\(\int \cot ^3(a+b x) \csc (a+b x) \, dx\) [163]

Optimal result

Integrand size = 15, antiderivative size = 26 \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\frac {\csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2686} \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\frac {\csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b} \]

[In]

Int[Cot[a + b*x]^3*Csc[a + b*x],x]

[Out]

Csc[a + b*x]/b - Csc[a + b*x]^3/(3*b)

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\frac {\csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b} \]

[In]

Integrate[Cot[a + b*x]^3*Csc[a + b*x],x]

[Out]

Csc[a + b*x]/b - Csc[a + b*x]^3/(3*b)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08

method	result	size
risch	\(\frac {2 i \left (3 \,{\mathrm e}^{5 i \left (b x +a \right )}-2 \,{\mathrm e}^{3 i \left (b x +a \right )}+3 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}\)	\(54\)
parallelrisch	\(\frac {-\left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\cot ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+9 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+9 \cot \left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}\)	\(55\)
derivativedivides	\(\frac {-\frac {\cos ^{4}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {\cos ^{4}\left (b x +a \right )}{3 \sin \left (b x +a \right )}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}}{b}\)	\(60\)
default	\(\frac {-\frac {\cos ^{4}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {\cos ^{4}\left (b x +a \right )}{3 \sin \left (b x +a \right )}+\frac {\left (2+\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right )}{3}}{b}\)	\(60\)
norman	\(\frac {-\frac {1}{24 b}+\frac {3 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}+\frac {3 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}-\frac {\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )}{24 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}\)	\(67\)

[In]

int(cos(b*x+a)^3/sin(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

2/3*I/b/(exp(2*I*(b*x+a))-1)^3*(3*exp(5*I*(b*x+a))-2*exp(3*I*(b*x+a))+3*exp(I*(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{2} - 2}{3 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).

Time = 0.51 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\begin {cases} \frac {2}{3 b \sin {\left (a + b x \right )}} - \frac {\cos ^{2}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{3}{\left (a \right )}}{\sin ^{4}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**3/sin(b*x+a)**4,x)

[Out]

Piecewise((2/(3*b*sin(a + b*x)) - cos(a + b*x)**2/(3*b*sin(a + b*x)**3), Ne(b, 0)), (x*cos(a)**3/sin(a)**4, Tr
ue))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\frac {3 \, \sin \left (b x + a\right )^{2} - 1}{3 \, b \sin \left (b x + a\right )^{3}} \]

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\frac {3 \, \sin \left (b x + a\right )^{2} - 1}{3 \, b \sin \left (b x + a\right )^{3}} \]

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^4,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \cot ^3(a+b x) \csc (a+b x) \, dx=\frac {{\sin \left (a+b\,x\right )}^2-\frac {1}{3}}{b\,{\sin \left (a+b\,x\right )}^3} \]

[In]

int(cos(a + b*x)^3/sin(a + b*x)^4,x)

[Out]

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