\(\int \cot ^3(a+b x) \csc ^2(a+b x) \, dx\) [177]

Optimal result

Integrand size = 17, antiderivative size = 15 \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=-\frac {\cot ^4(a+b x)}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2687, 30} \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=-\frac {\cot ^4(a+b x)}{4 b} \]

[In]

Int[Cot[a + b*x]^3*Csc[a + b*x]^2,x]

[Out]

-1/4*Cot[a + b*x]^4/b

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2687

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int x^3 \, dx,x,-\cot (a+b x)\right )}{b} \\ & = -\frac {\cot ^4(a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=-\frac {\cot ^4(a+b x)}{4 b} \]

[In]

Integrate[Cot[a + b*x]^3*Csc[a + b*x]^2,x]

[Out]

-1/4*Cot[a + b*x]^4/b

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47

method	result	size
derivativedivides	\(-\frac {\cos ^{4}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4} b}\)	\(22\)
default	\(-\frac {\cos ^{4}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4} b}\)	\(22\)
risch	\(-\frac {2 \left ({\mathrm e}^{6 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4}}\)	\(38\)
parallelrisch	\(\frac {-\left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\left (\cot ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+4 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+4 \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}\)	\(59\)
norman	\(\frac {-\frac {1}{64 b}+\frac {\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}{16 b}+\frac {\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )}{16 b}-\frac {\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}\)	\(67\)

[In]

int(cos(b*x+a)^3/sin(b*x+a)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (13) = 26\).

Time = 0.43 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.60 \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=-\frac {2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (12) = 24\).

Time = 0.55 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.93 \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=\begin {cases} \frac {1}{4 b \sin ^{2}{\left (a + b x \right )}} - \frac {\cos ^{2}{\left (a + b x \right )}}{4 b \sin ^{4}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{3}{\left (a \right )}}{\sin ^{5}{\left (a \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(b*x+a)**3/sin(b*x+a)**5,x)

[Out]

Piecewise((1/(4*b*sin(a + b*x)**2) - cos(a + b*x)**2/(4*b*sin(a + b*x)**4), Ne(b, 0)), (x*cos(a)**3/sin(a)**5,
True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.67 \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=\frac {2 \, \sin \left (b x + a\right )^{2} - 1}{4 \, b \sin \left (b x + a\right )^{4}} \]

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.67 \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=\frac {2 \, \sin \left (b x + a\right )^{2} - 1}{4 \, b \sin \left (b x + a\right )^{4}} \]

[In]

integrate(cos(b*x+a)^3/sin(b*x+a)^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.67 \[ \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx=-\frac {{\left ({\sin \left (a+b\,x\right )}^2-1\right )}^2}{4\,b\,{\sin \left (a+b\,x\right )}^4} \]

[In]

int(cos(a + b*x)^3/sin(a + b*x)^5,x)

[Out]

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