Integrand size = 18, antiderivative size = 83 \[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=\frac {35 \text {arctanh}(\sin (a+b x))}{256 b}-\frac {3 \csc (a+b x)}{32 b}-\frac {\csc ^3(a+b x)}{96 b}+\frac {13 \sec (a+b x) \tan (a+b x)}{256 b}+\frac {\sec (a+b x) \tan ^3(a+b x)}{128 b} \] Output:
35/256*arctanh(sin(b*x+a))/b-3/32*csc(b*x+a)/b-1/96*csc(b*x+a)^3/b+13/256* sec(b*x+a)*tan(b*x+a)/b+1/128*sec(b*x+a)*tan(b*x+a)^3/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.37 \[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=-\frac {\csc ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},3,-\frac {1}{2},\sin ^2(a+b x)\right )}{96 b} \] Input:
Integrate[Csc[2*a + 2*b*x]^5*Sin[a + b*x],x]
Output:
-1/96*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 3, -1/2, Sin[a + b*x]^2])/b
Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4776, 3042, 3101, 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.
| \(\displaystyle \int \sin (a+b x) \csc ^5(2 a+2 b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^5}dx\) |
| \(\Big \downarrow \) 4776 |
| \(\displaystyle \frac {1}{32} \int \csc ^4(a+b x) \sec ^5(a+b x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{32} \int \csc (a+b x)^4 \sec (a+b x)^5dx\) |
| \(\Big \downarrow \) 3101 |
| \(\displaystyle -\frac {\int -\frac {\csc ^8(a+b x)}{\left (1-\csc ^2(a+b x)\right )^3}d\csc (a+b x)}{32 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\csc ^8(a+b x)}{\left (1-\csc ^2(a+b x)\right )^3}d\csc (a+b x)}{32 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {7}{4} \int \frac {\csc ^6(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {7}{4} \left (\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\csc ^4(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)\right )-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {\frac {7}{4} \left (\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\csc ^2(a+b x)+\frac {1}{1-\csc ^2(a+b x)}-1\right )d\csc (a+b x)\right )-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {7}{4} \left (\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\csc (a+b x))-\frac {1}{3} \csc ^3(a+b x)-\csc (a+b x)\right )\right )-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{32 b}\) |
Input:
Int[Csc[2*a + 2*b*x]^5*Sin[a + b*x],x]
Output:
-1/32*(-1/4*Csc[a + b*x]^7/(1 - Csc[a + b*x]^2)^2 + (7*(Csc[a + b*x]^5/(2* (1 - Csc[a + b*x]^2)) - (5*(ArcTanh[Csc[a + b*x]] - Csc[a + b*x] - Csc[a + b*x]^3/3))/2))/4)/b
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_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[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 = 3.98 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05
| method | result | size |
| default | \(\frac {\frac {1}{4 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{4}}-\frac {7}{12 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {35}{24 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {35}{8 \sin \left (b x +a \right )}+\frac {35 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{32 b}\) | \(87\) |
| risch | \(-\frac {i \left (105 \,{\mathrm e}^{13 i \left (b x +a \right )}+70 \,{\mathrm e}^{11 i \left (b x +a \right )}-329 \,{\mathrm e}^{9 i \left (b x +a \right )}-204 \,{\mathrm e}^{7 i \left (b x +a \right )}-329 \,{\mathrm e}^{5 i \left (b x +a \right )}+70 \,{\mathrm e}^{3 i \left (b x +a \right )}+105 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{384 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}+\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{256 b}-\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{256 b}\) | \(148\) |
Input:
int(csc(2*b*x+2*a)^5*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
1/32/b*(1/4/sin(b*x+a)^3/cos(b*x+a)^4-7/12/sin(b*x+a)^3/cos(b*x+a)^2+35/24 /sin(b*x+a)/cos(b*x+a)^2-35/8/sin(b*x+a)+35/8*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.69 \[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=-\frac {210 \, \cos \left (b x + a\right )^{6} - 280 \, \cos \left (b x + a\right )^{4} - 105 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 105 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 42 \, \cos \left (b x + a\right )^{2} + 12}{1536 \, {\left (b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )} \sin \left (b x + a\right )} \] Input:
integrate(csc(2*b*x+2*a)^5*sin(b*x+a),x, algorithm="fricas")
Output:
-1/1536*(210*cos(b*x + a)^6 - 280*cos(b*x + a)^4 - 105*(cos(b*x + a)^6 - c os(b*x + a)^4)*log(sin(b*x + a) + 1)*sin(b*x + a) + 105*(cos(b*x + a)^6 - cos(b*x + a)^4)*log(-sin(b*x + a) + 1)*sin(b*x + a) + 42*cos(b*x + a)^2 + 12)/((b*cos(b*x + a)^6 - b*cos(b*x + a)^4)*sin(b*x + a))
Timed out. \[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=\text {Timed out} \] Input:
integrate(csc(2*b*x+2*a)**5*sin(b*x+a),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 3088 vs. \(2 (73) = 146\).
