Integrand size = 35, antiderivative size = 68 \[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\frac {33}{32} \text {arctanh}\left (\frac {1}{2} \sec (x) \sqrt {\sin (2 x)}\right )-\frac {9 \cos (x)}{16 \sqrt {\sin (2 x)}}-\frac {5 \cos (x) \cot (x)}{24 \sqrt {\sin (2 x)}}+\frac {\cos (x) \cot ^2(x)}{20 \sqrt {\sin (2 x)}} \] Output:
33/32*arctanh(1/2*sin(2*x)^(1/2)/cos(x))-9/16*cos(x)/sin(2*x)^(1/2)-5/24*c os(x)*cot(x)/sin(2*x)^(1/2)+1/20*cos(x)*cot(x)^2/sin(2*x)^(1/2)
Time = 7.08 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.63 \[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\frac {\cos (x) \sqrt {\sin (2 x)} \left (\frac {1}{15} \csc (x) \left (-147-50 \cot (x)+12 \csc ^2(x)\right )+\frac {33 \arctan \left (\frac {\sqrt {\tan \left (\frac {x}{2}\right )}}{\sqrt {-1+\tan ^2\left (\frac {x}{2}\right )}}\right ) \sqrt {-\frac {\cos (x)}{2+2 \cos (x)}} \sec (x)}{\sqrt {\tan \left (\frac {x}{2}\right )}}\right ) (\cos (2 x)-3 \tan (x))}{16 (\cos (x)+\cos (3 x)-6 \sin (x))} \] Input:
Integrate[(Cos[x]^3*(Cos[2*x] - 3*Tan[x]))/((Sin[x]^2 - Sin[2*x])*Sin[2*x] ^(5/2)),x]
Output:
(Cos[x]*Sqrt[Sin[2*x]]*((Csc[x]*(-147 - 50*Cot[x] + 12*Csc[x]^2))/15 + (33 *ArcTan[Sqrt[Tan[x/2]]/Sqrt[-1 + Tan[x/2]^2]]*Sqrt[-(Cos[x]/(2 + 2*Cos[x]) )]*Sec[x])/Sqrt[Tan[x/2]])*(Cos[2*x] - 3*Tan[x]))/(16*(Cos[x] + Cos[3*x] - 6*Sin[x]))
Time = 1.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4890, 4889, 25, 2035, 2333, 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.
| \(\displaystyle \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (x)^3 (\cos (2 x)-3 \tan (x))}{\left (\sin (x)^2-\sin (2 x)\right ) \sin (2 x)^{5/2}}dx\) |
| \(\Big \downarrow \) 4890 |
| \(\displaystyle \frac {\sin ^5(x) \int \frac {\cos (x)^3 \csc ^5(x) (\cos (2 x)-3 \tan (x)) \tan ^{\frac {5}{2}}(x)}{\sin (x)^2-\sin (2 x)}dx}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 4889 |
| \(\displaystyle \frac {\sin ^5(x) \int -\frac {-3 \tan ^3(x)-\tan ^2(x)-3 \tan (x)+1}{(2-\tan (x)) \tan ^{\frac {7}{2}}(x)}d\tan (x)}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sin ^5(x) \int \frac {-3 \tan ^3(x)-\tan ^2(x)-3 \tan (x)+1}{(2-\tan (x)) \tan ^{\frac {7}{2}}(x)}d\tan (x)}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {2 \sin ^5(x) \int \frac {\cot ^6(x) \left (-3 \tan ^3(x)-\tan ^2(x)-3 \tan (x)+1\right )}{2-\tan (x)}d\sqrt {\tan (x)}}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle -\frac {2 \sin ^5(x) \int \left (\frac {\cot ^6(x)}{2}-\frac {5 \cot ^4(x)}{4}-\frac {9 \cot ^2(x)}{8}+\frac {33}{8 (\tan (x)-2)}\right )d\sqrt {\tan (x)}}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 \sin ^5(x) \left (-\frac {33 \text {arctanh}\left (\frac {\sqrt {\tan (x)}}{\sqrt {2}}\right )}{8 \sqrt {2}}-\frac {1}{10} \cot ^5(x)+\frac {5 \cot ^3(x)}{12}+\frac {9 \cot (x)}{8}\right )}{\sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}\) |
Input:
Int[(Cos[x]^3*(Cos[2*x] - 3*Tan[x]))/((Sin[x]^2 - Sin[2*x])*Sin[2*x]^(5/2) ),x]
Output:
(-2*((-33*ArcTanh[Sqrt[Tan[x]]/Sqrt[2]])/(8*Sqrt[2]) + (9*Cot[x])/8 + (5*C ot[x]^3)/12 - Cot[x]^5/10)*Sin[x]^5)/(Sin[2*x]^(5/2)*Tan[x]^(5/2))
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -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 4 vs. order 3.
