Integrand size = 20, antiderivative size = 102 \[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=-\frac {5 \cot (a+b x)}{16 b}-\frac {5 \cot ^3(a+b x)}{64 b}-\frac {3 \cot ^5(a+b x)}{160 b}-\frac {\cot ^7(a+b x)}{448 b}+\frac {15 \tan (a+b x)}{64 b}+\frac {\tan ^3(a+b x)}{32 b}+\frac {\tan ^5(a+b x)}{320 b} \] Output:
-5/16*cot(b*x+a)/b-5/64*cot(b*x+a)^3/b-3/160*cot(b*x+a)^5/b-1/448*cot(b*x+ a)^7/b+15/64*tan(b*x+a)/b+1/32*tan(b*x+a)^3/b+1/320*tan(b*x+a)^5/b
Time = 0.27 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.29 \[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=-\frac {281 \cot (a+b x)}{1120 b}-\frac {53 \cot (a+b x) \csc ^2(a+b x)}{1120 b}-\frac {27 \cot (a+b x) \csc ^4(a+b x)}{2240 b}-\frac {\cot (a+b x) \csc ^6(a+b x)}{448 b}+\frac {33 \tan (a+b x)}{160 b}+\frac {\sec ^2(a+b x) \tan (a+b x)}{40 b}+\frac {\sec ^4(a+b x) \tan (a+b x)}{320 b} \] Input:
Integrate[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^6,x]
Output:
(-281*Cot[a + b*x])/(1120*b) - (53*Cot[a + b*x]*Csc[a + b*x]^2)/(1120*b) - (27*Cot[a + b*x]*Csc[a + b*x]^4)/(2240*b) - (Cot[a + b*x]*Csc[a + b*x]^6) /(448*b) + (33*Tan[a + b*x])/(160*b) + (Sec[a + b*x]^2*Tan[a + b*x])/(40*b ) + (Sec[a + b*x]^4*Tan[a + b*x])/(320*b)
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 4776, 3042, 3100, 244, 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.
\(\displaystyle \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^2 \sin (2 a+2 b x)^6}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{64} \int \csc ^8(a+b x) \sec ^6(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{64} \int \csc (a+b x)^8 \sec (a+b x)^6dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {\int \cot ^8(a+b x) \left (\tan ^2(a+b x)+1\right )^6d\tan (a+b x)}{64 b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\cot ^8(a+b x)+6 \cot ^6(a+b x)+15 \cot ^4(a+b x)+20 \cot ^2(a+b x)+\tan ^4(a+b x)+6 \tan ^2(a+b x)+15\right )d\tan (a+b x)}{64 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{5} \tan ^5(a+b x)+2 \tan ^3(a+b x)+15 \tan (a+b x)-\frac {1}{7} \cot ^7(a+b x)-\frac {6}{5} \cot ^5(a+b x)-5 \cot ^3(a+b x)-20 \cot (a+b x)}{64 b}\) |
Input:
Int[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^6,x]
Output:
(-20*Cot[a + b*x] - 5*Cot[a + b*x]^3 - (6*Cot[a + b*x]^5)/5 - Cot[a + b*x] ^7/7 + 15*Tan[a + b*x] + 2*Tan[a + b*x]^3 + Tan[a + b*x]^5/5)/(64*b)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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]
Result contains complex when optimal does not.
Time = 0.82 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {32 i \left (20 \,{\mathrm e}^{10 i \left (b x +a \right )}-5 \,{\mathrm e}^{8 i \left (b x +a \right )}-10 \,{\mathrm e}^{6 i \left (b x +a \right )}+4 \,{\mathrm e}^{4 i \left (b x +a \right )}+2 \,{\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{35 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{7} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{5}}\) | \(90\) |
default | \(\frac {-\frac {1}{7 \sin \left (b x +a \right )^{7} \cos \left (b x +a \right )^{5}}+\frac {12}{35 \sin \left (b x +a \right )^{5} \cos \left (b x +a \right )^{5}}-\frac {24}{35 \sin \left (b x +a \right )^{5} \cos \left (b x +a \right )^{3}}+\frac {64}{35 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{3}}-\frac {128}{35 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )}+\frac {512}{35 \sin \left (b x +a \right ) \cos \left (b x +a \right )}-\frac {1024 \cot \left (b x +a \right )}{35}}{64 b}\) | \(123\) |
Input:
int(csc(b*x+a)^2*csc(2*b*x+2*a)^6,x,method=_RETURNVERBOSE)
Output:
32/35*I*(20*exp(10*I*(b*x+a))-5*exp(8*I*(b*x+a))-10*exp(6*I*(b*x+a))+4*exp (4*I*(b*x+a))+2*exp(2*I*(b*x+a))-1)/b/(exp(2*I*(b*x+a))-1)^7/(exp(2*I*(b*x +a))+1)^5
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.16 \[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=-\frac {1024 \, \cos \left (b x + a\right )^{12} - 3584 \, \cos \left (b x + a\right )^{10} + 4480 \, \cos \left (b x + a\right )^{8} - 2240 \, \cos \left (b x + a\right )^{6} + 280 \, \cos \left (b x + a\right )^{4} + 28 \, \cos \left (b x + a\right )^{2} + 7}{2240 \, {\left (b \cos \left (b x + a\right )^{11} - 3 \, b \cos \left (b x + a\right )^{9} + 3 \, b \cos \left (b x + a\right )^{7} - b \cos \left (b x + a\right )^{5}\right )} \sin \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^6,x, algorithm="fricas")
Output:
-1/2240*(1024*cos(b*x + a)^12 - 3584*cos(b*x + a)^10 + 4480*cos(b*x + a)^8 - 2240*cos(b*x + a)^6 + 280*cos(b*x + a)^4 + 28*cos(b*x + a)^2 + 7)/((b*c os(b*x + a)^11 - 3*b*cos(b*x + a)^9 + 3*b*cos(b*x + a)^7 - b*cos(b*x + a)^ 5)*sin(b*x + a))
Timed out. \[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)**2*csc(2*b*x+2*a)**6,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 2710 vs. \(2 (88) = 176\).
