\(\int \frac {\cos (x) \cos (2 x) \sin (3 x)}{(-5+4 \sin ^2(x))^{5/2}} \, dx\) [425]

Optimal result

Integrand size = 23, antiderivative size = 49 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=-\frac {1}{4 \left (-5+4 \sin ^2(x)\right )^{3/2}}-\frac {5}{8 \sqrt {-5+4 \sin ^2(x)}}+\frac {1}{8} \sqrt {-5+4 \sin ^2(x)} \]

[Out]

-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)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4441, 1261, 712} \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {1}{8} \sqrt {4 \sin ^2(x)-5}-\frac {5}{8 \sqrt {4 \sin ^2(x)-5}}-\frac {1}{4 \left (4 \sin ^2(x)-5\right )^{3/2}} \]

[In]

Int[(Cos[x]*Cos[2*x]*Sin[3*x])/(-5 + 4*Sin[x]^2)^(5/2),x]

[Out]

-1/4*1/(-5 + 4*Sin[x]^2)^(3/2) - 5/(8*Sqrt[-5 + 4*Sin[x]^2]) + Sqrt[-5 + 4*Sin[x]^2]/8

Rule 712

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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1261

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[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]

Rule 4441

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist[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])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x \left (3-10 x^2+8 x^4\right )}{\left (-5+4 x^2\right )^{5/2}} \, dx,x,\sin (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {3-10 x+8 x^2}{(-5+4 x)^{5/2}} \, dx,x,\sin ^2(x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {3}{(-5+4 x)^{5/2}}+\frac {5}{2 (-5+4 x)^{3/2}}+\frac {1}{2 \sqrt {-5+4 x}}\right ) \, dx,x,\sin ^2(x)\right ) \\ & = -\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)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.57 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {12+11 \cos (2 x)+\cos (4 x)}{4 \left (-5+4 \sin ^2(x)\right )^{3/2}} \]

[In]

Integrate[(Cos[x]*Cos[2*x]*Sin[3*x])/(-5 + 4*Sin[x]^2)^(5/2),x]

[Out]

(12 + 11*Cos[2*x] + Cos[4*x])/(4*(-5 + 4*Sin[x]^2)^(3/2))

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\sqrt {-4 \left (\cos ^{2}\left (x \right )\right )-1}}{8}-\frac {1}{4 {\left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )}^{\frac {3}{2}}}-\frac {5}{8 \sqrt {-4 \left (\cos ^{2}\left (x \right )\right )-1}}\)	\(38\)
default	\(\frac {\sqrt {-4 \left (\cos ^{2}\left (x \right )\right )-1}}{8}-\frac {1}{4 {\left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )}^{\frac {3}{2}}}-\frac {5}{8 \sqrt {-4 \left (\cos ^{2}\left (x \right )\right )-1}}\)	\(38\)

[In]

int(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/8*(-4*cos(x)^2-1)^(1/2)-1/4/(-4*cos(x)^2-1)^(3/2)-5/8/(-4*cos(x)^2-1)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {{\left (4 \, \cos \left (x\right )^{4} + 7 \, \cos \left (x\right )^{2} + 1\right )} \sqrt {-4 \, \cos \left (x\right )^{2} - 1}}{2 \, {\left (16 \, \cos \left (x\right )^{4} + 8 \, \cos \left (x\right )^{2} + 1\right )}} \]

[In]

integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x, algorithm="fricas")

[Out]

1/2*(4*cos(x)^4 + 7*cos(x)^2 + 1)*sqrt(-4*cos(x)^2 - 1)/(16*cos(x)^4 + 8*cos(x)^2 + 1)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)**2)**(5/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (37) = 74\).

Time = 0.23 (sec) , antiderivative size = 192, normalized size of antiderivative = 3.92 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=-\frac {{\left (\cos \left (11 \, x\right ) + 14 \, \cos \left (9 \, x\right ) + 58 \, \cos \left (7 \, x\right ) + 94 \, \cos \left (5 \, x\right ) + 58 \, \cos \left (3 \, x\right ) + 15 \, \cos \left (x\right )\right )} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 3 \, \sin \left (2 \, x\right ), -\cos \left (4 \, x\right ) - 3 \, \cos \left (2 \, x\right ) - 1\right )\right ) - {\left (\sin \left (11 \, x\right ) + 14 \, \sin \left (9 \, x\right ) + 58 \, \sin \left (7 \, x\right ) + 94 \, \sin \left (5 \, x\right ) + 58 \, \sin \left (3 \, x\right ) + 13 \, \sin \left (x\right )\right )} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (4 \, x\right ) + 3 \, \sin \left (2 \, x\right ), -\cos \left (4 \, x\right ) - 3 \, \cos \left (2 \, x\right ) - 1\right )\right )}{8 \, {\left (2 \, {\left (3 \, \cos \left (2 \, x\right ) + 1\right )} \cos \left (4 \, x\right ) + \cos \left (4 \, x\right )^{2} + 9 \, \cos \left (2 \, x\right )^{2} + \sin \left (4 \, x\right )^{2} + 6 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) + 9 \, \sin \left (2 \, x\right )^{2} + 6 \, \cos \left (2 \, x\right ) + 1\right )}^{\frac {5}{4}}} \]

[In]

integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.67 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {1}{8} \, \sqrt {4 \, \sin \left (x\right )^{2} - 5} - \frac {20 \, \sin \left (x\right )^{2} - 23}{8 \, {\left (4 \, \sin \left (x\right )^{2} - 5\right )}^{\frac {3}{2}}} \]

[In]

integrate(cos(x)*cos(2*x)*sin(3*x)/(-5+4*sin(x)^2)^(5/2),x, algorithm="giac")

[Out]

1/8*sqrt(4*sin(x)^2 - 5) - 1/8*(20*sin(x)^2 - 23)/(4*sin(x)^2 - 5)^(3/2)

Mupad [B] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.57 \[ \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx=\frac {2\,{\cos \left (2\,x\right )}^2+11\,\cos \left (2\,x\right )+11}{4\,{\left (-2\,\cos \left (2\,x\right )-3\right )}^{3/2}} \]

[In]

int((cos(2*x)*sin(3*x)*cos(x))/(4*sin(x)^2 - 5)^(5/2),x)

[Out]

(11*cos(2*x) + 2*cos(2*x)^2 + 11)/(4*(- 2*cos(2*x) - 3)^(3/2))