3.2.75 \(\int \cot ^5(a+b x) \, dx\) [175]

Optimal. Leaf size=42 \[ \frac {\cot ^2(a+b x)}{2 b}-\frac {\cot ^4(a+b x)}{4 b}+\frac {\log (\sin (a+b x))}{b} \]

[Out]

1/2*cot(b*x+a)^2/b-1/4*cot(b*x+a)^4/b+ln(sin(b*x+a))/b

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Rubi [A]
time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3554, 3556} \begin {gather*} -\frac {\cot ^4(a+b x)}{4 b}+\frac {\cot ^2(a+b x)}{2 b}+\frac {\log (\sin (a+b x))}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[a + b*x]^5,x]

[Out]

Cot[a + b*x]^2/(2*b) - Cot[a + b*x]^4/(4*b) + Log[Sin[a + b*x]]/b

Rule 3554

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 3556

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

Rubi steps

\begin {align*} \int \cot ^5(a+b x) \, dx &=-\frac {\cot ^4(a+b x)}{4 b}-\int \cot ^3(a+b x) \, dx\\ &=\frac {\cot ^2(a+b x)}{2 b}-\frac {\cot ^4(a+b x)}{4 b}+\int \cot (a+b x) \, dx\\ &=\frac {\cot ^2(a+b x)}{2 b}-\frac {\cot ^4(a+b x)}{4 b}+\frac {\log (\sin (a+b x))}{b}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 46, normalized size = 1.10 \begin {gather*} \frac {2 \cot ^2(a+b x)-\cot ^4(a+b x)+4 \log (\cos (a+b x))+4 \log (\tan (a+b x))}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[a + b*x]^5,x]

[Out]

(2*Cot[a + b*x]^2 - Cot[a + b*x]^4 + 4*Log[Cos[a + b*x]] + 4*Log[Tan[a + b*x]])/(4*b)

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Maple [A]
time = 0.04, size = 33, normalized size = 0.79

method	result	size
derivativedivides	\(\frac {-\frac {\left (\cot ^{4}\left (b x +a \right )\right )}{4}+\frac {\left (\cot ^{2}\left (b x +a \right )\right )}{2}+\ln \left (\sin \left (b x +a \right )\right )}{b}\)	\(33\)
default	\(\frac {-\frac {\left (\cot ^{4}\left (b x +a \right )\right )}{4}+\frac {\left (\cot ^{2}\left (b x +a \right )\right )}{2}+\ln \left (\sin \left (b x +a \right )\right )}{b}\)	\(33\)
risch	\(-i x -\frac {2 i a}{b}-\frac {4 \left ({\mathrm e}^{6 i \left (b x +a \right )}-{\mathrm e}^{4 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}{b}\)	\(77\)
norman	\(\frac {-\frac {1}{64 b}+\frac {3 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{16 b}+\frac {3 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{16 b}-\frac {\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}-\frac {\ln \left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/4*cot(b*x+a)^4+1/2*cot(b*x+a)^2+ln(sin(b*x+a)))

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Maxima [A]
time = 0.28, size = 38, normalized size = 0.90 \begin {gather*} \frac {\frac {4 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4}} + 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((4*sin(b*x + a)^2 - 1)/sin(b*x + a)^4 + 2*log(sin(b*x + a)^2))/b

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Fricas [A]
time = 0.41, size = 70, normalized size = 1.67 \begin {gather*} -\frac {4 \, \cos \left (b x + a\right )^{2} - 4 \, {\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right ) - 3}{4 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.61, size = 61, normalized size = 1.45 \begin {gather*} \begin {cases} \frac {\log {\left (\sin {\left (a + b x \right )} \right )}}{b} + \frac {\cos ^{2}{\left (a + b x \right )}}{2 b \sin ^{2}{\left (a + b x \right )}} - \frac {\cos ^{4}{\left (a + b x \right )}}{4 b \sin ^{4}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{5}{\left (a \right )}}{\sin ^{5}{\left (a \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**5/sin(b*x+a)**5,x)

[Out]

Piecewise((log(sin(a + b*x))/b + cos(a + b*x)**2/(2*b*sin(a + b*x)**2) - cos(a + b*x)**4/(4*b*sin(a + b*x)**4)
, Ne(b, 0)), (x*cos(a)**5/sin(a)**5, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (38) = 76\).
time = 4.90, size = 164, normalized size = 3.90 \begin {gather*} -\frac {\frac {{\left (\frac {12 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {48 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} + \frac {12 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 32 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 64 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{64 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*((12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 48*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)*(cos(b*x
+ a) + 1)^2/(cos(b*x + a) - 1)^2 + 12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x +
a) + 1)^2 - 32*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 64*log(abs(-(cos(b*x + a) - 1)/(cos(b*x + a
) + 1) + 1)))/b

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Mupad [B]
time = 0.42, size = 52, normalized size = 1.24 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (a+b\,x\right )\right )}{b}-\frac {\ln \left ({\mathrm {tan}\left (a+b\,x\right )}^2+1\right )}{2\,b}+\frac {\frac {{\mathrm {tan}\left (a+b\,x\right )}^2}{2}-\frac {1}{4}}{b\,{\mathrm {tan}\left (a+b\,x\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^5/sin(a + b*x)^5,x)

[Out]

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

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