Time = 0.41 (sec) , antiderivative size = 3088, normalized size of antiderivative = 37.20 \[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(2*b*x+2*a)^5*sin(b*x+a),x, algorithm="maxima")
Output:
1/1536*(4*(105*sin(13*b*x + 13*a) + 70*sin(11*b*x + 11*a) - 329*sin(9*b*x + 9*a) - 204*sin(7*b*x + 7*a) - 329*sin(5*b*x + 5*a) + 70*sin(3*b*x + 3*a) + 105*sin(b*x + a))*cos(14*b*x + 14*a) - 420*(sin(12*b*x + 12*a) - 3*sin( 10*b*x + 10*a) - 3*sin(8*b*x + 8*a) + 3*sin(6*b*x + 6*a) + 3*sin(4*b*x + 4 *a) - sin(2*b*x + 2*a))*cos(13*b*x + 13*a) + 4*(70*sin(11*b*x + 11*a) - 32 9*sin(9*b*x + 9*a) - 204*sin(7*b*x + 7*a) - 329*sin(5*b*x + 5*a) + 70*sin( 3*b*x + 3*a) + 105*sin(b*x + a))*cos(12*b*x + 12*a) + 280*(3*sin(10*b*x + 10*a) + 3*sin(8*b*x + 8*a) - 3*sin(6*b*x + 6*a) - 3*sin(4*b*x + 4*a) + sin (2*b*x + 2*a))*cos(11*b*x + 11*a) + 12*(329*sin(9*b*x + 9*a) + 204*sin(7*b *x + 7*a) + 329*sin(5*b*x + 5*a) - 70*sin(3*b*x + 3*a) - 105*sin(b*x + a)) *cos(10*b*x + 10*a) - 1316*(3*sin(8*b*x + 8*a) - 3*sin(6*b*x + 6*a) - 3*si n(4*b*x + 4*a) + sin(2*b*x + 2*a))*cos(9*b*x + 9*a) + 12*(204*sin(7*b*x + 7*a) + 329*sin(5*b*x + 5*a) - 70*sin(3*b*x + 3*a) - 105*sin(b*x + a))*cos( 8*b*x + 8*a) + 816*(3*sin(6*b*x + 6*a) + 3*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(7*b*x + 7*a) - 84*(47*sin(5*b*x + 5*a) - 10*sin(3*b*x + 3*a) - 1 5*sin(b*x + a))*cos(6*b*x + 6*a) + 1316*(3*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(5*b*x + 5*a) + 420*(2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*cos(4*b *x + 4*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 + 4*a) - cos(2*b*x + 2*a) - 1) *cos(14*b*x + 14*a) + cos(14*b*x + 14*a)^2 - 2*(3*cos(10*b*x + 10*a) + ...
