Integrand size = 19, antiderivative size = 84 \[ \int \cos (e+f x) (a+a \sec (e+f x))^m \, dx=-\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2}+m,\frac {1}{2},2,\frac {3}{2}+m,\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1-\sec (e+f x)}} \]
-AppellF1(1/2+m,2,1/2,3/2+m,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*(a+a*sec(f*x+ e))^m*2^(1/2)*tan(f*x+e)/f/(1+2*m)/(1-sec(f*x+e))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3781\) vs. \(2(84)=168\).
Time = 17.47 (sec) , antiderivative size = 3781, normalized size of antiderivative = 45.01 \[ \int \cos (e+f x) (a+a \sec (e+f x))^m \, dx=\text {Result too large to show} \]
(2^(1 + m)*Cos[(e + f*x)/2]^3*Cos[e + f*x]*(Cos[(e + f*x)/2]^2*Sec[e + f*x ])^m*(a*(1 + Sec[e + f*x]))^m*Sin[(e + f*x)/2]*((-3*AppellF1[1/2, m, 1, 3/ 2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2)/(3*AppellF 1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*(AppellF1[3 /2, m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^ 2) + (2*AppellF1[1/2, m, 2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]) /(AppellF1[1/2, m, 2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*( -2*AppellF1[3/2, m, 3, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*A ppellF1[3/2, 1 + m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[ (e + f*x)/2]^2)/3)))/(f*(2^m*Cos[(e + f*x)/2]^4*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^m*((-3*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x) /2]^2]*Sec[(e + f*x)/2]^2)/(3*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*(AppellF1[3/2, m, 2, 5/2, Tan[(e + f*x)/2]^2, -T an[(e + f*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(e + f*x)/2]^2, -T an[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) + (2*AppellF1[1/2, m, 2, 3/2, Tan[ (e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])/(AppellF1[1/2, m, 2, 3/2, Tan[(e + f *x)/2]^2, -Tan[(e + f*x)/2]^2] + (2*(-2*AppellF1[3/2, m, 3, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + m*AppellF1[3/2, 1 + m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)/3)) - 3*2^m*Cos[(e...
Time = 0.35 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 4315, 3042, 4314, 153}
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.
\(\displaystyle \int \cos (e+f x) (a \sec (e+f x)+a)^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^m}{\csc \left (e+f x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4315 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \cos (e+f x) (\sec (e+f x)+1)^mdx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (\sec (e+f x)+1)^{-m} (a \sec (e+f x)+a)^m \int \frac {\left (\csc \left (e+f x+\frac {\pi }{2}\right )+1\right )^m}{\csc \left (e+f x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4314 |
\(\displaystyle -\frac {\tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \int \frac {\cos ^2(e+f x) (\sec (e+f x)+1)^{m-\frac {1}{2}}}{\sqrt {1-\sec (e+f x)}}d\sec (e+f x)}{f \sqrt {1-\sec (e+f x)}}\) |
\(\Big \downarrow \) 153 |
\(\displaystyle -\frac {\sqrt {2} \tan (e+f x) (a \sec (e+f x)+a)^m \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},2,m+\frac {3}{2},\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right )}{f (2 m+1) \sqrt {1-\sec (e+f x)}}\) |
-((Sqrt[2]*AppellF1[1/2 + m, 1/2, 2, 3/2 + m, (1 + Sec[e + f*x])/2, 1 + Se c[e + f*x]]*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(1 + 2*m)*Sqrt[1 - Sec [e + f*x]]))
3.4.46.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(b*e - a*f)^p*((a + b*x)^(m + 1)/(b^(p + 1)*(m + 1)*Simp lify[b/(b*c - a*d)]^n))*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}, x] && !IntegerQ[m] && !IntegerQ[n] && IntegerQ[p] && GtQ[Simplify[b/( b*c - a*d)], 0] && !(GtQ[Simplify[d/(d*a - c*b)], 0] && SimplerQ[c + d*x, a + b*x])
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]
\[\int \cos \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m}d x\]
\[ \int \cos (e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right ) \,d x } \]
\[ \int \cos (e+f x) (a+a \sec (e+f x))^m \, dx=\int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \cos {\left (e + f x \right )}\, dx \]
\[ \int \cos (e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right ) \,d x } \]
\[ \int \cos (e+f x) (a+a \sec (e+f x))^m \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right ) \,d x } \]
Timed out. \[ \int \cos (e+f x) (a+a \sec (e+f x))^m \, dx=\int \cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]