3.1.79 \(\int \cos ^2(x) \cot ^3(x) \, dx\) [79]

Optimal. Leaf size=22 \[ -\frac {1}{2} \csc ^2(x)-2 \log (\sin (x))+\frac {\sin ^2(x)}{2} \]

[Out]

-1/2*csc(x)^2-2*ln(sin(x))+1/2*sin(x)^2

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Rubi [A]
time = 0.02, antiderivative size = 22, 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 = {2670, 272, 45} \begin {gather*} \frac {\sin ^2(x)}{2}-\frac {1}{2} \csc ^2(x)-2 \log (\sin (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2*Cot[x]^3,x]

[Out]

-1/2*Csc[x]^2 - 2*Log[Sin[x]] + Sin[x]^2/2

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rubi steps

\begin {align*} \int \cos ^2(x) \cot ^3(x) \, dx &=\text {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^3} \, dx,x,-\sin (x)\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \frac {(1-x)^2}{x^2} \, dx,x,\sin ^2(x)\right )\\ &=\frac {1}{2} \text {Subst}\left (\int \left (1+\frac {1}{x^2}-\frac {2}{x}\right ) \, dx,x,\sin ^2(x)\right )\\ &=-\frac {1}{2} \csc ^2(x)-2 \log (\sin (x))+\frac {\sin ^2(x)}{2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 20, normalized size = 0.91 \begin {gather*} \frac {1}{2} \left (-\csc ^2(x)-4 \log (\sin (x))+\sin ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2*Cot[x]^3,x]

[Out]

(-Csc[x]^2 - 4*Log[Sin[x]] + Sin[x]^2)/2

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Maple [A]
time = 0.03, size = 29, normalized size = 1.32

method	result	size
default	\(-\frac {\cos ^{6}\left (x \right )}{2 \sin \left (x \right )^{2}}-\frac {\left (\cos ^{4}\left (x \right )\right )}{2}-\left (\cos ^{2}\left (x \right )\right )-2 \ln \left (\sin \left (x \right )\right )\)	\(29\)
risch	\(2 i x -\frac {{\mathrm e}^{2 i x}}{8}-\frac {{\mathrm e}^{-2 i x}}{8}+\frac {2 \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2}}-2 \ln \left ({\mathrm e}^{2 i x}-1\right )\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^5*sin(x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/2/sin(x)^2*cos(x)^6-1/2*cos(x)^4-cos(x)^2-2*ln(sin(x))

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Maxima [A]
time = 2.56, size = 20, normalized size = 0.91 \begin {gather*} \frac {1}{2} \, \sin \left (x\right )^{2} - \frac {1}{2 \, \sin \left (x\right )^{2}} - \log \left (\sin \left (x\right )^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^5*sin(x)^2,x, algorithm="maxima")

[Out]

1/2*sin(x)^2 - 1/2/sin(x)^2 - log(sin(x)^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (18) = 36\).
time = 0.66, size = 37, normalized size = 1.68 \begin {gather*} -\frac {2 \, \cos \left (x\right )^{4} - 3 \, \cos \left (x\right )^{2} + 8 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \sin \left (x\right )\right ) - 1}{4 \, {\left (\cos \left (x\right )^{2} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^5*sin(x)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.03, size = 20, normalized size = 0.91 \begin {gather*} - 2 \log {\left (\sin {\left (x \right )} \right )} + \frac {\sin ^{2}{\left (x \right )}}{2} - \frac {1}{2 \sin ^{2}{\left (x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**5*sin(x)**2,x)

[Out]

-2*log(sin(x)) + sin(x)**2/2 - 1/(2*sin(x)**2)

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Giac [A]
time = 1.21, size = 28, normalized size = 1.27 \begin {gather*} -\frac {1}{2} \, \cos \left (x\right )^{2} + \frac {1}{2 \, {\left (\cos \left (x\right )^{2} - 1\right )}} - \log \left (-\cos \left (x\right )^{2} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^5*sin(x)^2,x, algorithm="giac")

[Out]

-1/2*cos(x)^2 + 1/2/(cos(x)^2 - 1) - log(-cos(x)^2 + 1)

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Mupad [B]
time = 0.23, size = 32, normalized size = 1.45 \begin {gather*} \ln \left ({\mathrm {tan}\left (x\right )}^2+1\right )-2\,\ln \left (\mathrm {tan}\left (x\right )\right )-\frac {{\mathrm {tan}\left (x\right )}^2+\frac {1}{2}}{{\mathrm {tan}\left (x\right )}^4+{\mathrm {tan}\left (x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^5*sin(x)^2,x)

[Out]

log(tan(x)^2 + 1) - 2*log(tan(x)) - (tan(x)^2 + 1/2)/(tan(x)^2 + tan(x)^4)

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