Integrand size = 21, antiderivative size = 127 \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {a (4-n) \operatorname {Hypergeometric2F1}\left (1,-1+n,n,\frac {1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^{-1+n}}{4 d (1-n)}+\frac {a \operatorname {Hypergeometric2F1}(1,-1+n,n,1+\sec (c+d x)) (a+a \sec (c+d x))^{-1+n}}{d (1-n)}+\frac {a (a+a \sec (c+d x))^{-1+n}}{2 d (1-\sec (c+d x))} \]
-1/4*a*(4-n)*hypergeom([1, -1+n],[n],1/2+1/2*sec(d*x+c))*(a+a*sec(d*x+c))^ (-1+n)/d/(1-n)+a*hypergeom([1, -1+n],[n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(- 1+n)/d/(1-n)+1/2*a*(a+a*sec(d*x+c))^(-1+n)/d/(1-sec(d*x+c))
Time = 0.20 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {a \left (-2+2 n+(-4+n) \operatorname {Hypergeometric2F1}\left (1,-1+n,n,\frac {1}{2} (1+\sec (c+d x))\right ) (-1+\sec (c+d x))+4 \operatorname {Hypergeometric2F1}(1,-1+n,n,1+\sec (c+d x)) (-1+\sec (c+d x))\right ) (a (1+\sec (c+d x)))^{-1+n}}{4 d (-1+n) (-1+\sec (c+d x))} \]
-1/4*(a*(-2 + 2*n + (-4 + n)*Hypergeometric2F1[1, -1 + n, n, (1 + Sec[c + d*x])/2]*(-1 + Sec[c + d*x]) + 4*Hypergeometric2F1[1, -1 + n, n, 1 + Sec[c + d*x]]*(-1 + Sec[c + d*x]))*(a*(1 + Sec[c + d*x]))^(-1 + n))/(d*(-1 + n) *(-1 + Sec[c + d*x]))
Time = 0.31 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 25, 4368, 27, 114, 25, 27, 174, 75, 78}
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.
\(\displaystyle \int \cot ^3(c+d x) (a \sec (c+d x)+a)^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^n}{\cot \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\left (\csc \left (\frac {1}{2} (2 c+\pi )+d x\right ) a+a\right )^n}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}dx\) |
\(\Big \downarrow \) 4368 |
\(\displaystyle \frac {a^4 \int \frac {\cos (c+d x) (\sec (c+d x) a+a)^{n-2}}{a^2 (1-\sec (c+d x))^2}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \int \frac {\cos (c+d x) (\sec (c+d x) a+a)^{n-2}}{(1-\sec (c+d x))^2}d\sec (c+d x)}{d}\) |
\(\Big \downarrow \) 114 |
\(\displaystyle \frac {a^2 \left (\frac {(a \sec (c+d x)+a)^{n-1}}{2 a (1-\sec (c+d x))}-\frac {\int -\frac {a \cos (c+d x) (\sec (c+d x) a+a)^{n-2} ((2-n) \sec (c+d x)+2)}{1-\sec (c+d x)}d\sec (c+d x)}{2 a}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {a^2 \left (\frac {\int \frac {a \cos (c+d x) (\sec (c+d x) a+a)^{n-2} ((2-n) \sec (c+d x)+2)}{1-\sec (c+d x)}d\sec (c+d x)}{2 a}+\frac {(a \sec (c+d x)+a)^{n-1}}{2 a (1-\sec (c+d x))}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \left (\frac {1}{2} \int \frac {\cos (c+d x) (\sec (c+d x) a+a)^{n-2} ((2-n) \sec (c+d x)+2)}{1-\sec (c+d x)}d\sec (c+d x)+\frac {(a \sec (c+d x)+a)^{n-1}}{2 a (1-\sec (c+d x))}\right )}{d}\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {a^2 \left (\frac {1}{2} \left ((4-n) \int \frac {(\sec (c+d x) a+a)^{n-2}}{1-\sec (c+d x)}d\sec (c+d x)+2 \int \cos (c+d x) (\sec (c+d x) a+a)^{n-2}d\sec (c+d x)\right )+\frac {(a \sec (c+d x)+a)^{n-1}}{2 a (1-\sec (c+d x))}\right )}{d}\) |
\(\Big \downarrow \) 75 |
\(\displaystyle \frac {a^2 \left (\frac {1}{2} \left ((4-n) \int \frac {(\sec (c+d x) a+a)^{n-2}}{1-\sec (c+d x)}d\sec (c+d x)+\frac {2 (a \sec (c+d x)+a)^{n-1} \operatorname {Hypergeometric2F1}(1,n-1,n,\sec (c+d x)+1)}{a (1-n)}\right )+\frac {(a \sec (c+d x)+a)^{n-1}}{2 a (1-\sec (c+d x))}\right )}{d}\) |
\(\Big \downarrow \) 78 |
\(\displaystyle \frac {a^2 \left (\frac {1}{2} \left (\frac {2 (a \sec (c+d x)+a)^{n-1} \operatorname {Hypergeometric2F1}(1,n-1,n,\sec (c+d x)+1)}{a (1-n)}-\frac {(4-n) (a \sec (c+d x)+a)^{n-1} \operatorname {Hypergeometric2F1}\left (1,n-1,n,\frac {1}{2} (\sec (c+d x)+1)\right )}{2 a (1-n)}\right )+\frac {(a \sec (c+d x)+a)^{n-1}}{2 a (1-\sec (c+d x))}\right )}{d}\) |
(a^2*((a + a*Sec[c + d*x])^(-1 + n)/(2*a*(1 - Sec[c + d*x])) + (-1/2*((4 - n)*Hypergeometric2F1[1, -1 + n, n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d *x])^(-1 + n))/(a*(1 - n)) + (2*Hypergeometric2F1[1, -1 + n, n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(-1 + n))/(a*(1 - n)))/2))/d
3.3.24.3.1 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_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b *c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] && !IntegerQ[m] && IntegerQ[n]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(d*b^(m - 1))^(-1) Subst[Int[(-a + b*x)^((m - 1)/2 )*((a + b*x)^((m - 1)/2 + n)/x), x], x, Csc[c + d*x]], x] /; FreeQ[{a, b, c , d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && !IntegerQ[n]
\[\int \cot \left (d x +c \right )^{3} \left (a +a \sec \left (d x +c \right )\right )^{n}d x\]
\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]
\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \cot ^{3}{\left (c + d x \right )}\, dx \]
\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]
\[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3} \,d x } \]
Timed out. \[ \int \cot ^3(c+d x) (a+a \sec (c+d x))^n \, dx=\int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]