Integrand size = 23, antiderivative size = 49 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=-\frac {1}{4 \left (-5+4 \sin ^2(x)\right )^{3/2}}-\frac {5}{8 \sqrt {-5+4 \sin ^2(x)}}+\frac {1}{8} \sqrt {-5+4 \sin ^2(x)} \] Output:
-1/4/(-5+4*sin(x)^2)^(3/2)-5/8/(-5+4*sin(x)^2)^(1/2)+1/8*(-5+4*sin(x)^2)^( 1/2)
Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.57 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {12+11 \cos (2 x)+\cos (4 x)}{4 \left (-5+4 \sin ^2(x)\right )^{3/2}} \] Input:
Integrate[(Cos[x]*Cos[2*x]*Sin[3*x])/(-5 + 4*Sin[x]^2)^(5/2),x]
Output:
(12 + 11*Cos[2*x] + Cos[4*x])/(4*(-5 + 4*Sin[x]^2)^(3/2))
Time = 0.31 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4856, 1576, 1140, 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.
\(\displaystyle \int \frac {\sin (3 x) \cos (x) \cos (2 x)}{\left (4 \sin ^2(x)-5\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (3 x) \cos (x) \cos (2 x)}{\left (4 \sin (x)^2-5\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4856 |
\(\displaystyle \int \frac {\sin (x) \left (8 \sin ^4(x)-10 \sin ^2(x)+3\right )}{\left (4 \sin ^2(x)-5\right )^{5/2}}d\sin (x)\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \frac {1}{2} \int \frac {8 \sin ^4(x)-10 \sin ^2(x)+3}{\left (4 \sin ^2(x)-5\right )^{5/2}}d\sin ^2(x)\) |
\(\Big \downarrow \) 1140 |
\(\displaystyle \frac {1}{2} \int \left (\frac {1}{2 \sqrt {4 \sin ^2(x)-5}}+\frac {5}{2 \left (4 \sin ^2(x)-5\right )^{3/2}}+\frac {3}{\left (4 \sin ^2(x)-5\right )^{5/2}}\right )d\sin ^2(x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{4} \sqrt {4 \sin ^2(x)-5}-\frac {5}{4 \sqrt {4 \sin ^2(x)-5}}-\frac {1}{2 \left (4 \sin ^2(x)-5\right )^{3/2}}\right )\) |
Input:
Int[(Cos[x]*Cos[2*x]*Sin[3*x])/(-5 + 4*Sin[x]^2)^(5/2),x]
Output:
(-1/2*1/(-5 + 4*Sin[x]^2)^(3/2) - 5/(4*Sqrt[-5 + 4*Sin[x]^2]) + Sqrt[-5 + 4*Sin[x]^2]/4)/2
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x _Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] , x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Sin[c*(a + b*x)], x]}, Simp[d/(b*c) Subst[Int[SubstFor[1, 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, Cos] || EqQ[F, cos])
Time = 0.35 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {1}{2 \left (-4 \cos \left (x \right )^{2}-1\right )^{\frac {3}{2}}}+\frac {7 \cos \left (x \right )^{2}}{2 \left (-4 \cos \left (x \right )^{2}-1\right )^{\frac {3}{2}}}+\frac {2 \cos \left (x \right )^{4}}{\left (-4 \cos \left (x \right )^{2}-1\right )^{\frac {3}{2}}}\) | \(46\) |
default | \(\frac {1}{2 \left (-4 \cos \left (x \right )^{2}-1\right )^{\frac {3}{2}}}+\frac {7 \cos \left (x \right )^{2}}{2 \left (-4 \cos \left (x \right )^{2}-1\right )^{\frac {3}{2}}}+\frac {2 \cos \left (x \right )^{4}}{\left (-4 \cos \left (x \right )^{2}-1\right )^{\frac {3}{2}}}\) | \(46\) |
Input:
int(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE )
Output:
1/2/(-4*cos(x)^2-1)^(3/2)+7/2*cos(x)^2/(-4*cos(x)^2-1)^(3/2)+2*cos(x)^4/(- 4*cos(x)^2-1)^(3/2)
Time = 0.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {{\left (4 \, \cos \left (x\right )^{4} + 7 \, \cos \left (x\right )^{2} + 1\right )} \sqrt {-4 \, \cos \left (x\right )^{2} - 1}}{2 \, {\left (16 \, \cos \left (x\right )^{4} + 8 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x, algorithm="fri cas")
Output:
1/2*(4*cos(x)^4 + 7*cos(x)^2 + 1)*sqrt(-4*cos(x)^2 - 1)/(16*cos(x)^4 + 8*c os(x)^2 + 1)
Timed out. \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)**2)**(5/2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (37) = 74\).
