Integrand size = 18, antiderivative size = 89 \[ \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx=-\frac {35 \text {arctanh}(\cos (a+b x))}{256 b}+\frac {35 \sec (a+b x)}{256 b}+\frac {35 \sec ^3(a+b x)}{768 b}-\frac {7 \csc ^2(a+b x) \sec ^3(a+b x)}{256 b}-\frac {\csc ^4(a+b x) \sec ^3(a+b x)}{128 b} \]
-35/256*arctanh(cos(b*x+a))/b+35/256*sec(b*x+a)/b+35/768*sec(b*x+a)^3/b-7/ 256*csc(b*x+a)^2*sec(b*x+a)^3/b-1/128*csc(b*x+a)^4*sec(b*x+a)^3/b
Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(89)=178\).
Time = 0.73 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.01 \[ \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx=-\frac {\csc ^{10}(a+b x) \left (-204+658 \cos (2 (a+b x))-228 \cos (3 (a+b x))+140 \cos (4 (a+b x))-76 \cos (5 (a+b x))-210 \cos (6 (a+b x))+76 \cos (7 (a+b x))-315 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-105 \cos (5 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+105 \cos (7 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+3 \cos (a+b x) \left (76+105 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-105 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )+315 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+105 \cos (5 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )-105 \cos (7 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )}{768 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )^3} \]
-1/768*(Csc[a + b*x]^10*(-204 + 658*Cos[2*(a + b*x)] - 228*Cos[3*(a + b*x) ] + 140*Cos[4*(a + b*x)] - 76*Cos[5*(a + b*x)] - 210*Cos[6*(a + b*x)] + 76 *Cos[7*(a + b*x)] - 315*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 105*Cos[5 *(a + b*x)]*Log[Cos[(a + b*x)/2]] + 105*Cos[7*(a + b*x)]*Log[Cos[(a + b*x) /2]] + 3*Cos[a + b*x]*(76 + 105*Log[Cos[(a + b*x)/2]] - 105*Log[Sin[(a + b *x)/2]]) + 315*Cos[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + 105*Cos[5*(a + b*x )]*Log[Sin[(a + b*x)/2]] - 105*Cos[7*(a + b*x)]*Log[Sin[(a + b*x)/2]]))/(b *(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2)^3)
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4775, 3042, 3102, 25, 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.
\(\displaystyle \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^5}dx\) |
\(\Big \downarrow \) 4775 |
\(\displaystyle \frac {1}{32} \int \csc ^5(a+b x) \sec ^4(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{32} \int \csc (a+b x)^5 \sec (a+b x)^4dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {\sec ^8(a+b x)}{\left (1-\sec ^2(a+b x)\right )^3}d\sec (a+b x)}{32 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sec ^8(a+b x)}{\left (1-\sec ^2(a+b x)\right )^3}d\sec (a+b x)}{32 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {7}{4} \int \frac {\sec ^6(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{32 b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\sec ^4(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)\right )-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{32 b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\sec ^2(a+b x)+\frac {1}{1-\sec ^2(a+b x)}-1\right )d\sec (a+b x)\right )-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{32 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sec (a+b x))-\frac {1}{3} \sec ^3(a+b x)-\sec (a+b x)\right )\right )-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{32 b}\) |
(-1/4*Sec[a + b*x]^7/(1 - Sec[a + b*x]^2)^2 + (7*(Sec[a + b*x]^5/(2*(1 - S ec[a + b*x]^2)) - (5*(ArcTanh[Sec[a + b*x]] - Sec[a + b*x] - Sec[a + b*x]^ 3/3))/2))/4)/(32*b)
3.2.40.3.1 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[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_ Symbol] :> Simp[2^p/e^p Int[(e*Cos[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && I ntegerQ[p]
Time = 5.29 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {-\frac {1}{4 \sin \left (x b +a \right )^{4} \cos \left (x b +a \right )^{3}}+\frac {7}{12 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{3}}-\frac {35}{24 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )}+\frac {35}{8 \cos \left (x b +a \right )}+\frac {35 \ln \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}{8}}{32 b}\) | \(89\) |
risch | \(\frac {105 \,{\mathrm e}^{13 i \left (x b +a \right )}-70 \,{\mathrm e}^{11 i \left (x b +a \right )}-329 \,{\mathrm e}^{9 i \left (x b +a \right )}+204 \,{\mathrm e}^{7 i \left (x b +a \right )}-329 \,{\mathrm e}^{5 i \left (x b +a \right )}-70 \,{\mathrm e}^{3 i \left (x b +a \right )}+105 \,{\mathrm e}^{i \left (x b +a \right )}}{384 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{3}}+\frac {35 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{256 b}-\frac {35 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{256 b}\) | \(145\) |
1/32/b*(-1/4/sin(b*x+a)^4/cos(b*x+a)^3+7/12/sin(b*x+a)^2/cos(b*x+a)^3-35/2 4/sin(b*x+a)^2/cos(b*x+a)+35/8/cos(b*x+a)+35/8*ln(csc(b*x+a)-cot(b*x+a)))
Time = 0.25 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.66 \[ \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx=\frac {210 \, \cos \left (b x + a\right )^{6} - 350 \, \cos \left (b x + a\right )^{4} + 112 \, \cos \left (b x + a\right )^{2} - 105 \, {\left (\cos \left (b x + a\right )^{7} - 2 \, \cos \left (b x + a\right )^{5} + \cos \left (b x + a\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 105 \, {\left (\cos \left (b x + a\right )^{7} - 2 \, \cos \left (b x + a\right )^{5} + \cos \left (b x + a\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 16}{1536 \, {\left (b \cos \left (b x + a\right )^{7} - 2 \, b \cos \left (b x + a\right )^{5} + b \cos \left (b x + a\right )^{3}\right )}} \]
1/1536*(210*cos(b*x + a)^6 - 350*cos(b*x + a)^4 + 112*cos(b*x + a)^2 - 105 *(cos(b*x + a)^7 - 2*cos(b*x + a)^5 + cos(b*x + a)^3)*log(1/2*cos(b*x + a) + 1/2) + 105*(cos(b*x + a)^7 - 2*cos(b*x + a)^5 + cos(b*x + a)^3)*log(-1/ 2*cos(b*x + a) + 1/2) + 16)/(b*cos(b*x + a)^7 - 2*b*cos(b*x + a)^5 + b*cos (b*x + a)^3)
Timed out. \[ \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3846 vs. \(2 (79) = 158\).
