Integrand size = 18, antiderivative size = 66 \[ \int \csc ^4(2 a+2 b x) \sin (a+b x) \, dx=-\frac {5 \text {arctanh}(\cos (a+b x))}{32 b}+\frac {5 \sec (a+b x)}{32 b}+\frac {5 \sec ^3(a+b x)}{96 b}-\frac {\csc ^2(a+b x) \sec ^3(a+b x)}{32 b} \]
-5/32*arctanh(cos(b*x+a))/b+5/32*sec(b*x+a)/b+5/96*sec(b*x+a)^3/b-1/32*csc (b*x+a)^2*sec(b*x+a)^3/b
Leaf count is larger than twice the leaf count of optimal. \(205\) vs. \(2(66)=132\).
Time = 0.66 (sec) , antiderivative size = 205, normalized size of antiderivative = 3.11 \[ \int \csc ^4(2 a+2 b x) \sin (a+b x) \, dx=\frac {\csc ^8(a+b x) \left (22-40 \cos (2 (a+b x))+13 \cos (3 (a+b x))-30 \cos (4 (a+b x))+13 \cos (5 (a+b x))+15 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+15 \cos (5 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-15 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )-15 \cos (5 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (-26-30 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+30 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )\right )}{24 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )^3} \]
(Csc[a + b*x]^8*(22 - 40*Cos[2*(a + b*x)] + 13*Cos[3*(a + b*x)] - 30*Cos[4 *(a + b*x)] + 13*Cos[5*(a + b*x)] + 15*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/ 2]] + 15*Cos[5*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 15*Cos[3*(a + b*x)]*Log[ Sin[(a + b*x)/2]] - 15*Cos[5*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b* x]*(-26 - 30*Log[Cos[(a + b*x)/2]] + 30*Log[Sin[(a + b*x)/2]])))/(24*b*(Cs c[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2)^3)
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3042, 4776, 3042, 3102, 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.
| \(\displaystyle \int \sin (a+b x) \csc ^4(2 a+2 b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^4}dx\) |
| \(\Big \downarrow \) 4776 |
| \(\displaystyle \frac {1}{16} \int \csc ^3(a+b x) \sec ^4(a+b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{16} \int \csc (a+b x)^3 \sec (a+b x)^4dx\) |
| \(\Big \downarrow \) 3102 |
| \(\displaystyle \frac {\int \frac {\sec ^6(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)}{16 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\sec ^4(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)}{16 b}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle \frac {\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\sec ^2(a+b x)+\frac {1}{1-\sec ^2(a+b x)}-1\right )d\sec (a+b x)}{16 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sec (a+b x))-\frac {1}{3} \sec ^3(a+b x)-\sec (a+b x)\right )}{16 b}\) |
(Sec[a + b*x]^5/(2*(1 - Sec[a + b*x]^2)) - (5*(ArcTanh[Sec[a + b*x]] - Sec [a + b*x] - Sec[a + b*x]^3/3))/2)/(16*b)
3.1.11.3.1 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 = 2.23 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.08
| method | result | size |
| default | \(\frac {\frac {1}{3 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )^{3}}-\frac {5}{6 \sin \left (x b +a \right )^{2} \cos \left (x b +a \right )}+\frac {5}{2 \cos \left (x b +a \right )}+\frac {5 \ln \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}{2}}{16 b}\) | \(71\) |
| risch | \(\frac {15 \,{\mathrm e}^{9 i \left (x b +a \right )}+20 \,{\mathrm e}^{7 i \left (x b +a \right )}-22 \,{\mathrm e}^{5 i \left (x b +a \right )}+20 \,{\mathrm e}^{3 i \left (x b +a \right )}+15 \,{\mathrm e}^{i \left (x b +a \right )}}{48 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{32 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{32 b}\) | \(123\) |
1/16/b*(1/3/sin(b*x+a)^2/cos(b*x+a)^3-5/6/sin(b*x+a)^2/cos(b*x+a)+5/2/cos( b*x+a)+5/2*ln(csc(b*x+a)-cot(b*x+a)))
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.70 \[ \int \csc ^4(2 a+2 b x) \sin (a+b x) \, dx=\frac {30 \, \cos \left (b x + a\right )^{4} - 20 \, \cos \left (b x + a\right )^{2} - 15 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (b x + a\right )^{5} - \cos \left (b x + a\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 4}{192 \, {\left (b \cos \left (b x + a\right )^{5} - b \cos \left (b x + a\right )^{3}\right )}} \]
1/192*(30*cos(b*x + a)^4 - 20*cos(b*x + a)^2 - 15*(cos(b*x + a)^5 - cos(b* x + a)^3)*log(1/2*cos(b*x + a) + 1/2) + 15*(cos(b*x + a)^5 - cos(b*x + a)^ 3)*log(-1/2*cos(b*x + a) + 1/2) - 4)/(b*cos(b*x + a)^5 - b*cos(b*x + a)^3)
Timed out. \[ \int \csc ^4(2 a+2 b x) \sin (a+b x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 2174 vs. \(2 (58) = 116\).