Time = 0.05 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.92 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=-\frac {{\left (\cos \left (11 \, x\right ) + 14 \, \cos \left (9 \, x\right ) + 58 \, \cos \left (7 \, x\right ) + 94 \, \cos \left (5 \, x\right ) + 58 \, \cos \left (3 \, x\right ) + 15 \, \cos \left (x\right )\right )} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 3 \, \sin \left (2 \, x\right ), -\cos \left (4 \, x\right ) - 3 \, \cos \left (2 \, x\right ) - 1\right )\right ) - {\left (\sin \left (11 \, x\right ) + 14 \, \sin \left (9 \, x\right ) + 58 \, \sin \left (7 \, x\right ) + 94 \, \sin \left (5 \, x\right ) + 58 \, \sin \left (3 \, x\right ) + 13 \, \sin \left (x\right )\right )} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 3 \, \sin \left (2 \, x\right ), -\cos \left (4 \, x\right ) - 3 \, \cos \left (2 \, x\right ) - 1\right )\right )}{8 \, {\left (2 \, {\left (3 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 9 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 6 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 9 \, \sin \left (2 \, x\right )^{2} + 6 \, \cos \left (2 \, x\right ) + 1\right )}^{\frac {5}{4}}} \] Input:
integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x, algorithm="max ima")
Output:
-1/8*((cos(11*x) + 14*cos(9*x) + 58*cos(7*x) + 94*cos(5*x) + 58*cos(3*x) + 15*cos(x))*cos(5/2*arctan2(sin(4*x) + 3*sin(2*x), -cos(4*x) - 3*cos(2*x) - 1)) - (sin(11*x) + 14*sin(9*x) + 58*sin(7*x) + 94*sin(5*x) + 58*sin(3*x) + 13*sin(x))*sin(5/2*arctan2(sin(4*x) + 3*sin(2*x), -cos(4*x) - 3*cos(2*x ) - 1)))/(2*(3*cos(2*x) + 1)*cos(4*x) + cos(4*x)^2 + 9*cos(2*x)^2 + sin(4* x)^2 + 6*sin(4*x)*sin(2*x) + 9*sin(2*x)^2 + 6*cos(2*x) + 1)^(5/4)
Time = 0.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.67 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {1}{8} \, \sqrt {4 \, \sin \left (x\right )^{2} - 5} - \frac {20 \, \sin \left (x\right )^{2} - 23}{8 \, {\left (4 \, \sin \left (x\right )^{2} - 5\right )}^{\frac {3}{2}}} \] Input:
integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x, algorithm="gia c")
Output:
1/8*sqrt(4*sin(x)^2 - 5) - 1/8*(20*sin(x)^2 - 23)/(4*sin(x)^2 - 5)^(3/2)
Time = 0.34 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.57 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {2\,{\cos \left (2\,x\right )}^2+11\,\cos \left (2\,x\right )+11}{4\,{\left (-2\,\cos \left (2\,x\right )-3\right )}^{3/2}} \] Input:
int((cos(2*x)*sin(3*x)*cos(x))/(4*sin(x)^2 - 5)^(5/2),x)
Output:
(11*cos(2*x) + 2*cos(2*x)^2 + 11)/(4*(- 2*cos(2*x) - 3)^(3/2))
\[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {\sqrt {4 \sin \left (x \right )^{2}-5}\, \cos \left (2 x \right ) \cos \left (x \right ) \sin \left (3 x \right )}{64 \sin \left (x \right )^{6}-240 \sin \left (x \right )^{4}+300 \sin \left (x \right )^{2}-125}d x \] Input:
int(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x)
Output:
int((sqrt(4*sin(x)**2 - 5)*cos(2*x)*cos(x)*sin(3*x))/(64*sin(x)**6 - 240*s in(x)**4 + 300*sin(x)**2 - 125),x)