Integrand size = 21, antiderivative size = 114 \[ \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {10 a^3 \log (1-\sin (c+d x))}{d}-\frac {6 a^3 \sin (c+d x)}{d}-\frac {3 a^3 \sin ^2(c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {a^5}{2 d (a-a \sin (c+d x))^2}-\frac {5 a^4}{d (a-a \sin (c+d x))} \]
-10*a^3*ln(1-sin(d*x+c))/d-6*a^3*sin(d*x+c)/d-3/2*a^3*sin(d*x+c)^2/d-1/3*a ^3*sin(d*x+c)^3/d+1/2*a^5/d/(a-a*sin(d*x+c))^2-5*a^4/d/(a-a*sin(d*x+c))
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.64 \[ \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {a^3 \left (60 \log (1-\sin (c+d x))+\frac {27-30 \sin (c+d x)}{(-1+\sin (c+d x))^2}+36 \sin (c+d x)+9 \sin ^2(c+d x)+2 \sin ^3(c+d x)\right )}{6 d} \]
-1/6*(a^3*(60*Log[1 - Sin[c + d*x]] + (27 - 30*Sin[c + d*x])/(-1 + Sin[c + d*x])^2 + 36*Sin[c + d*x] + 9*Sin[c + d*x]^2 + 2*Sin[c + d*x]^3))/d
Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 49, 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.
| \(\displaystyle \int \tan ^5(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \tan (c+d x)^5 (a \sin (c+d x)+a)^3dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x)}{(a-a \sin (c+d x))^3}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\int \left (\frac {a^5}{(a-a \sin (c+d x))^3}-\frac {5 a^4}{(a-a \sin (c+d x))^2}+\frac {10 a^3}{a-a \sin (c+d x)}-\sin ^2(c+d x) a^2-3 \sin (c+d x) a^2-6 a^2\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {a^5}{2 (a-a \sin (c+d x))^2}-\frac {5 a^4}{a-a \sin (c+d x)}-\frac {1}{3} a^3 \sin ^3(c+d x)-\frac {3}{2} a^3 \sin ^2(c+d x)-6 a^3 \sin (c+d x)-10 a^3 \log (a-a \sin (c+d x))}{d}\) |
(-10*a^3*Log[a - a*Sin[c + d*x]] - 6*a^3*Sin[c + d*x] - (3*a^3*Sin[c + d*x ]^2)/2 - (a^3*Sin[c + d*x]^3)/3 + a^5/(2*(a - a*Sin[c + d*x])^2) - (5*a^4) /(a - a*Sin[c + d*x]))/d
3.9.70.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.42 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.32
| method | result | size |
| parallelrisch | \(\frac {10 \left (\left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right ) \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 \cos \left (2 d x +2 c \right )+6-8 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\frac {17 \cos \left (2 d x +2 c \right )}{12}+\frac {\cos \left (4 d x +4 c \right )}{48}+\frac {61 \sin \left (d x +c \right )}{24}-\frac {3 \sin \left (3 d x +3 c \right )}{16}+\frac {\sin \left (5 d x +5 c \right )}{240}-\frac {23}{16}\right ) a^{3}}{d \left (\cos \left (2 d x +2 c \right )-3+4 \sin \left (d x +c \right )\right )}\) | \(150\) |
| risch | \(10 i a^{3} x -\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {3 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {25 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {25 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {i a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {20 i a^{3} c}{d}+\frac {2 i \left (-5 a^{3} {\mathrm e}^{i \left (d x +c \right )}-9 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+5 a^{3} {\mathrm e}^{3 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{4} d}-\frac {20 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(210\) |
| derivativedivides | \(\frac {a^{3} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(302\) |
| default | \(\frac {a^{3} \left (\frac {\sin ^{9}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {5 \left (\sin ^{9}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {5 \left (\sin ^{7}\left (d x +c \right )\right )}{8}-\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}-\frac {35 \sin \left (d x +c \right )}{8}+\frac {35 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin ^{8}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}-\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{2}-\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{4}-\frac {3 \left (\sin ^{2}\left (d x +c \right )\right )}{2}-3 \ln \left (\cos \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {\sin ^{7}\left (d x +c \right )}{4 \cos \left (d x +c \right )^{4}}-\frac {3 \left (\sin ^{7}\left (d x +c \right )\right )}{8 \cos \left (d x +c \right )^{2}}-\frac {3 \left (\sin ^{5}\left (d x +c \right )\right )}{8}-\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{8}-\frac {15 \sin \left (d x +c \right )}{8}+\frac {15 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\) | \(302\) |
| norman | \(\frac {\frac {20 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {64 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {64 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {20 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {40 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {188 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {16 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {188 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {40 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {20 a^{3} \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {20 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {20 a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {20 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {10 a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(321\) |
