3.76 \(\int \csc ^5(x) \, dx\)

Optimal. Leaf size=26 \[ -\frac {3}{8} \tanh ^{-1}(\cos (x))-\frac {1}{4} \cot (x) \csc ^3(x)-\frac {3}{8} \cot (x) \csc (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3768, 3770} \[ -\frac {3}{8} \tanh ^{-1}(\cos (x))-\frac {1}{4} \cot (x) \csc ^3(x)-\frac {3}{8} \cot (x) \csc (x) \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^5,x]

[Out]

(-3*ArcTanh[Cos[x]])/8 - (3*Cot[x]*Csc[x])/8 - (Cot[x]*Csc[x]^3)/4

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \csc ^5(x) \, dx &=-\frac {1}{4} \cot (x) \csc ^3(x)+\frac {3}{4} \int \csc ^3(x) \, dx\\ &=-\frac {3}{8} \cot (x) \csc (x)-\frac {1}{4} \cot (x) \csc ^3(x)+\frac {3}{8} \int \csc (x) \, dx\\ &=-\frac {3}{8} \tanh ^{-1}(\cos (x))-\frac {3}{8} \cot (x) \csc (x)-\frac {1}{4} \cot (x) \csc ^3(x)\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 71, normalized size = 2.73 \[ -\frac {1}{64} \csc ^4\left (\frac {x}{2}\right )-\frac {3}{32} \csc ^2\left (\frac {x}{2}\right )+\frac {1}{64} \sec ^4\left (\frac {x}{2}\right )+\frac {3}{32} \sec ^2\left (\frac {x}{2}\right )+\frac {3}{8} \log \left (\sin \left (\frac {x}{2}\right )\right )-\frac {3}{8} \log \left (\cos \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^5,x]

[Out]

(-3*Csc[x/2]^2)/32 - Csc[x/2]^4/64 - (3*Log[Cos[x/2]])/8 + (3*Log[Sin[x/2]])/8 + (3*Sec[x/2]^2)/32 + Sec[x/2]^
4/64

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

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Csc[x]^5,x]

[Out]

Could not integrate

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fricas [B]  time = 0.97, size = 69, normalized size = 2.65 \[ \frac {6 \, \cos \relax (x)^{3} - 3 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 3 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 10 \, \cos \relax (x)}{16 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)^5,x, algorithm="fricas")

[Out]

1/16*(6*cos(x)^3 - 3*(cos(x)^4 - 2*cos(x)^2 + 1)*log(1/2*cos(x) + 1/2) + 3*(cos(x)^4 - 2*cos(x)^2 + 1)*log(-1/
2*cos(x) + 1/2) - 10*cos(x))/(cos(x)^4 - 2*cos(x)^2 + 1)

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giac [A]  time = 0.99, size = 38, normalized size = 1.46 \[ \frac {3 \, \cos \relax (x)^{3} - 5 \, \cos \relax (x)}{8 \, {\left (\cos \relax (x)^{2} - 1\right )}^{2}} - \frac {3}{16} \, \log \left (\cos \relax (x) + 1\right ) + \frac {3}{16} \, \log \left (-\cos \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)^5,x, algorithm="giac")

[Out]

1/8*(3*cos(x)^3 - 5*cos(x))/(cos(x)^2 - 1)^2 - 3/16*log(cos(x) + 1) + 3/16*log(-cos(x) + 1)

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maple [A]  time = 0.31, size = 26, normalized size = 1.00

method	result	size
default	\(\left (-\frac {\left (\csc ^{3}\relax (x )\right )}{4}-\frac {3 \csc \relax (x )}{8}\right ) \cot \relax (x )+\frac {3 \ln \left (\csc \relax (x )-\cot \relax (x )\right )}{8}\)	\(26\)
norman	\(\frac {-\frac {1}{64}-\frac {\left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {\left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{8}+\frac {\left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{64}}{\tan \left (\frac {x}{2}\right )^{4}}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{8}\)	\(42\)
risch	\(\frac {3 \,{\mathrm e}^{7 i x}-11 \,{\mathrm e}^{5 i x}-11 \,{\mathrm e}^{3 i x}+3 \,{\mathrm e}^{i x}}{4 \left ({\mathrm e}^{2 i x}-1\right )^{4}}+\frac {3 \ln \left ({\mathrm e}^{i x}-1\right )}{8}-\frac {3 \ln \left ({\mathrm e}^{i x}+1\right )}{8}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(x)^5,x,method=_RETURNVERBOSE)

[Out]

(-1/4*csc(x)^3-3/8*csc(x))*cot(x)+3/8*ln(csc(x)-cot(x))

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maxima [B]  time = 0.43, size = 42, normalized size = 1.62 \[ \frac {3 \, \cos \relax (x)^{3} - 5 \, \cos \relax (x)}{8 \, {\left (\cos \relax (x)^{4} - 2 \, \cos \relax (x)^{2} + 1\right )}} - \frac {3}{16} \, \log \left (\cos \relax (x) + 1\right ) + \frac {3}{16} \, \log \left (\cos \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)^5,x, algorithm="maxima")

[Out]

1/8*(3*cos(x)^3 - 5*cos(x))/(cos(x)^4 - 2*cos(x)^2 + 1) - 3/16*log(cos(x) + 1) + 3/16*log(cos(x) - 1)

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mupad [B]  time = 0.29, size = 33, normalized size = 1.27 \[ -\frac {3\,\mathrm {atanh}\left (\cos \relax (x)\right )}{8}-\frac {\frac {5\,\cos \relax (x)}{8}-\frac {3\,{\cos \relax (x)}^3}{8}}{{\cos \relax (x)}^4-2\,{\cos \relax (x)}^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sin(x)^5,x)

[Out]

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

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sympy [A]  time = 0.15, size = 46, normalized size = 1.77 \[ \frac {3 \cos ^{3}{\relax (x )} - 5 \cos {\relax (x )}}{8 \cos ^{4}{\relax (x )} - 16 \cos ^{2}{\relax (x )} + 8} + \frac {3 \log {\left (\cos {\relax (x )} - 1 \right )}}{16} - \frac {3 \log {\left (\cos {\relax (x )} + 1 \right )}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sin(x)**5,x)

[Out]

(3*cos(x)**3 - 5*cos(x))/(8*cos(x)**4 - 16*cos(x)**2 + 8) + 3*log(cos(x) - 1)/16 - 3*log(cos(x) + 1)/16

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