\(\int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx\) [165]

Optimal result

Integrand size = 19, antiderivative size = 72 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=-\frac {a^4 (a \sin (e+f x))^{-4+m}}{f (4-m)}+\frac {2 a^2 (a \sin (e+f x))^{-2+m}}{f (2-m)}+\frac {(a \sin (e+f x))^m}{f m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2672, 276} \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=-\frac {a^4 (a \sin (e+f x))^{m-4}}{f (4-m)}+\frac {2 a^2 (a \sin (e+f x))^{m-2}}{f (2-m)}+\frac {(a \sin (e+f x))^m}{f m} \]

[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 276

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 2672

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int x^{-5+m} \left (a^2-x^2\right )^2 \, dx,x,a \sin (e+f x)\right )}{f} \\ & = \frac {\text {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] (verified)

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\frac {\left (8-6 m+m^2-2 (-4+m) m \csc ^2(e+f x)+(-2+m) m \csc ^4(e+f x)\right ) (a \sin (e+f x))^m}{f (-4+m) (-2+m) m} \]

[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)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.02 (sec) , antiderivative size = 6931, normalized size of antiderivative = 96.26

method	result	size
risch	\(\text {Expression too large to display}\)	\(6931\)

[In]

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

[Out]

result too large to display

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.56 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\frac {{\left ({\left (m^{2} - 6 \, m + 8\right )} \cos \left (f x + e\right )^{4} + 4 \, {\left (m - 4\right )} \cos \left (f x + e\right )^{2} + 8\right )} \left (a \sin \left (f x + e\right )\right )^{m}}{{\left (f m^{3} - 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{4} + f m^{3} - 6 \, f m^{2} - 2 \, {\left (f m^{3} - 6 \, f m^{2} + 8 \, f m\right )} \cos \left (f x + e\right )^{2} + 8 \, f m} \]

[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)

Sympy [F]

\[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\int \left (a \sin {\left (e + f x \right )}\right )^{m} \cot ^{5}{\left (e + f x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\frac {\frac {a^{m} \sin \left (f x + e\right )^{m}}{m} - \frac {2 \, a^{m} \sin \left (f x + e\right )^{m}}{{\left (m - 2\right )} \sin \left (f x + e\right )^{2}} + \frac {a^{m} \sin \left (f x + e\right )^{m}}{{\left (m - 4\right )} \sin \left (f x + e\right )^{4}}}{f} \]

[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

Giac [F]

\[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=\int { \left (a \sin \left (f x + e\right )\right )^{m} \cot \left (f x + e\right )^{5} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 8.69 (sec) , antiderivative size = 219, normalized size of antiderivative = 3.04 \[ \int \cot ^5(e+f x) (a \sin (e+f x))^m \, dx=-\frac {{\left (a\,\sin \left (e+f\,x\right )\right )}^m\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-1\right )\,\left (-\frac {2\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )}{f\,m}+\frac {\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (6\,m^2-4\,m+48\right )}{f\,m\,\left (m^2-6\,m+8\right )}+\frac {2\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,\left (-4\,m^2+8\,m+32\right )}{f\,m\,\left (m^2-6\,m+8\right )}\right )}{16\,{\sin \left (e+f\,x\right )}^4} \]

[In]

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

[Out]

-((a*sin(e + f*x))^m*(sin(4*e + 4*f*x)*1i + 2*sin(2*e + 2*f*x)^2 - 1)*(((sin(4*e + 4*f*x)*1i - 2*sin(2*e + 2*f
*x)^2 + 1)*(6*m^2 - 4*m + 48))/(f*m*(m^2 - 6*m + 8)) - (2*(2*sin(2*e + 2*f*x)^2 - 1)*(sin(4*e + 4*f*x)*1i - 2*
sin(2*e + 2*f*x)^2 + 1))/(f*m) + (2*(2*sin(e + f*x)^2 - 1)*(sin(4*e + 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1)*(8
*m - 4*m^2 + 32))/(f*m*(m^2 - 6*m + 8))))/(16*sin(e + f*x)^4)