\(\int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\) [273]

Optimal result

Integrand size = 27, antiderivative size = 280 \[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=\frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right ),-7-4 \sqrt {3}\right ) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {e^{2/3}+\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \tan (c+d x)}{d (a-a \sec (c+d x)) \sqrt {a+a \sec (c+d x)} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}} \] Output:

2*3^(3/4)*(1/2*6^(1/2)+1/2*2^(1/2))*a^2*EllipticF(((1-3^(1/2))*e^(1/3)-(e* 
sec(d*x+c))^(1/3))/((1+3^(1/2))*e^(1/3)-(e*sec(d*x+c))^(1/3)),I*3^(1/2)+2* 
I)*(e^(1/3)-(e*sec(d*x+c))^(1/3))*((e^(2/3)+e^(1/3)*(e*sec(d*x+c))^(1/3)+( 
e*sec(d*x+c))^(2/3))/((1+3^(1/2))*e^(1/3)-(e*sec(d*x+c))^(1/3))^2)^(1/2)*t 
an(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*s 
ec(d*x+c))^(1/3))/((1+3^(1/2))*e^(1/3)-(e*sec(d*x+c))^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.09 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.25 \[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {3}{2},1-\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)} \sqrt {a (1+\sec (c+d x))} \tan \left (\frac {1}{2} (c+d x)\right )}{d \sqrt [3]{\sec (c+d x)}} \] Input:

Integrate[(e*Sec[c + d*x])^(1/3)*Sqrt[a + a*Sec[c + d*x]],x]
 

Output:

(2*Hypergeometric2F1[1/2, 2/3, 3/2, 1 - Sec[c + d*x]]*(e*Sec[c + d*x])^(1/ 
3)*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c + d*x)/2])/(d*Sec[c + d*x]^(1/3))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 4293, 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.

\(\displaystyle \int \sqrt {a \sec (c+d x)+a} \sqrt [3]{e \sec (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \sqrt [3]{e \csc \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 4293

\(\displaystyle -\frac {a^2 e \tan (c+d x) \int \frac {1}{(e \sec (c+d x))^{2/3} \sqrt {a-a \sec (c+d x)}}d\sec (c+d x)}{d \sqrt {a-a \sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {3 a^2 \tan (c+d x) \int \frac {1}{\sqrt {a-a \sec (c+d x)}}d\sqrt [3]{e \sec (c+d x)}}{d \sqrt {a-a \sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 759

\(\displaystyle \frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right ),-7-4 \sqrt {3}\right )}{d (a-a \sec (c+d x)) \sqrt {a \sec (c+d x)+a} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}\)

Input:

Int[(e*Sec[c + d*x])^(1/3)*Sqrt[a + a*Sec[c + d*x]],x]
 

Output:

(2*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*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*Sec[c + d*x])^(1/3) + (e*Sec[c + d*x])^(2/3))/((1 + Sqrt[3])*e^(1/3) 
 - (e*Sec[c + d*x])^(1/3))^2]*Tan[c + d*x])/(d*(a - a*Sec[c + d*x])*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])
 

Defintions of rubi rules used

rule 73

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]
 

rule 759

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4293

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]
 

Maple [F]

\[\int \left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}} \sqrt {a +a \sec \left (d x +c \right )}d x\]

Input:

int((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x)
 

Output:

int((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x)
 

Fricas [F]

\[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas 
")
 

Output:

integral(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(1/3), x)
 

Sympy [F]

\[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sqrt [3]{e \sec {\left (c + d x \right )}}\, dx \] Input:

integrate((e*sec(d*x+c))**(1/3)*(a+a*sec(d*x+c))**(1/2),x)
 

Output:

Integral(sqrt(a*(sec(c + d*x) + 1))*(e*sec(c + d*x))**(1/3), x)
 

Maxima [F]

\[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima 
")
 

Output:

integrate(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(1/3), x)
 

Giac [F]

\[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}} \,d x } \] Input:

integrate((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=\int \sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{1/3} \,d x \] Input:

int((a + a/cos(c + d*x))^(1/2)*(e/cos(c + d*x))^(1/3),x)
 

Output:

int((a + a/cos(c + d*x))^(1/2)*(e/cos(c + d*x))^(1/3), x)
 

Reduce [F]

\[ \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx=e^{\frac {1}{3}} \sqrt {a}\, \left (\int \sec \left (d x +c \right )^{\frac {1}{3}} \sqrt {\sec \left (d x +c \right )+1}d x \right ) \] Input:

int((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x)
 

Output:

e**(1/3)*sqrt(a)*int(sec(c + d*x)**(1/3)*sqrt(sec(c + d*x) + 1),x)