3.5 \(\int \cot ^5(a+b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.027822, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3473, 3475} \[ -\frac{\cot ^4(a+b x)}{4 b}+\frac{\cot ^2(a+b x)}{2 b}+\frac{\log (\sin (a+b x))}{b} \]

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 3473

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 3475

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*}

Mathematica [A]  time = 0.106946, size = 46, normalized size = 1.1 \[ \frac{-\cot ^4(a+b x)+2 \cot ^2(a+b x)+4 \log (\tan (a+b x))+4 \log (\cos (a+b x))}{4 b} \]

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 = 0.012, size = 44, normalized size = 1.1 \begin{align*} -{\frac{ \left ( \cot \left ( bx+a \right ) \right ) ^{4}}{4\,b}}+{\frac{ \left ( \cot \left ( bx+a \right ) \right ) ^{2}}{2\,b}}-{\frac{\ln \left ( \left ( \cot \left ( bx+a \right ) \right ) ^{2}+1 \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(b*x+a)^5,x)

[Out]

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

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Maxima [A]  time = 1.00492, size = 51, normalized size = 1.21 \begin{align*} \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(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 [B]  time = 1.39003, size = 217, normalized size = 5.17 \begin{align*} \frac{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (-\frac{1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac{1}{2}\right ) - 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 2}{2 \,{\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(b*x+a)^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.923948, size = 66, normalized size = 1.57 \begin{align*} \begin{cases} \tilde{\infty } x & \text{for}\: a = 0 \wedge b = 0 \\x \cot ^{5}{\left (a \right )} & \text{for}\: b = 0 \\\tilde{\infty } x & \text{for}\: a = - b x \\- \frac{\log{\left (\tan ^{2}{\left (a + b x \right )} + 1 \right )}}{2 b} + \frac{\log{\left (\tan{\left (a + b x \right )} \right )}}{b} + \frac{1}{2 b \tan ^{2}{\left (a + b x \right )}} - \frac{1}{4 b \tan ^{4}{\left (a + b x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(b*x+a)**5,x)

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (x*cot(a)**5, Eq(b, 0)), (zoo*x, Eq(a, -b*x)), (-log(tan(a + b*x)**2 +
1)/(2*b) + log(tan(a + b*x))/b + 1/(2*b*tan(a + b*x)**2) - 1/(4*b*tan(a + b*x)**4), True))

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Giac [B]  time = 1.15783, size = 221, normalized size = 5.26 \begin{align*} -\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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(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