Integrand size = 14, antiderivative size = 75 \[ \int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {3 \sqrt {2} \operatorname {AppellF1}\left (\frac {1}{6},\frac {1}{2},1,\frac {7}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}} \]
3*AppellF1(1/6,1,1/2,7/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan(d*x+ c)/d/(a+a*sec(d*x+c))^(1/3)/(1-sec(d*x+c))^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(718\) vs. \(2(75)=150\).
Time = 4.78 (sec) , antiderivative size = 718, normalized size of antiderivative = 9.57 \[ \int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\frac {45 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},1,\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 (c+d x) (1+\sec (c+d x))^2 \tan \left (\frac {1}{2} (c+d x)\right ) \left (9 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},1,\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 \left (3 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{3},2,\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 {2}{3},1,\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 \sqrt [3]{a (1+\sec (c+d x))} \left (40 \left (3 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{3},2,\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 {2}{3},1,\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 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+6 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (-15 \operatorname {AppellF1}\left (\frac {3}{2},-\frac {1}{3},2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1-10 \cos (c+d x)+3 \cos (2 (c+d x)))-5 \operatorname {AppellF1}\left (\frac {3}{2},\frac {2}{3},1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1-10 \cos (c+d x)+3 \cos (2 (c+d x)))-24 \left (9 \operatorname {AppellF1}\left (\frac {5}{2},-\frac {1}{3},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 )+3 \operatorname {AppellF1}\left (\frac {5}{2},\frac {2}{3},2,\frac {7}{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 {5}{2},\frac {5}{3},1,\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) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+135 \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{3},1,\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 \left (3+3 \sec (c+d x)-3 \sin (c+d x) \tan (c+d x)-\tan ^2(c+d x)\right )\right )} \]
(45*AppellF1[1/2, -1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*C os[c + d*x]*(1 + Sec[c + d*x])^2*Tan[(c + d*x)/2]*(9*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, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))/(d*(a* (1 + Sec[c + d*x]))^(1/3)*(40*(3*AppellF1[3/2, -1/3, 2, 5/2, Tan[(c + d*x) /2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^ 2, -Tan[(c + d*x)/2]^2])^2*Sec[c + d*x]*Sin[(c + d*x)/2]^2*Tan[(c + d*x)/2 ]^2 + 6*AppellF1[1/2, -1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^ 2]*Sec[c + d*x]^2*Sin[(c + d*x)/2]^2*(-15*AppellF1[3/2, -1/3, 2, 5/2, Tan[ (c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 - 10*Cos[c + d*x] + 3*Cos[2*(c + d *x)]) - 5*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2] ^2]*(1 - 10*Cos[c + d*x] + 3*Cos[2*(c + d*x)]) - 24*(9*AppellF1[5/2, -1/3, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 3*AppellF1[5/2, 2/3, 2 , 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[5/2, 5/3, 1, 7/ 2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[c + d*x]*Tan[(c + d*x)/2] ^2) + 135*AppellF1[1/2, -1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2 ]^2]^2*(3 + 3*Sec[c + d*x] - 3*Sin[c + d*x]*Tan[c + d*x] - Tan[c + d*x]^2) ))
Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4266, 3042, 4265, 149, 25, 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.
| \(\displaystyle \int \frac {1}{\sqrt [3]{a \sec (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt [3]{a \csc \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 4266 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)+1} \int \frac {1}{\sqrt [3]{\sec (c+d x)+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt [3]{\sec (c+d x)+1} \int \frac {1}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 4265 |
| \(\displaystyle -\frac {\tan (c+d x) \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 149 |
| \(\displaystyle -\frac {6 \tan (c+d x) \int \frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {6 \tan (c+d x) \int -\frac {\cos (c+d x)}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} \sqrt [6]{\sec (c+d x)+1} \sqrt [3]{a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 936 |
| \(\displaystyle \frac {3 \sqrt {2} \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{2},\frac {7}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}}\) |
(3*Sqrt[2]*AppellF1[1/6, 1, 1/2, 7/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x]) /2]*Tan[c + d*x])/(d*Sqrt[1 - Sec[c + d*x]]*(a + a*Sec[c + d*x])^(1/3))
3.2.57.3.1 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[((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[(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]
\[\int \frac {1}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}d x\]
Timed out. \[ \int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {1}{\sqrt [3]{a \sec {\left (c + d x \right )} + a}}\, dx \]
\[ \int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
\[ \int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \]