Integrand size = 30, antiderivative size = 95 \[ \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\frac {\cos ^5(x)}{5 \sin ^{\frac {5}{2}}(2 x)}+\frac {\cos ^4(x) \sin (x)}{6 \sin ^{\frac {5}{2}}(2 x)}-\frac {3 \cos ^3(x) \sin ^2(x)}{4 \sin ^{\frac {5}{2}}(2 x)}+\frac {3 \text {arctanh}\left (\frac {\sqrt {\tan (x)}}{\sqrt {2}}\right ) \sin ^5(x)}{4 \sqrt {2} \sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)} \]
1/5*cos(x)^5/sin(2*x)^(5/2)+1/6*cos(x)^4*sin(x)/sin(2*x)^(5/2)-3/4*cos(x)^ 3*sin(x)^2/sin(2*x)^(5/2)+3/8*arctanh(1/2*tan(x)^(1/2)*2^(1/2))*sin(x)^5/s in(2*x)^(5/2)*2^(1/2)/tan(x)^(5/2)
Time = 4.51 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\frac {\sec (x) \sqrt {\sin (2 x)} \left (4 \cot (x) \csc ^2(x) (-33+57 \cos (2 x)+10 \sin (2 x))+\frac {180 \sqrt {2} \arctan \left (\frac {\sqrt {\tan \left (\frac {x}{2}\right )}}{\sqrt {-1+\tan ^2\left (\frac {x}{2}\right )}}\right ) \sqrt {-\frac {1}{1+\sec (x)}}}{\sqrt {\tan \left (\frac {x}{2}\right )}}\right )}{3840} \]
(Sec[x]*Sqrt[Sin[2*x]]*(4*Cot[x]*Csc[x]^2*(-33 + 57*Cos[2*x] + 10*Sin[2*x] ) + (180*Sqrt[2]*ArcTan[Sqrt[Tan[x/2]]/Sqrt[-1 + Tan[x/2]^2]]*Sqrt[-(1 + S ec[x])^(-1)])/Sqrt[Tan[x/2]]))/3840
Time = 0.77 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.68, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 4890, 4889, 25, 518, 1585, 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.
| \(\displaystyle \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (x)^3 \cos (2 x)}{\left (\sin (x)^2-\sin (2 x)\right ) \sin (2 x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4890 |
| \(\displaystyle \frac {\sin ^5(x) \int \frac {\cos (x)^3 \cos (2 x) \csc ^5(x) \tan ^{\frac {5}{2}}(x)}{\sin (x)^2-\sin (2 x)}dx}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 4889 |
| \(\displaystyle \frac {\sin ^5(x) \int -\frac {1-\tan ^2(x)}{(2-\tan (x)) \tan ^{\frac {7}{2}}(x)}d\tan (x)}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sin ^5(x) \int \frac {1-\tan ^2(x)}{(2-\tan (x)) \tan ^{\frac {7}{2}}(x)}d\tan (x)}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 518 |
| \(\displaystyle -\frac {2 \sin ^5(x) \int \frac {\cot ^6(x) \left (1-\tan ^2(x)\right )}{2-\tan (x)}d\sqrt {\tan (x)}}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 1585 |
| \(\displaystyle -\frac {2 \sin ^5(x) \int \left (\frac {\cot ^6(x)}{2}+\frac {\cot ^4(x)}{4}-\frac {3 \cot ^2(x)}{8}+\frac {3}{8 (\tan (x)-2)}\right )d\sqrt {\tan (x)}}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sin ^5(x) \left (-\frac {3 \text {arctanh}\left (\frac {\sqrt {\tan (x)}}{\sqrt {2}}\right )}{8 \sqrt {2}}-\frac {1}{10} \cot ^5(x)-\frac {\cot ^3(x)}{12}+\frac {3 \cot (x)}{8}\right )}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
(-2*((-3*ArcTanh[Sqrt[Tan[x]]/Sqrt[2]])/(8*Sqrt[2]) + (3*Cot[x])/8 - Cot[x ]^3/12 - Cot[x]^5/10)*Sin[x]^5)/(Sin[2*x]^(5/2)*Tan[x]^(5/2))
3.7.36.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^(2*m + 1)*(e*c + d*x^2)^ n*(a*e^2 + b*x^4)^p, x], x, Sqrt[e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && IGtQ[p, 0] && IGtQ[q, -2]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, With[{d = FreeFactors [Tan[v], x]}, Simp[d/Coefficient[v, x, 1] Subst[Int[SubstFor[1/(1 + d^2*x ^2), Tan[v]/d, u, x], x], x, Tan[v]/d], x]] /; !FalseQ[v] && FunctionOfQ[N onfreeFactors[Tan[v], x], u, x]] /; InverseFunctionFreeQ[u, x] && !MatchQ[ u, (v_.)*((c_.)*tan[w_]^(n_.)*tan[z_]^(n_.))^(p_.) /; FreeQ[{c, p}, x] && I ntegerQ[n] && LinearQ[w, x] && EqQ[z, 2*w]]
