Integrand size = 21, antiderivative size = 123 \[ \int \cot ^5(e+f x) (a+a \sin (e+f x))^m \, dx=\frac {(9-m) \csc ^3(e+f x) (a+a \sin (e+f x))^{3+m}}{12 a^3 f}-\frac {\csc ^4(e+f x) (a+a \sin (e+f x))^{3+m}}{4 a^3 f}-\frac {\left (12-9 m+m^2\right ) \operatorname {Hypergeometric2F1}(3,3+m,4+m,1+\sin (e+f x)) (a+a \sin (e+f x))^{3+m}}{12 a^3 f (3+m)} \]
1/12*(9-m)*csc(f*x+e)^3*(a+a*sin(f*x+e))^(3+m)/a^3/f-1/4*csc(f*x+e)^4*(a+a *sin(f*x+e))^(3+m)/a^3/f-1/12*(m^2-9*m+12)*hypergeom([3, 3+m],[4+m],1+sin( f*x+e))*(a+a*sin(f*x+e))^(3+m)/a^3/f/(3+m)
Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.67 \[ \int \cot ^5(e+f x) (a+a \sin (e+f x))^m \, dx=-\frac {\left ((3+m) \csc ^3(e+f x) (-9+m+3 \csc (e+f x))+\left (12-9 m+m^2\right ) \operatorname {Hypergeometric2F1}(3,3+m,4+m,1+\sin (e+f x))\right ) (1+\sin (e+f x))^3 (a (1+\sin (e+f x)))^m}{12 f (3+m)} \]
-1/12*(((3 + m)*Csc[e + f*x]^3*(-9 + m + 3*Csc[e + f*x]) + (12 - 9*m + m^2 )*Hypergeometric2F1[3, 3 + m, 4 + m, 1 + Sin[e + f*x]])*(1 + Sin[e + f*x]) ^3*(a*(1 + Sin[e + f*x]))^m)/(f*(3 + m))
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3186, 100, 25, 27, 87, 75}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cot ^5(e+f x) (a \sin (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m}{\tan (e+f x)^5}dx\) |
\(\Big \downarrow \) 3186 |
\(\displaystyle \frac {\int \frac {\csc ^5(e+f x) (a-a \sin (e+f x))^2 (\sin (e+f x) a+a)^{m+2}}{a^5}d(a \sin (e+f x))}{f}\) |
\(\Big \downarrow \) 100 |
\(\displaystyle \frac {\frac {\int -\frac {\csc ^4(e+f x) (a (9-m)-4 a \sin (e+f x)) (\sin (e+f x) a+a)^{m+2}}{a^3}d(a \sin (e+f x))}{4 a}-\frac {\csc ^4(e+f x) (a \sin (e+f x)+a)^{m+3}}{4 a^3}}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {-\frac {\int \frac {\csc ^4(e+f x) (a (9-m)-4 a \sin (e+f x)) (\sin (e+f x) a+a)^{m+2}}{a^3}d(a \sin (e+f x))}{4 a}-\frac {\csc ^4(e+f x) (a \sin (e+f x)+a)^{m+3}}{4 a^3}}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {1}{4} \int \frac {\csc ^4(e+f x) (a (9-m)-4 a \sin (e+f x)) (\sin (e+f x) a+a)^{m+2}}{a^4}d(a \sin (e+f x))-\frac {\csc ^4(e+f x) (a \sin (e+f x)+a)^{m+3}}{4 a^3}}{f}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (m^2-9 m+12\right ) \int \frac {\csc ^3(e+f x) (\sin (e+f x) a+a)^{m+2}}{a^3}d(a \sin (e+f x))+\frac {(9-m) \csc ^3(e+f x) (a \sin (e+f x)+a)^{m+3}}{3 a^3}\right )-\frac {\csc ^4(e+f x) (a \sin (e+f x)+a)^{m+3}}{4 a^3}}{f}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {(9-m) \csc ^3(e+f x) (a \sin (e+f x)+a)^{m+3}}{3 a^3}-\frac {\left (m^2-9 m+12\right ) (a \sin (e+f x)+a)^{m+3} \operatorname {Hypergeometric2F1}(3,m+3,m+4,\sin (e+f x)+1)}{3 a^3 (m+3)}\right )-\frac {\csc ^4(e+f x) (a \sin (e+f x)+a)^{m+3}}{4 a^3}}{f}\) |
(-1/4*(Csc[e + f*x]^4*(a + a*Sin[e + f*x])^(3 + m))/a^3 + (((9 - m)*Csc[e + f*x]^3*(a + a*Sin[e + f*x])^(3 + m))/(3*a^3) - ((12 - 9*m + m^2)*Hyperge ometric2F1[3, 3 + m, 4 + m, 1 + Sin[e + f*x]]*(a + a*Sin[e + f*x])^(3 + m) )/(3*a^3*(3 + m)))/4)/f
3.2.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x )^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[n] && (IntegerQ[m] || GtQ[-d/(b*c), 0])
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[x^p*((a + x)^(m - (p + 1)/2)/(a - x) ^((p + 1)/2)), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && E qQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]
\[\int \left (\cot ^{5}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
\[ \int \cot ^5(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cot \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \cot ^5(e+f x) (a+a \sin (e+f x))^m \, dx=\text {Timed out} \]
\[ \int \cot ^5(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cot \left (f x + e\right )^{5} \,d x } \]
\[ \int \cot ^5(e+f x) (a+a \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cot \left (f x + e\right )^{5} \,d x } \]
Timed out. \[ \int \cot ^5(e+f x) (a+a \sin (e+f x))^m \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^5\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m \,d x \]