Integrand size = 21, antiderivative size = 126 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5 \text {arctanh}(\sin (c+d x))}{32 a^3 d}-\frac {a}{16 d (a+a \sin (c+d x))^4}-\frac {1}{12 d (a+a \sin (c+d x))^3}-\frac {3}{32 a d (a+a \sin (c+d x))^2}+\frac {1}{32 d \left (a^3-a^3 \sin (c+d x)\right )}-\frac {1}{8 d \left (a^3+a^3 \sin (c+d x)\right )} \]
5/32*arctanh(sin(d*x+c))/a^3/d-1/16*a/d/(a+a*sin(d*x+c))^4-1/12/d/(a+a*sin (d*x+c))^3-3/32/a/d/(a+a*sin(d*x+c))^2+1/32/d/(a^3-a^3*sin(d*x+c))-1/8/d/( a^3+a^3*sin(d*x+c))
Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.75 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\sec ^2(c+d x) \left (32+15 \sin (c+d x)-35 \sin ^2(c+d x)-45 \sin ^3(c+d x)-15 \sin ^4(c+d x)+15 \text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x)) (1+\sin (c+d x))^4\right )}{96 a^3 d (1+\sin (c+d x))^3} \]
-1/96*(Sec[c + d*x]^2*(32 + 15*Sin[c + d*x] - 35*Sin[c + d*x]^2 - 45*Sin[c + d*x]^3 - 15*Sin[c + d*x]^4 + 15*ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x ])*(1 + Sin[c + d*x])^4))/(a^3*d*(1 + Sin[c + d*x])^3)
Time = 0.31 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.94, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a \sin (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a^3 \int \frac {1}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^5}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^3 \int \left (\frac {1}{32 a^5 (a-a \sin (c+d x))^2}+\frac {1}{8 a^5 (\sin (c+d x) a+a)^2}+\frac {3}{16 a^4 (\sin (c+d x) a+a)^3}+\frac {1}{4 a^3 (\sin (c+d x) a+a)^4}+\frac {1}{4 a^2 (\sin (c+d x) a+a)^5}+\frac {5}{32 a^5 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \left (\frac {5 \text {arctanh}(\sin (c+d x))}{32 a^6}+\frac {1}{32 a^5 (a-a \sin (c+d x))}-\frac {1}{8 a^5 (a \sin (c+d x)+a)}-\frac {3}{32 a^4 (a \sin (c+d x)+a)^2}-\frac {1}{12 a^3 (a \sin (c+d x)+a)^3}-\frac {1}{16 a^2 (a \sin (c+d x)+a)^4}\right )}{d}\) |
(a^3*((5*ArcTanh[Sin[c + d*x]])/(32*a^6) + 1/(32*a^5*(a - a*Sin[c + d*x])) - 1/(16*a^2*(a + a*Sin[c + d*x])^4) - 1/(12*a^3*(a + a*Sin[c + d*x])^3) - 3/(32*a^4*(a + a*Sin[c + d*x])^2) - 1/(8*a^5*(a + a*Sin[c + d*x]))))/d
3.1.85.3.1 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 = 1.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.72
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{64}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{64}}{d \,a^{3}}\) | \(91\) |
| default | \(\frac {-\frac {1}{32 \left (\sin \left (d x +c \right )-1\right )}-\frac {5 \ln \left (\sin \left (d x +c \right )-1\right )}{64}-\frac {1}{16 \left (1+\sin \left (d x +c \right )\right )^{4}}-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {3}{32 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )}+\frac {5 \ln \left (1+\sin \left (d x +c \right )\right )}{64}}{d \,a^{3}}\) | \(91\) |
| risch | \(-\frac {i \left (90 i {\mathrm e}^{8 i \left (d x +c \right )}+15 \,{\mathrm e}^{9 i \left (d x +c \right )}-150 i {\mathrm e}^{6 i \left (d x +c \right )}-200 \,{\mathrm e}^{7 i \left (d x +c \right )}+150 i {\mathrm e}^{4 i \left (d x +c \right )}-142 \,{\mathrm e}^{5 i \left (d x +c \right )}-90 i {\mathrm e}^{2 i \left (d x +c \right )}-200 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{48 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{8} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{2} a^{3} d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{32 a^{3} d}-\frac {5 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{32 a^{3} d}\) | \(185\) |
| norman | \(\frac {\frac {27 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d a}+\frac {27 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{16 d a}+\frac {31 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {31 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {33 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}+\frac {33 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {9 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {9 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {89 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d a}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{8}}-\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32 a^{3} d}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32 a^{3} d}\) | \(242\) |
