Integrand size = 21, antiderivative size = 84 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b \csc (c+d x)}{a^2 d}-\frac {\csc ^2(c+d x)}{2 a d}-\frac {\left (a^2-b^2\right ) \log (\sin (c+d x))}{a^3 d}+\frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^3 d} \]
b*csc(d*x+c)/a^2/d-1/2*csc(d*x+c)^2/a/d-(a^2-b^2)*ln(sin(d*x+c))/a^3/d+(a^ 2-b^2)*ln(a+b*sin(d*x+c))/a^3/d
Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {-2 a b \csc (c+d x)+a^2 \csc ^2(c+d x)+2 \left (a^2-b^2\right ) (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))}{2 a^3 d} \]
-1/2*(-2*a*b*Csc[c + d*x] + a^2*Csc[c + d*x]^2 + 2*(a^2 - b^2)*(Log[Sin[c + d*x]] - Log[a + b*Sin[c + d*x]]))/(a^3*d)
Time = 0.26 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3200, 522, 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.
\(\displaystyle \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^3 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{b^3 (a+b \sin (c+d x))}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\frac {\csc ^3(c+d x)}{a b}-\frac {\csc ^2(c+d x)}{a^2}+\frac {\left (b^2-a^2\right ) \csc (c+d x)}{a^3 b}+\frac {a^2-b^2}{a^3 (a+b \sin (c+d x))}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b \csc (c+d x)}{a^2}-\frac {\left (a^2-b^2\right ) \log (b \sin (c+d x))}{a^3}+\frac {\left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^3}-\frac {\csc ^2(c+d x)}{2 a}}{d}\) |
((b*Csc[c + d*x])/a^2 - Csc[c + d*x]^2/(2*a) - ((a^2 - b^2)*Log[b*Sin[c + d*x]])/a^3 + ((a^2 - b^2)*Log[a + b*Sin[c + d*x]])/a^3)/d
3.13.100.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.40 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(\frac {-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(76\) |
default | \(\frac {-\frac {1}{2 a \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}+\frac {b}{a^{2} \sin \left (d x +c \right )}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(76\) |
norman | \(\frac {-\frac {1}{8 a d}-\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2} d}+\frac {b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{a^{3} d}-\frac {\left (a^{2}-b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\) | \(151\) |
parallelrisch | \(\frac {-a^{2} \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 a^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-8 \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) b^{2}-8 a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+4 a b \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+4 a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a^{3} d}\) | \(162\) |
risch | \(\frac {2 i \left (-i a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{3 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) b^{2}}{a^{3} d}\) | \(177\) |
1/d*(-1/2/a/sin(d*x+c)^2+1/a^3*(-a^2+b^2)*ln(sin(d*x+c))+1/a^2*b/sin(d*x+c )+(a^2-b^2)/a^3*ln(a+b*sin(d*x+c)))
Time = 0.39 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.40 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 \, a b \sin \left (d x + c\right ) - a^{2} - 2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} + b^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right )}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - a^{3} d\right )}} \]
-1/2*(2*a*b*sin(d*x + c) - a^2 - 2*((a^2 - b^2)*cos(d*x + c)^2 - a^2 + b^2 )*log(b*sin(d*x + c) + a) + 2*((a^2 - b^2)*cos(d*x + c)^2 - a^2 + b^2)*log (-1/2*sin(d*x + c)))/(a^3*d*cos(d*x + c)^2 - a^3*d)
Timed out. \[ \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{3}} - \frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{3}} + \frac {2 \, b \sin \left (d x + c\right ) - a}{a^{2} \sin \left (d x + c\right )^{2}}}{2 \, d} \]
1/2*(2*(a^2 - b^2)*log(b*sin(d*x + c) + a)/a^3 - 2*(a^2 - b^2)*log(sin(d*x + c))/a^3 + (2*b*sin(d*x + c) - a)/(a^2*sin(d*x + c)^2))/d
Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.05 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {2 \, {\left (a^{2} - b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3}} - \frac {2 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{3} b} - \frac {2 \, a b \sin \left (d x + c\right ) - a^{2}}{a^{3} \sin \left (d x + c\right )^{2}}}{2 \, d} \]
-1/2*(2*(a^2 - b^2)*log(abs(sin(d*x + c)))/a^3 - 2*(a^2*b - b^3)*log(abs(b *sin(d*x + c) + a))/(a^3*b) - (2*a*b*sin(d*x + c) - a^2)/(a^3*sin(d*x + c) ^2))/d
Time = 11.54 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.71 \[ \int \frac {\cot ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-b^2\right )}{a^3\,d}-\frac {\frac {a}{2}-2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,a^2\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^2-b^2\right )}{a^3\,d} \]