Integrand size = 21, antiderivative size = 105 \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {\sqrt {2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right ) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \tan (c+d x)}{d \sqrt {1+\sec (c+d x)} \sqrt [3]{a+b \sec (c+d x)}} \]
AppellF1(1/2,1/3,1/2,3/2,b*(1-sec(d*x+c))/(a+b),1/2-1/2*sec(d*x+c))*((a+b* sec(d*x+c))/(a+b))^(1/3)*2^(1/2)*tan(d*x+c)/d/(a+b*sec(d*x+c))^(1/3)/(1+se c(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(310\) vs. \(2(105)=210\).
Time = 14.39 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.95 \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\frac {15 (a-b)^2 (a+b) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) \cos (c+d x) \cot ^3(c+d x) (1+\sec (c+d x)) (b-b \sec (c+d x)) (a+b \sec (c+d x))^{2/3}}{b^2 (-a+b) d \left (3 (a-b) \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {3}{2},\frac {8}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) (b+a \cos (c+d x))+(a+b) \left (10 (a-b) \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) \cos (c+d x)+3 \operatorname {AppellF1}\left (\frac {5}{3},\frac {3}{2},\frac {1}{2},\frac {8}{3},\frac {a+b \sec (c+d x)}{a-b},\frac {a+b \sec (c+d x)}{a+b}\right ) (b+a \cos (c+d x))\right )\right )} \]
(15*(a - b)^2*(a + b)*AppellF1[2/3, 1/2, 1/2, 5/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d*x])/(a + b)]*Cos[c + d*x]*Cot[c + d*x]^3*(1 + Sec[ c + d*x])*(b - b*Sec[c + d*x])*(a + b*Sec[c + d*x])^(2/3))/(b^2*(-a + b)*d *(3*(a - b)*AppellF1[5/3, 1/2, 3/2, 8/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d*x])/(a + b)]*(b + a*Cos[c + d*x]) + (a + b)*(10*(a - b)*Appe llF1[2/3, 1/2, 1/2, 5/3, (a + b*Sec[c + d*x])/(a - b), (a + b*Sec[c + d*x] )/(a + b)]*Cos[c + d*x] + 3*AppellF1[5/3, 3/2, 1/2, 8/3, (a + b*Sec[c + d* x])/(a - b), (a + b*Sec[c + d*x])/(a + b)]*(b + a*Cos[c + d*x]))))
Time = 0.28 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4321, 156, 155}
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.
\(\displaystyle \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt [3]{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4321 |
\(\displaystyle -\frac {\tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1}}\) |
\(\Big \downarrow \) 156 |
\(\displaystyle -\frac {\tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \int \frac {1}{\sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{\frac {a}{a+b}+\frac {b \sec (c+d x)}{a+b}}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 155 |
\(\displaystyle \frac {\sqrt {2} \tan (c+d x) \sqrt [3]{\frac {a+b \sec (c+d x)}{a+b}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2},\frac {1}{3},\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),\frac {b (1-\sec (c+d x))}{a+b}\right )}{d \sqrt {\sec (c+d x)+1} \sqrt [3]{a+b \sec (c+d x)}}\) |
(Sqrt[2]*AppellF1[1/2, 1/2, 1/3, 3/2, (1 - Sec[c + d*x])/2, (b*(1 - Sec[c + d*x]))/(a + b)]*((a + b*Sec[c + d*x])/(a + b))^(1/3)*Tan[c + d*x])/(d*Sq rt[1 + Sec[c + d*x]]*(a + b*Sec[c + d*x])^(1/3))
3.8.2.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* Simplify[b/(b*e - a*f)]^p))*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, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] && GtQ[Sim plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] && !(GtQ[Simpl ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d *x, a + b*x]) && !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c - e*d)], 0] && SimplerQ[e + f*x, a + b*x])
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) ^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p ]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p]) Int[(a + b*x)^m*(c + d*x)^n*Si mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[p] & & GtQ[Simplify[b/(b*c - a*d)], 0] && !GtQ[Simplify[b/(b*e - a*f)], 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[Cot[e + f*x]/(f*Sqrt[1 + Csc[e + f*x]]*Sqrt[1 - Csc[e + f*x ]]) Subst[Int[(a + b*x)^m/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Csc[e + f*x]] , x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !IntegerQ[2*m]
\[\int \frac {\sec \left (d x +c \right )}{\left (a +b \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\sqrt [3]{a + b \sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int { \frac {\sec \left (d x + c\right )}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {\sec (c+d x)}{\sqrt [3]{a+b \sec (c+d x)}} \, dx=\int \frac {1}{\cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]