Integrand size = 20, antiderivative size = 60 \[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {3 \cot ^2(a+b x)}{16 b}-\frac {\cot ^4(a+b x)}{32 b}+\frac {3 \log (\tan (a+b x))}{8 b}+\frac {\tan ^2(a+b x)}{16 b} \] Output:
-3/16*cot(b*x+a)^2/b-1/32*cot(b*x+a)^4/b+3/8*ln(tan(b*x+a))/b+1/16*tan(b*x +a)^2/b
Time = 0.24 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {4 \csc ^2(a+b x)+\csc ^4(a+b x)+12 \log (\cos (a+b x))-12 \log (\sin (a+b x))-2 \sec ^2(a+b x)}{32 b} \] Input:
Integrate[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^3,x]
Output:
-1/32*(4*Csc[a + b*x]^2 + Csc[a + b*x]^4 + 12*Log[Cos[a + b*x]] - 12*Log[S in[a + b*x]] - 2*Sec[a + b*x]^2)/b
Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.78, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 4776, 3042, 3100, 243, 49, 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.
\(\displaystyle \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (a+b x)^2 \sin (2 a+2 b x)^3}dx\) |
\(\Big \downarrow \) 4776 |
\(\displaystyle \frac {1}{8} \int \csc ^5(a+b x) \sec ^3(a+b x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{8} \int \csc (a+b x)^5 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3100 |
\(\displaystyle \frac {\int \cot ^5(a+b x) \left (\tan ^2(a+b x)+1\right )^3d\tan (a+b x)}{8 b}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\int \cot ^3(a+b x) \left (\tan ^2(a+b x)+1\right )^3d\tan ^2(a+b x)}{16 b}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (\cot ^3(a+b x)+3 \cot ^2(a+b x)+3 \cot (a+b x)+1\right )d\tan ^2(a+b x)}{16 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\tan ^2(a+b x)-\frac {1}{2} \cot ^2(a+b x)-3 \cot (a+b x)+3 \log \left (\tan ^2(a+b x)\right )}{16 b}\) |
Input:
Int[Csc[a + b*x]^2*Csc[2*a + 2*b*x]^3,x]
Output:
(-3*Cot[a + b*x] - Cot[a + b*x]^2/2 + 3*Log[Tan[a + b*x]^2] + Tan[a + b*x] ^2)/(16*b)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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]
Time = 0.33 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.03
method | result | size |
default | \(\frac {-\frac {1}{4 \sin \left (b x +a \right )^{4} \cos \left (b x +a \right )^{2}}+\frac {3}{4 \sin \left (b x +a \right )^{2} \cos \left (b x +a \right )^{2}}-\frac {3}{2 \sin \left (b x +a \right )^{2}}+3 \ln \left (\tan \left (b x +a \right )\right )}{8 b}\) | \(62\) |
risch | \(\frac {3 \,{\mathrm e}^{10 i \left (b x +a \right )}-6 \,{\mathrm e}^{8 i \left (b x +a \right )}-2 \,{\mathrm e}^{6 i \left (b x +a \right )}-6 \,{\mathrm e}^{4 i \left (b x +a \right )}+3 \,{\mathrm e}^{2 i \left (b x +a \right )}}{4 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{8 b}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{8 b}\) | \(123\) |
Input:
int(csc(b*x+a)^2*csc(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)
Output:
1/8/b*(-1/4/sin(b*x+a)^4/cos(b*x+a)^2+3/4/sin(b*x+a)^2/cos(b*x+a)^2-3/2/si n(b*x+a)^2+3*ln(tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 138 vs. \(2 (52) = 104\).
Time = 0.08 (sec) , antiderivative size = 138, normalized size of antiderivative = 2.30 \[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {6 \, \cos \left (b x + a\right )^{4} - 9 \, \cos \left (b x + a\right )^{2} - 6 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) + 6 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (b x + a\right )^{2} + \frac {1}{4}\right ) + 2}{32 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )}} \] Input:
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^3,x, algorithm="fricas")
Output:
1/32*(6*cos(b*x + a)^4 - 9*cos(b*x + a)^2 - 6*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*log(cos(b*x + a)^2) + 6*(cos(b*x + a)^6 - 2*cos(b *x + a)^4 + cos(b*x + a)^2)*log(-1/4*cos(b*x + a)^2 + 1/4) + 2)/(b*cos(b*x + a)^6 - 2*b*cos(b*x + a)^4 + b*cos(b*x + a)^2)
\[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=\int \csc ^{2}{\left (a + b x \right )} \csc ^{3}{\left (2 a + 2 b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**2*csc(2*b*x+2*a)**3,x)
Output:
Integral(csc(a + b*x)**2*csc(2*a + 2*b*x)**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 3188 vs. \(2 (52) = 104\).
