\(\int \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx\) [287]

Optimal result

Integrand size = 25, antiderivative size = 80 \[ \int \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx=\frac {2\ 2^{5/6} a \operatorname {AppellF1}\left (\frac {1}{2},\frac {4}{3},-\frac {5}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{5/6}} \] Output:

2*2^(5/6)*a*AppellF1(1/2,4/3,-5/6,3/2,1-sec(d*x+c),1/2-1/2*sec(d*x+c))*(a+ 
a*sec(d*x+c))^(1/3)*tan(d*x+c)/d/(1+sec(d*x+c))^(5/6)
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 34.83 (sec) , antiderivative size = 2325, normalized size of antiderivative = 29.06 \[ \int \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx=\text {Result too large to show} \] Input:

Integrate[(a + a*Sec[c + d*x])^(4/3)/Sec[c + d*x]^(1/3),x]
 

Output:

(-3*(a*(1 + Sec[c + d*x]))^(4/3)*((1 + Sec[c + d*x])^(1/3)/Sec[c + d*x]^(1 
/3) + Sec[c + d*x]^(2/3)*(1 + Sec[c + d*x])^(1/3))*(-8*Tan[(c + d*x)/2] + 
(AppellF1[-4/3, -2/3, -2/3, -1/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + 
I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(((-I + Tan[(c + d*x)/2])/ 
(-1 + Tan[(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x) 
/2]))^(2/3)) - AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x) 
/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c 
 + d*x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*( 
1 + Tan[(c + d*x)/2])^2))/(4*2^(2/3)*d*(Sec[(c + d*x)/2]^2)^(1/3)*(1 + Sec 
[c + d*x])^(4/3)*((Tan[(c + d*x)/2]*(-8*Tan[(c + d*x)/2] + (AppellF1[-4/3, 
 -2/3, -2/3, -1/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c 
 + d*x)/2])]*Sec[(c + d*x)/2]^2)/(((-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + 
d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) - 
AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/( 
1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1 
/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d* 
x)/2])^2))/(4*2^(2/3)*(Sec[(c + d*x)/2]^2)^(1/3)) - (3*(-4*Sec[(c + d*x)/2 
]^2 + (Sec[(c + d*x)/2]^2*(((-4/3 + (4*I)/3)*AppellF1[-1/3, -2/3, 1/3, 2/3 
, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[ 
(c + d*x)/2]^2)/(-1 + Tan[(c + d*x)/2])^2 - ((4/3 + (4*I)/3)*AppellF1[-...
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4315, 3042, 4312, 150}

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.

\(\displaystyle \int \frac {(a \sec (c+d x)+a)^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{4/3}}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4315

\(\displaystyle \frac {a \sqrt [3]{a \sec (c+d x)+a} \int \frac {(\sec (c+d x)+1)^{4/3}}{\sqrt [3]{\sec (c+d x)}}dx}{\sqrt [3]{\sec (c+d x)+1}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \sqrt [3]{a \sec (c+d x)+a} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{4/3}}{\sqrt [3]{\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{\sqrt [3]{\sec (c+d x)+1}}\)

\(\Big \downarrow \) 4312

\(\displaystyle \frac {a \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \int \frac {(\sec (c+d x)+1)^{5/6}}{\sqrt {1-\sec (c+d x)} \sec ^{\frac {4}{3}}(c+d x)}d(1-\sec (c+d x))}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{5/6}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {2\ 2^{5/6} a \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} \operatorname {AppellF1}\left (\frac {1}{2},\frac {4}{3},-\frac {5}{6},\frac {3}{2},1-\sec (c+d x),\frac {1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}}\)

Input:

Int[(a + a*Sec[c + d*x])^(4/3)/Sec[c + d*x]^(1/3),x]
 

Output:

(2*2^(5/6)*a*AppellF1[1/2, 4/3, -5/6, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + 
d*x])/2]*(a + a*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(d*(1 + Sec[c + d*x])^(5 
/6))
 

Defintions of rubi rules used

rule 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 3042

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

rule 4312

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_), x_Symbol] :> Simp[(-(a*(d/b))^n)*(Cot[e + f*x]/(a^(n - 2)*f*Sqrt 
[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]))   Subst[Int[(a - x)^(n - 1) 
*((2*a - x)^(m - 1/2)/Sqrt[x]), x], x, a - b*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] & 
&  !IntegerQ[n] && GtQ[a*(d/b), 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]

\[\int \frac {\left (a +a \sec \left (d x +c \right )\right )^{\frac {4}{3}}}{\sec \left (d x +c \right )^{\frac {1}{3}}}d x\]

Input:

int((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x)
 

Output:

int((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))**(4/3)/sec(d*x+c)**(1/3),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^(4/3)/sec(d*x + c)^(1/3), x)
 

Giac [F]

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

integrate((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^(4/3)/sec(d*x + c)^(1/3), x)
 

Mupad [F(-1)]

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

int((a + a/cos(c + d*x))^(4/3)/(1/cos(c + d*x))^(1/3),x)
 

Output:

int((a + a/cos(c + d*x))^(4/3)/(1/cos(c + d*x))^(1/3), x)
 

Reduce [F]

\[ \int \frac {(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx=a^{\frac {4}{3}} \left (\int \frac {\left (\sec \left (d x +c \right )+1\right )^{\frac {1}{3}}}{\sec \left (d x +c \right )^{\frac {1}{3}}}d x +\int \left (\sec \left (d x +c \right )+1\right )^{\frac {1}{3}} \sec \left (d x +c \right )^{\frac {2}{3}}d x \right ) \] Input:

int((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x)
 

Output:

a**(1/3)*a*(int((sec(c + d*x) + 1)**(1/3)/sec(c + d*x)**(1/3),x) + int(((s 
ec(c + d*x) + 1)**(1/3)*sec(c + d*x))/sec(c + d*x)**(1/3),x))