736*a^2*cos(d*x + c) - 240*a^2)*sin(d*x + c))/( d*cos(d*x + c))
Timed out. \[ \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.28 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.11 \[ \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx=-\frac {64 \, {\left (6 \, \sin \left (d x + c\right )^{5} + 10 \, \sin \left (d x + c\right )^{3} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 30 \, \sin \left (d x + c\right )\right )} a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} + 120 \, {\left (15 \, d x + 15 \, c - \frac {9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{2}}{960 \, d} \]
-1/960*(64*(6*sin(d*x + c)^5 + 10*sin(d*x + c)^3 - 15*log(sin(d*x + c) + 1 ) + 15*log(sin(d*x + c) - 1) + 30*sin(d*x + c))*a^2 - 5*(4*sin(2*d*x + 2*c )^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^2 + 120* (15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*tan(d*x + c))/(tan(d*x + c)^4 + 2*t an(d*x + c)^2 + 1) - 8*tan(d*x + c))*a^2)/d
Time = 0.35 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.23 \[ \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx=-\frac {375 \, {\left (d x + c\right )} a^{2} - 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 480 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {480 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (615 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 3485 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 7926 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 8586 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2595 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 345 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]
-1/240*(375*(d*x + c)*a^2 - 480*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 4 80*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 480*a^2*tan(1/2*d*x + 1/2*c)/( tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(615*a^2*tan(1/2*d*x + 1/2*c)^11 + 3485*a^ 2*tan(1/2*d*x + 1/2*c)^9 + 7926*a^2*tan(1/2*d*x + 1/2*c)^7 + 8586*a^2*tan( 1/2*d*x + 1/2*c)^5 + 2595*a^2*tan(1/2*d*x + 1/2*c)^3 + 345*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
Time = 14.71 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.50 \[ \int (a+a \sec (c+d x))^2 \sin ^6(c+d x) \, dx=\frac {\frac {57\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}}{8}+\frac {431\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{12}+\frac {8041\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{120}+\frac {91\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}-\frac {797\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}-\frac {27\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}-\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {25\,a^2\,x}{16}+\frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]
((91*a^2*tan(c/2 + (d*x)/2)^7)/2 - (797*a^2*tan(c/2 + (d*x)/2)^5)/40 - (27 *a^2*tan(c/2 + (d*x)/2)^3)/4 + (8041*a^2*tan(c/2 + (d*x)/2)^9)/120 + (431* a^2*tan(c/2 + (d*x)/2)^11)/12 + (57*a^2*tan(c/2 + (d*x)/2)^13)/8 - (7*a^2* tan(c/2 + (d*x)/2))/8)/(d*(5*tan(c/2 + (d*x)/2)^2 + 9*tan(c/2 + (d*x)/2)^4 + 5*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 - 9*tan(c/2 + (d*x)/2)^ 10 - 5*tan(c/2 + (d*x)/2)^12 - tan(c/2 + (d*x)/2)^14 + 1)) - (25*a^2*x)/16 + (4*a^2*atanh(tan(c/2 + (d*x)/2)))/d