Time = 0.25 (sec) , antiderivative size = 2710, normalized size of antiderivative = 26.57 \[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^6,x, algorithm="maxima")
Output:
-32/35*((20*sin(10*b*x + 10*a) - 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) + 4*sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*cos(24*b*x + 24*a) - 2*(20*sin( 10*b*x + 10*a) - 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) + 4*sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*cos(22*b*x + 22*a) - 4*(20*sin(10*b*x + 10*a) - 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) + 4*sin(4*b*x + 4*a) + 2*sin(2*b *x + 2*a))*cos(20*b*x + 20*a) + 10*(20*sin(10*b*x + 10*a) - 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) + 4*sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*cos( 18*b*x + 18*a) + 5*(20*sin(10*b*x + 10*a) - 5*sin(8*b*x + 8*a) - 10*sin(6* b*x + 6*a) + 4*sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*cos(16*b*x + 16*a) - 20*(20*sin(10*b*x + 10*a) - 5*sin(8*b*x + 8*a) - 10*sin(6*b*x + 6*a) + 4* sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*cos(14*b*x + 14*a) - (20*cos(10*b*x + 10*a) - 5*cos(8*b*x + 8*a) - 10*cos(6*b*x + 6*a) + 4*cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*sin(24*b*x + 24*a) + 2*(20*cos(10*b*x + 10*a) - 5 *cos(8*b*x + 8*a) - 10*cos(6*b*x + 6*a) + 4*cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*sin(22*b*x + 22*a) + 4*(20*cos(10*b*x + 10*a) - 5*cos(8*b*x + 8*a) - 10*cos(6*b*x + 6*a) + 4*cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1) *sin(20*b*x + 20*a) - 10*(20*cos(10*b*x + 10*a) - 5*cos(8*b*x + 8*a) - 10* cos(6*b*x + 6*a) + 4*cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*sin(18*b*x + 18*a) - 5*(20*cos(10*b*x + 10*a) - 5*cos(8*b*x + 8*a) - 10*cos(6*b*x + 6*a) + 4*cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*sin(16*b*x + 16*a) ...
Time = 0.16 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.75 \[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=\frac {7 \, \tan \left (b x + a\right )^{5} + 70 \, \tan \left (b x + a\right )^{3} - \frac {700 \, \tan \left (b x + a\right )^{6} + 175 \, \tan \left (b x + a\right )^{4} + 42 \, \tan \left (b x + a\right )^{2} + 5}{\tan \left (b x + a\right )^{7}} + 525 \, \tan \left (b x + a\right )}{2240 \, b} \] Input:
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^6,x, algorithm="giac")
Output:
1/2240*(7*tan(b*x + a)^5 + 70*tan(b*x + a)^3 - (700*tan(b*x + a)^6 + 175*t an(b*x + a)^4 + 42*tan(b*x + a)^2 + 5)/tan(b*x + a)^7 + 525*tan(b*x + a))/ b
Time = 18.49 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.81 \[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=\frac {15\,\mathrm {tan}\left (a+b\,x\right )}{64\,b}+\frac {{\mathrm {tan}\left (a+b\,x\right )}^3}{32\,b}+\frac {{\mathrm {tan}\left (a+b\,x\right )}^5}{320\,b}-\frac {{\mathrm {cot}\left (a+b\,x\right )}^7\,\left (\frac {5\,{\mathrm {tan}\left (a+b\,x\right )}^6}{16}+\frac {5\,{\mathrm {tan}\left (a+b\,x\right )}^4}{64}+\frac {3\,{\mathrm {tan}\left (a+b\,x\right )}^2}{160}+\frac {1}{448}\right )}{b} \] Input:
int(1/(sin(a + b*x)^2*sin(2*a + 2*b*x)^6),x)
Output:
(15*tan(a + b*x))/(64*b) + tan(a + b*x)^3/(32*b) + tan(a + b*x)^5/(320*b) - (cot(a + b*x)^7*((3*tan(a + b*x)^2)/160 + (5*tan(a + b*x)^4)/64 + (5*tan (a + b*x)^6)/16 + 1/448))/b
\[ \int \csc ^2(a+b x) \csc ^6(2 a+2 b x) \, dx=\int \csc \left (2 b x +2 a \right )^{6} \csc \left (b x +a \right )^{2}d x \] Input:
int(csc(b*x+a)^2*csc(2*b*x+2*a)^6,x)
Output:
int(csc(2*a + 2*b*x)**6*csc(a + b*x)**2,x)