\(\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx\) [282]

Optimal result

Integrand size = 27, antiderivative size = 76 \[ \int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {3 \operatorname {AppellF1}\left (-\frac {1}{3},\frac {1}{2},1,\frac {2}{3},\sec (c+d x),-\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \] Output:

3*AppellF1(-1/3,1,1/2,2/3,-sec(d*x+c),sec(d*x+c))*tan(d*x+c)/d/(1-sec(d*x+ 
c))^(1/2)/(e*sec(d*x+c))^(1/3)/(a+a*sec(d*x+c))^(1/2)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(3346\) vs. \(2(76)=152\).

Time = 17.01 (sec) , antiderivative size = 3346, normalized size of antiderivative = 44.03 \[ \int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\text {Result too large to show} \] Input:

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

Output:

-(((Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/6)*Tan[(c + d*x)/2]*(-1 + Tan[(c + 
 d*x)/2]^2)*((2*AppellF1[3/2, 1/6, 1/3, 5/2, Tan[(c + d*x)/2]^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) + 
(3*(1 + (3*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x) 
/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(9*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c 
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[( 
c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c 
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))))/(Sec[(c + d*x)/2 
]^2)^(1/3)))/(d*(e*Sec[c + d*x])^(1/3)*Sqrt[a*(1 + Sec[c + d*x])]*(-(Sec[( 
c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/6)*Tan[(c + d*x)/2]^2*( 
(2*AppellF1[3/2, 1/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*T 
an[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/6) + (3*(1 + (3*Ap 
pellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 
+ Tan[(c + d*x)/2]^2)*(9*AppellF1[1/2, 1/6, 1/3, 3/2, Tan[(c + d*x)/2]^2, 
-Tan[(c + d*x)/2]^2] + (-2*AppellF1[3/2, 1/6, 4/3, 5/2, Tan[(c + d*x)/2]^2 
, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 7/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, 
-Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))))/(Sec[(c + d*x)/2]^2)^(1/3))) 
- (Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/6)*(-1 + Tan[(c 
 + d*x)/2]^2)*((2*AppellF1[3/2, 1/6, 1/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c 
 + d*x)/2]^2]*Tan[(c + d*x)/2]^2)/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/...
 

Rubi [A] (warning: unable to verify)

Time = 0.49 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84, 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.

\(\displaystyle \int \frac {1}{\sqrt {a \sec (c+d x)+a} \sqrt [3]{e \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \sqrt [3]{e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4315

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

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt {\sec (c+d x)+1} \int \frac {1}{\sqrt [3]{e \csc \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )+1}}dx}{\sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 4314

\(\displaystyle -\frac {e \tan (c+d x) \int \frac {1}{\sqrt {1-\sec (c+d x)} (e \sec (c+d x))^{4/3} (\sec (c+d x)+1)}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\)

\(\Big \downarrow \) 148

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {3 \sin (c+d x) \operatorname {AppellF1}\left (-\frac {1}{3},1,\frac {1}{2},\frac {2}{3},-\sec (c+d x),\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\)

Input:

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

Output:

(3*AppellF1[-1/3, 1, 1/2, 2/3, -Sec[c + d*x], Sec[c + d*x]]*Sin[c + d*x])/ 
(d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 148

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]
 

rule 1012

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])
 

rule 3042

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

rule 4314

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]
 

rule 4315

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{\frac {4}{3}}+\sec \left (d x +c \right )^{\frac {1}{3}}}d x \right )}{e^{\frac {1}{3}} a} \] Input:

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

Output:

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