3.161 \(\int \cos (a+b x) \cot ^4(a+b x) \, dx\)

Optimal. Leaf size=37 \[ \frac{\sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{2 \csc (a+b x)}{b} \]

[Out]

(2*Csc[a + b*x])/b - Csc[a + b*x]^3/(3*b) + Sin[a + b*x]/b

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Rubi [A]  time = 0.0254015, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2590, 270} \[ \frac{\sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{2 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Cot[a + b*x]^4,x]

[Out]

(2*Csc[a + b*x])/b - Csc[a + b*x]^3/(3*b) + Sin[a + b*x]/b

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos (a+b x) \cot ^4(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^4} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}-\frac{2}{x^2}\right ) \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{2 \csc (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{\sin (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0187425, size = 37, normalized size = 1. \[ \frac{\sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{2 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Cot[a + b*x]^4,x]

[Out]

(2*Csc[a + b*x])/b - Csc[a + b*x]^3/(3*b) + Sin[a + b*x]/b

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Maple [A]  time = 0.012, size = 68, normalized size = 1.8 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}}+{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{\sin \left ( bx+a \right ) }}+ \left ({\frac{8}{3}}+ \left ( \cos \left ( bx+a \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) \sin \left ( bx+a \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^5/sin(b*x+a)^4,x)

[Out]

1/b*(-1/3*cos(b*x+a)^6/sin(b*x+a)^3+cos(b*x+a)^6/sin(b*x+a)+(8/3+cos(b*x+a)^4+4/3*cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 0.989545, size = 47, normalized size = 1.27 \begin{align*} \frac{\frac{6 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{3}} + 3 \, \sin \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/3*((6*sin(b*x + a)^2 - 1)/sin(b*x + a)^3 + 3*sin(b*x + a))/b

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Fricas [A]  time = 2.17163, size = 117, normalized size = 3.16 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} + 8}{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)^5/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/3*(3*cos(b*x + a)^4 - 12*cos(b*x + a)^2 + 8)/((b*cos(b*x + a)^2 - b)*sin(b*x + a))

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Sympy [A]  time = 2.75586, size = 63, normalized size = 1.7 \begin{align*} \begin{cases} \frac{8 \sin{\left (a + b x \right )}}{3 b} + \frac{4 \cos ^{2}{\left (a + b x \right )}}{3 b \sin{\left (a + b x \right )}} - \frac{\cos ^{4}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{5}{\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)**5/sin(b*x+a)**4,x)

[Out]

Piecewise((8*sin(a + b*x)/(3*b) + 4*cos(a + b*x)**2/(3*b*sin(a + b*x)) - cos(a + b*x)**4/(3*b*sin(a + b*x)**3)
, Ne(b, 0)), (x*cos(a)**5/sin(a)**4, True))

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Giac [A]  time = 1.16345, size = 47, normalized size = 1.27 \begin{align*} \frac{\frac{6 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{3}} + 3 \, \sin \left (b x + a\right )}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5/sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/3*((6*sin(b*x + a)^2 - 1)/sin(b*x + a)^3 + 3*sin(b*x + a))/b