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(b*x+a)/b-3*sin(b*x+a)/b+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 = 50, 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} \[ -\frac {\sin ^5(a+b x)}{5 b}+\frac {\sin ^3(a+b x)}{b}-\frac {3 \sin (a+b x)}{b}-\frac {\csc (a+b x)}{b} \]

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 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 ^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*}

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Mathematica [A]  time = 0.02, size = 50, normalized size = 1.00 \[ -\frac {\sin ^5(a+b x)}{5 b}+\frac {\sin ^3(a+b x)}{b}-\frac {3 \sin (a+b x)}{b}-\frac {\csc (a+b x)}{b} \]

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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fricas [A]  time = 0.42, size = 43, normalized size = 0.86 \[ \frac {\cos \left (b x + a\right )^{6} + 2 \, \cos \left (b x + a\right )^{4} + 8 \, \cos \left (b x + a\right )^{2} - 16}{5 \, b \sin \left (b x + a\right )} \]

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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giac [A]  time = 0.26, size = 42, normalized size = 0.84 \[ -\frac {\sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3} + \frac {5}{\sin \left (b x + a\right )} + 15 \, \sin \left (b x + a\right )}{5 \, b} \]

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

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maple [A]  time = 0.02, size = 62, normalized size = 1.24 \[ \frac {-\frac {\cos ^{8}\left (b x +a \right )}{\sin \left (b x +a \right )}-\left (\frac {16}{5}+\cos ^{6}\left (b x +a \right )+\frac {6 \left (\cos ^{4}\left (b x +a \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (b x +a \right )\right )}{5}\right ) \sin \left (b x +a \right )}{b} \]

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 = 0.38, size = 42, normalized size = 0.84 \[ -\frac {\sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3} + \frac {5}{\sin \left (b x + a\right )} + 15 \, \sin \left (b x + a\right )}{5 \, b} \]

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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mupad [B]  time = 0.48, size = 43, normalized size = 0.86 \[ -\frac {{\sin \left (a+b\,x\right )}^6-5\,{\sin \left (a+b\,x\right )}^4+15\,{\sin \left (a+b\,x\right )}^2+5}{5\,b\,\sin \left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^7/sin(a + b*x)^2,x)

[Out]

-(15*sin(a + b*x)^2 - 5*sin(a + b*x)^4 + sin(a + b*x)^6 + 5)/(5*b*sin(a + b*x))

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sympy [A]  time = 8.67, size = 82, normalized size = 1.64 \[ \begin {cases} - \frac {16 \sin ^{5}{\left (a + b x \right )}}{5 b} - \frac {8 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{b} - \frac {6 \sin {\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{b} - \frac {\cos ^{6}{\left (a + b x \right )}}{b \sin {\left (a + b x \right )}} & \text {for}\: b \neq 0 \\\frac {x \cos ^{7}{\relax (a )}}{\sin ^{2}{\relax (a )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**7/sin(b*x+a)**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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