Integrand size = 21, antiderivative size = 146 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{64 a^2 d}+\frac {1}{64 d (a-a \sin (c+d x))^2}+\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {7 a}{48 d (a+a \sin (c+d x))^3}+\frac {1}{4 d (a+a \sin (c+d x))^2}-\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )} \]
5/64*arctanh(sin(d*x+c))/a^2/d+1/64/d/(a-a*sin(d*x+c))^2+1/32*a^2/d/(a+a*s in(d*x+c))^4-7/48*a/d/(a+a*sin(d*x+c))^3+1/4/d/(a+a*sin(d*x+c))^2-5/64/d/( a^2-a^2*sin(d*x+c))-5/32/d/(a^2+a^2*sin(d*x+c))
Time = 0.31 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.62 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {15 \text {arctanh}(\sin (c+d x))+\frac {-16-47 \sin (c+d x)-14 \sin ^2(c+d x)+74 \sin ^3(c+d x)+66 \sin ^4(c+d x)-15 \sin ^5(c+d x)}{(-1+\sin (c+d x))^2 (1+\sin (c+d x))^4}}{192 a^2 d} \]
(15*ArcTanh[Sin[c + d*x]] + (-16 - 47*Sin[c + d*x] - 14*Sin[c + d*x]^2 + 7 4*Sin[c + d*x]^3 + 66*Sin[c + d*x]^4 - 15*Sin[c + d*x]^5)/((-1 + Sin[c + d *x])^2*(1 + Sin[c + d*x])^4))/(192*a^2*d)
Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3186, 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.
| \(\displaystyle \int \frac {\tan ^5(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (c+d x)^5}{(a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3186 |
| \(\displaystyle \frac {\int \frac {a^5 \sin ^5(c+d x)}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (-\frac {a^2}{8 (\sin (c+d x) a+a)^5}+\frac {7 a}{16 (\sin (c+d x) a+a)^4}+\frac {1}{32 (a-a \sin (c+d x))^3}-\frac {1}{2 (\sin (c+d x) a+a)^3}+\frac {5}{64 \left (a^2-a^2 \sin ^2(c+d x)\right ) a}-\frac {5}{64 (a-a \sin (c+d x))^2 a}+\frac {5}{32 (\sin (c+d x) a+a)^2 a}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {5 \text {arctanh}(\sin (c+d x))}{64 a^2}+\frac {a^2}{32 (a \sin (c+d x)+a)^4}-\frac {7 a}{48 (a \sin (c+d x)+a)^3}+\frac {1}{64 (a-a \sin (c+d x))^2}+\frac {1}{4 (a \sin (c+d x)+a)^2}-\frac {5}{64 a (a-a \sin (c+d x))}-\frac {5}{32 a (a \sin (c+d x)+a)}}{d}\) |
((5*ArcTanh[Sin[c + d*x]])/(64*a^2) + 1/(64*(a - a*Sin[c + d*x])^2) - 5/(6 4*a*(a - a*Sin[c + d*x])) + a^2/(32*(a + a*Sin[c + d*x])^4) - (7*a)/(48*(a + a*Sin[c + d*x])^3) + 1/(4*(a + a*Sin[c + d*x])^2) - 5/(32*a*(a + a*Sin[ c + d*x])))/d
3.1.62.3.1 Defintions of rubi rules used
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[((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 = 2.00 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.71
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{128}+\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {7}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{128}}{d \,a^{2}}\) | \(103\) |
| default | \(\frac {\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}+\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{128}+\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {7}{48 \left (1+\sin \left (d x +c \right )\right )^{3}}+\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{128}}{d \,a^{2}}\) | \(103\) |
| risch | \(-\frac {i \left (416 i {\mathrm e}^{8 i \left (d x +c \right )}-221 \,{\mathrm e}^{3 i \left (d x +c \right )}-132 i {\mathrm e}^{2 i \left (d x +c \right )}-14 \,{\mathrm e}^{5 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}-56 i {\mathrm e}^{6 i \left (d x +c \right )}+14 \,{\mathrm e}^{7 i \left (d x +c \right )}-132 i {\mathrm e}^{10 i \left (d x +c \right )}+221 \,{\mathrm e}^{9 i \left (d x +c \right )}+416 i {\mathrm e}^{4 i \left (d x +c \right )}+15 \,{\mathrm e}^{11 i \left (d x +c \right )}\right )}{96 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4} d \,a^{2}}-\frac {5 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{64 a^{2} d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{64 a^{2} d}\) | \(208\) |
1/d/a^2*(1/64/(sin(d*x+c)-1)^2+5/64/(sin(d*x+c)-1)-5/128*ln(sin(d*x+c)-1)+ 1/32/(1+sin(d*x+c))^4-7/48/(1+sin(d*x+c))^3+1/4/(1+sin(d*x+c))^2-5/32/(1+s in(d*x+c))+5/128*ln(1+sin(d*x+c)))
