Integrand size = 21, antiderivative size = 112 \[ \int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx=\frac {a^2 \cos (c+d x)}{d}+\frac {2 a^2 \cos ^2(c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos ^4(c+d x)}{2 d}-\frac {a^2 \cos ^5(c+d x)}{5 d}-\frac {2 a^2 \log (\cos (c+d x))}{d}+\frac {a^2 \sec (c+d x)}{d} \]
a^2*cos(d*x+c)/d+2*a^2*cos(d*x+c)^2/d+1/3*a^2*cos(d*x+c)^3/d-1/2*a^2*cos(d *x+c)^4/d-1/5*a^2*cos(d*x+c)^5/d-2*a^2*ln(cos(d*x+c))/d+a^2*sec(d*x+c)/d
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.78 \[ \int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {a^2 (-750-275 \cos (2 (c+d x))-165 \cos (3 (c+d x))-2 \cos (4 (c+d x))+15 \cos (5 (c+d x))+3 \cos (6 (c+d x))+30 \cos (c+d x) (-3+32 \log (\cos (c+d x)))) \sec (c+d x)}{480 d} \]
-1/480*(a^2*(-750 - 275*Cos[2*(c + d*x)] - 165*Cos[3*(c + d*x)] - 2*Cos[4* (c + d*x)] + 15*Cos[5*(c + d*x)] + 3*Cos[6*(c + d*x)] + 30*Cos[c + d*x]*(- 3 + 32*Log[Cos[c + d*x]]))*Sec[c + d*x])/d
Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.92, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4360, 3042, 25, 3315, 27, 99, 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.
| \(\displaystyle \int \sin ^5(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos \left (c+d x-\frac {\pi }{2}\right )^5 \left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^2dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int \sin ^3(c+d x) \tan ^2(c+d x) (a (-\cos (c+d x))-a)^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\cos \left (c+d x+\frac {\pi }{2}\right )^5 \left (a \left (-\sin \left (c+d x+\frac {\pi }{2}\right )\right )-a\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\cos \left (\frac {1}{2} (2 c+\pi )+d x\right )^5 \left (\sin \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^2}{\sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2}dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle -\frac {\int (a-a \cos (c+d x))^2 (\cos (c+d x) a+a)^4 \sec ^2(c+d x)d(a \cos (c+d x))}{a^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {(a-a \cos (c+d x))^2 (\cos (c+d x) a+a)^4 \sec ^2(c+d x)}{a^2}d(a \cos (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (\cos ^4(c+d x) a^4+2 \cos ^3(c+d x) a^4-\cos ^2(c+d x) a^4+\sec ^2(c+d x) a^4-4 \cos (c+d x) a^4+2 \sec (c+d x) a^4-a^4\right )d(a \cos (c+d x))}{a^3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {1}{5} a^5 \cos ^5(c+d x)+\frac {1}{2} a^5 \cos ^4(c+d x)-\frac {1}{3} a^5 \cos ^3(c+d x)-2 a^5 \cos ^2(c+d x)-a^5 \cos (c+d x)-a^5 \sec (c+d x)+2 a^5 \log (a \cos (c+d x))}{a^3 d}\) |
-((-(a^5*Cos[c + d*x]) - 2*a^5*Cos[c + d*x]^2 - (a^5*Cos[c + d*x]^3)/3 + ( a^5*Cos[c + d*x]^4)/2 + (a^5*Cos[c + d*x]^5)/5 + 2*a^5*Log[a*Cos[c + d*x]] - a^5*Sec[c + d*x])/(a^3*d))
3.1.21.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^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 = 1.89 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.08
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}}{d}\) | \(121\) |
| default | \(\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )-\frac {a^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}}{d}\) | \(121\) |
| parts | \(-\frac {a^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5 d}+\frac {a^{2} \left (\frac {\sin \left (d x +c \right )^{6}}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )\right )}{d}+\frac {2 a^{2} \left (-\frac {\sin \left (d x +c \right )^{4}}{4}-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(126\) |
| parallelrisch | \(-\frac {\left (-320 \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \cos \left (d x +c \right )+320 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )+320 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )+\cos \left (6 d x +6 c \right )-\frac {874 \cos \left (d x +c \right )}{3}-\frac {275 \cos \left (2 d x +2 c \right )}{3}-55 \cos \left (3 d x +3 c \right )-\frac {2 \cos \left (4 d x +4 c \right )}{3}+5 \cos \left (5 d x +5 c \right )-250\right ) a^{2}}{160 d \cos \left (d x +c \right )}\) | \(140\) |
| risch | \(2 i a^{2} x +\frac {3 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {9 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{16 d}+\frac {9 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{16 d}+\frac {3 a^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {4 i a^{2} c}{d}+\frac {2 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}-\frac {a^{2} \cos \left (5 d x +5 c \right )}{80 d}-\frac {a^{2} \cos \left (4 d x +4 c \right )}{16 d}+\frac {a^{2} \cos \left (3 d x +3 c \right )}{48 d}\) | \(188\) |
