Integrand size = 21, antiderivative size = 104 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{4 a^2 d}-\frac {a}{12 d (a+a \sin (c+d x))^3}-\frac {1}{8 d (a+a \sin (c+d x))^2}+\frac {1}{16 d \left (a^2-a^2 \sin (c+d x)\right )}-\frac {3}{16 d \left (a^2+a^2 \sin (c+d x)\right )} \]
1/4*arctanh(sin(d*x+c))/a^2/d-1/12*a/d/(a+a*sin(d*x+c))^3-1/8/d/(a+a*sin(d *x+c))^2+1/16/d/(a^2-a^2*sin(d*x+c))-3/16/d/(a^2+a^2*sin(d*x+c))
Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.82 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\sec ^2(c+d x) \left (4-\sin (c+d x)-6 \sin ^2(c+d x)-3 \sin ^3(c+d x)+3 \text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x)) (1+\sin (c+d x))^3\right )}{12 a^2 d (1+\sin (c+d x))^2} \]
-1/12*(Sec[c + d*x]^2*(4 - Sin[c + d*x] - 6*Sin[c + d*x]^2 - 3*Sin[c + d*x ]^3 + 3*ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x])*(1 + Sin[c + d*x])^3))/( a^2*d*(1 + Sin[c + d*x])^2)
Time = 0.29 (sec) , antiderivative size = 99, 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.
| \(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a \sin (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a^3 \int \frac {1}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^4}d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {a^3 \int \left (\frac {1}{16 a^4 (a-a \sin (c+d x))^2}+\frac {3}{16 a^4 (\sin (c+d x) a+a)^2}+\frac {1}{4 a^3 (\sin (c+d x) a+a)^3}+\frac {1}{4 a^2 (\sin (c+d x) a+a)^4}+\frac {1}{4 a^4 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \left (\frac {\text {arctanh}(\sin (c+d x))}{4 a^5}+\frac {1}{16 a^4 (a-a \sin (c+d x))}-\frac {3}{16 a^4 (a \sin (c+d x)+a)}-\frac {1}{8 a^3 (a \sin (c+d x)+a)^2}-\frac {1}{12 a^2 (a \sin (c+d x)+a)^3}\right )}{d}\) |
(a^3*(ArcTanh[Sin[c + d*x]]/(4*a^5) + 1/(16*a^4*(a - a*Sin[c + d*x])) - 1/ (12*a^2*(a + a*Sin[c + d*x])^3) - 1/(8*a^3*(a + a*Sin[c + d*x])^2) - 3/(16 *a^4*(a + a*Sin[c + d*x]))))/d
3.1.72.3.1 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 = 0.87 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.76
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}-\frac {1}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}}{d \,a^{2}}\) | \(79\) |
| default | \(\frac {-\frac {1}{12 \left (1+\sin \left (d x +c \right )\right )^{3}}-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}-\frac {1}{16 \left (\sin \left (d x +c \right )-1\right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}}{d \,a^{2}}\) | \(79\) |
| risch | \(-\frac {i \left (12 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}+8 i {\mathrm e}^{4 i \left (d x +c \right )}-13 \,{\mathrm e}^{5 i \left (d x +c \right )}+12 i {\mathrm e}^{2 i \left (d x +c \right )}+13 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{6 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{6} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{2} a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 a^{2} d}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{4 a^{2} d}\) | \(162\) |
| norman | \(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {3 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}-\frac {4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d a}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}+\frac {7 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 a^{2} d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 a^{2} d}\) | \(204\) |
| parallelrisch | \(\frac {\left (-12 \cos \left (2 d x +2 c \right )+3 \cos \left (4 d x +4 c \right )-12 \sin \left (d x +c \right )-12 \sin \left (3 d x +3 c \right )-15\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-3 \cos \left (4 d x +4 c \right )+15+12 \sin \left (3 d x +3 c \right )+12 \sin \left (d x +c \right )+12 \cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-8 \cos \left (2 d x +2 c \right )-4 \cos \left (4 d x +4 c \right )+42 \sin \left (d x +c \right )+10 \sin \left (3 d x +3 c \right )+12}{12 a^{2} d \left (-\cos \left (4 d x +4 c \right )+5+4 \sin \left (3 d x +3 c \right )+4 \sin \left (d x +c \right )+4 \cos \left (2 d x +2 c \right )\right )}\) | \(209\) |
1/d/a^2*(-1/12/(1+sin(d*x+c))^3-1/8/(1+sin(d*x+c))^2-3/16/(1+sin(d*x+c))+1 /8*ln(1+sin(d*x+c))-1/16/(sin(d*x+c)-1)-1/8*ln(sin(d*x+c)-1))
Time = 0.30 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.71 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {12 \, \cos \left (d x + c\right )^{2} + 3 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 4}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )^{2}\right )}} \]
1/24*(12*cos(d*x + c)^2 + 3*(cos(d*x + c)^4 - 2*cos(d*x + c)^2*sin(d*x + c ) - 2*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*(cos(d*x + c)^4 - 2*cos(d* x + c)^2*sin(d*x + c) - 2*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(3*co s(d*x + c)^2 - 4)*sin(d*x + c) - 4)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d* x + c)^2*sin(d*x + c) - 2*a^2*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 4\right )}}{a^{2} \sin \left (d x + c\right )^{4} + 2 \, a^{2} \sin \left (d x + c\right )^{3} - 2 \, a^{2} \sin \left (d x + c\right ) - a^{2}} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{24 \, d} \]
-1/24*(2*(3*sin(d*x + c)^3 + 6*sin(d*x + c)^2 + sin(d*x + c) - 4)/(a^2*sin (d*x + c)^4 + 2*a^2*sin(d*x + c)^3 - 2*a^2*sin(d*x + c) - a^2) - 3*log(sin (d*x + c) + 1)/a^2 + 3*log(sin(d*x + c) - 1)/a^2)/d
Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} + \frac {3 \, {\left (2 \, \sin \left (d x + c\right ) - 3\right )}}{a^{2} {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {11 \, \sin \left (d x + c\right )^{3} + 42 \, \sin \left (d x + c\right )^{2} + 57 \, \sin \left (d x + c\right ) + 30}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{3}}}{48 \, d} \]
1/48*(6*log(abs(sin(d*x + c) + 1))/a^2 - 6*log(abs(sin(d*x + c) - 1))/a^2 + 3*(2*sin(d*x + c) - 3)/(a^2*(sin(d*x + c) - 1)) - (11*sin(d*x + c)^3 + 4 2*sin(d*x + c)^2 + 57*sin(d*x + c) + 30)/(a^2*(sin(d*x + c) + 1)^3))/d
Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^3}{4}+\frac {{\sin \left (c+d\,x\right )}^2}{2}+\frac {\sin \left (c+d\,x\right )}{12}-\frac {1}{3}}{d\,\left (-a^2\,{\sin \left (c+d\,x\right )}^4-2\,a^2\,{\sin \left (c+d\,x\right )}^3+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )}+\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{4\,a^2\,d} \]