Integrand size = 21, antiderivative size = 240 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx=\frac {a^2 \left (12+9 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (1,-2+n,-1+n,\frac {1}{2} (1+\sec (c+d x))\right ) (a+a \sec (c+d x))^{-2+n}}{16 d (2-n)}+\frac {a^2 (3+n) \sec ^2(c+d x) (a+a \sec (c+d x))^{-2+n}}{4 d (1-n) (1-\sec (c+d x))^2}-\frac {a^2 \sec ^3(c+d x) (a+a \sec (c+d x))^{-2+n}}{d (1-n) (1-\sec (c+d x))^2}-\frac {a^2 (a+a \sec (c+d x))^{-2+n} \left (12+4 n-7 n^2-n^3-2 (1-n) (6+n) \sec (c+d x)\right )}{8 d \left (2-3 n+n^2\right ) (1-\sec (c+d x))} \]
1/16*a^2*(n^2+9*n+12)*hypergeom([1, -2+n],[-1+n],1/2+1/2*sec(d*x+c))*(a+a* sec(d*x+c))^(-2+n)/d/(2-n)+1/4*a^2*(3+n)*sec(d*x+c)^2*(a+a*sec(d*x+c))^(-2 +n)/d/(1-n)/(1-sec(d*x+c))^2-a^2*sec(d*x+c)^3*(a+a*sec(d*x+c))^(-2+n)/d/(1 -n)/(1-sec(d*x+c))^2-1/8*a^2*(a+a*sec(d*x+c))^(-2+n)*(12+4*n-7*n^2-n^3-2*( 1-n)*(6+n)*sec(d*x+c))/d/(n^2-3*n+2)/(1-sec(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(492\) vs. \(2(240)=480\).
Time = 3.89 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.05 \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx=-\frac {\cos (c+d x) (1+\sec (c+d x))^{-n} (a (1+\sec (c+d x)))^n \left (2^{1+n} \cot ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{-1+n}-3\ 2^n n \cot ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{-1+n}+2^n n^2 \cot ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{-1+n}-3\ 2^{2+n} (-2+n) \operatorname {Hypergeometric2F1}\left (1,1-n,2-n,\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{-1+n}-2^n \left (-18+7 n+n^2\right ) \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec (c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{-1+n}+32 \sec (c+d x) (1+\sec (c+d x))^n-12 n \sec (c+d x) (1+\sec (c+d x))^n-12 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (1+\sec (c+d x))^n+2 n \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (1+\sec (c+d x))^n-2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (1+\sec (c+d x))^n+2 n \sec ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) (1+\sec (c+d x))^n\right )}{64 d (-2+n) (-1+n)} \]
-1/64*(Cos[c + d*x]*(a*(1 + Sec[c + d*x]))^n*(2^(1 + n)*Cot[(c + d*x)/2]^4 *Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) - 3*2^n*n*Cot[(c + d*x)/2]^4*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) + 2^n* n^2*Cot[(c + d*x)/2]^4*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) - 3*2^(2 + n)*(-2 + n)*Hypergeometric2F1[1, 1 - n, 2 - n, Cos[c + d*x ]*Sec[(c + d*x)/2]^2]*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) - 2^n*(-18 + 7*n + n^2)*Hypergeometric2F1[2, 1 - n, 2 - n, Cos[c + d*x ]*Sec[(c + d*x)/2]^2]*Sec[c + d*x]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n) + 32*Sec[c + d*x]*(1 + Sec[c + d*x])^n - 12*n*Sec[c + d*x]*(1 + Sec[c + d*x])^n - 12*Sec[(c + d*x)/2]^2*Sec[c + d*x]*(1 + Sec[c + d*x])^n + 2*n* Sec[(c + d*x)/2]^2*Sec[c + d*x]*(1 + Sec[c + d*x])^n - 2*Sec[(c + d*x)/2]^ 4*Sec[c + d*x]*(1 + Sec[c + d*x])^n + 2*n*Sec[(c + d*x)/2]^4*Sec[c + d*x]* (1 + Sec[c + d*x])^n))/(d*(-2 + n)*(-1 + n)*(1 + Sec[c + d*x])^n)
Time = 0.43 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4361, 25, 27, 111, 25, 27, 166, 27, 163, 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.
