\(\int \cot ^4(a+b x) \, dx\) [4]

Optimal result

Integrand size = 8, antiderivative size = 27 \[ \int \cot ^4(a+b x) \, dx=x+\frac {\cot (a+b x)}{b}-\frac {\cot ^3(a+b x)}{3 b} \]

[Out]

x+cot(b*x+a)/b-1/3*cot(b*x+a)^3/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3554, 8} \[ \int \cot ^4(a+b x) \, dx=-\frac {\cot ^3(a+b x)}{3 b}+\frac {\cot (a+b x)}{b}+x \]

[In]

Int[Cot[a + b*x]^4,x]

[Out]

x + Cot[a + b*x]/b - Cot[a + b*x]^3/(3*b)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cot ^3(a+b x)}{3 b}-\int \cot ^2(a+b x) \, dx \\ & = \frac {\cot (a+b x)}{b}-\frac {\cot ^3(a+b x)}{3 b}+\int 1 \, dx \\ & = x+\frac {\cot (a+b x)}{b}-\frac {\cot ^3(a+b x)}{3 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \cot ^4(a+b x) \, dx=-\frac {\cot ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\tan ^2(a+b x)\right )}{3 b} \]

[In]

Integrate[Cot[a + b*x]^4,x]

[Out]

-1/3*(Cot[a + b*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, -Tan[a + b*x]^2])/b

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {-\cot \left (b x +a \right )^{3}+3 b x +3 \cot \left (b x +a \right )}{3 b}\)	\(29\)
derivativedivides	\(\frac {-\frac {\cot \left (b x +a \right )^{3}}{3}+\cot \left (b x +a \right )-\frac {\pi }{2}+\operatorname {arccot}\left (\cot \left (b x +a \right )\right )}{b}\)	\(32\)
default	\(\frac {-\frac {\cot \left (b x +a \right )^{3}}{3}+\cot \left (b x +a \right )-\frac {\pi }{2}+\operatorname {arccot}\left (\cot \left (b x +a \right )\right )}{b}\)	\(32\)
norman	\(\frac {\frac {\tan \left (b x +a \right )^{2}}{b}+x \tan \left (b x +a \right )^{3}-\frac {1}{3 b}}{\tan \left (b x +a \right )^{3}}\)	\(38\)
risch	\(x +\frac {4 i \left (3 \,{\mathrm e}^{4 i \left (b x +a \right )}-3 \,{\mathrm e}^{2 i \left (b x +a \right )}+2\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}\)	\(46\)

[In]

int(cot(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/3*(-cot(b*x+a)^3+3*b*x+3*cot(b*x+a))/b

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.11 \[ \int \cot ^4(a+b x) \, dx=\frac {4 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 3 \, {\left (b x \cos \left (2 \, b x + 2 \, a\right ) - b x\right )} \sin \left (2 \, b x + 2 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) - 2}{3 \, {\left (b \cos \left (2 \, b x + 2 \, a\right ) - b\right )} \sin \left (2 \, b x + 2 \, a\right )} \]

[In]

integrate(cot(b*x+a)^4,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \cot ^4(a+b x) \, dx=\begin {cases} x - \frac {\cot ^{3}{\left (a + b x \right )}}{3 b} + \frac {\cot {\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \cot ^{4}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(cot(b*x+a)**4,x)

[Out]

Piecewise((x - cot(a + b*x)**3/(3*b) + cot(a + b*x)/b, Ne(b, 0)), (x*cot(a)**4, True))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \cot ^4(a+b x) \, dx=\frac {3 \, b x + 3 \, a + \frac {3 \, \tan \left (b x + a\right )^{2} - 1}{\tan \left (b x + a\right )^{3}}}{3 \, b} \]

[In]

integrate(cot(b*x+a)^4,x, algorithm="maxima")

[Out]

1/3*(3*b*x + 3*a + (3*tan(b*x + a)^2 - 1)/tan(b*x + a)^3)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (25) = 50\).

Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.30 \[ \int \cot ^4(a+b x) \, dx=\frac {\tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3} + 24 \, b x + 24 \, a + \frac {15 \, \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{2} - 1}{\tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )^{3}} - 15 \, \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}{24 \, b} \]

[In]

integrate(cot(b*x+a)^4,x, algorithm="giac")

[Out]

1/24*(tan(1/2*b*x + 1/2*a)^3 + 24*b*x + 24*a + (15*tan(1/2*b*x + 1/2*a)^2 - 1)/tan(1/2*b*x + 1/2*a)^3 - 15*tan
(1/2*b*x + 1/2*a))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \cot ^4(a+b x) \, dx=x+\frac {\mathrm {cot}\left (a+b\,x\right )-\frac {{\mathrm {cot}\left (a+b\,x\right )}^3}{3}}{b} \]

[In]

int(cot(a + b*x)^4,x)

[Out]

x + (cot(a + b*x) - cot(a + b*x)^3/3)/b