| norman | \(\frac {-\frac {64 a^{2}}{15 d}-\frac {64 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {16 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}-\frac {16 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d}-\frac {196 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {2 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}+\frac {2 a^{2} \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d}\) | \(199\) |
1/d*(a^2*(sin(d*x+c)^6/cos(d*x+c)+(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos( d*x+c))+2*a^2*(-1/4*sin(d*x+c)^4-1/2*sin(d*x+c)^2-ln(cos(d*x+c)))-1/5*a^2* (8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c))
Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03 \[ \int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {48 \, a^{2} \cos \left (d x + c\right )^{6} + 120 \, a^{2} \cos \left (d x + c\right )^{5} - 80 \, a^{2} \cos \left (d x + c\right )^{4} - 480 \, a^{2} \cos \left (d x + c\right )^{3} - 240 \, a^{2} \cos \left (d x + c\right )^{2} + 480 \, a^{2} \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + 195 \, a^{2} \cos \left (d x + c\right ) - 240 \, a^{2}}{240 \, d \cos \left (d x + c\right )} \]
-1/240*(48*a^2*cos(d*x + c)^6 + 120*a^2*cos(d*x + c)^5 - 80*a^2*cos(d*x + c)^4 - 480*a^2*cos(d*x + c)^3 - 240*a^2*cos(d*x + c)^2 + 480*a^2*cos(d*x + c)*log(-cos(d*x + c)) + 195*a^2*cos(d*x + c) - 240*a^2)/(d*cos(d*x + c))
\[ \int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx=a^{2} \left (\int 2 \sin ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \sin ^{5}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{5}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(2*sin(c + d*x)**5*sec(c + d*x), x) + Integral(sin(c + d*x)* *5*sec(c + d*x)**2, x) + Integral(sin(c + d*x)**5, x))
Time = 0.20 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84 \[ \int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx=-\frac {6 \, a^{2} \cos \left (d x + c\right )^{5} + 15 \, a^{2} \cos \left (d x + c\right )^{4} - 10 \, a^{2} \cos \left (d x + c\right )^{3} - 60 \, a^{2} \cos \left (d x + c\right )^{2} - 30 \, a^{2} \cos \left (d x + c\right ) + 60 \, a^{2} \log \left (\cos \left (d x + c\right )\right ) - \frac {30 \, a^{2}}{\cos \left (d x + c\right )}}{30 \, d} \]
-1/30*(6*a^2*cos(d*x + c)^5 + 15*a^2*cos(d*x + c)^4 - 10*a^2*cos(d*x + c)^ 3 - 60*a^2*cos(d*x + c)^2 - 30*a^2*cos(d*x + c) + 60*a^2*log(cos(d*x + c)) - 30*a^2/cos(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 270 vs. \(2 (106) = 212\).
Time = 0.36 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.41 \[ \int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx=\frac {60 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - 60 \, a^{2} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {60 \, {\left (2 \, a^{2} + \frac {a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1}\right )}}{\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1} + \frac {69 \, a^{2} - \frac {525 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {1650 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {1610 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {745 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {137 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}}{30 \, d} \]
1/30*(60*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - 60*a^2 *log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) + 60*(2*a^2 + a^2*(c os(d*x + c) - 1)/(cos(d*x + c) + 1))/((cos(d*x + c) - 1)/(cos(d*x + c) + 1 ) + 1) + (69*a^2 - 525*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1650*a^ 2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 - 1610*a^2*(cos(d*x + c) - 1)^ 3/(cos(d*x + c) + 1)^3 + 745*a^2*(cos(d*x + c) - 1)^4/(cos(d*x + c) + 1)^4 - 137*a^2*(cos(d*x + c) - 1)^5/(cos(d*x + c) + 1)^5)/((cos(d*x + c) - 1)/ (cos(d*x + c) + 1) - 1)^5)/d
Time = 14.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int (a+a \sec (c+d x))^2 \sin ^5(c+d x) \, dx=\frac {a^2\,\cos \left (c+d\,x\right )+\frac {a^2}{\cos \left (c+d\,x\right )}+2\,a^2\,{\cos \left (c+d\,x\right )}^2+\frac {a^2\,{\cos \left (c+d\,x\right )}^3}{3}-\frac {a^2\,{\cos \left (c+d\,x\right )}^4}{2}-\frac {a^2\,{\cos \left (c+d\,x\right )}^5}{5}-2\,a^2\,\ln \left (\cos \left (c+d\,x\right )\right )}{d} \]