Time = 0.31 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.36 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {132 \, \cos \left (d x + c\right )^{4} - 236 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} + 44 \, \cos \left (d x + c\right )^{2} - 12\right )} \sin \left (d x + c\right ) + 72}{384 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]
-1/384*(132*cos(d*x + c)^4 - 236*cos(d*x + c)^2 - 15*(cos(d*x + c)^6 - 2*c os(d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*log(sin(d*x + c) + 1) + 15* (cos(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*log(-s in(d*x + c) + 1) - 2*(15*cos(d*x + c)^4 + 44*cos(d*x + c)^2 - 12)*sin(d*x + c) + 72)/(a^2*d*cos(d*x + c)^6 - 2*a^2*d*cos(d*x + c)^4*sin(d*x + c) - 2 *a^2*d*cos(d*x + c)^4)
\[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\tan ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 66 \, \sin \left (d x + c\right )^{4} - 74 \, \sin \left (d x + c\right )^{3} + 14 \, \sin \left (d x + c\right )^{2} + 47 \, \sin \left (d x + c\right ) + 16\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{384 \, d} \]
-1/384*(2*(15*sin(d*x + c)^5 - 66*sin(d*x + c)^4 - 74*sin(d*x + c)^3 + 14* sin(d*x + c)^2 + 47*sin(d*x + c) + 16)/(a^2*sin(d*x + c)^6 + 2*a^2*sin(d*x + c)^5 - a^2*sin(d*x + c)^4 - 4*a^2*sin(d*x + c)^3 - a^2*sin(d*x + c)^2 + 2*a^2*sin(d*x + c) + a^2) - 15*log(sin(d*x + c) + 1)/a^2 + 15*log(sin(d*x + c) - 1)/a^2)/d
Time = 1.54 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, {\left (15 \, \sin \left (d x + c\right )^{2} - 10 \, \sin \left (d x + c\right ) - 1\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 740 \, \sin \left (d x + c\right )^{3} + 1086 \, \sin \left (d x + c\right )^{2} + 676 \, \sin \left (d x + c\right ) + 157}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{1536 \, d} \]
1/1536*(60*log(abs(sin(d*x + c) + 1))/a^2 - 60*log(abs(sin(d*x + c) - 1))/ a^2 + 6*(15*sin(d*x + c)^2 - 10*sin(d*x + c) - 1)/(a^2*(sin(d*x + c) - 1)^ 2) - (125*sin(d*x + c)^4 + 740*sin(d*x + c)^3 + 1086*sin(d*x + c)^2 + 676* sin(d*x + c) + 157)/(a^2*(sin(d*x + c) + 1)^4))/d
Time = 9.93 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.47 \[ \int \frac {\tan ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,a^2\,d}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}}{32}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}}{8}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{96}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{3}-\frac {121\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{48}-\frac {119\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{12}-\frac {121\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{48}-\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{3}+\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{96}+\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{32}}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-17\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+28\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-17\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-12\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\right )} \]
(5*atanh(tan(c/2 + (d*x)/2)))/(32*a^2*d) - ((5*tan(c/2 + (d*x)/2))/32 + (5 *tan(c/2 + (d*x)/2)^2)/8 + (35*tan(c/2 + (d*x)/2)^3)/96 - (5*tan(c/2 + (d* x)/2)^4)/3 - (121*tan(c/2 + (d*x)/2)^5)/48 - (119*tan(c/2 + (d*x)/2)^6)/12 - (121*tan(c/2 + (d*x)/2)^7)/48 - (5*tan(c/2 + (d*x)/2)^8)/3 + (35*tan(c/ 2 + (d*x)/2)^9)/96 + (5*tan(c/2 + (d*x)/2)^10)/8 + (5*tan(c/2 + (d*x)/2)^1 1)/32)/(d*(2*a^2*tan(c/2 + (d*x)/2)^2 - 12*a^2*tan(c/2 + (d*x)/2)^3 - 17*a ^2*tan(c/2 + (d*x)/2)^4 + 8*a^2*tan(c/2 + (d*x)/2)^5 + 28*a^2*tan(c/2 + (d *x)/2)^6 + 8*a^2*tan(c/2 + (d*x)/2)^7 - 17*a^2*tan(c/2 + (d*x)/2)^8 - 12*a ^2*tan(c/2 + (d*x)/2)^9 + 2*a^2*tan(c/2 + (d*x)/2)^10 + 4*a^2*tan(c/2 + (d *x)/2)^11 + a^2*tan(c/2 + (d*x)/2)^12 + a^2 + 4*a^2*tan(c/2 + (d*x)/2)))