Integrand size = 28, antiderivative size = 172 \[ \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^n \, dx=\frac {6 a^3 (c-c \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {2 a^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+n,\frac {3}{2}+n,1-\sec (e+f x)\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a+a \sec (e+f x)}}-\frac {2 a^3 (c-c \sec (e+f x))^{1+n} \tan (e+f x)}{c f (3+2 n) \sqrt {a+a \sec (e+f x)}} \]
6*a^3*(c-c*sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n)/(a+a*sec(f*x+e))^(1/2)+2*a^3 *hypergeom([1, 1/2+n],[3/2+n],1-sec(f*x+e))*(c-c*sec(f*x+e))^n*tan(f*x+e)/ f/(1+2*n)/(a+a*sec(f*x+e))^(1/2)-2*a^3*(c-c*sec(f*x+e))^(1+n)*tan(f*x+e)/c /f/(3+2*n)/(a+a*sec(f*x+e))^(1/2)
Time = 0.60 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.59 \[ \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^n \, dx=\frac {2 a^3 (c-c \sec (e+f x))^n \left (4 (2+n)+(3+2 n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+n,\frac {3}{2}+n,1-\sec (e+f x)\right )+(1+2 n) \sec (e+f x)\right ) \tan (e+f x)}{f (1+2 n) (3+2 n) \sqrt {a (1+\sec (e+f x))}} \]
(2*a^3*(c - c*Sec[e + f*x])^n*(4*(2 + n) + (3 + 2*n)*Hypergeometric2F1[1, 1/2 + n, 3/2 + n, 1 - Sec[e + f*x]] + (1 + 2*n)*Sec[e + f*x])*Tan[e + f*x] )/(f*(1 + 2*n)*(3 + 2*n)*Sqrt[a*(1 + Sec[e + f*x])])
Time = 0.38 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 4400, 27, 99, 2009}
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.
\(\displaystyle \int (a \sec (e+f x)+a)^{5/2} (c-c \sec (e+f x))^n \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
\(\Big \downarrow \) 4400 |
\(\displaystyle -\frac {a c \tan (e+f x) \int a^2 \cos (e+f x) (\sec (e+f x)+1)^2 (c-c \sec (e+f x))^{n-\frac {1}{2}}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^3 c \tan (e+f x) \int \cos (e+f x) (\sec (e+f x)+1)^2 (c-c \sec (e+f x))^{n-\frac {1}{2}}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^3 c \tan (e+f x) \int \left (\cos (e+f x) (c-c \sec (e+f x))^{n-\frac {1}{2}}+3 (c-c \sec (e+f x))^{n-\frac {1}{2}}-\frac {(c-c \sec (e+f x))^{n+\frac {1}{2}}}{c}\right )d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^3 c \tan (e+f x) \left (\frac {2 (c-c \sec (e+f x))^{n+\frac {3}{2}}}{c^2 (2 n+3)}-\frac {2 (c-c \sec (e+f x))^{n+\frac {1}{2}} \operatorname {Hypergeometric2F1}\left (1,n+\frac {1}{2},n+\frac {3}{2},1-\sec (e+f x)\right )}{c (2 n+1)}-\frac {6 (c-c \sec (e+f x))^{n+\frac {1}{2}}}{c (2 n+1)}\right )}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
-((a^3*c*((-6*(c - c*Sec[e + f*x])^(1/2 + n))/(c*(1 + 2*n)) - (2*Hypergeom etric2F1[1, 1/2 + n, 3/2 + n, 1 - Sec[e + f*x]]*(c - c*Sec[e + f*x])^(1/2 + n))/(c*(1 + 2*n)) + (2*(c - c*Sec[e + f*x])^(3/2 + n))/(c^2*(3 + 2*n)))* Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))
3.2.38.3.1 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_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[a*c*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d *x)^(n - 1/2)/x), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
\[\int \left (a +a \sec \left (f x +e \right )\right )^{\frac {5}{2}} \left (c -c \sec \left (f x +e \right )\right )^{n}d x\]
\[ \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^n \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} \,d x } \]
integral((a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2)*sqrt(a*sec(f*x + e) + a)*(-c*sec(f*x + e) + c)^n, x)
Timed out. \[ \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^n \, dx=\text {Timed out} \]
\[ \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^n \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} \,d x } \]
\[ \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^n \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{n} \,d x } \]
Timed out. \[ \int (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^n \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]