3.157 \(\int \cos ^5(a+b x) \cot ^4(a+b x) \, dx\)

Optimal. Leaf size=68 \[ \frac{\sin ^5(a+b x)}{5 b}-\frac{4 \sin ^3(a+b x)}{3 b}+\frac{6 \sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{4 \csc (a+b x)}{b} \]

[Out]

(4*Csc[a + b*x])/b - Csc[a + b*x]^3/(3*b) + (6*Sin[a + b*x])/b - (4*Sin[a + b*x]^3)/(3*b) + Sin[a + b*x]^5/(5*
b)

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Rubi [A]  time = 0.0440119, antiderivative size = 68, 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} \[ \frac{\sin ^5(a+b x)}{5 b}-\frac{4 \sin ^3(a+b x)}{3 b}+\frac{6 \sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{4 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^5*Cot[a + b*x]^4,x]

[Out]

(4*Csc[a + b*x])/b - Csc[a + b*x]^3/(3*b) + (6*Sin[a + b*x])/b - (4*Sin[a + b*x]^3)/(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 ^4(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{x^4} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \left (6+\frac{1}{x^4}-\frac{4}{x^2}-4 x^2+x^4\right ) \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{4 \csc (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{6 \sin (a+b x)}{b}-\frac{4 \sin ^3(a+b x)}{3 b}+\frac{\sin ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0376901, size = 68, normalized size = 1. \[ \frac{\sin ^5(a+b x)}{5 b}-\frac{4 \sin ^3(a+b x)}{3 b}+\frac{6 \sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{4 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^5*Cot[a + b*x]^4,x]

[Out]

(4*Csc[a + b*x])/b - Csc[a + b*x]^3/(3*b) + (6*Sin[a + b*x])/b - (4*Sin[a + b*x]^3)/(3*b) + Sin[a + b*x]^5/(5*
b)

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Maple [A]  time = 0.043, size = 90, normalized size = 1.3 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{10}}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}}+{\frac{7\, \left ( \cos \left ( bx+a \right ) \right ) ^{10}}{3\,\sin \left ( bx+a \right ) }}+{\frac{7\,\sin \left ( bx+a \right ) }{3} \left ({\frac{128}{35}}+ \left ( \cos \left ( bx+a \right ) \right ) ^{8}+{\frac{8\, \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{7}}+{\frac{48\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{35}}+{\frac{64\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{35}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^9/sin(b*x+a)^4,x)

[Out]

1/b*(-1/3/sin(b*x+a)^3*cos(b*x+a)^10+7/3/sin(b*x+a)*cos(b*x+a)^10+7/3*(128/35+cos(b*x+a)^8+8/7*cos(b*x+a)^6+48
/35*cos(b*x+a)^4+64/35*cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 0.990353, size = 76, normalized size = 1.12 \begin{align*} \frac{3 \, \sin \left (b x + a\right )^{5} - 20 \, \sin \left (b x + a\right )^{3} + \frac{5 \,{\left (12 \, \sin \left (b x + a\right )^{2} - 1\right )}}{\sin \left (b x + a\right )^{3}} + 90 \, \sin \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^9/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/15*(3*sin(b*x + a)^5 - 20*sin(b*x + a)^3 + 5*(12*sin(b*x + a)^2 - 1)/sin(b*x + a)^3 + 90*sin(b*x + a))/b

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Fricas [A]  time = 2.27207, size = 176, normalized size = 2.59 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{8} + 8 \, \cos \left (b x + a\right )^{6} + 48 \, \cos \left (b x + a\right )^{4} - 192 \, \cos \left (b x + a\right )^{2} + 128}{15 \,{\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)^9/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/15*(3*cos(b*x + a)^8 + 8*cos(b*x + a)^6 + 48*cos(b*x + a)^4 - 192*cos(b*x + a)^2 + 128)/((b*cos(b*x + a)^2
- b)*sin(b*x + a))

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Sympy [A]  time = 24.745, size = 105, normalized size = 1.54 \begin{align*} \begin{cases} \frac{128 \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac{64 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac{16 \sin{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{b} + \frac{8 \cos ^{6}{\left (a + b x \right )}}{3 b \sin{\left (a + b x \right )}} - \frac{\cos ^{8}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{9}{\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)**9/sin(b*x+a)**4,x)

[Out]

Piecewise((128*sin(a + b*x)**5/(15*b) + 64*sin(a + b*x)**3*cos(a + b*x)**2/(3*b) + 16*sin(a + b*x)*cos(a + b*x
)**4/b + 8*cos(a + b*x)**6/(3*b*sin(a + b*x)) - cos(a + b*x)**8/(3*b*sin(a + b*x)**3), Ne(b, 0)), (x*cos(a)**9
/sin(a)**4, True))

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Giac [A]  time = 1.19047, size = 76, normalized size = 1.12 \begin{align*} \frac{3 \, \sin \left (b x + a\right )^{5} - 20 \, \sin \left (b x + a\right )^{3} + \frac{5 \,{\left (12 \, \sin \left (b x + a\right )^{2} - 1\right )}}{\sin \left (b x + a\right )^{3}} + 90 \, \sin \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^9/sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/15*(3*sin(b*x + a)^5 - 20*sin(b*x + a)^3 + 5*(12*sin(b*x + a)^2 - 1)/sin(b*x + a)^3 + 90*sin(b*x + a))/b