Integrand size = 25, antiderivative size = 62 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \]
Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {a^3 \csc (c+d x)}{d}+\frac {3 a^3 \log (\sin (c+d x))}{d}+\frac {3 a^3 \sin (c+d x)}{d}+\frac {a^3 \sin ^2(c+d x)}{2 d} \]
-((a^3*Csc[c + d*x])/d) + (3*a^3*Log[Sin[c + d*x]])/d + (3*a^3*Sin[c + d*x ])/d + (a^3*Sin[c + d*x]^2)/(2*d)
Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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.
\(\displaystyle \int \cot (c+d x) \csc (c+d x) (a \sin (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x) (a \sin (c+d x)+a)^3}{\sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \csc ^2(c+d x) (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a \int \frac {\csc ^2(c+d x) (\sin (c+d x) a+a)^3}{a^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {a \int \left (a \csc ^2(c+d x)+3 a \csc (c+d x)+3 a+a \sin (c+d x)\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \left (\frac {1}{2} a^2 \sin ^2(c+d x)+3 a^2 \sin (c+d x)-a^2 \csc (c+d x)+3 a^2 \log (a \sin (c+d x))\right )}{d}\) |
(a*(-(a^2*Csc[c + d*x]) + 3*a^2*Log[a*Sin[c + d*x]] + 3*a^2*Sin[c + d*x] + (a^2*Sin[c + d*x]^2)/2))/d
3.3.10.3.1 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 = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.73
method | result | size |
derivativedivides | \(-\frac {a^{3} \left (\csc \left (d x +c \right )+3 \ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\) | \(45\) |
default | \(-\frac {a^{3} \left (\csc \left (d x +c \right )+3 \ln \left (\csc \left (d x +c \right )\right )-\frac {3}{\csc \left (d x +c \right )}-\frac {1}{2 \csc \left (d x +c \right )^{2}}\right )}{d}\) | \(45\) |
parallelrisch | \(-\frac {a^{3} \left (12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right ) \cos \left (d x +c \right )+2 \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\cos \left (2 d x +2 c \right )-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+12 \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}\) | \(94\) |
risch | \(-3 i a^{3} x -\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {3 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {3 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {6 i a^{3} c}{d}-\frac {2 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {3 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d}\) | \(140\) |
norman | \(\frac {\frac {2 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3}}{2 d}+\frac {4 a^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {9 a^{3} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 a^{3} \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {3 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 a^{3} \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}\) | \(192\) |
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.26 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=-\frac {12 \, a^{3} \cos \left (d x + c\right )^{2} - 12 \, a^{3} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) \sin \left (d x + c\right ) - 8 \, a^{3} + {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} - a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \sin \left (d x + c\right )} \]
-1/4*(12*a^3*cos(d*x + c)^2 - 12*a^3*log(1/2*sin(d*x + c))*sin(d*x + c) - 8*a^3 + (2*a^3*cos(d*x + c)^2 - a^3)*sin(d*x + c))/(d*sin(d*x + c))
\[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=a^{3} \left (\int \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin {\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )} \csc ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(cos(c + d*x)*csc(c + d*x)**2, x) + Integral(3*sin(c + d*x)* cos(c + d*x)*csc(c + d*x)**2, x) + Integral(3*sin(c + d*x)**2*cos(c + d*x) *csc(c + d*x)**2, x) + Integral(sin(c + d*x)**3*cos(c + d*x)*csc(c + d*x)* *2, x))
Time = 0.19 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left (\sin \left (d x + c\right )\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, a^{3}}{\sin \left (d x + c\right )}}{2 \, d} \]
1/2*(a^3*sin(d*x + c)^2 + 6*a^3*log(sin(d*x + c)) + 6*a^3*sin(d*x + c) - 2 *a^3/sin(d*x + c))/d
Time = 0.34 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.89 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {a^{3} \sin \left (d x + c\right )^{2} + 6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right ) + 6 \, a^{3} \sin \left (d x + c\right ) - \frac {2 \, a^{3}}{\sin \left (d x + c\right )}}{2 \, d} \]
1/2*(a^3*sin(d*x + c)^2 + 6*a^3*log(abs(sin(d*x + c))) + 6*a^3*sin(d*x + c ) - 2*a^3/sin(d*x + c))/d
Time = 9.34 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.52 \[ \int \cot (c+d x) \csc (c+d x) (a+a \sin (c+d x))^3 \, dx=\frac {3\,a^3\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {11\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^3}{d\,\left (2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d}-\frac {3\,a^3\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]