3.165 \(\int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx\)

Optimal. Leaf size=72 \[ -\frac{a^4 (a \sin (e+f x))^{m-4}}{f (4-m)}+\frac{2 a^2 (a \sin (e+f x))^{m-2}}{f (2-m)}+\frac{(a \sin (e+f x))^m}{f m} \]

[Out]

-((a^4*(a*Sin[e + f*x])^(-4 + m))/(f*(4 - m))) + (2*a^2*(a*Sin[e + f*x])^(-2 + m))/(f*(2 - m)) + (a*Sin[e + f*
x])^m/(f*m)

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Rubi [A]  time = 0.0609768, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2592, 270} \[ -\frac{a^4 (a \sin (e+f x))^{m-4}}{f (4-m)}+\frac{2 a^2 (a \sin (e+f x))^{m-2}}{f (2-m)}+\frac{(a \sin (e+f x))^m}{f m} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5*(a*Sin[e + f*x])^m,x]

[Out]

-((a^4*(a*Sin[e + f*x])^(-4 + m))/(f*(4 - m))) + (2*a^2*(a*Sin[e + f*x])^(-2 + m))/(f*(2 - m)) + (a*Sin[e + f*
x])^m/(f*m)

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(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 \cot ^5(e+f x) (a \sin (e+f x))^m \, dx &=\frac{\operatorname{Subst}\left (\int x^{-5+m} \left (a^2-x^2\right )^2 \, dx,x,a \sin (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^4 x^{-5+m}-2 a^2 x^{-3+m}+x^{-1+m}\right ) \, dx,x,a \sin (e+f x)\right )}{f}\\ &=-\frac{a^4 (a \sin (e+f x))^{-4+m}}{f (4-m)}+\frac{2 a^2 (a \sin (e+f x))^{-2+m}}{f (2-m)}+\frac{(a \sin (e+f x))^m}{f m}\\ \end{align*}

Mathematica [A]  time = 0.324903, size = 62, normalized size = 0.86 \[ \frac{\left ((m-2) m \csc ^4(e+f x)-2 (m-4) m \csc ^2(e+f x)+m^2-6 m+8\right ) (a \sin (e+f x))^m}{f (m-4) (m-2) m} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5*(a*Sin[e + f*x])^m,x]

[Out]

((8 - 6*m + m^2 - 2*(-4 + m)*m*Csc[e + f*x]^2 + (-2 + m)*m*Csc[e + f*x]^4)*(a*Sin[e + f*x])^m)/(f*(-4 + m)*(-2
+ m)*m)

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Maple [C]  time = 0.691, size = 7964, normalized size = 110.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5*(a*sin(f*x+e))^m,x)

[Out]

result too large to display

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Maxima [A]  time = 0.959883, size = 96, normalized size = 1.33 \begin{align*} \frac{\frac{a^{m} \sin \left (f x + e\right )^{m}}{m} - \frac{2 \, a^{m} \sin \left (f x + e\right )^{m}}{{\left (m - 2\right )} \sin \left (f x + e\right )^{2}} + \frac{a^{m} \sin \left (f x + e\right )^{m}}{{\left (m - 4\right )} \sin \left (f x + e\right )^{4}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a*sin(f*x+e))^m,x, algorithm="maxima")

[Out]

(a^m*sin(f*x + e)^m/m - 2*a^m*sin(f*x + e)^m/((m - 2)*sin(f*x + e)^2) + a^m*sin(f*x + e)^m/((m - 4)*sin(f*x +
e)^4))/f

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Fricas [A]  time = 1.65627, size = 267, normalized size = 3.71 \begin{align*} \frac{{\left ({\left (m^{2} - 6 \, m + 8\right )} \cos \left (f x + e\right )^{4} + 4 \,{\left (m - 4\right )} \cos \left (f x + e\right )^{2} + 8\right )} \left (a \sin \left (f x + e\right )\right )^{m}}{{\left (f m^{3} - 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{4} + f m^{3} - 6 \, f m^{2} - 2 \,{\left (f m^{3} - 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{2} + 8 \, f m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a*sin(f*x+e))^m,x, algorithm="fricas")

[Out]

((m^2 - 6*m + 8)*cos(f*x + e)^4 + 4*(m - 4)*cos(f*x + e)^2 + 8)*(a*sin(f*x + e))^m/((f*m^3 - 6*f*m^2 + 8*f*m)*
cos(f*x + e)^4 + f*m^3 - 6*f*m^2 - 2*(f*m^3 - 6*f*m^2 + 8*f*m)*cos(f*x + e)^2 + 8*f*m)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**5*(a*sin(f*x+e))**m,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \sin \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a*sin(f*x+e))^m,x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e))^m*cot(f*x + e)^5, x)