Integrand size = 27, antiderivative size = 101 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 \csc (c+d x)}{a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {\csc ^3(c+d x)}{3 a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (1+\sin (c+d x))}{a^2 d}-\frac {1}{d \left (a^2+a^2 \sin (c+d x)\right )} \]
-3*csc(d*x+c)/a^2/d+csc(d*x+c)^2/a^2/d-1/3*csc(d*x+c)^3/a^2/d-4*ln(sin(d*x +c))/a^2/d+4*ln(1+sin(d*x+c))/a^2/d-1/d/(a^2+a^2*sin(d*x+c))
Time = 0.66 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.97 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {3 \csc (c+d x)}{a^2 d}+\frac {\csc ^2(c+d x)}{a^2 d}-\frac {\csc ^3(c+d x)}{3 a^2 d}-\frac {4 \log (\sin (c+d x))}{a^2 d}+\frac {4 \log (1+\sin (c+d x))}{a^2 d}-\frac {1}{a^2 d (1+\sin (c+d x))} \]
(-3*Csc[c + d*x])/(a^2*d) + Csc[c + d*x]^2/(a^2*d) - Csc[c + d*x]^3/(3*a^2 *d) - (4*Log[Sin[c + d*x]])/(a^2*d) + (4*Log[1 + Sin[c + d*x]])/(a^2*d) - 1/(a^2*d*(1 + Sin[c + d*x]))
Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3312, 27, 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.
\(\displaystyle \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a \sin (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)}{\sin (c+d x)^4 (a \sin (c+d x)+a)^2}dx\) |
\(\Big \downarrow \) 3312 |
\(\displaystyle \frac {\int \frac {\csc ^4(c+d x)}{(\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {\csc ^4(c+d x)}{a^4 (\sin (c+d x) a+a)^2}d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {a^3 \int \left (\frac {\csc ^4(c+d x)}{a^6}-\frac {2 \csc ^3(c+d x)}{a^6}+\frac {3 \csc ^2(c+d x)}{a^6}-\frac {4 \csc (c+d x)}{a^6}+\frac {4}{a^5 (\sin (c+d x) a+a)}+\frac {1}{a^4 (\sin (c+d x) a+a)^2}\right )d(a \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (-\frac {\csc ^3(c+d x)}{3 a^5}+\frac {\csc ^2(c+d x)}{a^5}-\frac {3 \csc (c+d x)}{a^5}-\frac {4 \log (a \sin (c+d x))}{a^5}+\frac {4 \log (a \sin (c+d x)+a)}{a^5}-\frac {1}{a^4 (a \sin (c+d x)+a)}\right )}{d}\) |
(a^3*((-3*Csc[c + d*x])/a^5 + Csc[c + d*x]^2/a^5 - Csc[c + d*x]^3/(3*a^5) - (4*Log[a*Sin[c + d*x]])/a^5 + (4*Log[a + a*Sin[c + d*x]])/a^5 - 1/(a^4*( a + a*Sin[c + d*x]))))/d
3.3.39.3.1 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[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_)]*((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 = 0.26 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.60
method | result | size |
derivativedivides | \(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\left (\csc ^{2}\left (d x +c \right )\right )+3 \csc \left (d x +c \right )-\frac {1}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\) | \(61\) |
default | \(-\frac {\frac {\left (\csc ^{3}\left (d x +c \right )\right )}{3}-\left (\csc ^{2}\left (d x +c \right )\right )+3 \csc \left (d x +c \right )-\frac {1}{\csc \left (d x +c \right )+1}-4 \ln \left (\csc \left (d x +c \right )+1\right )}{d \,a^{2}}\) | \(61\) |
risch | \(-\frac {8 i \left (3 i {\mathrm e}^{6 i \left (d x +c \right )}+3 \,{\mathrm e}^{7 i \left (d x +c \right )}-8 i {\mathrm e}^{4 i \left (d x +c \right )}-7 \,{\mathrm e}^{5 i \left (d x +c \right )}+3 i {\mathrm e}^{2 i \left (d x +c \right )}+7 \,{\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{2} d \,a^{2}}+\frac {8 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {4 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{2}}\) | \(160\) |
parallelrisch | \(\frac {192 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-96 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-28 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-28 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+114 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d \,a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}\) | \(162\) |
