Integrand size = 27, antiderivative size = 36 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 x}{a^2}-\frac {\text {arctanh}(\cos (c+d x))}{a^2 d}-\frac {\cos (c+d x)}{a^2 d} \]
Time = 0.63 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {2 c+2 d x+\cos (c+d x)+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 d} \]
Time = 0.43 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 3348, 3042, 3225, 3042, 3214, 3042, 4257}
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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x) (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3348 |
\(\displaystyle \frac {\int \csc (c+d x) (a-a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^2}{\sin (c+d x)}dx}{a^4}\) |
\(\Big \downarrow \) 3225 |
\(\displaystyle \frac {\int \csc (c+d x) \left (a^2-2 a^2 \sin (c+d x)\right )dx-\frac {a^2 \cos (c+d x)}{d}}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a^2-2 a^2 \sin (c+d x)}{\sin (c+d x)}dx-\frac {a^2 \cos (c+d x)}{d}}{a^4}\) |
\(\Big \downarrow \) 3214 |
\(\displaystyle \frac {a^2 \int \csc (c+d x)dx-\frac {a^2 \cos (c+d x)}{d}-2 a^2 x}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a^2 \int \csc (c+d x)dx-\frac {a^2 \cos (c+d x)}{d}-2 a^2 x}{a^4}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {-\frac {a^2 \text {arctanh}(\cos (c+d x))}{d}-\frac {a^2 \cos (c+d x)}{d}-2 a^2 x}{a^4}\) |
3.5.25.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f _.)*(x_)]), x_Symbol] :> Simp[(-b^2)*(Cos[e + f*x]/(d*f)), x] + Simp[1/d Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a^(2*m) Int[(d* Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \(\frac {-2 d x +1+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\cos \left (d x +c \right )}{d \,a^{2}}\) | \(32\) |
derivativedivides | \(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(48\) |
default | \(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(48\) |
risch | \(-\frac {2 x}{a^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,a^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d \,a^{2}}\) | \(81\) |
norman | \(\frac {-\frac {2}{a d}-\frac {2 x}{a}-\frac {6 x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {12 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {20 x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {24 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {24 x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {20 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {12 x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {6 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {2 x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {10 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {2 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {6 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {14 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}\) | \(348\) |
Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {4 \, d x + 2 \, \cos \left (d x + c\right ) + \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right )}{2 \, a^{2} d} \]
-1/2*(4*d*x + 2*cos(d*x + c) + log(1/2*cos(d*x + c) + 1/2) - log(-1/2*cos( d*x + c) + 1/2))/(a^2*d)
\[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos ^{4}{\left (c + d x \right )} \csc {\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (36) = 72\).
Time = 0.33 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.28 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2}{a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {4 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{d} \]
-(2/(a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) + 4*arctan(sin(d*x + c )/(cos(d*x + c) + 1))/a^2 - log(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d
Time = 0.31 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.44 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (d x + c\right )}}{a^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} + \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}}}{d} \]
-(2*(d*x + c)/a^2 - log(abs(tan(1/2*d*x + 1/2*c)))/a^2 + 2/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2))/d
Time = 10.27 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.69 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {4\,\mathrm {atan}\left (\frac {16}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8}-\frac {8\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{16\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+8}\right )}{a^2\,d}-\frac {2}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d} \]