Time = 0.24 (sec) , antiderivative size = 3188, normalized size of antiderivative = 53.13 \[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=\text {Too large to display} \] Input:
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^3,x, algorithm="maxima")
Output:
1/16*(4*(3*cos(10*b*x + 10*a) - 6*cos(8*b*x + 8*a) - 2*cos(6*b*x + 6*a) - 6*cos(4*b*x + 4*a) + 3*cos(2*b*x + 2*a))*cos(12*b*x + 12*a) + 4*(9*cos(8*b *x + 8*a) + 16*cos(6*b*x + 6*a) + 9*cos(4*b*x + 4*a) - 12*cos(2*b*x + 2*a) + 3)*cos(10*b*x + 10*a) - 24*cos(10*b*x + 10*a)^2 - 4*(22*cos(6*b*x + 6*a ) - 12*cos(4*b*x + 4*a) - 9*cos(2*b*x + 2*a) + 6)*cos(8*b*x + 8*a) + 24*co s(8*b*x + 8*a)^2 - 8*(11*cos(4*b*x + 4*a) - 8*cos(2*b*x + 2*a) + 1)*cos(6* b*x + 6*a) - 32*cos(6*b*x + 6*a)^2 + 12*(3*cos(2*b*x + 2*a) - 2)*cos(4*b*x + 4*a) + 24*cos(4*b*x + 4*a)^2 - 24*cos(2*b*x + 2*a)^2 + 3*(2*(2*cos(10*b *x + 10*a) + cos(8*b*x + 8*a) - 4*cos(6*b*x + 6*a) + cos(4*b*x + 4*a) + 2* cos(2*b*x + 2*a) - 1)*cos(12*b*x + 12*a) - cos(12*b*x + 12*a)^2 - 4*(cos(8 *b*x + 8*a) - 4*cos(6*b*x + 6*a) + cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(10*b*x + 10*a) - 4*cos(10*b*x + 10*a)^2 + 2*(4*cos(6*b*x + 6*a) - cos(4*b*x + 4*a) - 2*cos(2*b*x + 2*a) + 1)*cos(8*b*x + 8*a) - cos(8*b*x + 8*a)^2 + 8*(cos(4*b*x + 4*a) + 2*cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - 16*cos(6*b*x + 6*a)^2 - 2*(2*cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos( 4*b*x + 4*a)^2 - 4*cos(2*b*x + 2*a)^2 + 2*(2*sin(10*b*x + 10*a) + sin(8*b* x + 8*a) - 4*sin(6*b*x + 6*a) + sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin (12*b*x + 12*a) - sin(12*b*x + 12*a)^2 - 4*(sin(8*b*x + 8*a) - 4*sin(6*b*x + 6*a) + sin(4*b*x + 4*a) + 2*sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - 4*si n(10*b*x + 10*a)^2 + 2*(4*sin(6*b*x + 6*a) - sin(4*b*x + 4*a) - 2*sin(2...
Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23 \[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {\frac {6 \, \sin \left (b x + a\right )^{4} - 3 \, \sin \left (b x + a\right )^{2} - 1}{{\left (\sin \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )^{4}} + 6 \, \log \left (-\sin \left (b x + a\right )^{2} + 1\right ) - 12 \, \log \left ({\left | \sin \left (b x + a\right ) \right |}\right )}{32 \, b} \] Input:
integrate(csc(b*x+a)^2*csc(2*b*x+2*a)^3,x, algorithm="giac")
Output:
-1/32*((6*sin(b*x + a)^4 - 3*sin(b*x + a)^2 - 1)/((sin(b*x + a)^2 - 1)*sin (b*x + a)^4) + 6*log(-sin(b*x + a)^2 + 1) - 12*log(abs(sin(b*x + a))))/b
Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.37 \[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {3\,\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{16\,b}-\frac {3\,\ln \left (\cos \left (a+b\,x\right )\right )}{8\,b}+\frac {\frac {3\,{\cos \left (a+b\,x\right )}^4}{16}-\frac {9\,{\cos \left (a+b\,x\right )}^2}{32}+\frac {1}{16}}{b\,\left ({\cos \left (a+b\,x\right )}^6-2\,{\cos \left (a+b\,x\right )}^4+{\cos \left (a+b\,x\right )}^2\right )} \] Input:
int(1/(sin(a + b*x)^2*sin(2*a + 2*b*x)^3),x)
Output:
(3*log(sin(a + b*x)^2))/(16*b) - (3*log(cos(a + b*x)))/(8*b) + ((3*cos(a + b*x)^4)/16 - (9*cos(a + b*x)^2)/32 + 1/16)/(b*(cos(a + b*x)^2 - 2*cos(a + b*x)^4 + cos(a + b*x)^6))
\[ \int \csc ^2(a+b x) \csc ^3(2 a+2 b x) \, dx=\int \csc \left (2 b x +2 a \right )^{3} \csc \left (b x +a \right )^{2}d x \] Input:
int(csc(b*x+a)^2*csc(2*b*x+2*a)^3,x)
Output:
int(csc(2*a + 2*b*x)**3*csc(a + b*x)**2,x)