Integrand size = 21, antiderivative size = 146 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {15 \text {arctanh}(\sin (c+d x))}{64 a^2 d}+\frac {1}{64 d (a-a \sin (c+d x))^2}-\frac {a^2}{32 d (a+a \sin (c+d x))^4}-\frac {a}{16 d (a+a \sin (c+d x))^3}-\frac {3}{32 d (a+a \sin (c+d x))^2}+\frac {5}{64 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {5}{32 d \left (a^2+a^2 \sin (c+d x)\right )} \]
15/64*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* sin(d*x+c))^4-1/16*a/d/(a+a*sin(d*x+c))^3-3/32/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 = 0.22 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\sec ^4(c+d x) (1-\sin (c+d x))^2 (1+\sin (c+d x))^2 \left (\frac {15}{64} \text {arctanh}(\sin (c+d x))+\frac {1}{64 (1-\sin (c+d x))^2}+\frac {5}{64 (1-\sin (c+d x))}-\frac {1}{32 (1+\sin (c+d x))^4}-\frac {1}{16 (1+\sin (c+d x))^3}-\frac {3}{32 (1+\sin (c+d x))^2}-\frac {5}{32 (1+\sin (c+d x))}\right )}{a^2 d} \]
(Sec[c + d*x]^4*(1 - Sin[c + d*x])^2*(1 + Sin[c + d*x])^2*((15*ArcTanh[Sin [c + d*x]])/64 + 1/(64*(1 - Sin[c + d*x])^2) + 5/(64*(1 - Sin[c + d*x])) - 1/(32*(1 + Sin[c + d*x])^4) - 1/(16*(1 + Sin[c + d*x])^3) - 3/(32*(1 + Si n[c + d*x])^2) - 5/(32*(1 + Sin[c + d*x]))))/(a^2*d)
Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3146, 54, 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.
| \(\displaystyle \int \frac {\sec ^5(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^5 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a^5 \int \frac {1}{(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^5 \int \left (\frac {5}{64 a^6 (a-a \sin (c+d x))^2}+\frac {5}{32 a^6 (\sin (c+d x) a+a)^2}+\frac {1}{32 a^5 (a-a \sin (c+d x))^3}+\frac {3}{16 a^5 (\sin (c+d x) a+a)^3}+\frac {3}{16 a^4 (\sin (c+d x) a+a)^4}+\frac {1}{8 a^3 (\sin (c+d x) a+a)^5}+\frac {15}{64 a^6 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^5 \left (\frac {15 \text {arctanh}(\sin (c+d x))}{64 a^7}+\frac {5}{64 a^6 (a-a \sin (c+d x))}-\frac {5}{32 a^6 (a \sin (c+d x)+a)}+\frac {1}{64 a^5 (a-a \sin (c+d x))^2}-\frac {3}{32 a^5 (a \sin (c+d x)+a)^2}-\frac {1}{16 a^4 (a \sin (c+d x)+a)^3}-\frac {1}{32 a^3 (a \sin (c+d x)+a)^4}\right )}{d}\) |
(a^5*((15*ArcTanh[Sin[c + d*x]])/(64*a^7) + 1/(64*a^5*(a - a*Sin[c + d*x]) ^2) + 5/(64*a^6*(a - a*Sin[c + d*x])) - 1/(32*a^3*(a + a*Sin[c + d*x])^4) - 1/(16*a^4*(a + a*Sin[c + d*x])^3) - 3/(32*a^5*(a + a*Sin[c + d*x])^2) - 5/(32*a^6*(a + a*Sin[c + d*x]))))/d
3.1.74.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(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] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 1.38 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.71
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right )}{128}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right )}{128}}{d \,a^{2}}\) | \(103\) |
| default | \(\frac {-\frac {1}{32 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {5}{32 \left (1+\sin \left (d x +c \right )\right )}+\frac {15 \ln \left (1+\sin \left (d x +c \right )\right )}{128}+\frac {1}{64 \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {5}{64 \left (\sin \left (d x +c \right )-1\right )}-\frac {15 \ln \left (\sin \left (d x +c \right )-1\right )}{128}}{d \,a^{2}}\) | \(103\) |
| risch | \(-\frac {i \left (60 i {\mathrm e}^{10 i \left (d x +c \right )}+15 \,{\mathrm e}^{11 i \left (d x +c \right )}+160 i {\mathrm e}^{8 i \left (d x +c \right )}-35 \,{\mathrm e}^{9 i \left (d x +c \right )}+72 i {\mathrm e}^{6 i \left (d x +c \right )}-242 \,{\mathrm e}^{7 i \left (d x +c \right )}+160 i {\mathrm e}^{4 i \left (d x +c \right )}+242 \,{\mathrm e}^{5 i \left (d x +c \right )}+60 i {\mathrm e}^{2 i \left (d x +c \right )}+35 \,{\mathrm e}^{3 i \left (d x +c \right )}-15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{32 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{4} a^{2} d}+\frac {15 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{64 a^{2} d}-\frac {15 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{64 a^{2} d}\) | \(208\) |
| norman | \(\frac {\frac {17 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {17 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {23 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{32 d a}+\frac {49 \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {57 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {57 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {29 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}+\frac {29 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32 d a}}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 a^{2} d}+\frac {15 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 a^{2} d}\) | \(280\) |
