Integrand size = 21, antiderivative size = 220 \[ \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx=\frac {\sqrt {2} (a+b) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-1-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt {1+\sec (e+f x)}}-\frac {\sqrt {2} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right ) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \tan (e+f x)}{b f \sqrt {1+\sec (e+f x)}} \]
(a+b)*AppellF1(1/2,-1-m,1/2,3/2,b*(1-sec(f*x+e))/(a+b),1/2-1/2*sec(f*x+e)) *(a+b*sec(f*x+e))^m*2^(1/2)*tan(f*x+e)/b/f/(((a+b*sec(f*x+e))/(a+b))^m)/(1 +sec(f*x+e))^(1/2)-a*AppellF1(1/2,-m,1/2,3/2,b*(1-sec(f*x+e))/(a+b),1/2-1/ 2*sec(f*x+e))*(a+b*sec(f*x+e))^m*2^(1/2)*tan(f*x+e)/b/f/(((a+b*sec(f*x+e)) /(a+b))^m)/(1+sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(5564\) vs. \(2(220)=440\).
Time = 23.09 (sec) , antiderivative size = 5564, normalized size of antiderivative = 25.29 \[ \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx=\text {Result too large to show} \]
Time = 0.48 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4325, 3042, 4321, 156, 155}
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.
\(\displaystyle \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right )^2 \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx\) |
\(\Big \downarrow \) 4325 |
\(\displaystyle \frac {\int \sec (e+f x) (a+b \sec (e+f x))^{m+1}dx}{b}-\frac {a \int \sec (e+f x) (a+b \sec (e+f x))^mdx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{m+1}dx}{b}-\frac {a \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^mdx}{b}\) |
\(\Big \downarrow \) 4321 |
\(\displaystyle \frac {a \tan (e+f x) \int \frac {(a+b \sec (e+f x))^m}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}-\frac {\tan (e+f x) \int \frac {(a+b \sec (e+f x))^{m+1}}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle \frac {a \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (e+f x)}{a+b}\right )^m}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}-\frac {(a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \int \frac {\left (\frac {a}{a+b}+\frac {b \sec (e+f x)}{a+b}\right )^{m+1}}{\sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}d\sec (e+f x)}{b f \sqrt {1-\sec (e+f x)} \sqrt {\sec (e+f x)+1}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\sqrt {2} (a+b) \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m-1,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt {\sec (e+f x)+1}}-\frac {\sqrt {2} a \tan (e+f x) (a+b \sec (e+f x))^m \left (\frac {a+b \sec (e+f x)}{a+b}\right )^{-m} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},-m,\frac {3}{2},\frac {1}{2} (1-\sec (e+f x)),\frac {b (1-\sec (e+f x))}{a+b}\right )}{b f \sqrt {\sec (e+f x)+1}}\) |
(Sqrt[2]*(a + b)*AppellF1[1/2, 1/2, -1 - m, 3/2, (1 - Sec[e + f*x])/2, (b* (1 - Sec[e + f*x]))/(a + b)]*(a + b*Sec[e + f*x])^m*Tan[e + f*x])/(b*f*Sqr t[1 + Sec[e + f*x]]*((a + b*Sec[e + f*x])/(a + b))^m) - (Sqrt[2]*a*AppellF 1[1/2, 1/2, -m, 3/2, (1 - Sec[e + f*x])/2, (b*(1 - Sec[e + f*x]))/(a + b)] *(a + b*Sec[e + f*x])^m*Tan[e + f*x])/(b*f*Sqrt[1 + Sec[e + f*x]]*((a + b* Sec[e + f*x])/(a + b))^m)
3.8.89.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*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, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x ]]) Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f*x]] , x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[-a/b Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] + Simp[1/b Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{ a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]
\[\int \sec \left (f x +e \right )^{2} \left (a +b \sec \left (f x +e \right )\right )^{m}d x\]
\[ \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
\[ \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx=\int \left (a + b \sec {\left (e + f x \right )}\right )^{m} \sec ^{2}{\left (e + f x \right )}\, dx \]
\[ \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
\[ \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx=\int { {\left (b \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )^{2} \,d x } \]
Timed out. \[ \int \sec ^2(e+f x) (a+b \sec (e+f x))^m \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^m}{{\cos \left (e+f\,x\right )}^2} \,d x \]