Integrand size = 27, antiderivative size = 60 \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=-\frac {5}{3} \csc ^2(3 x)+\frac {5}{12} \csc ^4(3 x)-\frac {1}{18} \csc ^6(3 x)-\frac {10}{3} \log (\sin (3 x))+\frac {5}{6} \sin ^2(3 x)-\frac {1}{12} \sin ^4(3 x) \] Output:
-5/3*csc(3*x)^2+5/12*csc(3*x)^4-1/18*csc(3*x)^6-10/3*ln(sin(3*x))+5/6*sin( 3*x)^2-1/12*sin(3*x)^4
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=-\frac {5}{3} \csc ^2(3 x)+\frac {5}{12} \csc ^4(3 x)-\frac {1}{18} \csc ^6(3 x)-\frac {10}{3} \log (\sin (3 x))+\frac {5}{6} \sin ^2(3 x)-\frac {1}{12} \sin ^4(3 x) \] Input:
Integrate[Cot[3*x]*(-1 + Csc[3*x]^2)^3*(1 - Sin[3*x]^2)^2,x]
Output:
(-5*Csc[3*x]^2)/3 + (5*Csc[3*x]^4)/12 - Csc[3*x]^6/18 - (10*Log[Sin[3*x]]) /3 + (5*Sin[3*x]^2)/6 - Sin[3*x]^4/12
Time = 0.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3654, 25, 3042, 4860, 25, 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.
\(\displaystyle \int \left (1-\sin ^2(3 x)\right )^2 \cot (3 x) \left (\csc ^2(3 x)-1\right )^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (1-\sin (3 x)^2\right )^2 \cot (3 x) \left (\csc (3 x)^2-1\right )^3dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \int -\cos ^4(3 x) \cot (3 x) \left (1-\csc ^2(3 x)\right )^3dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \cos ^4(3 x) \cot (3 x) \left (1-\csc ^2(3 x)\right )^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \cos (3 x)^4 \cot (3 x) \left (1-\csc (3 x)^2\right )^3dx\) |
\(\Big \downarrow \) 4860 |
\(\displaystyle -\frac {1}{3} \int -\csc ^7(3 x) \left (1-\sin ^2(3 x)\right )^5d\sin (3 x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \int \csc ^7(3 x) \left (1-\sin ^2(3 x)\right )^5d\sin (3 x)\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{6} \int \csc ^4(3 x) \left (1-\sin ^2(3 x)\right )^5d\sin ^2(3 x)\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{6} \int \left (\csc ^4(3 x)-5 \csc ^3(3 x)+10 \csc ^2(3 x)-10 \csc (3 x)-\sin ^2(3 x)+5\right )d\sin ^2(3 x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (-\frac {1}{2} \sin ^4(3 x)+5 \sin ^2(3 x)-\frac {1}{3} \csc ^3(3 x)+\frac {5}{2} \csc ^2(3 x)-10 \csc (3 x)-10 \log \left (\sin ^2(3 x)\right )\right )\) |
Input:
Int[Cot[3*x]*(-1 + Csc[3*x]^2)^3*(1 - Sin[3*x]^2)^2,x]
Output:
(-10*Csc[3*x] + (5*Csc[3*x]^2)/2 - Csc[3*x]^3/3 - 10*Log[Sin[3*x]^2] + 5*S in[3*x]^2 - Sin[3*x]^4/2)/6
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[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ a^p Int[ActivateTrig[u*cos[e + f*x]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Sin[c*(a + b*x)], x]}, Simp[1/(b*c) Subst[Int[SubstFor[1/x, Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d], x] /; FunctionOfQ[Sin[c*(a + b *x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Cot] || EqQ[F, cot])
