Integrand size = 20, antiderivative size = 112 \[ \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {105 \text {arctanh}(\cos (a+b x))}{256 b}+\frac {105 \sec (a+b x)}{256 b}+\frac {35 \sec ^3(a+b x)}{256 b}-\frac {21 \csc ^2(a+b x) \sec ^3(a+b x)}{256 b}-\frac {3 \csc ^4(a+b x) \sec ^3(a+b x)}{128 b}-\frac {\csc ^6(a+b x) \sec ^3(a+b x)}{96 b} \]
-105/256*arctanh(cos(b*x+a))/b+105/256*sec(b*x+a)/b+35/256*sec(b*x+a)^3/b- 21/256*csc(b*x+a)^2*sec(b*x+a)^3/b-3/128*csc(b*x+a)^4*sec(b*x+a)^3/b-1/96* csc(b*x+a)^6*sec(b*x+a)^3/b
Leaf count is larger than twice the leaf count of optimal. \(278\) vs. \(2(112)=224\).
Time = 1.15 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.48 \[ \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {\csc ^{12}(a+b x) \left (1150-4752 \cos (2 (a+b x))+1600 \cos (3 (a+b x))+504 \cos (4 (a+b x))+1680 \cos (6 (a+b x))-600 \cos (7 (a+b x))-630 \cos (8 (a+b x))+200 \cos (9 (a+b x))+2520 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-945 \cos (7 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+315 \cos (9 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-30 \cos (a+b x) \left (40+63 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-63 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )-2520 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+945 \cos (7 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )-315 \cos (9 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )}{3072 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )^3} \]
(Csc[a + b*x]^12*(1150 - 4752*Cos[2*(a + b*x)] + 1600*Cos[3*(a + b*x)] + 5 04*Cos[4*(a + b*x)] + 1680*Cos[6*(a + b*x)] - 600*Cos[7*(a + b*x)] - 630*C os[8*(a + b*x)] + 200*Cos[9*(a + b*x)] + 2520*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 945*Cos[7*(a + b*x)]*Log[Cos[(a + b*x)/2]] + 315*Cos[9*(a + b *x)]*Log[Cos[(a + b*x)/2]] - 30*Cos[a + b*x]*(40 + 63*Log[Cos[(a + b*x)/2] ] - 63*Log[Sin[(a + b*x)/2]]) - 2520*Cos[3*(a + b*x)]*Log[Sin[(a + b*x)/2] ] + 945*Cos[7*(a + b*x)]*Log[Sin[(a + b*x)/2]] - 315*Cos[9*(a + b*x)]*Log[ Sin[(a + b*x)/2]]))/(3072*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2)^3)
Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3042, 4776, 3042, 3102, 252, 252, 252, 254, 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.
| \(\displaystyle \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin (a+b x)^3 \sin (2 a+2 b x)^4}dx\) |
| \(\Big \downarrow \) 4776 |
| \(\displaystyle \frac {1}{16} \int \csc ^7(a+b x) \sec ^4(a+b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{16} \int \csc (a+b x)^7 \sec (a+b x)^4dx\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int \frac {\sec ^{10}(a+b x)}{\left (1-\sec ^2(a+b x)\right )^4}d\sec (a+b x)}{16 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {\sec ^9(a+b x)}{6 \left (1-\sec ^2(a+b x)\right )^3}-\frac {3}{2} \int \frac {\sec ^8(a+b x)}{\left (1-\sec ^2(a+b x)\right )^3}d\sec (a+b x)}{16 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {\sec ^9(a+b x)}{6 \left (1-\sec ^2(a+b x)\right )^3}-\frac {3}{2} \left (\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}-\frac {7}{4} \int \frac {\sec ^6(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)\right )}{16 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {\sec ^9(a+b x)}{6 \left (1-\sec ^2(a+b x)\right )^3}-\frac {3}{2} \left (\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}-\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\sec ^4(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)\right )\right )}{16 b}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {\frac {\sec ^9(a+b x)}{6 \left (1-\sec ^2(a+b x)\right )^3}-\frac {3}{2} \left (\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}-\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\sec ^2(a+b x)+\frac {1}{1-\sec ^2(a+b x)}-1\right )d\sec (a+b x)\right )\right )}{16 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\sec ^9(a+b x)}{6 \left (1-\sec ^2(a+b x)\right )^3}-\frac {3}{2} \left (\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}-\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sec (a+b x))-\frac {1}{3} \sec ^3(a+b x)-\sec (a+b x)\right )\right )\right )}{16 b}\) |