| parallelrisch | \(\frac {\left (-60 \sin \left (5 d x +5 c \right )-255 \cos \left (2 d x +2 c \right )-30 \cos \left (4 d x +4 c \right )+15 \cos \left (6 d x +6 c \right )-120 \sin \left (d x +c \right )-180 \sin \left (3 d x +3 c \right )-210\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (60 \sin \left (5 d x +5 c \right )+255 \cos \left (2 d x +2 c \right )+30 \cos \left (4 d x +4 c \right )-15 \cos \left (6 d x +6 c \right )+120 \sin \left (d x +c \right )+180 \sin \left (3 d x +3 c \right )+210\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+34 \sin \left (5 d x +5 c \right )-48 \cos \left (2 d x +2 c \right )-88 \cos \left (4 d x +4 c \right )-16 \cos \left (6 d x +6 c \right )+612 \sin \left (d x +c \right )+262 \sin \left (3 d x +3 c \right )+152}{64 a^{2} d \left (17 \cos \left (2 d x +2 c \right )-\cos \left (6 d x +6 c \right )+4 \sin \left (5 d x +5 c \right )+12 \sin \left (3 d x +3 c \right )+2 \cos \left (4 d x +4 c \right )+14+8 \sin \left (d x +c \right )\right )}\) | \(297\) |
1/d/a^2*(-1/32/(1+sin(d*x+c))^4-1/16/(1+sin(d*x+c))^3-3/32/(1+sin(d*x+c))^ 2-5/32/(1+sin(d*x+c))+15/128*ln(1+sin(d*x+c))+1/64/(sin(d*x+c)-1)^2-5/64/( sin(d*x+c)-1)-15/128*ln(sin(d*x+c)-1))
Time = 0.37 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {60 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{6} - 2 \, \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 12\right )} \sin \left (d x + c\right ) - 8}{128 \, {\left (a^{2} d \cos \left (d x + c\right )^{6} - 2 \, a^{2} d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{4}\right )}} \]
1/128*(60*cos(d*x + c)^4 - 20*cos(d*x + c)^2 + 15*(cos(d*x + c)^6 - 2*cos( d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*log(sin(d*x + c) + 1) - 15*(co s(d*x + c)^6 - 2*cos(d*x + c)^4*sin(d*x + c) - 2*cos(d*x + c)^4)*log(-sin( d*x + c) + 1) + 2*(15*cos(d*x + c)^4 - 20*cos(d*x + c)^2 - 12)*sin(d*x + c ) - 8)/(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)
\[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sec ^{5}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} + 30 \, \sin \left (d x + c\right )^{4} - 10 \, \sin \left (d x + c\right )^{3} - 50 \, \sin \left (d x + c\right )^{2} - 17 \, \sin \left (d x + c\right ) + 16\right )}}{a^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{2} \sin \left (d x + c\right )^{5} - a^{2} \sin \left (d x + c\right )^{4} - 4 \, a^{2} \sin \left (d x + c\right )^{3} - a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{128 \, d} \]
-1/128*(2*(15*sin(d*x + c)^5 + 30*sin(d*x + c)^4 - 10*sin(d*x + c)^3 - 50* sin(d*x + c)^2 - 17*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 = 0.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.86 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (45 \, \sin \left (d x + c\right )^{2} - 110 \, \sin \left (d x + c\right ) + 69\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}^{2}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 580 \, \sin \left (d x + c\right )^{3} + 1038 \, \sin \left (d x + c\right )^{2} + 868 \, \sin \left (d x + c\right ) + 301}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{512 \, d} \]
1/512*(60*log(abs(sin(d*x + c) + 1))/a^2 - 60*log(abs(sin(d*x + c) - 1))/a ^2 + 2*(45*sin(d*x + c)^2 - 110*sin(d*x + c) + 69)/(a^2*(sin(d*x + c) - 1) ^2) - (125*sin(d*x + c)^4 + 580*sin(d*x + c)^3 + 1038*sin(d*x + c)^2 + 868 *sin(d*x + c) + 301)/(a^2*(sin(d*x + c) + 1)^4))/d
Time = 0.24 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^5(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {15\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{64\,a^2\,d}+\frac {-\frac {15\,{\sin \left (c+d\,x\right )}^5}{64}-\frac {15\,{\sin \left (c+d\,x\right )}^4}{32}+\frac {5\,{\sin \left (c+d\,x\right )}^3}{32}+\frac {25\,{\sin \left (c+d\,x\right )}^2}{32}+\frac {17\,\sin \left (c+d\,x\right )}{64}-\frac {1}{4}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^6+2\,a^2\,{\sin \left (c+d\,x\right )}^5-a^2\,{\sin \left (c+d\,x\right )}^4-4\,a^2\,{\sin \left (c+d\,x\right )}^3-a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )} \]
(15*atanh(sin(c + d*x)))/(64*a^2*d) + ((17*sin(c + d*x))/64 + (25*sin(c + d*x)^2)/32 + (5*sin(c + d*x)^3)/32 - (15*sin(c + d*x)^4)/32 - (15*sin(c + d*x)^5)/64 - 1/4)/(d*(2*a^2*sin(c + d*x) + a^2 - a^2*sin(c + d*x)^2 - 4*a^ 2*sin(c + d*x)^3 - a^2*sin(c + d*x)^4 + 2*a^2*sin(c + d*x)^5 + a^2*sin(c + d*x)^6))