Integrand size = 20, antiderivative size = 90 \[ \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx=-\frac {5 \cot ^2(a+b x)}{32 b}-\frac {5 \cot ^4(a+b x)}{128 b}-\frac {\cot ^6(a+b x)}{192 b}+\frac {5 \log (\tan (a+b x))}{16 b}+\frac {5 \tan ^2(a+b x)}{64 b}+\frac {\tan ^4(a+b x)}{128 b} \]
-5/32*cot(b*x+a)^2/b-5/128*cot(b*x+a)^4/b-1/192*cot(b*x+a)^6/b+5/16*ln(tan (b*x+a))/b+5/64*tan(b*x+a)^2/b+1/128*tan(b*x+a)^4/b
Time = 0.43 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx=-\frac {36 \csc ^2(a+b x)+9 \csc ^4(a+b x)+2 \csc ^6(a+b x)+120 \log (\cos (a+b x))-120 \log (\sin (a+b x))-24 \sec ^2(a+b x)-3 \sec ^4(a+b x)}{384 b} \]
-1/384*(36*Csc[a + b*x]^2 + 9*Csc[a + b*x]^4 + 2*Csc[a + b*x]^6 + 120*Log[ Cos[a + b*x]] - 120*Log[Sin[a + b*x]] - 24*Sec[a + b*x]^2 - 3*Sec[a + b*x] ^4)/b
Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.81, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 4776, 3042, 3100, 243, 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.
\(\displaystyle \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^2 \sin (2 a+2 b x)^5}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{32} \int \csc ^7(a+b x) \sec ^5(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{32} \int \csc (a+b x)^7 \sec (a+b x)^5dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {\int \cot ^7(a+b x) \left (\tan ^2(a+b x)+1\right )^5d\tan (a+b x)}{32 b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \cot ^4(a+b x) \left (\tan ^2(a+b x)+1\right )^5d\tan ^2(a+b x)}{64 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\cot ^4(a+b x)+5 \cot ^3(a+b x)+10 \cot ^2(a+b x)+10 \cot (a+b x)+\tan ^2(a+b x)+5\right )d\tan ^2(a+b x)}{64 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} \tan ^4(a+b x)+5 \tan ^2(a+b x)-\frac {1}{3} \cot ^3(a+b x)-\frac {5}{2} \cot ^2(a+b x)-10 \cot (a+b x)+10 \log \left (\tan ^2(a+b x)\right )}{64 b}\) |
(-10*Cot[a + b*x] - (5*Cot[a + b*x]^2)/2 - Cot[a + b*x]^3/3 + 10*Log[Tan[a + b*x]^2] + 5*Tan[a + b*x]^2 + Tan[a + b*x]^4/2)/(64*b)
3.1.57.3.1 Defintions of rubi rules used
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[1/f Subst[Int[(1 + x^2)^((m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]] , x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]
Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[2^p/f^p Int[Cos[a + b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && I ntegerQ[p]
Time = 1.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.09
method | result | size |
default | \(\frac {-\frac {1}{6 \sin \left (x b +a \right )^{6} \cos \left (x b +a \right )^{4}}+\frac {5}{12 \sin \left (x b +a \right )^{4} \cos \left (x b +a \right )^{4}}-\frac {5}{6 \sin \left (x b +a \right )^{4} \cos \left (x b +a \right )^{2}}+\frac {5}{2 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{2}}-\frac {5}{\sin \left (x b +a \right )^{2}}+10 \ln \left (\tan \left (x b +a \right )\right )}{32 b}\) | \(98\) |
risch | \(\frac {15 \,{\mathrm e}^{18 i \left (x b +a \right )}-30 \,{\mathrm e}^{16 i \left (x b +a \right )}-40 \,{\mathrm e}^{14 i \left (x b +a \right )}+110 \,{\mathrm e}^{12 i \left (x b +a \right )}+18 \,{\mathrm e}^{10 i \left (x b +a \right )}+110 \,{\mathrm e}^{8 i \left (x b +a \right )}-40 \,{\mathrm e}^{6 i \left (x b +a \right )}-30 \,{\mathrm e}^{4 i \left (x b +a \right )}+15 \,{\mathrm e}^{2 i \left (x b +a \right )}}{24 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{6} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{4}}-\frac {5 \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{16 b}+\frac {5 \ln \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{16 b}\) | \(167\) |
1/32/b*(-1/6/sin(b*x+a)^6/cos(b*x+a)^4+5/12/sin(b*x+a)^4/cos(b*x+a)^4-5/6/ sin(b*x+a)^4/cos(b*x+a)^2+5/2/sin(b*x+a)^2/cos(b*x+a)^2-5/sin(b*x+a)^2+10* ln(tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (78) = 156\).
Time = 0.26 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.16 \[ \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx=\frac {60 \, \cos \left (b x + a\right )^{8} - 150 \, \cos \left (b x + a\right )^{6} + 110 \, \cos \left (b x + a\right )^{4} - 15 \, \cos \left (b x + a\right )^{2} - 60 \, {\left (\cos \left (b x + a\right )^{10} - 3 \, \cos \left (b x + a\right )^{8} + 3 \, \cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 60 \, {\left (\cos \left (b x + a\right )^{10} - 3 \, \cos \left (b x + a\right )^{8} + 3 \, \cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) - 3}{384 \, {\left (b \cos \left (b x + a\right )^{10} - 3 \, b \cos \left (b x + a\right )^{8} + 3 \, b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )}} \]
1/384*(60*cos(b*x + a)^8 - 150*cos(b*x + a)^6 + 110*cos(b*x + a)^4 - 15*co s(b*x + a)^2 - 60*(cos(b*x + a)^10 - 3*cos(b*x + a)^8 + 3*cos(b*x + a)^6 - cos(b*x + a)^4)*log(cos(b*x + a)^2) + 60*(cos(b*x + a)^10 - 3*cos(b*x + a )^8 + 3*cos(b*x + a)^6 - cos(b*x + a)^4)*log(-1/4*cos(b*x + a)^2 + 1/4) - 3)/(b*cos(b*x + a)^10 - 3*b*cos(b*x + a)^8 + 3*b*cos(b*x + a)^6 - b*cos(b* x + a)^4)
\[ \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx=\int \csc ^{2}{\left (a + b x \right )} \csc ^{5}{\left (2 a + 2 b x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 7650 vs. \(2 (78) = 156\).
