Integrand size = 26, antiderivative size = 108 \[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx=\frac {2 a^2 (1+4 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},1+\sec (e+f x)\right ) \tan (e+f x)}{f (1+2 n) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^2 (-\sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {a-a \sec (e+f x)}} \]
2*a^2*(1+4*n)*hypergeom([1/2, 1-n],[3/2],1+sec(f*x+e))*tan(f*x+e)/f/(1+2*n )/(a-a*sec(f*x+e))^(1/2)+2*a^2*(-sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n)/(a-a*s ec(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 3.32 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.93 \[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx=\frac {2^{-\frac {3}{2}+n} e^{-\frac {1}{2} i (e+f x)} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{\frac {1}{2}+n} \left (1+e^{2 i (e+f x)}\right )^{\frac {1}{2}+n} \csc ^3\left (\frac {1}{2} (e+f x)\right ) \left (-\frac {\operatorname {Hypergeometric2F1}\left (\frac {n}{2},\frac {3}{2}+n,\frac {2+n}{2},-e^{2 i (e+f x)}\right )}{n}+\frac {3 e^{i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1+n}{2},\frac {3}{2}+n,\frac {3+n}{2},-e^{2 i (e+f x)}\right )}{1+n}-\frac {3 e^{2 i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,\frac {2+n}{2},\frac {4+n}{2},-e^{2 i (e+f x)}\right )}{2+n}+\frac {e^{3 i (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {3}{2}+n,\frac {3+n}{2},\frac {5+n}{2},-e^{2 i (e+f x)}\right )}{3+n}\right ) (-\sec (e+f x))^n \sec ^{-\frac {3}{2}-n}(e+f x) (a-a \sec (e+f x))^{3/2}}{f} \]
(2^(-3/2 + n)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^(1/2 + n)*(1 + E ^((2*I)*(e + f*x)))^(1/2 + n)*Csc[(e + f*x)/2]^3*(-(Hypergeometric2F1[n/2, 3/2 + n, (2 + n)/2, -E^((2*I)*(e + f*x))]/n) + (3*E^(I*(e + f*x))*Hyperge ometric2F1[(1 + n)/2, 3/2 + n, (3 + n)/2, -E^((2*I)*(e + f*x))])/(1 + n) - (3*E^((2*I)*(e + f*x))*Hypergeometric2F1[3/2 + n, (2 + n)/2, (4 + n)/2, - E^((2*I)*(e + f*x))])/(2 + n) + (E^((3*I)*(e + f*x))*Hypergeometric2F1[3/2 + n, (3 + n)/2, (5 + n)/2, -E^((2*I)*(e + f*x))])/(3 + n))*(-Sec[e + f*x] )^n*Sec[e + f*x]^(-3/2 - n)*(a - a*Sec[e + f*x])^(3/2))/(E^((I/2)*(e + f*x ))*f)
Time = 0.49 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3042, 4301, 27, 2011, 3042, 4293, 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.
| \(\displaystyle \int (a-a \sec (e+f x))^{3/2} (-\sec (e+f x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a-a \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2} \left (-\csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
| \(\Big \downarrow \) 4301 |
| \(\displaystyle \frac {2 a^2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}-\frac {2 a \int -\frac {(-\sec (e+f x))^n (a (4 n+1)-a (4 n+1) \sec (e+f x))}{2 \sqrt {a-a \sec (e+f x)}}dx}{2 n+1}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a \int \frac {(-\sec (e+f x))^n (a (4 n+1)-a (4 n+1) \sec (e+f x))}{\sqrt {a-a \sec (e+f x)}}dx}{2 n+1}+\frac {2 a^2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle \frac {a (4 n+1) \int (-\sec (e+f x))^n \sqrt {a-a \sec (e+f x)}dx}{2 n+1}+\frac {2 a^2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (4 n+1) \int \left (-\csc \left (e+f x+\frac {\pi }{2}\right )\right )^n \sqrt {a-a \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{2 n+1}+\frac {2 a^2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}\) |
| \(\Big \downarrow \) 4293 |
| \(\displaystyle \frac {a^3 (4 n+1) \tan (e+f x) \int \frac {(-\sec (e+f x))^{n-1}}{\sqrt {\sec (e+f x) a+a}}d\sec (e+f x)}{f (2 n+1) \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}+\frac {2 a^2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}\) |
| \(\Big \downarrow \) 75 |
| \(\displaystyle \frac {2 a^2 (4 n+1) \tan (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},\sec (e+f x)+1\right )}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}+\frac {2 a^2 \tan (e+f x) (-\sec (e+f x))^n}{f (2 n+1) \sqrt {a-a \sec (e+f x)}}\) |
(2*a^2*(1 + 4*n)*Hypergeometric2F1[1/2, 1 - n, 3/2, 1 + Sec[e + f*x]]*Tan[ e + f*x])/(f*(1 + 2*n)*Sqrt[a - a*Sec[e + f*x]]) + (2*a^2*(-Sec[e + f*x])^ n*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[a - a*Sec[e + f*x]])
3.4.21.3.1 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[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a^2*d*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]] *Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(d*x)^(n - 1)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[b/(m + n - 1) Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^ 2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]
\[\int \left (-\sec \left (f x +e \right )\right )^{n} \left (a -a \sec \left (f x +e \right )\right )^{\frac {3}{2}}d x\]
\[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx=\int { {\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (-\sec \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx=\int \left (- \sec {\left (e + f x \right )}\right )^{n} \left (- a \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}\, dx \]
\[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx=\int { {\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (-\sec \left (f x + e\right )\right )^{n} \,d x } \]
\[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx=\int { {\left (-a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (-\sec \left (f x + e\right )\right )^{n} \,d x } \]
Timed out. \[ \int (-\sec (e+f x))^n (a-a \sec (e+f x))^{3/2} \, dx=\int {\left (a-\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (-\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]