Integrand size = 25, antiderivative size = 98 \[ \int \frac {(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx=\frac {2 \operatorname {AppellF1}\left (-\frac {3}{2},\frac {1}{2},\frac {1}{2}-m,-\frac {1}{2},\sec (e+f x),-\sec (e+f x)\right ) (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{3 f \sqrt {1-\sec (e+f x)} (d \sec (e+f x))^{3/2}} \]
2/3*AppellF1(-3/2,1/2-m,1/2,-1/2,-sec(f*x+e),sec(f*x+e))*(1+sec(f*x+e))^(- 1/2-m)*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/(d*sec(f*x+e))^(3/2)/(1-sec(f*x+e)) ^(1/2)
Result contains complex when optimal does not.
Time = 22.33 (sec) , antiderivative size = 3349, normalized size of antiderivative = 34.17 \[ \int \frac {(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx=\text {Result too large to show} \]
(2^(1 + m)*Sec[e + f*x]^(3/2)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^(1/2 + m)* (a*(1 + Sec[e + f*x]))^m*((Cos[2*(e + f*x)]^3*Sqrt[Sec[e + f*x]]*(1 + Sec[ e + f*x])^m)/4 + Cos[2*(e + f*x)]^2*Sqrt[Sec[e + f*x]]*((1 + Sec[e + f*x]) ^m/2 + (I/4)*(1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]) + Cos[2*(e + f*x)]*Sqr t[Sec[e + f*x]]*((1 + Sec[e + f*x])^m/4 + ((1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]^2)/4) + Sqrt[Sec[e + f*x]]*((-1/4*I)*(1 + Sec[e + f*x])^m*Sin[2*(e + f*x)] + ((1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]^2)/2 + (I/4)*(1 + Sec[e + f*x])^m*Sin[2*(e + f*x)]^3))*Tan[(e + f*x)/2]*(-(AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(1/2 + m)*Tan[(e + f*x)/2]^2) - (9*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x])/((Sec[(e + f* x)/2]^2)^(3/2)*(-3*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, - Tan[(e + f*x)/2]^2] + (5*AppellF1[3/2, -1/2 + m, 7/2, 5/2, Tan[(e + f*x)/2 ]^2, -Tan[(e + f*x)/2]^2] + (1 - 2*m)*AppellF1[3/2, 1/2 + m, 5/2, 5/2, Tan [(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))))/(3*f*(d*Sec[ e + f*x])^(3/2)*(1 + Sec[e + f*x])^m*((2^m*Sec[(e + f*x)/2]^2*(Cos[(e + f* x)/2]^2*Sec[e + f*x])^(1/2 + m)*(-(AppellF1[3/2, -1/2 + m, 5/2, 5/2, Tan[( e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(Cos[e + f*x]*Sec[(e + f*x)/2]^2)^(1/2 + m)*Tan[(e + f*x)/2]^2) - (9*AppellF1[1/2, -1/2 + m, 5/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[e + f*x])/((Sec[(e + f*x)/2]^2)^(3/...
Time = 0.42 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4315, 3042, 4314, 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.
\(\displaystyle \int \frac {(a \sec (e+f x)+a)^m}{(d \sec (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^m}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \frac {(\sec (e+f x)+1)^m}{(d \sec (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \frac {\left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^m}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {d \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \int \frac {(\sec (e+f x)+1)^{m-\frac {1}{2}}}{\sqrt {1-\sec (e+f x)} (d \sec (e+f x))^{5/2}}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {2 \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (-\frac {3}{2},\frac {1}{2},\frac {1}{2}-m,-\frac {1}{2},\sec (e+f x),-\sec (e+f x)\right )}{3 f \sqrt {1-\sec (e+f x)} (d \sec (e+f x))^{3/2}}\) |
(2*AppellF1[-3/2, 1/2, 1/2 - m, -1/2, Sec[e + f*x], -Sec[e + f*x]]*(1 + Se c[e + f*x])^(-1/2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(3*f*Sqrt[1 - Sec[e + f*x]]*(d*Sec[e + f*x])^(3/2))
3.4.50.3.1 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^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]
\[\int \frac {\left (a +a \sec \left (f x +e \right )\right )^{m}}{\left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{m}}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^m}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]