3.371 \(\int \cos (5 x) \csc ^5(x) \, dx\)

Optimal. Leaf size=20 \[ -\frac {1}{4} \csc ^4(x)+6 \csc ^2(x)+16 \log (\sin (x)) \]

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4366, 1247, 698} \[ -\frac {1}{4} \csc ^4(x)+6 \csc ^2(x)+16 \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Cos[5*x]*Csc[x]^5,x]

[Out]

6*Csc[x]^2 - Csc[x]^4/4 + 16*Log[Sin[x]]

Rule 698

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 1247

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 4366

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[d/(b*c), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2), Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d]
, x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

\begin {align*} \int \cos (5 x) \csc ^5(x) \, dx &=-\operatorname {Subst}\left (\int \frac {x \left (5-20 x^2+16 x^4\right )}{\left (1-x^2\right )^3} \, dx,x,\cos (x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {5-20 x+16 x^2}{(1-x)^3} \, dx,x,\cos ^2(x)\right )\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \left (-\frac {1}{(-1+x)^3}-\frac {12}{(-1+x)^2}-\frac {16}{-1+x}\right ) \, dx,x,\cos ^2(x)\right )\right )\\ &=6 \csc ^2(x)-\frac {\csc ^4(x)}{4}+16 \log (\sin (x))\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \[ -\frac {1}{4} \csc ^4(x)+6 \csc ^2(x)+16 \log (\sin (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[5*x]*Csc[x]^5,x]

[Out]

6*Csc[x]^2 - Csc[x]^4/4 + 16*Log[Sin[x]]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos (5 x) \csc ^5(x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Cos[5*x]*Csc[x]^5,x]

[Out]

Could not integrate

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fricas [B]  time = 0.97, size = 43, normalized size = 2.15 \[ -\frac {24 \, \cos \relax (x)^{2} - 64 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \relax (x)\right ) - 23}{4 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/sin(x)^5,x, algorithm="fricas")

[Out]

-1/4*(24*cos(x)^2 - 64*(cos(x)^4 - 2*cos(x)^2 + 1)*log(1/2*sin(x)) - 23)/(cos(x)^4 - 2*cos(x)^2 + 1)

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giac [A]  time = 0.63, size = 21, normalized size = 1.05 \[ \frac {24 \, \sin \relax (x)^{2} - 1}{4 \, \sin \relax (x)^{4}} + 16 \, \log \left ({\left | \sin \relax (x) \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/sin(x)^5,x, algorithm="giac")

[Out]

1/4*(24*sin(x)^2 - 1)/sin(x)^4 + 16*log(abs(sin(x)))

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maple [A]  time = 0.14, size = 35, normalized size = 1.75

method	result	size
default	\(-\frac {5}{4 \sin \relax (x )^{4}}+\frac {5 \left (\cos ^{4}\relax (x )\right )}{\sin \relax (x )^{4}}-4 \left (\cot ^{4}\relax (x )\right )+8 \left (\cot ^{2}\relax (x )\right )+16 \ln \left (\sin \relax (x )\right )\)	\(35\)
risch	\(-16 i x -\frac {4 \left (6 \,{\mathrm e}^{6 i x}-11 \,{\mathrm e}^{4 i x}+6 \,{\mathrm e}^{2 i x}\right )}{\left ({\mathrm e}^{2 i x}-1\right )^{4}}+16 \ln \left ({\mathrm e}^{2 i x}-1\right )\)	\(49\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(5*x)/sin(x)^5,x,method=_RETURNVERBOSE)

[Out]

-5/4/sin(x)^4+5/sin(x)^4*cos(x)^4-4*cot(x)^4+8*cot(x)^2+16*ln(sin(x))

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maxima [A]  time = 0.43, size = 33, normalized size = 1.65 \[ \frac {5}{\sin \relax (x)^{2}} + \frac {4 \, \sin \relax (x)^{2} - 1}{4 \, \sin \relax (x)^{4}} + \frac {11}{2} \, \log \left (\sin \relax (x)^{2}\right ) + 5 \, \log \left (\sin \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/sin(x)^5,x, algorithm="maxima")

[Out]

5/sin(x)^2 + 1/4*(4*sin(x)^2 - 1)/sin(x)^4 + 11/2*log(sin(x)^2) + 5*log(sin(x))

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mupad [B]  time = 0.07, size = 21, normalized size = 1.05 \[ 8\,\ln \left ({\sin \relax (x)}^2\right )+\frac {6\,{\sin \relax (x)}^2-\frac {1}{4}}{{\sin \relax (x)}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(5*x)/sin(x)^5,x)

[Out]

8*log(sin(x)^2) + (6*sin(x)^2 - 1/4)/sin(x)^4

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sympy [A]  time = 24.84, size = 22, normalized size = 1.10 \[ 8 \log {\left (\sin ^{2}{\relax (x )} \right )} + \frac {6}{\sin ^{2}{\relax (x )}} - \frac {1}{4 \sin ^{4}{\relax (x )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(5*x)/sin(x)**5,x)

[Out]

8*log(sin(x)**2) + 6/sin(x)**2 - 1/(4*sin(x)**4)

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