Integrand size = 19, antiderivative size = 60 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{4 a^2 d}-\frac {1}{4 d (a+a \sin (c+d x))^2}-\frac {1}{4 d \left (a^2+a^2 \sin (c+d x)\right )} \]
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.63 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))-\frac {2+\sin (c+d x)}{(1+\sin (c+d x))^2}}{4 a^2 d} \]
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x) (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a \int \frac {1}{(a-a \sin (c+d x)) (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a \int \left (\frac {1}{2 (\sin (c+d x) a+a)^3 a}+\frac {1}{4 \left (a^2-a^2 \sin ^2(c+d x)\right ) a^2}+\frac {1}{4 (\sin (c+d x) a+a)^2 a^2}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \left (\frac {\text {arctanh}(\sin (c+d x))}{4 a^3}-\frac {1}{4 a^2 (a \sin (c+d x)+a)}-\frac {1}{4 a (a \sin (c+d x)+a)^2}\right )}{d}\) |
(a*(ArcTanh[Sin[c + d*x]]/(4*a^3) - 1/(4*a*(a + a*Sin[c + d*x])^2) - 1/(4* a^2*(a + a*Sin[c + d*x]))))/d
3.1.70.3.1 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 = 0.49 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}}{d \,a^{2}}\) | \(55\) |
default | \(\frac {-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{8}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{8}}{d \,a^{2}}\) | \(55\) |
risch | \(-\frac {i \left (4 i {\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{2 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4}}-\frac {\ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{4 a^{2} d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 a^{2} d}\) | \(100\) |
norman | \(\frac {\frac {2 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d a}}{a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 a^{2} d}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 a^{2} d}\) | \(115\) |
parallelrisch | \(\frac {\left (3-\cos \left (2 d x +2 c \right )+4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \cos \left (2 d x +2 c \right )-6 \sin \left (d x +c \right )-2}{4 a^{2} d \left (-3+\cos \left (2 d x +2 c \right )-4 \sin \left (d x +c \right )\right )}\) | \(117\) |
1/d/a^2*(-1/8*ln(sin(d*x+c)-1)-1/4/(1+sin(d*x+c))^2-1/4/(1+sin(d*x+c))+1/8 *ln(1+sin(d*x+c)))
Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.75 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (\cos \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) - 2\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, \sin \left (d x + c\right ) + 4}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - 2 \, a^{2} d \sin \left (d x + c\right ) - 2 \, a^{2} d\right )}} \]
1/8*((cos(d*x + c)^2 - 2*sin(d*x + c) - 2)*log(sin(d*x + c) + 1) - (cos(d* x + c)^2 - 2*sin(d*x + c) - 2)*log(-sin(d*x + c) + 1) + 2*sin(d*x + c) + 4 )/(a^2*d*cos(d*x + c)^2 - 2*a^2*d*sin(d*x + c) - 2*a^2*d)
\[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.19 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.20 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (\sin \left (d x + c\right ) + 2\right )}}{a^{2} \sin \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) + a^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2}}}{8 \, d} \]
-1/8*(2*(sin(d*x + c) + 2)/(a^2*sin(d*x + c)^2 + 2*a^2*sin(d*x + c) + a^2) - log(sin(d*x + c) + 1)/a^2 + log(sin(d*x + c) - 1)/a^2)/d
Time = 0.30 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2}} - \frac {2 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2}} - \frac {3 \, \sin \left (d x + c\right )^{2} + 10 \, \sin \left (d x + c\right ) + 11}{a^{2} {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{16 \, d} \]
1/16*(2*log(abs(sin(d*x + c) + 1))/a^2 - 2*log(abs(sin(d*x + c) - 1))/a^2 - (3*sin(d*x + c)^2 + 10*sin(d*x + c) + 11)/(a^2*(sin(d*x + c) + 1)^2))/d
Time = 6.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \frac {\sec (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{4\,a^2\,d}-\frac {\frac {\sin \left (c+d\,x\right )}{4}+\frac {1}{2}}{d\,\left (a^2\,{\sin \left (c+d\,x\right )}^2+2\,a^2\,\sin \left (c+d\,x\right )+a^2\right )} \]