Integrand size = 21, antiderivative size = 88 \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^m \, dx=\frac {2^{\frac {1}{2}+m} \operatorname {AppellF1}\left (\frac {1}{2},1-n,\frac {1}{2}-m,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{f} \]
2^(1/2+m)*AppellF1(1/2,1-n,1/2-m,3/2,1-sec(f*x+e),1/2-1/2*sec(f*x+e))*(1+s ec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e))^m*tan(f*x+e)/f
Leaf count is larger than twice the leaf count of optimal. \(2248\) vs. \(2(88)=176\).
Time = 6.30 (sec) , antiderivative size = 2248, normalized size of antiderivative = 25.55 \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^m \, dx=\text {Result too large to show} \]
(3*2^(1 + m)*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^(-1 + n)*Sec[e + f*x]^n*(Cos[(e + f*x)/2 ]^2*Sec[e + f*x])^(m + n)*(a*(1 + Sec[e + f*x]))^m*Tan[(e + f*x)/2])/(f*(3 *AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[( e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 5/2, Tan[(e + f*x )/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)*((3*2^m*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2 ]^2)^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(m + n))/(3*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF 1[3/2, m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x) /2]^2])*Tan[(e + f*x)/2]^2) + (3*2^(1 + m)*(-1 + n)*AppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Sec[(e + f*x)/2]^2)^( -1 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(m + n)*Tan[(e + f*x)/2]^2)/(3*A ppellF1[1/2, m + n, 1 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-1 + n)*AppellF1[3/2, m + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (m + n)*AppellF1[3/2, 1 + m + n, 1 - n, 5/2, Tan[(e + f*x)/ 2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (3*2^(1 + m)*(Sec[(e + f *x)/2]^2)^(-1 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(m + n)*Tan[(e + f...
Time = 0.35 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4315, 3042, 4312, 150}
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.
\(\displaystyle \int \sec ^n(e+f x) (a \sec (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right )^n \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^mdx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \sec ^n(e+f x) (\sec (e+f x)+1)^mdx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \csc \left (e+f x+\frac {\pi }{2}\right )^n \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^mdx\) |
\(\Big \downarrow \) 4312 |
\(\displaystyle \frac {\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \int \frac {\sec ^{n-1}(e+f x) (\sec (e+f x)+1)^{m-\frac {1}{2}}}{\sqrt {1-\sec (e+f x)}}d(1-\sec (e+f x))}{f \sqrt {1-\sec (e+f x)}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (\frac {1}{2},1-n,\frac {1}{2}-m,\frac {3}{2},1-\sec (e+f x),\frac {1}{2} (1-\sec (e+f x))\right )}{f}\) |
(2^(1/2 + m)*AppellF1[1/2, 1 - n, 1/2 - m, 3/2, 1 - Sec[e + f*x], (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
3.4.31.3.1 Defintions of rubi rules used
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 , (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-(a*(d/b))^n)*(Cot[e + f*x]/(a^(n - 2)*f*Sqrt [a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(a - x)^(n - 1) *((2*a - x)^(m - 1/2)/Sqrt[x]), x], x, a - b*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] & & !IntegerQ[n] && GtQ[a*(d/b), 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]
\[\int \sec \left (f x +e \right )^{n} \left (a +a \sec \left (f x +e \right )\right )^{m}d x\]
\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{n} \,d x } \]
\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^m \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sec ^{n}{\left (e + f x \right )}\, dx \]
\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{n} \,d x } \]
\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{n} \,d x } \]
Timed out. \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^m \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]