\(\int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx\) [281]

Optimal result

Integrand size = 27, antiderivative size = 76 \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {3 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},1,\frac {4}{3},\sec (c+d x),-\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)} \tan (c+d x)}{d \sqrt {1-\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \] Output:

-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)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(761\) vs. \(2(76)=152\).

Time = 6.79 (sec) , antiderivative size = 761, normalized size of antiderivative = 10.01 \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(90*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]*(e*Sec[c + d*x])^(1/3)*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*((1 + Cos[c + d*x])^(-1))^(2/3)*(Sec[c + d*x]/(1 + Sec[c + d*x])) 
^(1/3)*Sqrt[a*(1 + Sec[c + d*x])]*(270*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]^4*(-1 + 4*Cos[c + d 
*x])*Sec[c + d*x] + 10*(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*Sin[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2 - 3*AppellF1 
[1/2, -1/6, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sin[(c + d* 
x)/2]^2*(20*AppellF1[3/2, -1/6, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d* 
x)/2]^2]*(-9 + (1 + 2*Cos[2*(c + d*x)])*Sec[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]*(-9 + (1 + 2*Cos[2*( 
c + d*x)])*Sec[c + d*x]) + 6*(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])*Tan[(c + d*x)/2]^2)))
 

Rubi [A] (warning: unable to verify)

Time = 0.46 (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

\(\displaystyle -\frac {3 \tan (c+d x) \operatorname {AppellF1}\left (\frac {1}{3},1,\frac {1}{2},\frac {4}{3},-\sec (c+d x),\sec (c+d x)\right ) \sqrt [3]{e \sec (c+d x)}}{d \sqrt {1-\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\)

Input:

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

Output:

(-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]])
 

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 936

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])
 

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]

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

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(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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