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(b*x+a)/b-1/3*csc(b*x+a)^3/b+6*sin(b*x+a)/b-4/3*sin(b*x+a)^3/b+1/5*sin(b*x+a)^5/b

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Rubi [A]  time = 0.04, antiderivative size = 68, 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} \[ \frac {\sin ^5(a+b x)}{5 b}-\frac {4 \sin ^3(a+b x)}{3 b}+\frac {6 \sin (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}+\frac {4 \csc (a+b x)}{b} \]

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 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 ^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*}

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Mathematica [A]  time = 0.03, size = 68, normalized size = 1.00 \[ \frac {\sin ^5(a+b x)}{5 b}-\frac {4 \sin ^3(a+b x)}{3 b}+\frac {6 \sin (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}+\frac {4 \csc (a+b x)}{b} \]

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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fricas [A]  time = 0.43, size = 68, normalized size = 1.00 \[ -\frac {3 \, \cos \left (b x + a\right )^{8} + 8 \, \cos \left (b x + a\right )^{6} + 48 \, \cos \left (b x + a\right )^{4} - 192 \, \cos \left (b x + a\right )^{2} + 128}{15 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \]

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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giac [A]  time = 0.22, size = 56, normalized size = 0.82 \[ \frac {3 \, \sin \left (b x + a\right )^{5} - 20 \, \sin \left (b x + a\right )^{3} + \frac {5 \, {\left (12 \, \sin \left (b x + a\right )^{2} - 1\right )}}{\sin \left (b x + a\right )^{3}} + 90 \, \sin \left (b x + a\right )}{15 \, b} \]

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

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maple [A]  time = 0.06, size = 90, normalized size = 1.32 \[ \frac {-\frac {\cos ^{10}\left (b x +a \right )}{3 \sin \left (b x +a \right )^{3}}+\frac {7 \left (\cos ^{10}\left (b x +a \right )\right )}{3 \sin \left (b x +a \right )}+\frac {7 \left (\frac {128}{35}+\cos ^{8}\left (b x +a \right )+\frac {8 \left (\cos ^{6}\left (b x +a \right )\right )}{7}+\frac {48 \left (\cos ^{4}\left (b x +a \right )\right )}{35}+\frac {64 \left (\cos ^{2}\left (b x +a \right )\right )}{35}\right ) \sin \left (b x +a \right )}{3}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^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 = 0.36, size = 56, normalized size = 0.82 \[ \frac {3 \, \sin \left (b x + a\right )^{5} - 20 \, \sin \left (b x + a\right )^{3} + \frac {5 \, {\left (12 \, \sin \left (b x + a\right )^{2} - 1\right )}}{\sin \left (b x + a\right )^{3}} + 90 \, \sin \left (b x + a\right )}{15 \, b} \]

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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mupad [B]  time = 0.52, size = 55, normalized size = 0.81 \[ \frac {3\,{\sin \left (a+b\,x\right )}^8-20\,{\sin \left (a+b\,x\right )}^6+90\,{\sin \left (a+b\,x\right )}^4+60\,{\sin \left (a+b\,x\right )}^2-5}{15\,b\,{\sin \left (a+b\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^9/sin(a + b*x)^4,x)

[Out]

(60*sin(a + b*x)^2 + 90*sin(a + b*x)^4 - 20*sin(a + b*x)^6 + 3*sin(a + b*x)^8 - 5)/(15*b*sin(a + b*x)^3)

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sympy [A]  time = 22.23, size = 105, normalized size = 1.54 \[ \begin {cases} \frac {128 \sin ^{5}{\left (a + b x \right )}}{15 b} + \frac {64 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{3 b} + \frac {16 \sin {\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{b} + \frac {8 \cos ^{6}{\left (a + b x \right )}}{3 b \sin {\left (a + b x \right )}} - \frac {\cos ^{8}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{9}{\relax (a )}}{\sin ^{4}{\relax (a )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**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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