Time = 0.29 (sec) , antiderivative size = 2174, normalized size of antiderivative = 32.94 \[ \int \csc ^4(2 a+2 b x) \sin (a+b x) \, dx=\text {Too large to display} \]
1/192*(4*(15*cos(9*b*x + 9*a) + 20*cos(7*b*x + 7*a) - 22*cos(5*b*x + 5*a) + 20*cos(3*b*x + 3*a) + 15*cos(b*x + a))*cos(10*b*x + 10*a) + 60*(cos(8*b* x + 8*a) - 2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) + cos(2*b*x + 2*a) + 1) *cos(9*b*x + 9*a) + 4*(20*cos(7*b*x + 7*a) - 22*cos(5*b*x + 5*a) + 20*cos( 3*b*x + 3*a) + 15*cos(b*x + a))*cos(8*b*x + 8*a) - 80*(2*cos(6*b*x + 6*a) + 2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) - 1)*cos(7*b*x + 7*a) + 8*(22*cos( 5*b*x + 5*a) - 20*cos(3*b*x + 3*a) - 15*cos(b*x + a))*cos(6*b*x + 6*a) + 8 8*(2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) - 1)*cos(5*b*x + 5*a) - 40*(4*cos (3*b*x + 3*a) + 3*cos(b*x + a))*cos(4*b*x + 4*a) + 80*(cos(2*b*x + 2*a) + 1)*cos(3*b*x + 3*a) + 60*cos(2*b*x + 2*a)*cos(b*x + a) - 15*(2*(cos(8*b*x + 8*a) - 2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) + cos(2*b*x + 2*a) + 1)*c os(10*b*x + 10*a) + cos(10*b*x + 10*a)^2 - 2*(2*cos(6*b*x + 6*a) + 2*cos(4 *b*x + 4*a) - cos(2*b*x + 2*a) - 1)*cos(8*b*x + 8*a) + cos(8*b*x + 8*a)^2 + 4*(2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) + 4*cos(6 *b*x + 6*a)^2 - 4*(cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + 4*cos(4*b*x + 4*a)^2 + cos(2*b*x + 2*a)^2 + 2*(sin(8*b*x + 8*a) - 2*sin(6*b*x + 6*a) - 2 *sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(10*b*x + 10*a) + sin(10*b*x + 10 *a)^2 - 2*(2*sin(6*b*x + 6*a) + 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin (8*b*x + 8*a) + sin(8*b*x + 8*a)^2 + 4*(2*sin(4*b*x + 4*a) - sin(2*b*x + 2 *a))*sin(6*b*x + 6*a) + 4*sin(6*b*x + 6*a)^2 + 4*sin(4*b*x + 4*a)^2 - 4...
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (58) = 116\).
Time = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.42 \[ \int \csc ^4(2 a+2 b x) \sin (a+b x) \, dx=-\frac {\frac {3 \, {\left (\frac {10 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} + \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {16 \, {\left (\frac {12 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {9 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 7\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{3}} - 30 \, \log \left (-\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1}\right )}{384 \, b} \]
-1/384*(3*(10*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)*(cos(b*x + a) + 1 )/(cos(b*x + a) - 1) + 3*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 16*(12*(c os(b*x + a) - 1)/(cos(b*x + a) + 1) + 9*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 7)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)^3 - 30*log(-(cos( b*x + a) - 1)/(cos(b*x + a) + 1)))/b
Time = 20.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.91 \[ \int \csc ^4(2 a+2 b x) \sin (a+b x) \, dx=\frac {-\frac {5\,{\cos \left (a+b\,x\right )}^4}{32}+\frac {5\,{\cos \left (a+b\,x\right )}^2}{48}+\frac {1}{48}}{b\,\left ({\cos \left (a+b\,x\right )}^3-{\cos \left (a+b\,x\right )}^5\right )}-\frac {5\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{32\,b} \]