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

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Rubi [A]  time = 0.04, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 53, normalized size = 1.00 \[ -\frac {\sin ^3(a+b x)}{3 b}+\frac {3 \sin (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}+\frac {3 \csc (a+b x)}{b} \]

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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fricas [A]  time = 0.42, size = 56, normalized size = 1.06 \[ -\frac {\cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} + 16}{3 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

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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giac [A]  time = 0.21, size = 41, normalized size = 0.77 \[ -\frac {{\left (\frac {1}{\sin \left (b x + a\right )} + \sin \left (b x + a\right )\right )}^{3} - \frac {12}{\sin \left (b x + a\right )} - 12 \, \sin \left (b x + a\right )}{3 \, b} \]

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

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maple [A]  time = 0.02, size = 80, normalized size = 1.51 \[ \frac {-\frac {\cos ^{8}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {5 \left (\cos ^{8}\left (b x +a \right )\right )}{3 \sin \left (b x +a \right )}+\frac {5 \left (\frac {16}{5}+\cos ^{6}\left (b x +a \right )+\frac {6 \left (\cos ^{4}\left (b x +a \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (b x +a \right )\right )}{5}\right ) \sin \left (b x +a \right )}{3}}{b} \]

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 = 0.32, size = 44, normalized size = 0.83 \[ -\frac {\sin \left (b x + a\right )^{3} - \frac {9 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{3}} - 9 \, \sin \left (b x + a\right )}{3 \, b} \]

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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mupad [B]  time = 0.48, size = 45, normalized size = 0.85 \[ \frac {-{\sin \left (a+b\,x\right )}^6+9\,{\sin \left (a+b\,x\right )}^4+9\,{\sin \left (a+b\,x\right )}^2-1}{3\,b\,{\sin \left (a+b\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^7/sin(a + b*x)^4,x)

[Out]

(9*sin(a + b*x)^2 + 9*sin(a + b*x)^4 - sin(a + b*x)^6 - 1)/(3*b*sin(a + b*x)^3)

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sympy [A]  time = 9.59, size = 82, normalized size = 1.55 \[ \begin {cases} \frac {16 \sin ^{3}{\left (a + b x \right )}}{3 b} + \frac {8 \sin {\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} + \frac {2 \cos ^{4}{\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} - \frac {\cos ^{6}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{7}{\relax (a )}}{\sin ^{4}{\relax (a )}} & \text {otherwise} \end {cases} \]

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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