Integrand size = 21, antiderivative size = 108 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {51 x}{8 a^3}-\frac {7 \sin (c+d x)}{a^3 d}+\frac {19 \cos (c+d x) \sin (c+d x)}{8 a^3 d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a^3 d}-\frac {4 \sin (c+d x)}{a^3 d (1+\cos (c+d x))}+\frac {\sin ^3(c+d x)}{a^3 d} \]
51/8*x/a^3-7*sin(d*x+c)/a^3/d+19/8*cos(d*x+c)*sin(d*x+c)/a^3/d+1/4*cos(d*x +c)^3*sin(d*x+c)/a^3/d-4*sin(d*x+c)/a^3/d/(1+cos(d*x+c))+sin(d*x+c)^3/a^3/ d
Time = 0.61 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.60 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left (2040 d x \cos \left (\frac {d x}{2}\right )+2040 d x \cos \left (c+\frac {d x}{2}\right )-3563 \sin \left (\frac {d x}{2}\right )-997 \sin \left (c+\frac {d x}{2}\right )-800 \sin \left (c+\frac {3 d x}{2}\right )-800 \sin \left (2 c+\frac {3 d x}{2}\right )+160 \sin \left (2 c+\frac {5 d x}{2}\right )+160 \sin \left (3 c+\frac {5 d x}{2}\right )-35 \sin \left (3 c+\frac {7 d x}{2}\right )-35 \sin \left (4 c+\frac {7 d x}{2}\right )+5 \sin \left (4 c+\frac {9 d x}{2}\right )+5 \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{640 a^3 d} \]
(Sec[c/2]*Sec[(c + d*x)/2]*(2040*d*x*Cos[(d*x)/2] + 2040*d*x*Cos[c + (d*x) /2] - 3563*Sin[(d*x)/2] - 997*Sin[c + (d*x)/2] - 800*Sin[c + (3*d*x)/2] - 800*Sin[2*c + (3*d*x)/2] + 160*Sin[2*c + (5*d*x)/2] + 160*Sin[3*c + (5*d*x )/2] - 35*Sin[3*c + (7*d*x)/2] - 35*Sin[4*c + (7*d*x)/2] + 5*Sin[4*c + (9* d*x)/2] + 5*Sin[5*c + (9*d*x)/2]))/(640*a^3*d)
Time = 0.60 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4360, 25, 25, 3042, 3354, 3042, 25, 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.
| \(\displaystyle \int \frac {\sin ^4(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos \left (c+d x-\frac {\pi }{2}\right )^4}{\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4360 |
| \(\displaystyle \int -\frac {\sin ^4(c+d x) \cos ^3(c+d x)}{(a (-\cos (c+d x))-a)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int -\frac {\cos ^3(c+d x) \sin ^4(c+d x)}{(\cos (c+d x) a+a)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \int \frac {\sin ^4(c+d x) \cos ^3(c+d x)}{(a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \cos \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3354 |
| \(\displaystyle \frac {\int \cos (c+d x) (a-a \cos (c+d x))^3 \cot ^2(c+d x)dx}{a^6}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int -\frac {\sin \left (c+d x-\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x-\frac {\pi }{2}\right ) a+a\right )^3}{\cos \left (c+d x-\frac {\pi }{2}\right )^2}dx}{a^6}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \left (\sin \left (\frac {1}{2} (2 c-\pi )+d x\right ) a+a\right )^3}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^2}dx}{a^6}\) |
| \(\Big \downarrow \) 3351 |
| \(\displaystyle -\frac {\int \left (-a \cos ^4(c+d x)+3 a \cos ^3(c+d x)-4 a \cos ^2(c+d x)+4 a \cos (c+d x)-4 a+\frac {4 a}{\cos (c+d x)+1}\right )dx}{a^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {-\frac {a \sin ^3(c+d x)}{d}+\frac {7 a \sin (c+d x)}{d}-\frac {a \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {19 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {4 a \sin (c+d x)}{d (\cos (c+d x)+1)}-\frac {51 a x}{8}}{a^4}\) |
-(((-51*a*x)/8 + (7*a*Sin[c + d*x])/d - (19*a*Cos[c + d*x]*Sin[c + d*x])/( 8*d) - (a*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (4*a*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) - (a*Sin[c + d*x]^3)/d)/a^4)
3.2.2.3.1 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_)*((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 = 0.91 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.60
| method | result | size |
| parallelrisch | \(\frac {204 d x +\left (-295+\cos \left (4 d x +4 c \right )-6 \cos \left (3 d x +3 c \right )+26 \cos \left (2 d x +2 c \right )-134 \cos \left (d x +c \right )\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 a^{3} d}\) | \(65\) |
| derivativedivides | \(\frac {-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 \left (-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}-\frac {149 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16}-\frac {123 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}-\frac {35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {51 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{3} d}\) | \(100\) |