10*((cos(2*d*x+2*c)-3+4*sin(d*x+c))*ln(sec(1/2*d*x+1/2*c)^2)+(-2*cos(2*d*x +2*c)+6-8*sin(d*x+c))*ln(tan(1/2*d*x+1/2*c)-1)+17/12*cos(2*d*x+2*c)+1/48*c os(4*d*x+4*c)+61/24*sin(d*x+c)-3/16*sin(3*d*x+3*c)+1/240*sin(5*d*x+5*c)-23 /16)*a^3/d/(cos(2*d*x+2*c)-3+4*sin(d*x+c))
Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.24 \[ \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\frac {10 \, a^{3} \cos \left (d x + c\right )^{4} + 115 \, a^{3} \cos \left (d x + c\right )^{2} - 80 \, a^{3} - 120 \, {\left (a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \sin \left (d x + c\right ) - 2 \, a^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 24 \, a^{3} \cos \left (d x + c\right )^{2} + 37 \, a^{3}\right )} \sin \left (d x + c\right )}{12 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]
1/12*(10*a^3*cos(d*x + c)^4 + 115*a^3*cos(d*x + c)^2 - 80*a^3 - 120*(a^3*c os(d*x + c)^2 + 2*a^3*sin(d*x + c) - 2*a^3)*log(-sin(d*x + c) + 1) + 2*(2* a^3*cos(d*x + c)^4 - 24*a^3*cos(d*x + c)^2 + 37*a^3)*sin(d*x + c))/(d*cos( d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)
Timed out. \[ \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx=-\frac {2 \, a^{3} \sin \left (d x + c\right )^{3} + 9 \, a^{3} \sin \left (d x + c\right )^{2} + 60 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) + 36 \, a^{3} \sin \left (d x + c\right ) - \frac {3 \, {\left (10 \, a^{3} \sin \left (d x + c\right ) - 9 \, a^{3}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{6 \, d} \]
-1/6*(2*a^3*sin(d*x + c)^3 + 9*a^3*sin(d*x + c)^2 + 60*a^3*log(sin(d*x + c ) - 1) + 36*a^3*sin(d*x + c) - 3*(10*a^3*sin(d*x + c) - 9*a^3)/(sin(d*x + c)^2 - 2*sin(d*x + c) + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 242 vs. \(2 (110) = 220\).
Time = 0.49 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.12 \[ \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\frac {30 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 60 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {55 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 183 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 80 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 183 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 55 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}} + \frac {125 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 524 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 804 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 524 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 125 \, a^{3}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}}}{3 \, d} \]
1/3*(30*a^3*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 60*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (55*a^3*tan(1/2*d*x + 1/2*c)^6 + 36*a^3*tan(1/2*d*x + 1/2* c)^5 + 183*a^3*tan(1/2*d*x + 1/2*c)^4 + 80*a^3*tan(1/2*d*x + 1/2*c)^3 + 18 3*a^3*tan(1/2*d*x + 1/2*c)^2 + 36*a^3*tan(1/2*d*x + 1/2*c) + 55*a^3)/(tan( 1/2*d*x + 1/2*c)^2 + 1)^3 + (125*a^3*tan(1/2*d*x + 1/2*c)^4 - 524*a^3*tan( 1/2*d*x + 1/2*c)^3 + 804*a^3*tan(1/2*d*x + 1/2*c)^2 - 524*a^3*tan(1/2*d*x + 1/2*c) + 125*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^4)/d
Time = 12.34 (sec) , antiderivative size = 321, normalized size of antiderivative = 2.82 \[ \int (a+a \sin (c+d x))^3 \tan ^5(c+d x) \, dx=\frac {10\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {20\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-60\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+\frac {320\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {500\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{3}+184\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-\frac {500\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {320\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-60\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+20\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+22\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-16\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+9\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {20\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d} \]
(10*a^3*log(tan(c/2 + (d*x)/2)^2 + 1))/d - ((320*a^3*tan(c/2 + (d*x)/2)^3) /3 - 60*a^3*tan(c/2 + (d*x)/2)^2 - (500*a^3*tan(c/2 + (d*x)/2)^4)/3 + 184* a^3*tan(c/2 + (d*x)/2)^5 - (500*a^3*tan(c/2 + (d*x)/2)^6)/3 + (320*a^3*tan (c/2 + (d*x)/2)^7)/3 - 60*a^3*tan(c/2 + (d*x)/2)^8 + 20*a^3*tan(c/2 + (d*x )/2)^9 + 20*a^3*tan(c/2 + (d*x)/2))/(d*(9*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2) - 16*tan(c/2 + (d*x)/2)^3 + 22*tan(c/2 + (d*x)/2)^4 - 24*tan(c /2 + (d*x)/2)^5 + 22*tan(c/2 + (d*x)/2)^6 - 16*tan(c/2 + (d*x)/2)^7 + 9*ta n(c/2 + (d*x)/2)^8 - 4*tan(c/2 + (d*x)/2)^9 + tan(c/2 + (d*x)/2)^10 + 1)) - (20*a^3*log(tan(c/2 + (d*x)/2) - 1))/d