Time = 1.46 (sec) , antiderivative size = 653, normalized size of antiderivative = 9.60
Input:
int(cos(x)^3*(cos(2*x)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x,meth od=_RETURNVERBOSE)
Output:
1/480*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/tan(1/2*x)^3*(-153*(tan(1/2*x)* (tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan (1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),2/(3+5^(1/2)),1/2*2^(1/2))* 5^(1/2)*tan(1/2*x)^2+153*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x) )^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2* x))^(1/2),-2/(5^(1/2)-3),1/2*2^(1/2))*5^(1/2)*tan(1/2*x)^2+1992*(1+tan(1/2 *x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticE((1+tan(1/ 2*x))^(1/2),1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^2- 1286*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*Elli pticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2 )*tan(1/2*x)^2+765*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x))^(1/2 )*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1 /2),2/(3+5^(1/2)),1/2*2^(1/2))*tan(1/2*x)^2+765*(tan(1/2*x)*(tan(1/2*x)^2- 1))^(1/2)*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2) *EllipticPi((1+tan(1/2*x))^(1/2),-2/(5^(1/2)-3),1/2*2^(1/2))*tan(1/2*x)^2- 12*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^6-140*(tan(1/2*x)*(tan(1 /2*x)^2-1))^(1/2)*tan(1/2*x)^5+12*tan(1/2*x)^4*(tan(1/2*x)*(tan(1/2*x)^2-1 ))^(1/2)+996*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^4+12*(tan(1/2*x)*( tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)^2-996*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*ta n(1/2*x)^2+140*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*tan(1/2*x)-12*(tan(1...
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (52) = 104\).
Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=-\frac {495 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} {\left (4 \, \cos \left (x\right ) + 3 \, \sin \left (x\right )\right )} + \frac {1}{2} \, \cos \left (x\right )^{2} + \frac {7}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) - 495 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right )^{2} + \frac {1}{2} \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} \sin \left (x\right ) - \frac {1}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + 4 \, \sqrt {2} {\left (147 \, \cos \left (x\right )^{2} - 50 \, \cos \left (x\right ) \sin \left (x\right ) - 135\right )} \sqrt {\cos \left (x\right ) \sin \left (x\right )} + 388 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{1920 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \] Input:
integrate(cos(x)^3*(cos(2*x)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2), x, algorithm="fricas")
Output:
-1/1920*(495*(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) - 495*(cos(x )^2 - 1)*log(1/2*cos(x)^2 + 1/2*sqrt(2)*sqrt(cos(x)*sin(x))*sin(x) - 1/2*c os(x)*sin(x) + 1/2)*sin(x) + 4*sqrt(2)*(147*cos(x)^2 - 50*cos(x)*sin(x) - 135)*sqrt(cos(x)*sin(x)) + 388*(cos(x)^2 - 1)*sin(x))/((cos(x)^2 - 1)*sin( x))
Timed out. \[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\text {Timed out} \] Input:
integrate(cos(x)**3*(cos(2*x)-3*tan(x))/(sin(x)**2-sin(2*x))/sin(2*x)**(5/ 2),x)
Output:
Timed out
Timed out. \[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\text {Timed out} \] Input:
integrate(cos(x)^3*(cos(2*x)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2), x, algorithm="maxima")
Output:
Timed out
\[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\int { \frac {{\left (\cos \left (2 \, x\right ) - 3 \, \tan \left (x\right )\right )} \cos \left (x\right )^{3}}{{\left (\sin \left (x\right )^{2} - \sin \left (2 \, x\right )\right )} \sin \left (2 \, x\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate(cos(x)^3*(cos(2*x)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2), x, algorithm="giac")
Output:
integrate((cos(2*x) - 3*tan(x))*cos(x)^3/((sin(x)^2 - sin(2*x))*sin(2*x)^( 5/2)), x)
Timed out. \[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\int -\frac {{\cos \left (x\right )}^3\,\left (\cos \left (2\,x\right )-3\,\mathrm {tan}\left (x\right )\right )}{{\sin \left (2\,x\right )}^{5/2}\,\left (\sin \left (2\,x\right )-{\sin \left (x\right )}^2\right )} \,d x \] Input:
int(-(cos(x)^3*(cos(2*x) - 3*tan(x)))/(sin(2*x)^(5/2)*(sin(2*x) - sin(x)^2 )),x)
Output:
int(-(cos(x)^3*(cos(2*x) - 3*tan(x)))/(sin(2*x)^(5/2)*(sin(2*x) - sin(x)^2 )), x)
\[ \int \frac {\cos ^3(x) (\cos (2 x)-3 \tan (x))}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=-\left (\int \frac {\sqrt {\sin \left (2 x \right )}\, \cos \left (2 x \right ) \cos \left (x \right )^{3}}{\sin \left (2 x \right )^{4}-\sin \left (2 x \right )^{3} \sin \left (x \right )^{2}}d x \right )+3 \left (\int \frac {\sqrt {\sin \left (2 x \right )}\, \cos \left (x \right )^{3} \tan \left (x \right )}{\sin \left (2 x \right )^{4}-\sin \left (2 x \right )^{3} \sin \left (x \right )^{2}}d x \right ) \] Input:
int(cos(x)^3*(cos(2*x)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x)
Output:
- int((sqrt(sin(2*x))*cos(2*x)*cos(x)**3)/(sin(2*x)**4 - sin(2*x)**3*sin( x)**2),x) + 3*int((sqrt(sin(2*x))*cos(x)**3*tan(x))/(sin(2*x)**4 - sin(2*x )**3*sin(x)**2),x)