Integrand size = 21, antiderivative size = 77 \[ \int \frac {\sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3 \text {arctanh}(\sin (c+d x))}{8 a d}+\frac {1}{8 d (a-a \sin (c+d x))}-\frac {a}{8 d (a+a \sin (c+d x))^2}-\frac {1}{4 d (a+a \sin (c+d x))} \]
3/8*arctanh(sin(d*x+c))/a/d+1/8/d/(a-a*sin(d*x+c))-1/8*a/d/(a+a*sin(d*x+c) )^2-1/4/d/(a+a*sin(d*x+c))
Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\sec ^2(c+d x) \left (2-3 \sin (c+d x)-3 \sin ^2(c+d x)+3 \text {arctanh}(\sin (c+d x)) (-1+\sin (c+d x)) (1+\sin (c+d x))^2\right )}{8 a d (1+\sin (c+d x))} \]
-1/8*(Sec[c + d*x]^2*(2 - 3*Sin[c + d*x] - 3*Sin[c + d*x]^2 + 3*ArcTanh[Si n[c + d*x]]*(-1 + Sin[c + d*x])*(1 + Sin[c + d*x])^2))/(a*d*(1 + Sin[c + d *x]))
Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (c+d x)^3 (a \sin (c+d x)+a)}dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {a^3 \int \frac {1}{(a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a^3 \int \left (\frac {1}{8 a^3 (a-a \sin (c+d x))^2}+\frac {1}{4 a^3 (\sin (c+d x) a+a)^2}+\frac {1}{4 a^2 (\sin (c+d x) a+a)^3}+\frac {3}{8 a^3 \left (a^2-a^2 \sin ^2(c+d x)\right )}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (\frac {3 \text {arctanh}(\sin (c+d x))}{8 a^4}+\frac {1}{8 a^3 (a-a \sin (c+d x))}-\frac {1}{4 a^3 (a \sin (c+d x)+a)}-\frac {1}{8 a^2 (a \sin (c+d x)+a)^2}\right )}{d}\) |
(a^3*((3*ArcTanh[Sin[c + d*x]])/(8*a^4) + 1/(8*a^3*(a - a*Sin[c + d*x])) - 1/(8*a^2*(a + a*Sin[c + d*x])^2) - 1/(4*a^3*(a + a*Sin[c + d*x]))))/d
3.1.59.3.1 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 = 0.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(\frac {-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{16}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{16}}{d a}\) | \(67\) |
default | \(\frac {-\frac {1}{8 \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 \left (1+\sin \left (d x +c \right )\right )}+\frac {3 \ln \left (1+\sin \left (d x +c \right )\right )}{16}-\frac {1}{8 \left (\sin \left (d x +c \right )-1\right )}-\frac {3 \ln \left (\sin \left (d x +c \right )-1\right )}{16}}{d a}\) | \(67\) |
risch | \(-\frac {i \left (6 i {\mathrm e}^{4 i \left (d x +c \right )}+3 \,{\mathrm e}^{5 i \left (d x +c \right )}-6 i {\mathrm e}^{2 i \left (d x +c \right )}+2 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{4 \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{4} \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d a}-\frac {3 \ln \left (-i+{\mathrm e}^{i \left (d x +c \right )}\right )}{8 a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 a d}\) | \(139\) |
parallelrisch | \(\frac {\left (-6 \cos \left (2 d x +2 c \right )-3 \sin \left (d x +c \right )-3 \sin \left (3 d x +3 c \right )-6\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (6 \cos \left (2 d x +2 c \right )+3 \sin \left (d x +c \right )+3 \sin \left (3 d x +3 c \right )+6\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \cos \left (2 d x +2 c \right )+14 \sin \left (d x +c \right )+2 \sin \left (3 d x +3 c \right )+2}{8 a d \left (\sin \left (3 d x +3 c \right )+\sin \left (d x +c \right )+2 \cos \left (2 d x +2 c \right )+2\right )}\) | \(161\) |
norman | \(\frac {\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 a d}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 a d}\) | \(163\) |
1/d/a*(-1/8/(1+sin(d*x+c))^2-1/4/(1+sin(d*x+c))+3/16*ln(1+sin(d*x+c))-1/8/ (sin(d*x+c)-1)-3/16*ln(sin(d*x+c)-1))
Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.62 \[ \int \frac {\sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {6 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 6 \, \sin \left (d x + c\right ) - 2}{16 \, {\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \]
-1/16*(6*cos(d*x + c)^2 - 3*(cos(d*x + c)^2*sin(d*x + c) + cos(d*x + c)^2) *log(sin(d*x + c) + 1) + 3*(cos(d*x + c)^2*sin(d*x + c) + cos(d*x + c)^2)* log(-sin(d*x + c) + 1) - 6*sin(d*x + c) - 2)/(a*d*cos(d*x + c)^2*sin(d*x + c) + a*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
Time = 0.18 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 2\right )}}{a \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} - \frac {3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} + \frac {3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \]
-1/16*(2*(3*sin(d*x + c)^2 + 3*sin(d*x + c) - 2)/(a*sin(d*x + c)^3 + a*sin (d*x + c)^2 - a*sin(d*x + c) - a) - 3*log(sin(d*x + c) + 1)/a + 3*log(sin( d*x + c) - 1)/a)/d
Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.25 \[ \int \frac {\sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac {6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac {2 \, {\left (3 \, \sin \left (d x + c\right ) - 5\right )}}{a {\left (\sin \left (d x + c\right ) - 1\right )}} - \frac {9 \, \sin \left (d x + c\right )^{2} + 26 \, \sin \left (d x + c\right ) + 21}{a {\left (\sin \left (d x + c\right ) + 1\right )}^{2}}}{32 \, d} \]
1/32*(6*log(abs(sin(d*x + c) + 1))/a - 6*log(abs(sin(d*x + c) - 1))/a + 2* (3*sin(d*x + c) - 5)/(a*(sin(d*x + c) - 1)) - (9*sin(d*x + c)^2 + 26*sin(d *x + c) + 21)/(a*(sin(d*x + c) + 1)^2))/d
Time = 6.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.96 \[ \int \frac {\sec ^3(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {3\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{8\,a\,d}+\frac {\frac {3\,{\sin \left (c+d\,x\right )}^2}{8}+\frac {3\,\sin \left (c+d\,x\right )}{8}-\frac {1}{4}}{d\,\left (-a\,{\sin \left (c+d\,x\right )}^3-a\,{\sin \left (c+d\,x\right )}^2+a\,\sin \left (c+d\,x\right )+a\right )} \]