Time = 0.33 (sec) , antiderivative size = 3846, normalized size of antiderivative = 43.21 \[ \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx=\text {Too large to display} \]
1/1536*(4*(105*cos(13*b*x + 13*a) - 70*cos(11*b*x + 11*a) - 329*cos(9*b*x + 9*a) + 204*cos(7*b*x + 7*a) - 329*cos(5*b*x + 5*a) - 70*cos(3*b*x + 3*a) + 105*cos(b*x + a))*cos(14*b*x + 14*a) - 420*(cos(12*b*x + 12*a) + 3*cos( 10*b*x + 10*a) - 3*cos(8*b*x + 8*a) - 3*cos(6*b*x + 6*a) + 3*cos(4*b*x + 4 *a) + cos(2*b*x + 2*a) - 1)*cos(13*b*x + 13*a) + 4*(70*cos(11*b*x + 11*a) + 329*cos(9*b*x + 9*a) - 204*cos(7*b*x + 7*a) + 329*cos(5*b*x + 5*a) + 70* cos(3*b*x + 3*a) - 105*cos(b*x + a))*cos(12*b*x + 12*a) + 280*(3*cos(10*b* x + 10*a) - 3*cos(8*b*x + 8*a) - 3*cos(6*b*x + 6*a) + 3*cos(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(11*b*x + 11*a) + 12*(329*cos(9*b*x + 9*a) - 204 *cos(7*b*x + 7*a) + 329*cos(5*b*x + 5*a) + 70*cos(3*b*x + 3*a) - 105*cos(b *x + a))*cos(10*b*x + 10*a) - 1316*(3*cos(8*b*x + 8*a) + 3*cos(6*b*x + 6*a ) - 3*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(9*b*x + 9*a) + 12*(204* cos(7*b*x + 7*a) - 329*cos(5*b*x + 5*a) - 70*cos(3*b*x + 3*a) + 105*cos(b* x + a))*cos(8*b*x + 8*a) + 816*(3*cos(6*b*x + 6*a) - 3*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(7*b*x + 7*a) - 84*(47*cos(5*b*x + 5*a) + 10*cos( 3*b*x + 3*a) - 15*cos(b*x + a))*cos(6*b*x + 6*a) + 1316*(3*cos(4*b*x + 4*a ) + cos(2*b*x + 2*a) - 1)*cos(5*b*x + 5*a) + 420*(2*cos(3*b*x + 3*a) - 3*c os(b*x + a))*cos(4*b*x + 4*a) + 280*(cos(2*b*x + 2*a) - 1)*cos(3*b*x + 3*a ) - 420*cos(2*b*x + 2*a)*cos(b*x + a) + 105*(2*(cos(12*b*x + 12*a) + 3*cos (10*b*x + 10*a) - 3*cos(8*b*x + 8*a) - 3*cos(6*b*x + 6*a) + 3*cos(4*b*x...
Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.96 \[ \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx=\frac {\frac {6 \, {\left (11 \, \cos \left (b x + a\right )^{3} - 13 \, \cos \left (b x + a\right )\right )}}{{\left (\cos \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac {16 \, {\left (9 \, \cos \left (b x + a\right )^{2} + 1\right )}}{\cos \left (b x + a\right )^{3}} - 105 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 105 \, \log \left (-\cos \left (b x + a\right ) + 1\right )}{1536 \, b} \]
1/1536*(6*(11*cos(b*x + a)^3 - 13*cos(b*x + a))/(cos(b*x + a)^2 - 1)^2 + 1 6*(9*cos(b*x + a)^2 + 1)/cos(b*x + a)^3 - 105*log(cos(b*x + a) + 1) + 105* log(-cos(b*x + a) + 1))/b
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \cos (a+b x) \csc ^5(2 a+2 b x) \, dx=\frac {\frac {35\,{\cos \left (a+b\,x\right )}^6}{256}-\frac {175\,{\cos \left (a+b\,x\right )}^4}{768}+\frac {7\,{\cos \left (a+b\,x\right )}^2}{96}+\frac {1}{96}}{b\,\left ({\cos \left (a+b\,x\right )}^7-2\,{\cos \left (a+b\,x\right )}^5+{\cos \left (a+b\,x\right )}^3\right )}-\frac {35\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{256\,b} \]