Integrand size = 27, antiderivative size = 81 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {a \cot ^6(c+d x)}{6 d}-\frac {a \cot ^8(c+d x)}{8 d}-\frac {b \csc ^3(c+d x)}{3 d}+\frac {2 b \csc ^5(c+d x)}{5 d}-\frac {b \csc ^7(c+d x)}{7 d} \]
-1/6*a*cot(d*x+c)^6/d-1/8*a*cot(d*x+c)^8/d-1/3*b*csc(d*x+c)^3/d+2/5*b*csc( d*x+c)^5/d-1/7*b*csc(d*x+c)^7/d
Time = 0.02 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.20 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \csc ^3(c+d x)}{3 d}-\frac {a \csc ^4(c+d x)}{4 d}+\frac {2 b \csc ^5(c+d x)}{5 d}+\frac {a \csc ^6(c+d x)}{3 d}-\frac {b \csc ^7(c+d x)}{7 d}-\frac {a \csc ^8(c+d x)}{8 d} \]
-1/3*(b*Csc[c + d*x]^3)/d - (a*Csc[c + d*x]^4)/(4*d) + (2*b*Csc[c + d*x]^5 )/(5*d) + (a*Csc[c + d*x]^6)/(3*d) - (b*Csc[c + d*x]^7)/(7*d) - (a*Csc[c + d*x]^8)/(8*d)
Time = 0.43 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3313, 3042, 25, 3086, 244, 2009, 3087, 244, 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.
\(\displaystyle \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (a+b \sin (c+d x))}{\sin (c+d x)^9}dx\) |
\(\Big \downarrow \) 3313 |
\(\displaystyle a \int \cot ^5(c+d x) \csc ^4(c+d x)dx+b \int \cot ^5(c+d x) \csc ^3(c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int -\sec \left (c+d x-\frac {\pi }{2}\right )^4 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx+b \int -\sec \left (c+d x-\frac {\pi }{2}\right )^3 \tan \left (c+d x-\frac {\pi }{2}\right )^5dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-b \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^3 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-\frac {b \int \csc ^2(c+d x) \left (1-\csc ^2(c+d x)\right )^2d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-\frac {b \int \left (\csc ^6(c+d x)-2 \csc ^4(c+d x)+\csc ^2(c+d x)\right )d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -a \int \sec \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \tan \left (\frac {1}{2} (2 c-\pi )+d x\right )^5dx-\frac {b \left (\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle -\frac {a \int -\cot ^5(c+d x) \left (\cot ^2(c+d x)+1\right )d(-\cot (c+d x))}{d}-\frac {b \left (\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {a \int \left (-\cot ^7(c+d x)-\cot ^5(c+d x)\right )d(-\cot (c+d x))}{d}-\frac {b \left (\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \left (\frac {1}{8} \cot ^8(c+d x)+\frac {1}{6} \cot ^6(c+d x)\right )}{d}-\frac {b \left (\frac {1}{7} \csc ^7(c+d x)-\frac {2}{5} \csc ^5(c+d x)+\frac {1}{3} \csc ^3(c+d x)\right )}{d}\) |
-((a*(Cot[c + d*x]^6/6 + Cot[c + d*x]^8/8))/d) - (b*(Csc[c + d*x]^3/3 - (2 *Csc[c + d*x]^5)/5 + Csc[c + d*x]^7/7))/d
3.13.11.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, 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 = 0.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90
method | result | size |
derivativedivides | \(-\frac {\frac {\left (\csc ^{8}\left (d x +c \right )\right ) a}{8}+\frac {b \left (\csc ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\csc ^{6}\left (d x +c \right )\right ) a}{3}-\frac {2 b \left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right ) a}{4}+\frac {b \left (\csc ^{3}\left (d x +c \right )\right )}{3}}{d}\) | \(73\) |
default | \(-\frac {\frac {\left (\csc ^{8}\left (d x +c \right )\right ) a}{8}+\frac {b \left (\csc ^{7}\left (d x +c \right )\right )}{7}-\frac {\left (\csc ^{6}\left (d x +c \right )\right ) a}{3}-\frac {2 b \left (\csc ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\csc ^{4}\left (d x +c \right )\right ) a}{4}+\frac {b \left (\csc ^{3}\left (d x +c \right )\right )}{3}}{d}\) | \(73\) |
