Integrand size = 25, antiderivative size = 111 \[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {m}{2},\frac {1}{2},-p,\frac {2+m}{2},\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a}\right ) \cot (e+f x) (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p} \sqrt {-\tan ^2(e+f x)}}{f m} \]
AppellF1(1/2*m,1/2,-p,1+1/2*m,sec(f*x+e)^2,-b*sec(f*x+e)^2/a)*cos(f*x+e)*( d*sec(f*x+e))^m*(a+b*sec(f*x+e)^2)^p*(-tan(f*x+e)^2)^(1/2)/f/m/((1+b*sec(f *x+e)^2/a)^p)/sin(f*x+e)
Leaf count is larger than twice the leaf count of optimal. \(2195\) vs. \(2(111)=222\).
Time = 20.08 (sec) , antiderivative size = 2195, normalized size of antiderivative = 19.77 \[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Result too large to show} \]
(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f *x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(d*Sec[e + f*x])^m*(Sec[ e + f*x]^2)^(-1 + m/2 + p)*(a + b*Sec[e + f*x]^2)^p*Tan[e + f*x])/(f*(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2 )/(a + b))] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, - ((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p , 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2)*(( 3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f* x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(m/2 + p ))/(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f *x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f* x]^2) - (6*a*(a + b)*p*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -( (b*Tan[e + f*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^(-1 + p)*(Sec[ e + f*x]^2)^(-1 + m/2 + p)*Sin[2*(e + f*x)]*Tan[e + f*x])/(3*(a + b)*Appel lF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p, 5/2, -Tan [e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2) + (6*(a + ...
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.
\(\displaystyle \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sec (e+f x))^m \left (a+b \sec (e+f x)^2\right )^pdx\) |
\(\Big \downarrow \) 4638 |
\(\displaystyle \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^pdx\) |
3.3.98.3.1 Defintions of rubi rules used
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> Unintegrable[(d*Sec[e + f*x])^m*(a + b*(c*Sec[e + f*x])^n)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x]
\[\int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]
\[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{m} \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{p}\, dx \]
\[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
\[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int {\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]