Time = 0.48 (sec) , antiderivative size = 7650, normalized size of antiderivative = 85.00 \[ \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx=\text {Too large to display} \]
1/96*(4*(15*cos(18*b*x + 18*a) - 30*cos(16*b*x + 16*a) - 40*cos(14*b*x + 1 4*a) + 110*cos(12*b*x + 12*a) + 18*cos(10*b*x + 10*a) + 110*cos(8*b*x + 8* a) - 40*cos(6*b*x + 6*a) - 30*cos(4*b*x + 4*a) + 15*cos(2*b*x + 2*a))*cos( 20*b*x + 20*a) + 4*(15*cos(16*b*x + 16*a) + 200*cos(14*b*x + 14*a) - 190*c os(12*b*x + 12*a) - 216*cos(10*b*x + 10*a) - 190*cos(8*b*x + 8*a) + 200*co s(6*b*x + 6*a) + 15*cos(4*b*x + 4*a) - 60*cos(2*b*x + 2*a) + 15)*cos(18*b* x + 18*a) - 120*cos(18*b*x + 18*a)^2 - 12*(40*cos(14*b*x + 14*a) + 130*cos (12*b*x + 12*a) - 102*cos(10*b*x + 10*a) + 130*cos(8*b*x + 8*a) + 40*cos(6 *b*x + 6*a) - 60*cos(4*b*x + 4*a) - 5*cos(2*b*x + 2*a) + 10)*cos(16*b*x + 16*a) + 360*cos(16*b*x + 16*a)^2 + 32*(100*cos(12*b*x + 12*a) + 78*cos(10* b*x + 10*a) + 100*cos(8*b*x + 8*a) - 80*cos(6*b*x + 6*a) - 15*cos(4*b*x + 4*a) + 25*cos(2*b*x + 2*a) - 5)*cos(14*b*x + 14*a) - 1280*cos(14*b*x + 14* a)^2 - 8*(642*cos(10*b*x + 10*a) - 220*cos(8*b*x + 8*a) - 400*cos(6*b*x + 6*a) + 195*cos(4*b*x + 4*a) + 95*cos(2*b*x + 2*a) - 55)*cos(12*b*x + 12*a) + 880*cos(12*b*x + 12*a)^2 - 24*(214*cos(8*b*x + 8*a) - 104*cos(6*b*x + 6 *a) - 51*cos(4*b*x + 4*a) + 36*cos(2*b*x + 2*a) - 3)*cos(10*b*x + 10*a) - 864*cos(10*b*x + 10*a)^2 + 40*(80*cos(6*b*x + 6*a) - 39*cos(4*b*x + 4*a) - 19*cos(2*b*x + 2*a) + 11)*cos(8*b*x + 8*a) + 880*cos(8*b*x + 8*a)^2 - 160 *(3*cos(4*b*x + 4*a) - 5*cos(2*b*x + 2*a) + 1)*cos(6*b*x + 6*a) - 1280*cos (6*b*x + 6*a)^2 + 60*(cos(2*b*x + 2*a) - 2)*cos(4*b*x + 4*a) + 360*cos(...
Time = 0.27 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.04 \[ \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx=-\frac {\frac {60 \, \sin \left (b x + a\right )^{8} - 90 \, \sin \left (b x + a\right )^{6} + 20 \, \sin \left (b x + a\right )^{4} + 5 \, \sin \left (b x + a\right )^{2} + 2}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2} \sin \left (b x + a\right )^{6}} + 60 \, \log \left (-\sin \left (b x + a\right )^{2} + 1\right ) - 120 \, \log \left ({\left | \sin \left (b x + a\right ) \right |}\right )}{384 \, b} \]
-1/384*((60*sin(b*x + a)^8 - 90*sin(b*x + a)^6 + 20*sin(b*x + a)^4 + 5*sin (b*x + a)^2 + 2)/((sin(b*x + a)^2 - 1)^2*sin(b*x + a)^6) + 60*log(-sin(b*x + a)^2 + 1) - 120*log(abs(sin(b*x + a))))/b
Time = 19.56 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.27 \[ \int \csc ^2(a+b x) \csc ^5(2 a+2 b x) \, dx=\frac {5\,\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{32\,b}-\frac {5\,\ln \left (\cos \left (a+b\,x\right )\right )}{16\,b}+\frac {-\frac {5\,{\cos \left (a+b\,x\right )}^8}{32}+\frac {25\,{\cos \left (a+b\,x\right )}^6}{64}-\frac {55\,{\cos \left (a+b\,x\right )}^4}{192}+\frac {5\,{\cos \left (a+b\,x\right )}^2}{128}+\frac {1}{128}}{b\,\left (-{\cos \left (a+b\,x\right )}^{10}+3\,{\cos \left (a+b\,x\right )}^8-3\,{\cos \left (a+b\,x\right )}^6+{\cos \left (a+b\,x\right )}^4\right )} \]