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

Optimal. Leaf size=49 \[ -\frac {1}{4 \left (-5+4 \sin ^2(x)\right )^{3/2}}-\frac {5}{8 \sqrt {-5+4 \sin ^2(x)}}+\frac {1}{8} \sqrt {-5+4 \sin ^2(x)} \]

[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)

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Rubi [A]
time = 0.08, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4441, 1261, 712} \begin {gather*} \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}} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {\cos (x) \cos (2 x) \sin (3 x)}{\left (-5+4 \sin ^2(x)\right )^{5/2}} \, dx &=\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*}

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Mathematica [A]
time = 0.07, size = 28, normalized size = 0.57 \begin {gather*} \frac {12+11 \cos (2 x)+\cos (4 x)}{4 \left (-5+4 \sin ^2(x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[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))

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Maple [A]
time = 0.08, size = 46, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {2 \left (\cos ^{4}\left (x \right )\right )}{\left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )^{\frac {3}{2}}}+\frac {7 \left (\cos ^{2}\left (x \right )\right )}{2 \left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )^{\frac {3}{2}}}+\frac {1}{2 \left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )^{\frac {3}{2}}}\)	\(46\)
default	\(\frac {2 \left (\cos ^{4}\left (x \right )\right )}{\left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )^{\frac {3}{2}}}+\frac {7 \left (\cos ^{2}\left (x \right )\right )}{2 \left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )^{\frac {3}{2}}}+\frac {1}{2 \left (-4 \left (\cos ^{2}\left (x \right )\right )-1\right )^{\frac {3}{2}}}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*cos(x)^4/(-4*cos(x)^2-1)^(3/2)+7/2*cos(x)^2/(-4*cos(x)^2-1)^(3/2)+1/2/(-4*cos(x)^2-1)^(3/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (37) = 74\).
time = 1.06, size = 192, normalized size = 3.92 \begin {gather*} -\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6436 deep

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Giac [C] Result contains complex when optimal does not.
time = 1.23, size = 33, normalized size = 0.67 \begin {gather*} \frac {1}{8} i \, \sqrt {4 \, \cos \left (x\right )^{2} + 1} - \frac {-20 i \, \cos \left (x\right )^{2} - 3 i}{8 \, {\left (4 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*I*sqrt(4*cos(x)^2 + 1) - 1/8*(-20*I*cos(x)^2 - 3*I)/(4*cos(x)^2 + 1)^(3/2)

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Mupad [B]
time = 0.63, size = 28, normalized size = 0.57 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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