(Sec[a + b*x]^9/(6*(1 - Sec[a + b*x]^2)^3) - (3*(Sec[a + b*x]^7/(4*(1 - Se c[a + b*x]^2)^2) - (7*(Sec[a + b*x]^5/(2*(1 - Sec[a + b*x]^2)) - (5*(ArcTa nh[Sec[a + b*x]] - Sec[a + b*x] - Sec[a + b*x]^3/3))/2))/4))/2)/(16*b)
3.1.72.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
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 = 1.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(\frac {-\frac {1}{6 \sin \left (x b +a \right )^{6} \cos \left (x b +a \right )^{3}}-\frac {3}{8 \sin \left (x b +a \right )^{4} \cos \left (x b +a \right )^{3}}+\frac {7}{8 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{3}}-\frac {35}{16 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )}+\frac {105}{16 \cos \left (x b +a \right )}+\frac {105 \ln \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}{16}}{16 b}\) | \(107\) |
| risch | \(\frac {315 \,{\mathrm e}^{17 i \left (x b +a \right )}-840 \,{\mathrm e}^{15 i \left (x b +a \right )}-252 \,{\mathrm e}^{13 i \left (x b +a \right )}+2376 \,{\mathrm e}^{11 i \left (x b +a \right )}-1150 \,{\mathrm e}^{9 i \left (x b +a \right )}+2376 \,{\mathrm e}^{7 i \left (x b +a \right )}-252 \,{\mathrm e}^{5 i \left (x b +a \right )}-840 \,{\mathrm e}^{3 i \left (x b +a \right )}+315 \,{\mathrm e}^{i \left (x b +a \right )}}{384 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{6} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{3}}-\frac {105 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{256 b}+\frac {105 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{256 b}\) | \(167\) |
1/16/b*(-1/6/sin(b*x+a)^6/cos(b*x+a)^3-3/8/sin(b*x+a)^4/cos(b*x+a)^3+7/8/s in(b*x+a)^2/cos(b*x+a)^3-35/16/sin(b*x+a)^2/cos(b*x+a)+105/16/cos(b*x+a)+1 05/16*ln(csc(b*x+a)-cot(b*x+a)))
Time = 0.26 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.73 \[ \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {630 \, \cos \left (b x + a\right )^{8} - 1680 \, \cos \left (b x + a\right )^{6} + 1386 \, \cos \left (b x + a\right )^{4} - 288 \, \cos \left (b x + a\right )^{2} - 315 \, {\left (\cos \left (b x + a\right )^{9} - 3 \, \cos \left (b x + a\right )^{7} + 3 \, \cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 315 \, {\left (\cos \left (b x + a\right )^{9} - 3 \, \cos \left (b x + a\right )^{7} + 3 \, \cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 32}{1536 \, {\left (b \cos \left (b x + a\right )^{9} - 3 \, b \cos \left (b x + a\right )^{7} + 3 \, b \cos \left (b x + a\right )^{5} - b \cos \left (b x + a\right )^{3}\right )}} \]
1/1536*(630*cos(b*x + a)^8 - 1680*cos(b*x + a)^6 + 1386*cos(b*x + a)^4 - 2 88*cos(b*x + a)^2 - 315*(cos(b*x + a)^9 - 3*cos(b*x + a)^7 + 3*cos(b*x + a )^5 - cos(b*x + a)^3)*log(1/2*cos(b*x + a) + 1/2) + 315*(cos(b*x + a)^9 - 3*cos(b*x + a)^7 + 3*cos(b*x + a)^5 - cos(b*x + a)^3)*log(-1/2*cos(b*x + a ) + 1/2) - 32)/(b*cos(b*x + a)^9 - 3*b*cos(b*x + a)^7 + 3*b*cos(b*x + a)^5 - b*cos(b*x + a)^3)
\[ \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx=\int \csc ^{3}{\left (a + b x \right )} \csc ^{4}{\left (2 a + 2 b x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 4268 vs. \(2 (100) = 200\).