| parallelrisch | \(\frac {\left (-120 \cos \left (2 d x +2 c \right )+90 \cos \left (4 d x +4 c \right )-210 \sin \left (d x +c \right )-195 \sin \left (3 d x +3 c \right )+15 \sin \left (5 d x +5 c \right )-210\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (120 \cos \left (2 d x +2 c \right )-90 \cos \left (4 d x +4 c \right )+210 \sin \left (d x +c \right )+195 \sin \left (3 d x +3 c \right )-15 \sin \left (5 d x +5 c \right )+210\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-144 \cos \left (2 d x +2 c \right )-162 \cos \left (4 d x +4 c \right )+748 \sin \left (d x +c \right )+236 \sin \left (3 d x +3 c \right )-32 \sin \left (5 d x +5 c \right )+306}{96 a^{3} d \left (-6 \cos \left (4 d x +4 c \right )+14-\sin \left (5 d x +5 c \right )+14 \sin \left (d x +c \right )+13 \sin \left (3 d x +3 c \right )+8 \cos \left (2 d x +2 c \right )\right )}\) | \(253\) |
1/d/a^3*(-1/32/(sin(d*x+c)-1)-5/64*ln(sin(d*x+c)-1)-1/16/(1+sin(d*x+c))^4- 1/12/(1+sin(d*x+c))^3-3/32/(1+sin(d*x+c))^2-1/8/(1+sin(d*x+c))+5/64*ln(1+s in(d*x+c)))
Time = 0.30 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.81 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {30 \, \cos \left (d x + c\right )^{4} - 130 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, {\left (3 \, \cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{4} - 4 \, \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 30 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )} \sin \left (d x + c\right ) + 36}{192 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2} + {\left (a^{3} d \cos \left (d x + c\right )^{4} - 4 \, a^{3} d \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )}} \]
-1/192*(30*cos(d*x + c)^4 - 130*cos(d*x + c)^2 - 15*(3*cos(d*x + c)^4 - 4* cos(d*x + c)^2 + (cos(d*x + c)^4 - 4*cos(d*x + c)^2)*sin(d*x + c))*log(sin (d*x + c) + 1) + 15*(3*cos(d*x + c)^4 - 4*cos(d*x + c)^2 + (cos(d*x + c)^4 - 4*cos(d*x + c)^2)*sin(d*x + c))*log(-sin(d*x + c) + 1) - 30*(3*cos(d*x + c)^2 - 2)*sin(d*x + c) + 36)/(3*a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2 + (a^3*d*cos(d*x + c)^4 - 4*a^3*d*cos(d*x + c)^2)*sin(d*x + c))
\[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sin ^{3}{\left (c + d x \right )} + 3 \sin ^{2}{\left (c + d x \right )} + 3 \sin {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Time = 0.19 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.16 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{4} + 45 \, \sin \left (d x + c\right )^{3} + 35 \, \sin \left (d x + c\right )^{2} - 15 \, \sin \left (d x + c\right ) - 32\right )}}{a^{3} \sin \left (d x + c\right )^{5} + 3 \, a^{3} \sin \left (d x + c\right )^{4} + 2 \, a^{3} \sin \left (d x + c\right )^{3} - 2 \, a^{3} \sin \left (d x + c\right )^{2} - 3 \, a^{3} \sin \left (d x + c\right ) - a^{3}} - \frac {15 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3}} + \frac {15 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3}}}{192 \, d} \]
-1/192*(2*(15*sin(d*x + c)^4 + 45*sin(d*x + c)^3 + 35*sin(d*x + c)^2 - 15* sin(d*x + c) - 32)/(a^3*sin(d*x + c)^5 + 3*a^3*sin(d*x + c)^4 + 2*a^3*sin( d*x + c)^3 - 2*a^3*sin(d*x + c)^2 - 3*a^3*sin(d*x + c) - a^3) - 15*log(sin (d*x + c) + 1)/a^3 + 15*log(sin(d*x + c) - 1)/a^3)/d
Time = 0.48 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.92 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {\frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3}} + \frac {12 \, {\left (5 \, \sin \left (d x + c\right ) - 7\right )}}{a^{3} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {125 \, \sin \left (d x + c\right )^{4} + 596 \, \sin \left (d x + c\right )^{3} + 1110 \, \sin \left (d x + c\right )^{2} + 996 \, \sin \left (d x + c\right ) + 405}{a^{3} {\left (\sin \left (d x + c\right ) + 1\right )}^{4}}}{768 \, d} \]
1/768*(60*log(abs(sin(d*x + c) + 1))/a^3 - 60*log(abs(sin(d*x + c) - 1))/a ^3 + 12*(5*sin(d*x + c) - 7)/(a^3*(sin(d*x + c) - 1)) - (125*sin(d*x + c)^ 4 + 596*sin(d*x + c)^3 + 1110*sin(d*x + c)^2 + 996*sin(d*x + c) + 405)/(a^ 3*(sin(d*x + c) + 1)^4))/d
Time = 0.20 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^3} \, dx=\frac {5\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{32\,a^3\,d}+\frac {\frac {5\,{\sin \left (c+d\,x\right )}^4}{32}+\frac {15\,{\sin \left (c+d\,x\right )}^3}{32}+\frac {35\,{\sin \left (c+d\,x\right )}^2}{96}-\frac {5\,\sin \left (c+d\,x\right )}{32}-\frac {1}{3}}{d\,\left (-a^3\,{\sin \left (c+d\,x\right )}^5-3\,a^3\,{\sin \left (c+d\,x\right )}^4-2\,a^3\,{\sin \left (c+d\,x\right )}^3+2\,a^3\,{\sin \left (c+d\,x\right )}^2+3\,a^3\,\sin \left (c+d\,x\right )+a^3\right )} \]