Integrand size = 27, antiderivative size = 78 \[ \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {3 \operatorname {AppellF1}\left (-\frac {2}{3},\frac {1}{2},1,\frac {1}{3},\sec (c+d x),-\sec (c+d x)\right ) \tan (c+d x)}{2 d \sqrt {1-\sec (c+d x)} (e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \]
3/2*AppellF1(-2/3,1,1/2,1/3,-sec(d*x+c),sec(d*x+c))*tan(d*x+c)/d/(e*sec(d* x+c))^(2/3)/(1-sec(d*x+c))^(1/2)/(a+a*sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(585\) vs. \(2(78)=156\).
Time = 7.85 (sec) , antiderivative size = 585, normalized size of antiderivative = 7.50 \[ \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sec ^{\frac {7}{6}}(c+d x) \left (-\frac {3}{2} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{6}}(c+d x) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )+\frac {5 \sqrt {\frac {1}{1+\cos (c+d x)}} (-1+3 \cos (c+d x)) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{5/6} \sin \left (\frac {1}{2} (c+d x)\right ) \left (-3 \cos ^{\frac {5}{6}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {3}{2},2 \sin ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt [3]{\sec ^2\left (\frac {1}{2} (c+d x)\right )}+2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{6},\frac {2}{3},\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^{5/6} \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{-120 \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{6},\frac {2}{3},\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\frac {1}{1+\cos (c+d x)}\right )^{2/3} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{5/6} \sin \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )+32 \operatorname {AppellF1}\left (\frac {5}{2},\frac {5}{6},\frac {5}{3},\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\frac {1}{1+\cos (c+d x)}\right )^{2/3} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{5/6} \sin \left (\frac {1}{2} (c+d x)\right ) \tan ^3\left (\frac {1}{2} (c+d x)\right )+5 \sqrt {2} \cos \left (\frac {1}{2} (c+d x)\right ) \left (3-4 \sqrt {2} \operatorname {AppellF1}\left (\frac {5}{2},\frac {11}{6},\frac {2}{3},\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (\frac {1}{1+\cos (c+d x)}\right )^{2/3} \left (\frac {\cos (c+d x)}{1+\cos (c+d x)}\right )^{5/6} \tan ^4\left (\frac {1}{2} (c+d x)\right )\right )}\right )}{d (e \sec (c+d x))^{2/3} \sqrt {a (1+\sec (c+d x))}} \]
(Sec[c + d*x]^(7/6)*((-3*Cos[(c + d*x)/2]*Sec[c + d*x]^(5/6)*(Sin[(c + d*x )/2] - Sin[(3*(c + d*x))/2]))/2 + (5*Sqrt[(1 + Cos[c + d*x])^(-1)]*(-1 + 3 *Cos[c + d*x])*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(5/6)*Sin[(c + d*x)/2]*(- 3*Cos[c + d*x]^(5/6)*Hypergeometric2F1[1/2, 5/6, 3/2, 2*Sin[(c + d*x)/2]^2 ]*(Sec[(c + d*x)/2]^2)^(1/3) + 2*AppellF1[3/2, 5/6, 2/3, 5/2, Tan[(c + d*x )/2]^2, -Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6)*Tan[( c + d*x)/2]^2))/(-120*AppellF1[3/2, 5/6, 2/3, 5/2, Tan[(c + d*x)/2]^2, -Ta n[(c + d*x)/2]^2]*((1 + Cos[c + d*x])^(-1))^(2/3)*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(5/6)*Sin[(c + d*x)/2]*Tan[(c + d*x)/2] + 32*AppellF1[5/2, 5/6, 5/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*((1 + Cos[c + d*x])^(-1 ))^(2/3)*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(5/6)*Sin[(c + d*x)/2]*Tan[(c + d*x)/2]^3 + 5*Sqrt[2]*Cos[(c + d*x)/2]*(3 - 4*Sqrt[2]*AppellF1[5/2, 11/6, 2/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*((1 + Cos[c + d*x])^(- 1))^(2/3)*(Cos[c + d*x]/(1 + Cos[c + d*x]))^(5/6)*Tan[(c + d*x)/2]^4))))/( d*(e*Sec[c + d*x])^(2/3)*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.46 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {3042, 4315, 3042, 4314, 148, 27, 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.
| \(\displaystyle \int \frac {1}{\sqrt {a \sec (c+d x)+a} (e \sec (c+d x))^{2/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx\) |
| \(\Big \downarrow \) 4315 |
| \(\displaystyle \frac {\sqrt {\sec (c+d x)+1} \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {\sec (c+d x)+1}}dx}{\sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\sec (c+d x)+1} \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4314 |
| \(\displaystyle -\frac {e \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} (e \sec (c+d x))^{5/3} (\sec (c+d x)+1)}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 148 |
| \(\displaystyle -\frac {3 \tan (c+d x) \int \frac {e \cos ^3(c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x) e+e)}d\sqrt [3]{e \sec (c+d x)}}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 e \tan (c+d x) \int \frac {\cos ^3(c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x) e+e)}d\sqrt [3]{e \sec (c+d x)}}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle \frac {3 \sin (c+d x) \cos (c+d x) \operatorname {AppellF1}\left (-\frac {2}{3},1,\frac {1}{2},\frac {1}{3},-\sec (c+d x),\sec (c+d x)\right )}{2 d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
(3*AppellF1[-2/3, 1, 1/2, 1/3, -Sec[c + d*x], Sec[c + d*x]]*Cos[c + d*x]*S in[c + d*x])/(2*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])
3.3.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((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 + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && 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[(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]
\[\int \frac {1}{\left (e \sec \left (d x +c \right )\right )^{\frac {2}{3}} \sqrt {a +a \sec \left (d x +c \right )}}d x\]
Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (e \sec {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]
\[ \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
\[ \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {1}{\sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {2}{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {1}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{2/3}} \,d x \]