3.164 \(\int \cot ^2(a+b x) \csc ^2(a+b x) \, dx\)

Optimal. Leaf size=15 \[ -\frac{\cot ^3(a+b x)}{3 b} \]

[Out]

-Cot[a + b*x]^3/(3*b)

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Rubi [A]  time = 0.029278, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2607, 30} \[ -\frac{\cot ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[a + b*x]^2*Csc[a + b*x]^2,x]

[Out]

-Cot[a + b*x]^3/(3*b)

Rule 2607

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])

Rule 30

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

Rubi steps

\begin{align*} \int \cot ^2(a+b x) \csc ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^2 \, dx,x,-\cot (a+b x)\right )}{b}\\ &=-\frac{\cot ^3(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.0095005, size = 15, normalized size = 1. \[ -\frac{\cot ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cot[a + b*x]^3/(3*b)

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Maple [A]  time = 0.012, size = 22, normalized size = 1.5 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2/sin(b*x+a)^4,x)

[Out]

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

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Maxima [A]  time = 0.992593, size = 18, normalized size = 1.2 \begin{align*} -\frac{1}{3 \, b \tan \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/3/(b*tan(b*x + a)^3)

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Fricas [B]  time = 2.1128, size = 78, normalized size = 5.2 \begin{align*} \frac{\cos \left (b x + a\right )^{3}}{3 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.84079, size = 71, normalized size = 4.73 \begin{align*} \begin{cases} \frac{\tan ^{3}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{24 b} - \frac{\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{8 b} + \frac{1}{8 b \tan{\left (\frac{a}{2} + \frac{b x}{2} \right )}} - \frac{1}{24 b \tan ^{3}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{2}{\left (a \right )}}{\sin ^{4}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2/sin(b*x+a)**4,x)

[Out]

Piecewise((tan(a/2 + b*x/2)**3/(24*b) - tan(a/2 + b*x/2)/(8*b) + 1/(8*b*tan(a/2 + b*x/2)) - 1/(24*b*tan(a/2 +
b*x/2)**3), Ne(b, 0)), (x*cos(a)**2/sin(a)**4, True))

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Giac [A]  time = 1.17765, size = 18, normalized size = 1.2 \begin{align*} -\frac{1}{3 \, b \tan \left (b x + a\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2/sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/3/(b*tan(b*x + a)^3)