| \(\displaystyle \int \csc ^5(c+d x) (a \sec (c+d x)+a)^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^n}{\cos \left (c+d x-\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4361 |
| \(\displaystyle -\frac {a^6 \int -\frac {\sec ^4(c+d x) (\sec (c+d x) a+a)^{n-3}}{a^3 (1-\sec (c+d x))^3}d(-\sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {a^6 \int \frac {\sec ^4(c+d x) (\sec (c+d x) a+a)^{n-3}}{a^3 (1-\sec (c+d x))^3}d(-\sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^3 \int \frac {\sec ^4(c+d x) (\sec (c+d x) a+a)^{n-3}}{(1-\sec (c+d x))^3}d(-\sec (c+d x))}{d}\) |
| \(\Big \downarrow \) 111 |
| \(\displaystyle \frac {a^3 \left (\frac {\int -\frac {a \sec ^2(c+d x) (\sec (c+d x) a+a)^{n-3} (n \sec (c+d x)+3)}{(1-\sec (c+d x))^3}d(-\sec (c+d x))}{a (1-n)}-\frac {\sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{a (1-n) (1-\sec (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {a^3 \left (-\frac {\int \frac {a \sec ^2(c+d x) (\sec (c+d x) a+a)^{n-3} (n \sec (c+d x)+3)}{(1-\sec (c+d x))^3}d(-\sec (c+d x))}{a (1-n)}-\frac {\sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{a (1-n) (1-\sec (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^3 \left (-\frac {\int \frac {\sec ^2(c+d x) (\sec (c+d x) a+a)^{n-3} (n \sec (c+d x)+3)}{(1-\sec (c+d x))^3}d(-\sec (c+d x))}{1-n}-\frac {\sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{a (1-n) (1-\sec (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 166 |
| \(\displaystyle \frac {a^3 \left (-\frac {\frac {\int -\frac {a \sec (c+d x) (\sec (c+d x) a+a)^{n-3} (2 (n+3)-(1-n) (n+6) \sec (c+d x))}{(1-\sec (c+d x))^2}d(-\sec (c+d x))}{4 a}-\frac {(n+3) \sec ^2(c+d x) (a \sec (c+d x)+a)^{n-2}}{4 a (1-\sec (c+d x))^2}}{1-n}-\frac {\sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{a (1-n) (1-\sec (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^3 \left (-\frac {\frac {1}{4} \int -\frac {\sec (c+d x) (\sec (c+d x) a+a)^{n-3} (2 (n+3)-(1-n) (n+6) \sec (c+d x))}{(1-\sec (c+d x))^2}d(-\sec (c+d x))-\frac {(n+3) \sec ^2(c+d x) (a \sec (c+d x)+a)^{n-2}}{4 a (1-\sec (c+d x))^2}}{1-n}-\frac {\sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{a (1-n) (1-\sec (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 163 |
| \(\displaystyle \frac {a^3 \left (-\frac {\frac {1}{4} \left (\frac {\left (-2 (1-n) (n+6) \sec (c+d x)-n^3-7 n^2+4 n+12\right ) (a \sec (c+d x)+a)^{n-2}}{2 a (2-n) (1-\sec (c+d x))}-\frac {1}{2} (1-n) \left (n^2+9 n+12\right ) \int \frac {(\sec (c+d x) a+a)^{n-3}}{1-\sec (c+d x)}d(-\sec (c+d x))\right )-\frac {(n+3) \sec ^2(c+d x) (a \sec (c+d x)+a)^{n-2}}{4 a (1-\sec (c+d x))^2}}{1-n}-\frac {\sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{a (1-n) (1-\sec (c+d x))^2}\right )}{d}\) |
| \(\Big \downarrow \) 78 |
| \(\displaystyle \frac {a^3 \left (-\frac {\frac {1}{4} \left (\frac {\left (-2 (1-n) (n+6) \sec (c+d x)-n^3-7 n^2+4 n+12\right ) (a \sec (c+d x)+a)^{n-2}}{2 a (2-n) (1-\sec (c+d x))}-\frac {(1-n) \left (n^2+9 n+12\right ) (a \sec (c+d x)+a)^{n-2} \operatorname {Hypergeometric2F1}\left (1,n-2,n-1,\frac {1}{2} (\sec (c+d x)+1)\right )}{4 a (2-n)}\right )-\frac {(n+3) \sec ^2(c+d x) (a \sec (c+d x)+a)^{n-2}}{4 a (1-\sec (c+d x))^2}}{1-n}-\frac {\sec ^3(c+d x) (a \sec (c+d x)+a)^{n-2}}{a (1-n) (1-\sec (c+d x))^2}\right )}{d}\) |
(a^3*(-((Sec[c + d*x]^3*(a + a*Sec[c + d*x])^(-2 + n))/(a*(1 - n)*(1 - Sec [c + d*x])^2)) - (-1/4*((3 + n)*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^(-2 + n))/(a*(1 - Sec[c + d*x])^2) + (-1/4*((1 - n)*(12 + 9*n + n^2)*Hypergeomet ric2F1[1, -2 + n, -1 + n, (1 + Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(-2 + n))/(a*(2 - n)) + ((a + a*Sec[c + d*x])^(-2 + n)*(12 + 4*n - 7*n^2 - n^3 - 2*(1 - n)*(6 + n)*Sec[c + d*x]))/(2*a*(2 - n)*(1 - Sec[c + d*x])))/4)/(1 - n)))/d
3.2.51.3.1 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[((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 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) )*((g_.) + (h_.)*(x_)), x_] :> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(b*c - a*d)* (m + 1)*x)/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] - Simp[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f *h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c* d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2* d*(b*c - a*d)*(m + 1)*(m + n + 3)) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((GeQ[m, -2] && LtQ[m, - 1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h )*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m _), x_Symbol] :> Simp[-(f*b^(p - 1))^(-1) Subst[Int[(-a + b*x)^((p - 1)/2 )*((a + b*x)^(m + (p - 1)/2)/x^(p + 1)), x], x, Csc[e + f*x]], x] /; FreeQ[ {a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
\[\int \csc \left (d x +c \right )^{5} \left (a +a \sec \left (d x +c \right )\right )^{n}d x\]
\[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx=\text {Timed out} \]
\[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{5} \,d x } \]
\[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \csc \left (d x + c\right )^{5} \,d x } \]
Timed out. \[ \int \csc ^5(c+d x) (a+a \sec (c+d x))^n \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n}{{\sin \left (c+d\,x\right )}^5} \,d x \]