3.133 \(\int \cos ^5(a+b x) \cot ^2(a+b x) \, dx\)

Optimal. Leaf size=50 \[ -\frac{\sin ^5(a+b x)}{5 b}+\frac{\sin ^3(a+b x)}{b}-\frac{3 \sin (a+b x)}{b}-\frac{\csc (a+b x)}{b} \]

[Out]

-(Csc[a + b*x]/b) - (3*Sin[a + b*x])/b + Sin[a + b*x]^3/b - Sin[a + b*x]^5/(5*b)

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Rubi [A]  time = 0.038405, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2590, 270} \[ -\frac{\sin ^5(a+b x)}{5 b}+\frac{\sin ^3(a+b x)}{b}-\frac{3 \sin (a+b x)}{b}-\frac{\csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Cot[a + b*x]^2,x]

[Out]

-(Csc[a + b*x]/b) - (3*Sin[a + b*x])/b + Sin[a + b*x]^3/b - Sin[a + b*x]^5/(5*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 ^5(a+b x) \cot ^2(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^2} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-3+\frac{1}{x^2}+3 x^2-x^4\right ) \, dx,x,-\sin (a+b x)\right )}{b}\\ &=-\frac{\csc (a+b x)}{b}-\frac{3 \sin (a+b x)}{b}+\frac{\sin ^3(a+b x)}{b}-\frac{\sin ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.025517, size = 50, normalized size = 1. \[ -\frac{\sin ^5(a+b x)}{5 b}+\frac{\sin ^3(a+b x)}{b}-\frac{3 \sin (a+b x)}{b}-\frac{\csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5*Cot[a + b*x]^2,x]

[Out]

-(Csc[a + b*x]/b) - (3*Sin[a + b*x])/b + Sin[a + b*x]^3/b - Sin[a + b*x]^5/(5*b)

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Maple [A]  time = 0.013, size = 62, normalized size = 1.2 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{\sin \left ( bx+a \right ) }}- \left ({\frac{16}{5}}+ \left ( \cos \left ( bx+a \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{5}} \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)^7/sin(b*x+a)^2,x)

[Out]

1/b*(-1/sin(b*x+a)*cos(b*x+a)^8-(16/5+cos(b*x+a)^6+6/5*cos(b*x+a)^4+8/5*cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 0.985622, size = 57, normalized size = 1.14 \begin{align*} -\frac{\sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3} + \frac{5}{\sin \left (b x + a\right )} + 15 \, \sin \left (b x + a\right )}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7/sin(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/5*(sin(b*x + a)^5 - 5*sin(b*x + a)^3 + 5/sin(b*x + a) + 15*sin(b*x + a))/b

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Fricas [A]  time = 1.9589, size = 111, normalized size = 2.22 \begin{align*} \frac{\cos \left (b x + a\right )^{6} + 2 \, \cos \left (b x + a\right )^{4} + 8 \, \cos \left (b x + a\right )^{2} - 16}{5 \, b \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7/sin(b*x+a)^2,x, algorithm="fricas")

[Out]

1/5*(cos(b*x + a)^6 + 2*cos(b*x + a)^4 + 8*cos(b*x + a)^2 - 16)/(b*sin(b*x + a))

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Sympy [A]  time = 8.10902, size = 82, normalized size = 1.64 \begin{align*} \begin{cases} - \frac{16 \sin ^{5}{\left (a + b x \right )}}{5 b} - \frac{8 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} - \frac{6 \sin{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{b} - \frac{\cos ^{6}{\left (a + b x \right )}}{b \sin{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{7}{\left (a \right )}}{\sin ^{2}{\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)**7/sin(b*x+a)**2,x)

[Out]

Piecewise((-16*sin(a + b*x)**5/(5*b) - 8*sin(a + b*x)**3*cos(a + b*x)**2/b - 6*sin(a + b*x)*cos(a + b*x)**4/b
- cos(a + b*x)**6/(b*sin(a + b*x)), Ne(b, 0)), (x*cos(a)**7/sin(a)**2, True))

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Giac [A]  time = 1.19002, size = 57, normalized size = 1.14 \begin{align*} -\frac{\sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3} + \frac{5}{\sin \left (b x + a\right )} + 15 \, \sin \left (b x + a\right )}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7/sin(b*x+a)^2,x, algorithm="giac")

[Out]

-1/5*(sin(b*x + a)^5 - 5*sin(b*x + a)^3 + 5/sin(b*x + a) + 15*sin(b*x + a))/b