Int[(u_)*((c_.)*sin[v_])^(m_), x_Symbol] :> With[{w = FunctionOfTrig[u*(Sin [v/2]^(2*m)/(c*Tan[v/2])^m), x]}, Simp[(c*Sin[v])^m*((c*Tan[v/2])^m/Sin[v/2 ]^(2*m)) Int[u*(Sin[v/2]^(2*m)/(c*Tan[v/2])^m), x], x] /; !FalseQ[w] && FunctionOfQ[NonfreeFactors[Tan[w], x], u*(Sin[v/2]^(2*m)/(c*Tan[v/2])^m), x ]] /; FreeQ[c, x] && LinearQ[v, x] && IntegerQ[m + 1/2] && !SumQ[u] && Inv erseFunctionFreeQ[u, x]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.83 (sec) , antiderivative size = 761, normalized size of antiderivative = 8.01
1/3840*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/tan(1/2*x)^3*(-4464*(tan(1/2*x )+1)^(1/2)*(-tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*EllipticE((tan(1/2* x)+1)^(1/2),1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)-1)*(tan(1/2*x)+1))^(1/2)* (tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^2+1772*(tan(1/2*x)+1)^(1/2) *(-tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*EllipticF((tan(1/2*x)+1)^(1/2 ),1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)-1)*(tan(1/2*x)+1))^(1/2)*(tan(1/2*x )*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^2+24*(tan(1/2*x)*(tan(1/2*x)-1)*(tan( 1/2*x)+1))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^6+3*2^(1/2 )*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(tan(1/2*x)+1 ))^(1/2)*sum((6*_alpha^3+7*_alpha^2+6*_alpha+1)*(_alpha^3+2*_alpha-3)*(tan (1/2*x)+1)^(1/2)*(1-tan(1/2*x))^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((tan( 1/2*x)+1)^(1/2),-1/4*_alpha^3-1/2*_alpha+3/4,1/2*2^(1/2))/(tan(1/2*x)*(tan (1/2*x)^2-1))^(1/2),_alpha=RootOf(_Z^4+_Z^3+2*_Z^2-_Z+1))*(tan(1/2*x)*(tan (1/2*x)^2-1))^(1/2)*tan(1/2*x)^2-40*(tan(1/2*x)*(tan(1/2*x)-1)*(tan(1/2*x) +1))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^5-1272*tan(1/2*x )^4*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(tan(1/2*x) +1))^(1/2)-1920*tan(1/2*x)^4*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*( tan(1/2*x)^2-1))^(1/2)-24*tan(1/2*x)^4*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2) *(tan(1/2*x)*(tan(1/2*x)-1)*(tan(1/2*x)+1))^(1/2)+1272*(tan(1/2*x)^3-tan(1 /2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(tan(1/2*x)+1))^(1/2)*tan(1/2*x...
Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.43 \[ \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=-\frac {45 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} {\left (4 \, \cos \left (x\right ) + 3 \, \sin \left (x\right )\right )} + \frac {1}{2} \, \cos \left (x\right )^{2} + \frac {7}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) - 45 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right )^{2} + \frac {1}{2} \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} \sin \left (x\right ) - \frac {1}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + 4 \, \sqrt {2} {\left (57 \, \cos \left (x\right )^{2} + 10 \, \cos \left (x\right ) \sin \left (x\right ) - 45\right )} \sqrt {\cos \left (x\right ) \sin \left (x\right )} + 268 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{1920 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \]
-1/1920*(45*(cos(x)^2 - 1)*log(-1/2*sqrt(2)*sqrt(cos(x)*sin(x))*(4*cos(x) + 3*sin(x)) + 1/2*cos(x)^2 + 7/2*cos(x)*sin(x) + 1/2)*sin(x) - 45*(cos(x)^ 2 - 1)*log(1/2*cos(x)^2 + 1/2*sqrt(2)*sqrt(cos(x)*sin(x))*sin(x) - 1/2*cos (x)*sin(x) + 1/2)*sin(x) + 4*sqrt(2)*(57*cos(x)^2 + 10*cos(x)*sin(x) - 45) *sqrt(cos(x)*sin(x)) + 268*(cos(x)^2 - 1)*sin(x))/((cos(x)^2 - 1)*sin(x))
Timed out. \[ \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\int { \frac {\cos \left (2 \, x\right ) \cos \left (x\right )^{3}}{{\left (\sin \left (x\right )^{2} - \sin \left (2 \, x\right )\right )} \sin \left (2 \, x\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\cos ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\int -\frac {\cos \left (2\,x\right )\,{\cos \left (x\right )}^3}{{\sin \left (2\,x\right )}^{5/2}\,\left (\sin \left (2\,x\right )-{\sin \left (x\right )}^2\right )} \,d x \]