Integrand size = 27, antiderivative size = 65 \[ \int \cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {3 a^3 \csc (c+d x)}{d}-\frac {3 a^3 \csc ^2(c+d x)}{2 d}-\frac {a^3 \csc ^3(c+d x)}{3 d}+\frac {a^3 \log (\sin (c+d x))}{d} \]
Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int \cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx=a^3 \left (-\frac {3 \csc (c+d x)}{d}-\frac {3 \csc ^2(c+d x)}{2 d}-\frac {\csc ^3(c+d x)}{3 d}+\frac {\log (\sin (c+d x))}{d}\right ) \]
a^3*((-3*Csc[c + d*x])/d - (3*Csc[c + d*x]^2)/(2*d) - Csc[c + d*x]^3/(3*d) + Log[Sin[c + d*x]]/d)
Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 49, 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.
\(\displaystyle \int \cot (c+d x) \csc ^3(c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x) (a \sin (c+d x)+a)^3}{\sin (c+d x)^4}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \csc ^4(c+d x) (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {\csc ^4(c+d x) (\sin (c+d x) a+a)^3}{a^4}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a^3 \int \left (\frac {\csc ^4(c+d x)}{a}+\frac {3 \csc ^3(c+d x)}{a}+\frac {3 \csc ^2(c+d x)}{a}+\frac {\csc (c+d x)}{a}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (\log (a \sin (c+d x))-\frac {1}{3} \csc ^3(c+d x)-\frac {3}{2} \csc ^2(c+d x)-3 \csc (c+d x)\right )}{d}\) |
3.3.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*(( c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b*f) Su bst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.69
method | result | size |
derivativedivides | \(-\frac {a^{3} \left (\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\csc ^{2}\left (d x +c \right )\right )}{2}+3 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(45\) |
default | \(-\frac {a^{3} \left (\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\csc ^{2}\left (d x +c \right )\right )}{2}+3 \csc \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )\right )\right )}{d}\) | \(45\) |
parallelrisch | \(-\frac {a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )+\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )+9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+24 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+39 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+39 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}\) | \(106\) |
risch | \(-i a^{3} x -\frac {2 i a^{3} c}{d}-\frac {2 i a^{3} \left (9 \,{\mathrm e}^{5 i \left (d x +c \right )}-22 \,{\mathrm e}^{3 i \left (d x +c \right )}+9 i {\mathrm e}^{4 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}-9 i {\mathrm e}^{2 i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(118\) |
norman | \(\frac {-\frac {a^{3}}{24 d}-\frac {3 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {7 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {53 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {59 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {53 a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {7 a^{3} \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {3 a^{3} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {a^{3} \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d}+\frac {15 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}+\frac {15 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(265\) |
Time = 0.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.40 \[ \int \cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {18 \, a^{3} \cos \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) - 20 \, a^{3} - 6 \, {\left (a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right )}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]
-1/6*(18*a^3*cos(d*x + c)^2 - 9*a^3*sin(d*x + c) - 20*a^3 - 6*(a^3*cos(d*x + c)^2 - a^3)*log(1/2*sin(d*x + c))*sin(d*x + c))/((d*cos(d*x + c)^2 - d) *sin(d*x + c))
Timed out. \[ \int \cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) - \frac {18 \, a^{3} \sin \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3}}{\sin \left (d x + c\right )^{3}}}{6 \, d} \]
1/6*(6*a^3*log(sin(d*x + c)) - (18*a^3*sin(d*x + c)^2 + 9*a^3*sin(d*x + c) + 2*a^3)/sin(d*x + c)^3)/d
Time = 0.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.91 \[ \int \cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) - \frac {18 \, a^{3} \sin \left (d x + c\right )^{2} + 9 \, a^{3} \sin \left (d x + c\right ) + 2 \, a^{3}}{\sin \left (d x + c\right )^{3}}}{6 \, d} \]
1/6*(6*a^3*log(abs(sin(d*x + c))) - (18*a^3*sin(d*x + c)^2 + 9*a^3*sin(d*x + c) + 2*a^3)/sin(d*x + c)^3)/d
Time = 9.38 (sec) , antiderivative size = 147, normalized size of antiderivative = 2.26 \[ \int \cot (c+d x) \csc ^3(c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,d}-\frac {a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (13\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3}{3}\right )}{8\,d}-\frac {13\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d}-\frac {a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]