norman | \(\frac {-\frac {1}{24 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d a}+\frac {\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}+\frac {33 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}+\frac {33 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}\) | \(208\) |
Time = 0.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.93 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} - 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + 12 \, {\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 7}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d - {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d\right )} \sin \left (d x + c\right )\right )}} \]
1/3*(6*cos(d*x + c)^2 - 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(1/2*sin(d*x + c)) + 12*(cos(d*x + c)^4 - 2*cos(d*x + c)^2 - (cos(d*x + c)^2 - 1)*sin(d*x + c) + 1)*log(sin(d*x + c) + 1) + 2*(6*cos(d*x + c)^2 - 5)*sin(d*x + c) - 7)/(a^2*d*cos(d*x + c)^4 - 2*a^2*d*cos(d*x + c)^2 + a^2*d - (a^2*d*cos(d*x + c)^2 - a^2*d)*sin(d*x + c))
\[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos {\left (c + d x \right )} \csc ^{4}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.18 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.89 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, \sin \left (d x + c\right )^{3} + 6 \, \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}{a^{2} \sin \left (d x + c\right )^{4} + a^{2} \sin \left (d x + c\right )^{3}} - \frac {12 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {12 \, \log \left (\sin \left (d x + c\right )\right )}{a^{2}}}{3 \, d} \]
-1/3*((12*sin(d*x + c)^3 + 6*sin(d*x + c)^2 - 2*sin(d*x + c) + 1)/(a^2*sin (d*x + c)^4 + a^2*sin(d*x + c)^3) - 12*log(sin(d*x + c) + 1)/a^2 + 12*log( sin(d*x + c))/a^2)/d
Time = 0.35 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=-\frac {\frac {12 \, \log \left ({\left | -\frac {a}{a \sin \left (d x + c\right ) + a} + 1 \right |}\right )}{a^{2}} + \frac {3}{{\left (a \sin \left (d x + c\right ) + a\right )} a} + \frac {\frac {30 \, a}{a \sin \left (d x + c\right ) + a} - \frac {18 \, a^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}} - 13}{a^{2} {\left (\frac {a}{a \sin \left (d x + c\right ) + a} - 1\right )}^{3}}}{3 \, d} \]
-1/3*(12*log(abs(-a/(a*sin(d*x + c) + a) + 1))/a^2 + 3/((a*sin(d*x + c) + a)*a) + (30*a/(a*sin(d*x + c) + a) - 18*a^2/(a*sin(d*x + c) + a)^2 - 13)/( a^2*(a/(a*sin(d*x + c) + a) - 1)^3))/d
Time = 9.72 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.00 \[ \int \frac {\cot (c+d x) \csc ^3(c+d x)}{(a+a \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}-\frac {-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+24\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {28\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{3}-\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {1}{3}}{d\,\left (8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+16\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+8\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {4\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}+\frac {8\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{a^2\,d}-\frac {13\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,a^2\,d} \]
tan(c/2 + (d*x)/2)^2/(4*a^2*d) - tan(c/2 + (d*x)/2)^3/(24*a^2*d) - ((28*ta n(c/2 + (d*x)/2)^2)/3 - (4*tan(c/2 + (d*x)/2))/3 + 24*tan(c/2 + (d*x)/2)^3 - 3*tan(c/2 + (d*x)/2)^4 + 1/3)/(d*(8*a^2*tan(c/2 + (d*x)/2)^3 + 16*a^2*t an(c/2 + (d*x)/2)^4 + 8*a^2*tan(c/2 + (d*x)/2)^5)) - (4*log(tan(c/2 + (d*x )/2)))/(a^2*d) + (8*log(tan(c/2 + (d*x)/2) + 1))/(a^2*d) - (13*tan(c/2 + ( d*x)/2))/(8*a^2*d)