Time = 59.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82
method | result | size |
derivativedivides | \(-\frac {\sin \left (3 x \right )^{4}}{12}-\frac {5 \cos \left (3 x \right )^{2}}{6}-\frac {10 \ln \left (\sin \left (3 x \right )\right )}{3}-\frac {5}{3 \sin \left (3 x \right )^{2}}+\frac {5}{12 \sin \left (3 x \right )^{4}}-\frac {1}{18 \sin \left (3 x \right )^{6}}\) | \(49\) |
default | \(-\frac {\sin \left (3 x \right )^{4}}{12}-\frac {5 \cos \left (3 x \right )^{2}}{6}-\frac {10 \ln \left (\sin \left (3 x \right )\right )}{3}-\frac {5}{3 \sin \left (3 x \right )^{2}}+\frac {5}{12 \sin \left (3 x \right )^{4}}-\frac {1}{18 \sin \left (3 x \right )^{6}}\) | \(49\) |
risch | \(10 i x -\frac {{\mathrm e}^{12 i x}}{192}-\frac {3 \,{\mathrm e}^{6 i x}}{16}-\frac {3 \,{\mathrm e}^{-6 i x}}{16}-\frac {{\mathrm e}^{-12 i x}}{192}+\frac {\frac {20 \,{\mathrm e}^{30 i x}}{3}-20 \,{\mathrm e}^{24 i x}+\frac {272 \,{\mathrm e}^{18 i x}}{9}-20 \,{\mathrm e}^{12 i x}+\frac {20 \,{\mathrm e}^{6 i x}}{3}}{\left ({\mathrm e}^{6 i x}-1\right )^{6}}-\frac {10 \ln \left ({\mathrm e}^{6 i x}-1\right )}{3}\) | \(91\) |
Input:
int(cot(3*x)*(-1+csc(3*x)^2)^3*(1-sin(3*x)^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/12*sin(3*x)^4-5/6*cos(3*x)^2-10/3*ln(sin(3*x))-5/3/sin(3*x)^2+5/12/sin( 3*x)^4-1/18/sin(3*x)^6
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (48) = 96\).
Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.72 \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=-\frac {24 \, \cos \left (3 \, x\right )^{10} + 120 \, \cos \left (3 \, x\right )^{8} - 609 \, \cos \left (3 \, x\right )^{6} + 387 \, \cos \left (3 \, x\right )^{4} + 333 \, \cos \left (3 \, x\right )^{2} + 960 \, {\left (\cos \left (3 \, x\right )^{6} - 3 \, \cos \left (3 \, x\right )^{4} + 3 \, \cos \left (3 \, x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (3 \, x\right )\right ) - 271}{288 \, {\left (\cos \left (3 \, x\right )^{6} - 3 \, \cos \left (3 \, x\right )^{4} + 3 \, \cos \left (3 \, x\right )^{2} - 1\right )}} \] Input:
integrate(cot(3*x)*(-1+csc(3*x)^2)^3*(1-sin(3*x)^2)^2,x, algorithm="fricas ")
Output:
-1/288*(24*cos(3*x)^10 + 120*cos(3*x)^8 - 609*cos(3*x)^6 + 387*cos(3*x)^4 + 333*cos(3*x)^2 + 960*(cos(3*x)^6 - 3*cos(3*x)^4 + 3*cos(3*x)^2 - 1)*log( 1/2*sin(3*x)) - 271)/(cos(3*x)^6 - 3*cos(3*x)^4 + 3*cos(3*x)^2 - 1)
Timed out. \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=\text {Timed out} \] Input:
integrate(cot(3*x)*(-1+csc(3*x)**2)**3*(1-sin(3*x)**2)**2,x)
Output:
Timed out
