Integrand size = 14, antiderivative size = 90 \[ \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx=-\frac {3 \sqrt {2} \operatorname {AppellF1}\left (-\frac {7}{6},\frac {1}{2},1,-\frac {1}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{7 a d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) (a+a \sec (c+d x))^{2/3}} \]
-3/7*AppellF1(-7/6,1,1/2,-1/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan (d*x+c)/a/d/(1+sec(d*x+c))/(a+a*sec(d*x+c))^(2/3)/(1-sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(3007\) vs. \(2(90)=180\).
Time = 17.58 (sec) , antiderivative size = 3007, normalized size of antiderivative = 33.41 \[ \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx=\text {Result too large to show} \]
(((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(1 + Sec[c + d*x])^(5/3)*((27*Sin [c + d*x])/7 - (30*Tan[(c + d*x)/2])/7 + (3*Sec[(c + d*x)/2]^2*Tan[(c + d* x)/2])/14))/(d*(a*(1 + Sec[c + d*x]))^(5/3)) + (2^(1/3)*(1 + Sec[c + d*x]) ^(5/3)*((16*(1 + Sec[c + d*x])^(1/3))/7 - (27*Cos[c + d*x]*(1 + Sec[c + d* x])^(1/3))/7)*Tan[(c + d*x)/2]*((-3*AppellF1[3/2, 1/3, 1, 5/2, Tan[(c + d* x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d *x)/2]^2)^(2/3) + Cos[(c + d*x)/2]^2*(-27 - (5*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(Appe llF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*A ppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + Appel lF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d *x)/2]^2)/9)))))/(7*d*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(a*(1 + Sec[ c + d*x]))^(5/3)*((Sec[(c + d*x)/2]^2*((-3*AppellF1[3/2, 1/3, 1, 5/2, Tan[ (c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec [(c + d*x)/2]^2)^(2/3) + Cos[(c + d*x)/2]^2*(-27 - (5*AppellF1[1/2, 1/3, 1 , 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2 )*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + ( 2*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan [(c + d*x)/2]^2)/9)))))/(7*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2...
Time = 0.35 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4266, 3042, 4265, 149, 25, 1012}
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.
\(\displaystyle \int \frac {1}{(a \sec (c+d x)+a)^{5/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/3}}dx\) |
\(\Big \downarrow \) 4266 |
\(\displaystyle \frac {(\sec (c+d x)+1)^{2/3} \int \frac {1}{(\sec (c+d x)+1)^{5/3}}dx}{a (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(\sec (c+d x)+1)^{2/3} \int \frac {1}{\left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{5/3}}dx}{a (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 4265 |
\(\displaystyle -\frac {\tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{13/6}}d\sec (c+d x)}{a d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 149 |
\(\displaystyle -\frac {6 \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int \frac {\cos ^9(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{a d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {6 \tan (c+d x) \sqrt [6]{\sec (c+d x)+1} \int -\frac {\cos ^9(c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{a d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {3 \sqrt {2} \sin (c+d x) \cos ^6(c+d x) \sqrt [6]{\sec (c+d x)+1} \operatorname {AppellF1}\left (-\frac {7}{6},1,\frac {1}{2},-\frac {1}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 a d \sqrt {1-\sec (c+d x)} (a \sec (c+d x)+a)^{2/3}}\) |
(-3*Sqrt[2]*AppellF1[-7/6, 1, 1/2, -1/6, 1 + Sec[c + d*x], (1 + Sec[c + d* x])/2]*Cos[c + d*x]^6*(1 + Sec[c + d*x])^(1/6)*Sin[c + d*x])/(7*a*d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(2/3))
3.2.63.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1 ) - 1)*(c - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[2*n] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^n*(Cot [c + d*x]/(d*Sqrt[1 + Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]])) Subst[Int[(1 + b*(x/a))^(n - 1/2)/(x*Sqrt[1 - b*(x/a)]), x], x, Csc[c + d*x]], x] /; Fr eeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && GtQ[a, 0 ]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^IntPar t[n]*((a + b*Csc[c + d*x])^FracPart[n]/(1 + (b/a)*Csc[c + d*x])^FracPart[n] ) Int[(1 + (b/a)*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[2*n] && !GtQ[a, 0]
\[\int \frac {1}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}}d x\]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx=\int \frac {1}{\left (a \sec {\left (c + d x \right )} + a\right )^{\frac {5}{3}}}\, dx \]
\[ \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
\[ \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]