| default | \(\frac {-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 \left (-\frac {77 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{16}-\frac {149 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{16}-\frac {123 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{16}-\frac {35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}+\frac {51 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}}{a^{3} d}\) | \(100\) |
| risch | \(\frac {51 x}{8 a^{3}}+\frac {25 i {\mathrm e}^{i \left (d x +c \right )}}{8 a^{3} d}-\frac {25 i {\mathrm e}^{-i \left (d x +c \right )}}{8 a^{3} d}-\frac {8 i}{a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\sin \left (4 d x +4 c \right )}{32 a^{3} d}-\frac {\sin \left (3 d x +3 c \right )}{4 a^{3} d}+\frac {5 \sin \left (2 d x +2 c \right )}{4 a^{3} d}\) | \(117\) |
| norman | \(\frac {\frac {51 x}{8 a}-\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {187 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a d}-\frac {245 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 a d}-\frac {141 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 a d}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {51 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {153 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 a}+\frac {51 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}+\frac {51 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8 a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} a^{2}}\) | \(188\) |
1/32*(204*d*x+(-295+cos(4*d*x+4*c)-6*cos(3*d*x+3*c)+26*cos(2*d*x+2*c)-134* cos(d*x+c))*tan(1/2*d*x+1/2*c))/a^3/d
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.77 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {51 \, d x \cos \left (d x + c\right ) + 51 \, d x + {\left (2 \, \cos \left (d x + c\right )^{4} - 6 \, \cos \left (d x + c\right )^{3} + 11 \, \cos \left (d x + c\right )^{2} - 29 \, \cos \left (d x + c\right ) - 80\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/8*(51*d*x*cos(d*x + c) + 51*d*x + (2*cos(d*x + c)^4 - 6*cos(d*x + c)^3 + 11*cos(d*x + c)^2 - 29*cos(d*x + c) - 80)*sin(d*x + c))/(a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\sin ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Leaf count of result is larger than twice the leaf count of optimal. 227 vs. \(2 (102) = 204\).
Time = 0.29 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.10 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {123 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {149 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {77 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{3} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {6 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {4 \, a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {a^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac {51 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}} + \frac {16 \, \sin \left (d x + c\right )}{a^{3} {\left (\cos \left (d x + c\right ) + 1\right )}}}{4 \, d} \]
-1/4*((35*sin(d*x + c)/(cos(d*x + c) + 1) + 123*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 149*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 77*sin(d*x + c)^7/(c os(d*x + c) + 1)^7)/(a^3 + 4*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a ^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a^3*sin(d*x + c)^6/(cos(d*x + c ) + 1)^6 + a^3*sin(d*x + c)^8/(cos(d*x + c) + 1)^8) - 51*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3 + 16*sin(d*x + c)/(a^3*(cos(d*x + c) + 1)))/d
Time = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {51 \, {\left (d x + c\right )}}{a^{3}} - \frac {32 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} - \frac {2 \, {\left (77 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 149 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 123 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4} a^{3}}}{8 \, d} \]
1/8*(51*(d*x + c)/a^3 - 32*tan(1/2*d*x + 1/2*c)/a^3 - 2*(77*tan(1/2*d*x + 1/2*c)^7 + 149*tan(1/2*d*x + 1/2*c)^5 + 123*tan(1/2*d*x + 1/2*c)^3 + 35*ta n(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a^3))/d
Time = 14.96 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {51\,x}{8\,a^3}-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d}-\frac {\frac {77\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {149\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {123\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {35\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a^3\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]