Time = 0.02 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.87 \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=-\frac {1}{12} \, \sin \left (3 \, x\right )^{4} + \frac {5}{6} \, \sin \left (3 \, x\right )^{2} - \frac {60 \, \sin \left (3 \, x\right )^{4} - 15 \, \sin \left (3 \, x\right )^{2} + 2}{36 \, \sin \left (3 \, x\right )^{6}} - \frac {5}{3} \, \log \left (\sin \left (3 \, x\right )^{2}\right ) \] Input:
integrate(cot(3*x)*(-1+csc(3*x)^2)^3*(1-sin(3*x)^2)^2,x, algorithm="maxima ")
Output:
-1/12*sin(3*x)^4 + 5/6*sin(3*x)^2 - 1/36*(60*sin(3*x)^4 - 15*sin(3*x)^2 + 2)/sin(3*x)^6 - 5/3*log(sin(3*x)^2)
Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=-\frac {1}{12} \, \sin \left (3 \, x\right )^{4} + \frac {5}{6} \, \sin \left (3 \, x\right )^{2} + \frac {110 \, \sin \left (3 \, x\right )^{6} - 60 \, \sin \left (3 \, x\right )^{4} + 15 \, \sin \left (3 \, x\right )^{2} - 2}{36 \, \sin \left (3 \, x\right )^{6}} - \frac {5}{3} \, \log \left (\sin \left (3 \, x\right )^{2}\right ) \] Input:
integrate(cot(3*x)*(-1+csc(3*x)^2)^3*(1-sin(3*x)^2)^2,x, algorithm="giac")
Output:
-1/12*sin(3*x)^4 + 5/6*sin(3*x)^2 + 1/36*(110*sin(3*x)^6 - 60*sin(3*x)^4 + 15*sin(3*x)^2 - 2)/sin(3*x)^6 - 5/3*log(sin(3*x)^2)
Time = 18.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.40 \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=\frac {\ln \left ({\left ({\mathrm {tan}\left (3\,x\right )}^2+1\right )}^5\right )}{3}-\frac {10\,\ln \left (\mathrm {tan}\left (3\,x\right )\right )}{3}-\frac {5\,{\mathrm {tan}\left (3\,x\right )}^8+\frac {15\,{\mathrm {tan}\left (3\,x\right )}^6}{2}+\frac {5\,{\mathrm {tan}\left (3\,x\right )}^4}{3}-\frac {5\,{\mathrm {tan}\left (3\,x\right )}^2}{12}+\frac {1}{6}}{3\,\left ({\mathrm {tan}\left (3\,x\right )}^{10}+2\,{\mathrm {tan}\left (3\,x\right )}^8+{\mathrm {tan}\left (3\,x\right )}^6\right )} \] Input:
int(cot(3*x)*(1/sin(3*x)^2 - 1)^3*(sin(3*x)^2 - 1)^2,x)
Output:
log((tan(3*x)^2 + 1)^5)/3 - (10*log(tan(3*x)))/3 - ((5*tan(3*x)^4)/3 - (5* tan(3*x)^2)/12 + (15*tan(3*x)^6)/2 + 5*tan(3*x)^8 + 1/6)/(3*(tan(3*x)^6 + 2*tan(3*x)^8 + tan(3*x)^10))
Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.33 \[ \int \cot (3 x) \left (-1+\csc ^2(3 x)\right )^3 \left (1-\sin ^2(3 x)\right )^2 \, dx=\frac {240 \,\mathrm {log}\left (\tan \left (\frac {3 x}{2}\right )^{2}+1\right ) \sin \left (3 x \right )^{6}-240 \,\mathrm {log}\left (\tan \left (\frac {3 x}{2}\right )\right ) \sin \left (3 x \right )^{6}-6 \sin \left (3 x \right )^{10}+60 \sin \left (3 x \right )^{8}+135 \sin \left (3 x \right )^{6}-120 \sin \left (3 x \right )^{4}+30 \sin \left (3 x \right )^{2}-4}{72 \sin \left (3 x \right )^{6}} \] Input:
int(cot(3*x)*(-1+csc(3*x)^2)^3*(1-sin(3*x)^2)^2,x)
Output:
(240*log(tan((3*x)/2)**2 + 1)*sin(3*x)**6 - 240*log(tan((3*x)/2))*sin(3*x) **6 - 6*sin(3*x)**10 + 60*sin(3*x)**8 + 135*sin(3*x)**6 - 120*sin(3*x)**4 + 30*sin(3*x)**2 - 4)/(72*sin(3*x)**6)