Integrand size = 27, antiderivative size = 325 \[ \int (e \sec (c+d x))^{4/3} \sqrt {a+a \sec (c+d x)} \, dx=\frac {6 a e \sqrt [3]{e \sec (c+d x)} \tan (c+d x)}{5 d \sqrt {a+a \sec (c+d x)}}+\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 e \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right ),-7-4 \sqrt {3}\right ) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {e^{2/3}+\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \tan (c+d x)}{5 d (a-a \sec (c+d x)) \sqrt {a+a \sec (c+d x)} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}} \]
6/5*a*e*(e*sec(d*x+c))^(1/3)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+4/5*3^(3/ 4)*a^2*e*EllipticF((-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1-3^(1/2)))/(-(e*sec(d* x+c))^(1/3)+e^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(e^(1/3)-(e*sec(d*x+c))^(1 /3))*(1/2*6^(1/2)+1/2*2^(1/2))*((e^(2/3)+e^(1/3)*(e*sec(d*x+c))^(1/3)+(e*s ec(d*x+c))^(2/3))/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan (d*x+c)/d/(a-a*sec(d*x+c))/(a+a*sec(d*x+c))^(1/2)/(e^(1/3)*(e^(1/3)-(e*sec (d*x+c))^(1/3))/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.22 \[ \int (e \sec (c+d x))^{4/3} \sqrt {a+a \sec (c+d x)} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {3}{2},1-\sec (c+d x)\right ) (e \sec (c+d x))^{4/3} \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{d \sec ^{\frac {4}{3}}(c+d x)} \]
(2*Hypergeometric2F1[-1/3, 1/2, 3/2, 1 - Sec[c + d*x]]*(e*Sec[c + d*x])^(4 /3)*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(d*Sec[c + d*x]^(4/3))
Time = 0.41 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 4293, 60, 73, 759}
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.
| \(\displaystyle \int \sqrt {a \sec (c+d x)+a} (e \sec (c+d x))^{4/3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx\) |
| \(\Big \downarrow \) 4293 |
| \(\displaystyle -\frac {a^2 e \tan (c+d x) \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a-a \sec (c+d x)}}d\sec (c+d x)}{d \sqrt {a-a \sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle -\frac {a^2 e \tan (c+d x) \left (\frac {2}{5} e \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a-a \sec (c+d x)}}d\sec (c+d x)-\frac {6 \sqrt {a-a \sec (c+d x)} \sqrt [3]{e \sec (c+d x)}}{5 a}\right )}{d \sqrt {a-a \sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {a^2 e \tan (c+d x) \left (\frac {6}{5} \int \frac {1}{\sqrt {a-a \sec (c+d x)}}d\sqrt [3]{e \sec (c+d x)}-\frac {6 \sqrt {a-a \sec (c+d x)} \sqrt [3]{e \sec (c+d x)}}{5 a}\right )}{d \sqrt {a-a \sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
| \(\Big \downarrow \) 759 |
| \(\displaystyle -\frac {a^2 e \tan (c+d x) \left (-\frac {4\ 3^{3/4} \sqrt {2+\sqrt {3}} \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right ),-7-4 \sqrt {3}\right )}{5 \sqrt {a-a \sec (c+d x)} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}-\frac {6 \sqrt {a-a \sec (c+d x)} \sqrt [3]{e \sec (c+d x)}}{5 a}\right )}{d \sqrt {a-a \sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\) |
-((a^2*e*((-6*(e*Sec[c + d*x])^(1/3)*Sqrt[a - a*Sec[c + d*x]])/(5*a) - (4* 3^(3/4)*Sqrt[2 + Sqrt[3]]*EllipticF[ArcSin[((1 - Sqrt[3])*e^(1/3) - (e*Sec [c + d*x])^(1/3))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))], -7 - 4*Sqrt[3]]*(e^(1/3) - (e*Sec[c + d*x])^(1/3))*Sqrt[(e^(2/3) + e^(1/3)*(e*S ec[c + d*x])^(1/3) + (e*Sec[c + d*x])^(2/3))/((1 + Sqrt[3])*e^(1/3) - (e*S ec[c + d*x])^(1/3))^2])/(5*Sqrt[a - a*Sec[c + d*x]]*Sqrt[(e^(1/3)*(e^(1/3) - (e*Sec[c + d*x])^(1/3)))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3 ))^2]))*Tan[c + d*x])/(d*Sqrt[a - a*Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]] ))
3.3.72.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[s* ((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] & & PosQ[a]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], 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)/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0]
\[\int \left (e \sec \left (d x +c \right )\right )^{\frac {4}{3}} \sqrt {a +a \sec \left (d x +c \right )}d x\]
\[ \int (e \sec (c+d x))^{4/3} \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]
Timed out. \[ \int (e \sec (c+d x))^{4/3} \sqrt {a+a \sec (c+d x)} \, dx=\text {Timed out} \]
\[ \int (e \sec (c+d x))^{4/3} \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]
\[ \int (e \sec (c+d x))^{4/3} \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \,d x } \]
Timed out. \[ \int (e \sec (c+d x))^{4/3} \sqrt {a+a \sec (c+d x)} \, dx=\int \sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{4/3} \,d x \]