Integrand size = 19, antiderivative size = 73 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f} \]
2^(1/2+m)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2*sec(f*x+e))*(1+sec(f*x+e))^ (-1/2-m)*(a+a*sec(f*x+e))^m*tan(f*x+e)/f
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a (1+\sec (e+f x)))^m \tan (e+f x)}{f} \]
(2^(1/2 + m)*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(-1/2 - m)*(a*(1 + Sec[e + f*x]))^m*Tan[e + f*x])/f
Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 4315, 3042, 4314, 79}
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.
\(\displaystyle \int \sec (e+f x) (a \sec (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^mdx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \sec (e+f x) (\sec (e+f x)+1)^mdx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^mdx\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \int \frac {(\sec (e+f x)+1)^{m-\frac {1}{2}}}{\sqrt {1-\sec (e+f x)}}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x))\right )}{f}\) |
(2^(1/2 + m)*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(-1/2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/f
3.4.44.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), 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)*((a + b*x)^(m - 1/2 )/Sqrt[a - b*x]), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && GtQ[a, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[a^IntPart[m]*((a + b*Csc[e + f*x])^FracPart[m ]/(1 + (b/a)*Csc[e + f*x])^FracPart[m]) Int[(1 + (b/a)*Csc[e + f*x])^m*(d *Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^ 2, 0] && !IntegerQ[m] && !GtQ[a, 0]
\[\int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m}d x\]
\[ \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \]
\[ \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sec {\left (e + f x \right )}\, dx \]
\[ \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \]
\[ \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \]
Timed out. \[ \int \sec (e+f x) (a+a \sec (e+f x))^m \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{\cos \left (e+f\,x\right )} \,d x \]