Integrand size = 27, antiderivative size = 76 \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {3 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},\sec (c+d x),-\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)} \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \]
-3*AppellF1(1/3,1,1/2,4/3,-sec(d*x+c),sec(d*x+c))*(e*sec(d*x+c))^(1/3)*tan (d*x+c)/d/(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. \(749\) vs. \(2(76)=152\).
Time = 7.80 (sec) , antiderivative size = 749, normalized size of antiderivative = 9.86 \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {720 e \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},\frac {2}{3},\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos \left (\frac {1}{2} (c+d x)\right ) (1+\cos (c+d x))^2 \sin \left (\frac {1}{2} (c+d x)\right ) \left (9 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},\frac {2}{3},\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-\left (4 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{6},\frac {5}{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 )+\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 )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{d (e \sec (c+d x))^{2/3} \sqrt {a (1+\sec (c+d x))} \left (4320 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},\frac {2}{3},\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2 \cos ^6\left (\frac {1}{2} (c+d x)\right ) (-1+4 \cos (c+d x))+160 \left (4 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{6},\frac {5}{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 )+\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 )\right )^2 \cos (c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+12 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{6},\frac {2}{3},\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (20 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{6},\frac {5}{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 ) (7+14 \cos (c+d x)+5 \cos (2 (c+d x))-2 \cos (3 (c+d x)))+5 \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 ) (7+14 \cos (c+d x)+5 \cos (2 (c+d x))-2 \cos (3 (c+d x)))-24 \left (40 \operatorname {AppellF1}\left (\frac {5}{2},-\frac {1}{6},\frac {8}{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 )+8 \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 )-5 \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 )\right ) \cos (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )} \]
(720*e*AppellF1[1/2, -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2] ^2]*Cos[(c + d*x)/2]*(1 + Cos[c + d*x])^2*Sin[(c + d*x)/2]*(9*AppellF1[1/2 , -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - (4*AppellF1[3 /2, -1/6, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/ 2, 5/6, 2/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/ 2]^2))/(d*(e*Sec[c + d*x])^(2/3)*Sqrt[a*(1 + Sec[c + d*x])]*(4320*AppellF1 [1/2, -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]^2*Cos[(c + d*x)/2]^6*(-1 + 4*Cos[c + d*x]) + 160*(4*AppellF1[3/2, -1/6, 5/3, 5/2, Tan [(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 5/6, 2/3, 5/2, Tan[( c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Cos[c + d*x]*Sin[(c + d*x)/2]^4 + 1 2*AppellF1[1/2, -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*S in[(c + d*x)/2]^2*(20*AppellF1[3/2, -1/6, 5/3, 5/2, Tan[(c + d*x)/2]^2, -T an[(c + d*x)/2]^2]*(7 + 14*Cos[c + d*x] + 5*Cos[2*(c + d*x)] - 2*Cos[3*(c + d*x)]) + 5*AppellF1[3/2, 5/6, 2/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d* x)/2]^2]*(7 + 14*Cos[c + d*x] + 5*Cos[2*(c + d*x)] - 2*Cos[3*(c + d*x)]) - 24*(40*AppellF1[5/2, -1/6, 8/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2 ]^2] + 8*AppellF1[5/2, 5/6, 5/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2 ]^2] - 5*AppellF1[5/2, 11/6, 2/3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/ 2]^2])*Cos[c + d*x]*Sin[(c + d*x)/2]^2)))
Time = 0.43 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, 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, 936}
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.
| \(\displaystyle \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt [3]{e \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4315 |
| \(\displaystyle \frac {\sqrt {\sec (c+d x)+1} \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {\sec (c+d x)+1}}dx}{\sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\sec (c+d x)+1} \int \frac {\sqrt [3]{e \csc \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4314 |
| \(\displaystyle -\frac {e \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} (e \sec (c+d x))^{2/3} (\sec (c+d x)+1)}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 148 |
| \(\displaystyle -\frac {3 \tan (c+d x) \int \frac {e}{\sqrt {1-\sec (c+d x)} (\sec (c+d x) e+e)}d\sqrt [3]{e \sec (c+d x)}}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {3 e \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} (\sec (c+d x) e+e)}d\sqrt [3]{e \sec (c+d x)}}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 936 |
| \(\displaystyle -\frac {3 \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{3},1,\frac {1}{2},\frac {4}{3},-\sec (c+d x),\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)}}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
(-3*AppellF1[1/3, 1, 1/2, 4/3, -Sec[c + d*x], Sec[c + d*x]]*(e*Sec[c + d*x ])^(1/3)*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])
3.3.81.3.1 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[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) ], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[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]
\[\int \frac {\left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}}}{\sqrt {a +a \sec \left (d x +c \right )}}d x\]
Timed out. \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\sqrt [3]{e \sec {\left (c + d x \right )}}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
\[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{1/3}}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]