Integrand size = 12, antiderivative size = 83 \[ \int (a+a \sec (e+f x))^m \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2}+m,\frac {1}{2},1,\frac {3}{2}+m,\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1-\sec (e+f x)}} \]
AppellF1(1/2+m,1,1/2,3/2+m,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*(a+a*sec(f*x+e ))^m*2^(1/2)*tan(f*x+e)/f/(1+2*m)/(1-sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(711\) vs. \(2(83)=166\).
Time = 7.36 (sec) , antiderivative size = 711, normalized size of antiderivative = 8.57 \[ \int (a+a \sec (e+f x))^m \, dx=\frac {30 \operatorname {AppellF1}\left (\frac {1}{2},m,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \cos (e+f x) (a (1+\sec (e+f x)))^m \sin (e+f x) \left (3 \operatorname {AppellF1}\left (\frac {1}{2},m,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \left (\operatorname {AppellF1}\left (\frac {3}{2},m,2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-m \operatorname {AppellF1}\left (\frac {3}{2},1+m,1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )}{f \left (45 \operatorname {AppellF1}\left (\frac {1}{2},m,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2 \cos ^2\left (\frac {1}{2} (e+f x)\right ) (1+2 m-2 m \cos (e+f x)+\cos (2 (e+f x)))+6 \operatorname {AppellF1}\left (\frac {1}{2},m,1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \left (-5 \operatorname {AppellF1}\left (\frac {3}{2},m,2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+2 m-2 (2+m) \cos (e+f x)+\cos (2 (e+f x)))+5 m \operatorname {AppellF1}\left (\frac {3}{2},1+m,1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) (1+2 m-2 (2+m) \cos (e+f x)+\cos (2 (e+f x)))-48 \left (2 \operatorname {AppellF1}\left (\frac {5}{2},m,3,\frac {7}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 m \operatorname {AppellF1}\left (\frac {5}{2},1+m,2,\frac {7}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+m (1+m) \operatorname {AppellF1}\left (\frac {5}{2},2+m,1,\frac {7}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \cot (e+f x) \csc (e+f x) \sin ^4\left (\frac {1}{2} (e+f x)\right )\right )+40 \left (\operatorname {AppellF1}\left (\frac {3}{2},m,2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-m \operatorname {AppellF1}\left (\frac {3}{2},1+m,1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )^2 \cos (e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )} \]
(30*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[ (e + f*x)/2]^2*Cos[e + f*x]*(a*(1 + Sec[e + f*x]))^m*Sin[e + f*x]*(3*Appel lF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*(AppellF1 [3/2, m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - m*AppellF1[3/2 , 1 + m, 1, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2 ]^2))/(f*(45*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2 ]^2]^2*Cos[(e + f*x)/2]^2*(1 + 2*m - 2*m*Cos[e + f*x] + Cos[2*(e + f*x)]) + 6*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[ (e + f*x)/2]^2*(-5*AppellF1[3/2, m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[e + f*x] + Cos[2*(e + f*x)]) + 5*m*App ellF1[3/2, 1 + m, 1, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + 2* m - 2*(2 + m)*Cos[e + f*x] + Cos[2*(e + f*x)]) - 48*(2*AppellF1[5/2, m, 3, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*m*AppellF1[5/2, 1 + m, 2, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*(1 + m)*AppellF1[5/2, 2 + m, 1, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cot[e + f*x]*Csc [e + f*x]*Sin[(e + f*x)/2]^4) + 40*(AppellF1[3/2, m, 2, 5/2, Tan[(e + f*x) /2]^2, -Tan[(e + f*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(e + f*x) /2]^2, -Tan[(e + f*x)/2]^2])^2*Cos[e + f*x]*Sin[(e + f*x)/2]^2*Tan[(e + f* x)/2]^2))
Time = 0.31 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4266, 3042, 4265, 153}
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.
| \(\displaystyle \int (a \sec (e+f x)+a)^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^mdx\) |
| \(\Big \downarrow \) 4266 |
| \(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int (\sec (e+f x)+1)^mdx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^mdx\) |
| \(\Big \downarrow \) 4265 |
| \(\displaystyle -\frac {\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \int \frac {\cos (e+f x) (\sec (e+f x)+1)^{m-\frac {1}{2}}}{\sqrt {1-\sec (e+f x)}}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)}}\) |
| \(\Big \downarrow \) 153 |
| \(\displaystyle \frac {\sqrt {2} \tan (e+f x) (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},1,m+\frac {3}{2},\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right )}{f (2 m+1) \sqrt {1-\sec (e+f x)}}\) |
(Sqrt[2]*AppellF1[1/2 + m, 1/2, 1, 3/2 + m, (1 + Sec[e + f*x])/2, 1 + Sec[ e + f*x]]*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(1 + 2*m)*Sqrt[1 - Sec[e + f*x]])
3.4.45.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp lify[b/(b*c - a*d)]^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( b*c - a*d)], 0] && !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, a + b*x])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^n*(Cot [c + d*x]/(d*Sqrt[1 + Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]])) Subst[Int[(1 + b*(x/a))^(n - 1/2)/(x*Sqrt[1 - b*(x/a)]), x], x, Csc[c + d*x]], x] /; Fr eeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0 ]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^IntPar t[n]*((a + b*Csc[c + d*x])^FracPart[n]/(1 + (b/a)*Csc[c + d*x])^FracPart[n] ) Int[(1 + (b/a)*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
\[\int \left (a +a \sec \left (f x +e \right )\right )^{m}d x\]
\[ \int (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (a+a \sec (e+f x))^m \, dx=\int \left (a \sec {\left (e + f x \right )} + a\right )^{m}\, dx \]
\[ \int (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (a+a \sec (e+f x))^m \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]