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.02 \[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=-\frac {\frac {6 \, {\left (11 \, \sin \left (b x + a\right )^{3} - 13 \, \sin \left (b x + a\right )\right )}}{{\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2}} + \frac {16 \, {\left (9 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 105 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 105 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{1536 \, b} \] Input:
integrate(csc(2*b*x+2*a)^5*sin(b*x+a),x, algorithm="giac")
Output:
-1/1536*(6*(11*sin(b*x + a)^3 - 13*sin(b*x + a))/(sin(b*x + a)^2 - 1)^2 + 16*(9*sin(b*x + a)^2 + 1)/sin(b*x + a)^3 - 105*log(sin(b*x + a) + 1) + 105 *log(-sin(b*x + a) + 1))/b
Time = 18.41 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=\frac {35\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{256\,b}-\frac {\frac {35\,{\sin \left (a+b\,x\right )}^6}{256}-\frac {175\,{\sin \left (a+b\,x\right )}^4}{768}+\frac {7\,{\sin \left (a+b\,x\right )}^2}{96}+\frac {1}{96}}{b\,\left ({\sin \left (a+b\,x\right )}^7-2\,{\sin \left (a+b\,x\right )}^5+{\sin \left (a+b\,x\right )}^3\right )} \] Input:
int(sin(a + b*x)/sin(2*a + 2*b*x)^5,x)
Output:
(35*atanh(sin(a + b*x)))/(256*b) - ((7*sin(a + b*x)^2)/96 - (175*sin(a + b *x)^4)/768 + (35*sin(a + b*x)^6)/256 + 1/96)/(b*(sin(a + b*x)^3 - 2*sin(a + b*x)^5 + sin(a + b*x)^7))
\[ \int \csc ^5(2 a+2 b x) \sin (a+b x) \, dx=\frac {280 \cos \left (2 b x +2 a \right ) \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )+35 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{3}-490 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )+280 \cos \left (2 b x +2 a \right ) \sin \left (2 b x +2 a \right )-288 \cos \left (2 b x +2 a \right ) \sin \left (b x +a \right )-980 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{3}+232 \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )-630 \left (\int \frac {\tan \left (b x +a \right )^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}d x \right ) \sin \left (2 b x +2 a \right )^{4} b +420 \left (\int \frac {1}{\tan \left (b x +a \right )^{4} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+\tan \left (b x +a \right )^{4}}d x \right ) \sin \left (2 b x +2 a \right )^{4} b +105 \sin \left (2 b x +2 a \right )^{4} \sin \left (b x +a \right )+630 \sin \left (2 b x +2 a \right )^{4} \tan \left (b x +a \right )+105 \sin \left (2 b x +2 a \right )^{4} a -525 \sin \left (2 b x +2 a \right )^{4} b x -735 \sin \left (2 b x +2 a \right )^{3}-70 \sin \left (2 b x +2 a \right )^{2} \sin \left (b x +a \right )+280 \sin \left (2 b x +2 a \right )}{2304 \sin \left (2 b x +2 a \right )^{4} b} \] Input:
int(csc(2*b*x+2*a)^5*sin(b*x+a),x)
Output:
(280*cos(2*a + 2*b*x)*cos(a + b*x)*sin(2*a + 2*b*x) + 35*cos(2*a + 2*b*x)* sin(2*a + 2*b*x)**3 - 490*cos(2*a + 2*b*x)*sin(2*a + 2*b*x)**2*sin(a + b*x ) + 280*cos(2*a + 2*b*x)*sin(2*a + 2*b*x) - 288*cos(2*a + 2*b*x)*sin(a + b *x) - 980*cos(a + b*x)*sin(2*a + 2*b*x)**3 + 232*cos(a + b*x)*sin(2*a + 2* b*x) - 630*int((tan(a + b*x)**2*tan((a + b*x)/2)**2)/(tan((a + b*x)/2)**2 + 1),x)*sin(2*a + 2*b*x)**4*b + 420*int(1/(tan(a + b*x)**4*tan((a + b*x)/2 )**2 + tan(a + b*x)**4),x)*sin(2*a + 2*b*x)**4*b + 105*sin(2*a + 2*b*x)**4 *sin(a + b*x) + 630*sin(2*a + 2*b*x)**4*tan(a + b*x) + 105*sin(2*a + 2*b*x )**4*a - 525*sin(2*a + 2*b*x)**4*b*x - 735*sin(2*a + 2*b*x)**3 - 70*sin(2* a + 2*b*x)**2*sin(a + b*x) + 280*sin(2*a + 2*b*x))/(2304*sin(2*a + 2*b*x)* *4*b)