\(\int \cot ^4(a+b x) \, dx\) [162]

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 = 0.01 (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 = 0.01 (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 = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-\frac {\left (\cot ^{3}\left (b x +a \right )\right )}{3}+\cot \left (b x +a \right )+b x +a}{b}\)	\(26\)
default	\(\frac {-\frac {\left (\cot ^{3}\left (b x +a \right )\right )}{3}+\cot \left (b x +a \right )+b x +a}{b}\)	\(26\)
risch	\(x +\frac {4 i \left (3 \,{\mathrm e}^{4 i \left (b x +a \right )}-3 \,{\mathrm e}^{2 i \left (b x +a \right )}+2\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}\)	\(46\)
parallelrisch	\(\frac {-\left (\cot ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )+24 b x +15 \cot \left (\frac {b x}{2}+\frac {a}{2}\right )-15 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}\)	\(57\)
norman	\(\frac {x \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\frac {1}{24 b}+\frac {5 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {5 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}\)	\(80\)

[In]

int(cos(b*x+a)^4/sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (25) = 50\).

Time = 0.31 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.56 \[ \int \cot ^4(a+b x) \, dx=\frac {4 \, \cos \left (b x + a\right )^{3} + 3 \, {\left (b x \cos \left (b x + a\right )^{2} - b x\right )} \sin \left (b x + a\right ) - 3 \, \cos \left (b x + a\right )}{3 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

[In]

integrate(cos(b*x+a)^4/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).

Time = 0.51 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \cot ^4(a+b x) \, dx=\begin {cases} x + \frac {\cos {\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} - \frac {\cos ^{3}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{4}{\left (a \right )}}{\sin ^{4}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**4/sin(b*x+a)**4,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (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(cos(b*x+a)^4/sin(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 = 0.39 (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(cos(b*x+a)^4/sin(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 = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \cot ^4(a+b x) \, dx=x+\frac {{\mathrm {tan}\left (a+b\,x\right )}^2-\frac {1}{3}}{b\,{\mathrm {tan}\left (a+b\,x\right )}^3} \]

[In]

int(cos(a + b*x)^4/sin(a + b*x)^4,x)

[Out]

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