Integrand size = 27, antiderivative size = 61 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \cot ^6(c+d x)}{6 d}-\frac {b \csc (c+d x)}{d}+\frac {2 b \csc ^3(c+d x)}{3 d}-\frac {b \csc ^5(c+d x)}{5 d} \]
Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \cot ^6(c+d x)}{6 d}-\frac {b \csc (c+d x)}{d}+\frac {2 b \csc ^3(c+d x)}{3 d}-\frac {b \csc ^5(c+d x)}{5 d} \]
-1/6*(a*Cot[c + d*x]^6)/d - (b*Csc[c + d*x])/d + (2*b*Csc[c + d*x]^3)/(3*d ) - (b*Csc[c + d*x]^5)/(5*d)
Time = 0.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3313, 3042, 25, 3086, 210, 2009, 3087, 15}
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.
\(\displaystyle \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a+b \sin (c+d x))}{\sin (c+d x)^7}dx\) |
\(\Big \downarrow \) 3313 |
\(\displaystyle a \int \cot ^5(c+d x) \csc ^2(c+d x)dx+b \int \cot ^5(c+d x) \csc (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int -\sec \left (c+d x-\frac {\pi }{2}\right )^2 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx+b \int -\sec \left (c+d x-\frac {\pi }{2}\right ) \tan \left (c+d x-\frac {\pi }{2}\right )^5dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-b \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right ) \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-\frac {b \int \left (\csc ^2(c+d x)-1\right )^2d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-\frac {b \int \left (\csc ^4(c+d x)-2 \csc ^2(c+d x)+1\right )d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^2 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-\frac {b \left (\frac {1}{5} \csc ^5(c+d x)-\frac {2}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle -\frac {a \int -\cot ^5(c+d x)d(-\cot (c+d x))}{d}-\frac {b \left (\frac {1}{5} \csc ^5(c+d x)-\frac {2}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 15 |
\(\displaystyle -\frac {a \cot ^6(c+d x)}{6 d}-\frac {b \left (\frac {1}{5} \csc ^5(c+d x)-\frac {2}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
3.13.9.3.1 Defintions of rubi rules used
Int[(a_.)*(x_)^(m_.), x_Symbol] :> Simp[a*(x^(m + 1)/(m + 1)), x] /; FreeQ[ {a, m}, x] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[Cos[e + f*x]^ p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[Cos[e + f*x]^p*(d*Sin[e + f*x ])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n, p}, x] && IntegerQ[(p - 1)/2 ] && IntegerQ[n] && ((LtQ[p, 0] && NeQ[a^2 - b^2, 0]) || LtQ[0, n, p - 1] | | LtQ[p + 1, -n, 2*p + 1])
Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15
method | result | size |
derivativedivides | \(-\frac {\frac {\left (\csc ^{6}\left (d x +c \right )\right ) a}{6}+\frac {b \left (\csc ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\csc ^{4}\left (d x +c \right )\right ) a}{2}-\frac {2 b \left (\csc ^{3}\left (d x +c \right )\right )}{3}+\frac {a \left (\csc ^{2}\left (d x +c \right )\right )}{2}+\csc \left (d x +c \right ) b}{d}\) | \(70\) |
default | \(-\frac {\frac {\left (\csc ^{6}\left (d x +c \right )\right ) a}{6}+\frac {b \left (\csc ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\csc ^{4}\left (d x +c \right )\right ) a}{2}-\frac {2 b \left (\csc ^{3}\left (d x +c \right )\right )}{3}+\frac {a \left (\csc ^{2}\left (d x +c \right )\right )}{2}+\csc \left (d x +c \right ) b}{d}\) | \(70\) |
risch | \(-\frac {2 i {\mathrm e}^{i \left (d x +c \right )} \left (15 i a \,{\mathrm e}^{9 i \left (d x +c \right )}+15 b \,{\mathrm e}^{10 i \left (d x +c \right )}-35 b \,{\mathrm e}^{8 i \left (d x +c \right )}+50 i a \,{\mathrm e}^{5 i \left (d x +c \right )}+78 b \,{\mathrm e}^{6 i \left (d x +c \right )}-78 b \,{\mathrm e}^{4 i \left (d x +c \right )}+15 i a \,{\mathrm e}^{i \left (d x +c \right )}+35 b \,{\mathrm e}^{2 i \left (d x +c \right )}-15 b \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}\) | \(148\) |
parallelrisch | \(\frac {-5 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -5 a \left (\cot ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-12 b \left (\cot ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+30 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +30 a \left (\cot ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+100 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+100 b \left (\cot ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-75 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -75 a \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-600 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-600 b \cot \left (\frac {d x}{2}+\frac {c}{2}\right )}{1920 d}\) | \(171\) |
norman | \(\frac {-\frac {a}{384 d}+\frac {5 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d}-\frac {3 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d}-\frac {3 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{128 d}+\frac {5 a \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d}-\frac {a \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{384 d}-\frac {b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{160 d}+\frac {11 b \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}-\frac {25 b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}-\frac {5 b \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d}-\frac {25 b \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{96 d}+\frac {11 b \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{240 d}-\frac {b \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{160 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(237\) |
-1/d*(1/6*csc(d*x+c)^6*a+1/5*b*csc(d*x+c)^5-1/2*csc(d*x+c)^4*a-2/3*b*csc(d *x+c)^3+1/2*a*csc(d*x+c)^2+csc(d*x+c)*b)
Time = 0.37 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.64 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\frac {15 \, a \cos \left (d x + c\right )^{4} - 15 \, a \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, b \cos \left (d x + c\right )^{4} - 20 \, b \cos \left (d x + c\right )^{2} + 8 \, b\right )} \sin \left (d x + c\right ) + 5 \, a}{30 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \]
1/30*(15*a*cos(d*x + c)^4 - 15*a*cos(d*x + c)^2 + 2*(15*b*cos(d*x + c)^4 - 20*b*cos(d*x + c)^2 + 8*b)*sin(d*x + c) + 5*a)/(d*cos(d*x + c)^6 - 3*d*co s(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)
Timed out. \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {30 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 20 \, b \sin \left (d x + c\right )^{3} - 15 \, a \sin \left (d x + c\right )^{2} + 6 \, b \sin \left (d x + c\right ) + 5 \, a}{30 \, d \sin \left (d x + c\right )^{6}} \]
-1/30*(30*b*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*b*sin(d*x + c)^3 - 1 5*a*sin(d*x + c)^2 + 6*b*sin(d*x + c) + 5*a)/(d*sin(d*x + c)^6)
Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {30 \, b \sin \left (d x + c\right )^{5} + 15 \, a \sin \left (d x + c\right )^{4} - 20 \, b \sin \left (d x + c\right )^{3} - 15 \, a \sin \left (d x + c\right )^{2} + 6 \, b \sin \left (d x + c\right ) + 5 \, a}{30 \, d \sin \left (d x + c\right )^{6}} \]
-1/30*(30*b*sin(d*x + c)^5 + 15*a*sin(d*x + c)^4 - 20*b*sin(d*x + c)^3 - 1 5*a*sin(d*x + c)^2 + 6*b*sin(d*x + c) + 5*a)/(d*sin(d*x + c)^6)
Time = 11.56 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13 \[ \int \cot ^5(c+d x) \csc ^2(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b\,{\sin \left (c+d\,x\right )}^5+\frac {a\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,b\,{\sin \left (c+d\,x\right )}^3}{3}-\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {b\,\sin \left (c+d\,x\right )}{5}+\frac {a}{6}}{d\,{\sin \left (c+d\,x\right )}^6} \]