Integrand size = 25, antiderivative size = 98 \[ \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx=-\frac {2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},\frac {1}{2}-m,\frac {5}{2},\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^{3/2} (1+\sec (e+f x))^{-\frac {1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{3 f \sqrt {1-\sec (e+f x)}} \]
-2/3*AppellF1(3/2,1/2-m,1/2,5/2,-sec(f*x+e),sec(f*x+e))*(d*sec(f*x+e))^(3/ 2)*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/(1-sec(f*x+e))^ (1/2)
Leaf count is larger than twice the leaf count of optimal. \(2529\) vs. \(2(98)=196\).
Time = 17.67 (sec) , antiderivative size = 2529, normalized size of antiderivative = 25.81 \[ \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx=\text {Result too large to show} \]
(-3*2^(1 + m)*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[( e + f*x)/2]^2]*Sqrt[Sec[e + f*x]]*(d*Sec[e + f*x])^(3/2)*(Cos[(e + f*x)/2] ^2*Sec[e + f*x])^m*(a*(1 + Sec[e + f*x]))^m*Tan[(e + f*x)/2])/(f*(-1 + Tan [(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^ 2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[ (e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)*((3*2^(1 + m)*Ap pellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*S ec[(e + f*x)/2]^2*Sqrt[Sec[e + f*x]]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^m*T an[(e + f*x)/2]^2)/((-1 + Tan[(e + f*x)/2]^2)^2*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)*AppellF 1[3/2, 5/2 + m, -1/2, 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, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2 ]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Sqrt[Sec[e + f*x]]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^m)/((-1 + Tan[(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)*A ppellF1[3/2, 5/2 + m, -1/2, 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, 3/2 + m, -1/2, 3/2, Tan[(e...
Time = 0.40 (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.
\(\displaystyle \int (d \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2} \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^mdx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int (d \sec (e+f x))^{3/2} (\sec (e+f x)+1)^mdx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2} \left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^mdx\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {d \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \int \frac {\sqrt {d \sec (e+f x)} (\sec (e+f x)+1)^{m-\frac {1}{2}}}{\sqrt {1-\sec (e+f x)}}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)}}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle -\frac {2 \tan (e+f x) (d \sec (e+f x))^{3/2} (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{2},\frac {1}{2}-m,\frac {5}{2},\sec (e+f x),-\sec (e+f x)\right )}{3 f \sqrt {1-\sec (e+f x)}}\) |
(-2*AppellF1[3/2, 1/2, 1/2 - m, 5/2, Sec[e + f*x], -Sec[e + f*x]]*(d*Sec[e + f*x])^(3/2)*(1 + Sec[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]])
3.4.47.3.1 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]
\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sec \left (f x +e \right )\right )^{m}d x\]
\[ \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx=\text {Timed out} \]
\[ \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x } \]
\[ \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2} \,d x \]