Time = 0.38 (sec) , antiderivative size = 4268, normalized size of antiderivative = 38.11 \[ \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx=\text {Too large to display} \]
1/1536*(4*(315*cos(17*b*x + 17*a) - 840*cos(15*b*x + 15*a) - 252*cos(13*b* x + 13*a) + 2376*cos(11*b*x + 11*a) - 1150*cos(9*b*x + 9*a) + 2376*cos(7*b *x + 7*a) - 252*cos(5*b*x + 5*a) - 840*cos(3*b*x + 3*a) + 315*cos(b*x + a) )*cos(18*b*x + 18*a) - 1260*(3*cos(16*b*x + 16*a) - 8*cos(12*b*x + 12*a) + 6*cos(10*b*x + 10*a) + 6*cos(8*b*x + 8*a) - 8*cos(6*b*x + 6*a) + 3*cos(2* b*x + 2*a) - 1)*cos(17*b*x + 17*a) + 12*(840*cos(15*b*x + 15*a) + 252*cos( 13*b*x + 13*a) - 2376*cos(11*b*x + 11*a) + 1150*cos(9*b*x + 9*a) - 2376*co s(7*b*x + 7*a) + 252*cos(5*b*x + 5*a) + 840*cos(3*b*x + 3*a) - 315*cos(b*x + a))*cos(16*b*x + 16*a) - 3360*(8*cos(12*b*x + 12*a) - 6*cos(10*b*x + 10 *a) - 6*cos(8*b*x + 8*a) + 8*cos(6*b*x + 6*a) - 3*cos(2*b*x + 2*a) + 1)*co s(15*b*x + 15*a) - 1008*(8*cos(12*b*x + 12*a) - 6*cos(10*b*x + 10*a) - 6*c os(8*b*x + 8*a) + 8*cos(6*b*x + 6*a) - 3*cos(2*b*x + 2*a) + 1)*cos(13*b*x + 13*a) + 32*(2376*cos(11*b*x + 11*a) - 1150*cos(9*b*x + 9*a) + 2376*cos(7 *b*x + 7*a) - 252*cos(5*b*x + 5*a) - 840*cos(3*b*x + 3*a) + 315*cos(b*x + a))*cos(12*b*x + 12*a) - 9504*(6*cos(10*b*x + 10*a) + 6*cos(8*b*x + 8*a) - 8*cos(6*b*x + 6*a) + 3*cos(2*b*x + 2*a) - 1)*cos(11*b*x + 11*a) + 24*(115 0*cos(9*b*x + 9*a) - 2376*cos(7*b*x + 7*a) + 252*cos(5*b*x + 5*a) + 840*co s(3*b*x + 3*a) - 315*cos(b*x + a))*cos(10*b*x + 10*a) + 4600*(6*cos(8*b*x + 8*a) - 8*cos(6*b*x + 6*a) + 3*cos(2*b*x + 2*a) - 1)*cos(9*b*x + 9*a) - 7 2*(792*cos(7*b*x + 7*a) - 84*cos(5*b*x + 5*a) - 280*cos(3*b*x + 3*a) + ...
Leaf count of result is larger than twice the leaf count of optimal. 268 vs. \(2 (100) = 200\).
Time = 0.31 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.39 \[ \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx=-\frac {\frac {285 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {21 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac {\frac {18 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {225 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - \frac {2966 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac {3513 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - \frac {660 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} + \frac {1155 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}} - 1}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}\right )}^{3}} - 1260 \, \log \left (-\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}{6144 \, b} \]
-1/6144*(285*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 21*(cos(b*x + a) - 1) ^2/(cos(b*x + a) + 1)^2 + (cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 + (18* (cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 225*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 - 2966*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 - 3513*(cos(b *x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 660*(cos(b*x + a) - 1)^5/(cos(b*x + a) + 1)^5 + 1155*(cos(b*x + a) - 1)^6/(cos(b*x + a) + 1)^6 - 1)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2)^ 3 - 1260*log(-(cos(b*x + a) - 1)/(cos(b*x + a) + 1)))/b
Time = 19.62 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.89 \[ \int \csc ^3(a+b x) \csc ^4(2 a+2 b x) \, dx=\frac {-\frac {105\,{\cos \left (a+b\,x\right )}^8}{256}+\frac {35\,{\cos \left (a+b\,x\right )}^6}{32}-\frac {231\,{\cos \left (a+b\,x\right )}^4}{256}+\frac {3\,{\cos \left (a+b\,x\right )}^2}{16}+\frac {1}{48}}{b\,\left (-{\cos \left (a+b\,x\right )}^9+3\,{\cos \left (a+b\,x\right )}^7-3\,{\cos \left (a+b\,x\right )}^5+{\cos \left (a+b\,x\right )}^3\right )}-\frac {105\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{256\,b} \]