3.177 \(\int \cot ^3(a+b x) \csc ^2(a+b x) \, dx\)

Optimal. Leaf size=15 \[ -\frac {\cot ^4(a+b x)}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, 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} \[ -\frac {\cot ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[a + b*x]^3*Csc[a + b*x]^2,x]

[Out]

-Cot[a + b*x]^4/(4*b)

Rule 30

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

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])

Rubi steps

\begin {align*} \int \cot ^3(a+b x) \csc ^2(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int x^3 \, dx,x,-\cot (a+b x)\right )}{b}\\ &=-\frac {\cot ^4(a+b x)}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ -\frac {\cot ^4(a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*Cot[a + b*x]^4/b

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fricas [B]  time = 0.44, size = 39, normalized size = 2.60 \[ -\frac {2 \, \cos \left (b x + a\right )^{2} - 1}{4 \, {\left (b \cos \left (b x + a\right )^{4} - 2 \, b \cos \left (b x + a\right )^{2} + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.28, size = 25, normalized size = 1.67 \[ \frac {2 \, \sin \left (b x + a\right )^{2} - 1}{4 \, b \sin \left (b x + a\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 22, normalized size = 1.47 \[ -\frac {\cos ^{4}\left (b x +a \right )}{4 \sin \left (b x +a \right )^{4} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.78, size = 25, normalized size = 1.67 \[ \frac {2 \, \sin \left (b x + a\right )^{2} - 1}{4 \, b \sin \left (b x + a\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.39, size = 25, normalized size = 1.67 \[ -\frac {{\left ({\sin \left (a+b\,x\right )}^2-1\right )}^2}{4\,b\,{\sin \left (a+b\,x\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 2.90, size = 44, normalized size = 2.93 \[ \begin {cases} \frac {1}{4 b \sin ^{2}{\left (a + b x \right )}} - \frac {\cos ^{2}{\left (a + b x \right )}}{4 b \sin ^{4}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{3}{\relax (a )}}{\sin ^{5}{\relax (a )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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