Integrand size = 23, antiderivative size = 117 \[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\frac {2 (d \sec (e+f x))^n \tan (e+f x)}{f (1+2 n) \sqrt {1+\sec (e+f x)}}-\frac {(1+4 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,1+n,\sec (e+f x)\right ) (d \sec (e+f x))^n \tan (e+f x)}{f n (1+2 n) \sqrt {1-\sec (e+f x)} \sqrt {1+\sec (e+f x)}} \]
2*(d*sec(f*x+e))^n*tan(f*x+e)/f/(1+2*n)/(1+sec(f*x+e))^(1/2)-(1+4*n)*hyper geom([1/2, n],[1+n],sec(f*x+e))*(d*sec(f*x+e))^n*tan(f*x+e)/f/n/(1+2*n)/(1 -sec(f*x+e))^(1/2)/(1+sec(f*x+e))^(1/2)
Time = 0.50 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73 \[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\frac {\left (-1+(1+4 n) \cos ^{\frac {1}{2}+n}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{2}+n,\frac {3}{2},2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) (d \sec (e+f x))^n \sqrt {1+\sec (e+f x)} \tan \left (\frac {1}{2} (e+f x)\right )}{f n} \]
((-1 + (1 + 4*n)*Cos[e + f*x]^(1/2 + n)*Hypergeometric2F1[1/2, 3/2 + n, 3/ 2, 2*Sin[(e + f*x)/2]^2])*(d*Sec[e + f*x])^n*Sqrt[1 + Sec[e + f*x]]*Tan[(e + f*x)/2])/(f*n)
Time = 0.44 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4301, 27, 2011, 3042, 4293, 74}
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.
| \(\displaystyle \int (\sec (e+f x)+1)^{3/2} (d \sec (e+f x))^n \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^{3/2} \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^ndx\) |
| \(\Big \downarrow \) 4301 |
| \(\displaystyle \frac {2 \int \frac {(d \sec (e+f x))^n (4 n+(4 n+1) \sec (e+f x)+1)}{2 \sqrt {\sec (e+f x)+1}}dx}{2 n+1}+\frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(d \sec (e+f x))^n (4 n+(4 n+1) \sec (e+f x)+1)}{\sqrt {\sec (e+f x)+1}}dx}{2 n+1}+\frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\) |
| \(\Big \downarrow \) 2011 |
| \(\displaystyle \frac {(4 n+1) \int (d \sec (e+f x))^n \sqrt {\sec (e+f x)+1}dx}{2 n+1}+\frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(4 n+1) \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^n \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right )+1}dx}{2 n+1}+\frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}\) |
| \(\Big \downarrow \) 4293 |
| \(\displaystyle \frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}-\frac {d (4 n+1) \tan (e+f x) \int \frac {(d \sec (e+f x))^{n-1}}{\sqrt {1-\sec (e+f x)}}d\sec (e+f x)}{f (2 n+1) \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
| \(\Big \downarrow \) 74 |
| \(\displaystyle \frac {2 \tan (e+f x) (d \sec (e+f x))^n}{f (2 n+1) \sqrt {\sec (e+f x)+1}}-\frac {(4 n+1) \tan (e+f x) (d \sec (e+f x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},n,n+1,\sec (e+f x)\right )}{f n (2 n+1) \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
(2*(d*Sec[e + f*x])^n*Tan[e + f*x])/(f*(1 + 2*n)*Sqrt[1 + Sec[e + f*x]]) - ((1 + 4*n)*Hypergeometric2F1[1/2, n, 1 + n, Sec[e + f*x]]*(d*Sec[e + f*x] )^n*Tan[e + f*x])/(f*n*(1 + 2*n)*Sqrt[1 - Sec[e + f*x]]*Sqrt[1 + Sec[e + f *x]])
3.4.3.3.1 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^n*((b*x )^(m + 1)/(b*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] && !IntegerQ[m] && (IntegerQ[n] || (GtQ[c, 0] && !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && 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]
\[\int \left (d \sec \left (f x +e \right )\right )^{n} \left (\sec \left (f x +e \right )+1\right )^{\frac {3}{2}}d x\]
\[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{n} \left (\sec {\left (e + f x \right )} + 1\right )^{\frac {3}{2}}\, dx \]
\[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]
\[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{n} {\left (\sec \left (f x + e\right ) + 1\right )}^{\frac {3}{2}} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^n (1+\sec (e+f x))^{3/2} \, dx=\int {\left (\frac {1}{\cos \left (e+f\,x\right )}+1\right )}^{3/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]