parallelrisch | \(-\frac {853 \left (a \sec \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\cos \left (2 d x +2 c \right )+\frac {2561 \cos \left (4 d x +4 c \right )}{5118}+\frac {73 \cos \left (6 d x +6 c \right )}{2559}-\frac {73 \cos \left (8 d x +8 c \right )}{20472}+\frac {5975}{6824}\right ) \csc \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {16384 b \left (\cos \left (2 d x +2 c \right )+\frac {5 \cos \left (4 d x +4 c \right )}{4}+\frac {57}{28}\right )}{12795}\right ) \left (\csc ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\sec ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4194304 d}\) | \(118\) |
risch | \(\frac {4 i \left (105 i a \,{\mathrm e}^{12 i \left (d x +c \right )}+70 b \,{\mathrm e}^{13 i \left (d x +c \right )}+140 i a \,{\mathrm e}^{10 i \left (d x +c \right )}-14 b \,{\mathrm e}^{11 i \left (d x +c \right )}+350 i a \,{\mathrm e}^{8 i \left (d x +c \right )}+172 b \,{\mathrm e}^{9 i \left (d x +c \right )}+140 i a \,{\mathrm e}^{6 i \left (d x +c \right )}-172 b \,{\mathrm e}^{7 i \left (d x +c \right )}+105 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+14 b \,{\mathrm e}^{5 i \left (d x +c \right )}-70 b \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{8}}\) | \(158\) |
-1/d*(1/8*csc(d*x+c)^8*a+1/7*b*csc(d*x+c)^7-1/3*csc(d*x+c)^6*a-2/5*b*csc(d *x+c)^5+1/4*csc(d*x+c)^4*a+1/3*b*csc(d*x+c)^3)
Time = 0.39 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.35 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {210 \, a \cos \left (d x + c\right )^{4} - 140 \, a \cos \left (d x + c\right )^{2} + 8 \, {\left (35 \, b \cos \left (d x + c\right )^{4} - 28 \, b \cos \left (d x + c\right )^{2} + 8 \, b\right )} \sin \left (d x + c\right ) + 35 \, a}{840 \, {\left (d \cos \left (d x + c\right )^{8} - 4 \, d \cos \left (d x + c\right )^{6} + 6 \, d \cos \left (d x + c\right )^{4} - 4 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]
-1/840*(210*a*cos(d*x + c)^4 - 140*a*cos(d*x + c)^2 + 8*(35*b*cos(d*x + c) ^4 - 28*b*cos(d*x + c)^2 + 8*b)*sin(d*x + c) + 35*a)/(d*cos(d*x + c)^8 - 4 *d*cos(d*x + c)^6 + 6*d*cos(d*x + c)^4 - 4*d*cos(d*x + c)^2 + d)
Timed out. \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {280 \, b \sin \left (d x + c\right )^{5} + 210 \, a \sin \left (d x + c\right )^{4} - 336 \, b \sin \left (d x + c\right )^{3} - 280 \, a \sin \left (d x + c\right )^{2} + 120 \, b \sin \left (d x + c\right ) + 105 \, a}{840 \, d \sin \left (d x + c\right )^{8}} \]
-1/840*(280*b*sin(d*x + c)^5 + 210*a*sin(d*x + c)^4 - 336*b*sin(d*x + c)^3 - 280*a*sin(d*x + c)^2 + 120*b*sin(d*x + c) + 105*a)/(d*sin(d*x + c)^8)
Time = 0.40 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {280 \, b \sin \left (d x + c\right )^{5} + 210 \, a \sin \left (d x + c\right )^{4} - 336 \, b \sin \left (d x + c\right )^{3} - 280 \, a \sin \left (d x + c\right )^{2} + 120 \, b \sin \left (d x + c\right ) + 105 \, a}{840 \, d \sin \left (d x + c\right )^{8}} \]
-1/840*(280*b*sin(d*x + c)^5 + 210*a*sin(d*x + c)^4 - 336*b*sin(d*x + c)^3 - 280*a*sin(d*x + c)^2 + 120*b*sin(d*x + c) + 105*a)/(d*sin(d*x + c)^8)
Time = 11.53 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int \cot ^5(c+d x) \csc ^4(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {280\,b\,{\sin \left (c+d\,x\right )}^5+210\,a\,{\sin \left (c+d\,x\right )}^4-336\,b\,{\sin \left (c+d\,x\right )}^3-280\,a\,{\sin \left (c+d\,x\right )}^2+120\,b\,\sin \left (c+d\,x\right )+105\,a}{840\,d\,{\sin \left (c+d\,x\right )}^8} \]