3.159 \(\int \cos ^3(a+b x) \cot ^4(a+b x) \, dx\)

Optimal. Leaf size=53 \[ -\frac{\sin ^3(a+b x)}{3 b}+\frac{3 \sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{3 \csc (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0402648, antiderivative size = 53, 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} \[ -\frac{\sin ^3(a+b x)}{3 b}+\frac{3 \sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{3 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Cot[a + b*x]^4,x]

[Out]

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

Mathematica [A]  time = 0.0255077, size = 53, normalized size = 1. \[ -\frac{\sin ^3(a+b x)}{3 b}+\frac{3 \sin (a+b x)}{b}-\frac{\csc ^3(a+b x)}{3 b}+\frac{3 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Cot[a + b*x]^4,x]

[Out]

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

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Maple [A]  time = 0.012, size = 80, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}}+{\frac{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{3\,\sin \left ( bx+a \right ) }}+{\frac{5\,\sin \left ( bx+a \right ) }{3} \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 ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^7/sin(b*x+a)^4,x)

[Out]

1/b*(-1/3/sin(b*x+a)^3*cos(b*x+a)^8+5/3/sin(b*x+a)*cos(b*x+a)^8+5/3*(16/5+cos(b*x+a)^6+6/5*cos(b*x+a)^4+8/5*co
s(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 0.98447, size = 59, normalized size = 1.11 \begin{align*} -\frac{\sin \left (b x + a\right )^{3} - \frac{9 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{3}} - 9 \, \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)^7/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

-1/3*(sin(b*x + a)^3 - (9*sin(b*x + a)^2 - 1)/sin(b*x + a)^3 - 9*sin(b*x + a))/b

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Fricas [A]  time = 2.34696, size = 142, normalized size = 2.68 \begin{align*} -\frac{\cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} + 16}{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)^7/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/3*(cos(b*x + a)^6 + 6*cos(b*x + a)^4 - 24*cos(b*x + a)^2 + 16)/((b*cos(b*x + a)^2 - b)*sin(b*x + a))

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Sympy [A]  time = 8.07703, size = 82, normalized size = 1.55 \begin{align*} \begin{cases} \frac{16 \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac{8 \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac{2 \cos ^{4}{\left (a + b x \right )}}{b \sin{\left (a + b x \right )}} - \frac{\cos ^{6}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{7}{\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)**7/sin(b*x+a)**4,x)

[Out]

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

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Giac [A]  time = 1.16567, size = 55, normalized size = 1.04 \begin{align*} -\frac{{\left (\frac{1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{3} - \frac{12}{\sin \left (b x + a\right )} - 12 \, \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)^7/sin(b*x+a)^4,x, algorithm="giac")

[Out]

-1/3*((1/sin(b*x + a) + sin(b*x + a))^3 - 12/sin(b*x + a) - 12*sin(b*x + a))/b