\(\int (a+a \sec (c+d x))^{5/3} \, dx\) [151]

Optimal result

Integrand size = 14, antiderivative size = 78 \[ \int (a+a \sec (c+d x))^{5/3} \, dx=\frac {4 \sqrt [6]{2} a \operatorname {AppellF1}\left (\frac {1}{2},-\frac {7}{6},1,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x)),1-\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{d (1+\sec (c+d x))^{7/6}} \] Output:

4*2^(1/6)*a*AppellF1(1/2,1,-7/6,3/2,1-sec(d*x+c),1/2-1/2*sec(d*x+c))*(a+a* 
sec(d*x+c))^(2/3)*tan(d*x+c)/d/(1+sec(d*x+c))^(7/6)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2694\) vs. \(2(78)=156\).

Time = 14.39 (sec) , antiderivative size = 2694, normalized size of antiderivative = 34.54 \[ \int (a+a \sec (c+d x))^{5/3} \, dx=\text {Result too large to show} \] Input:

Integrate[(a + a*Sec[c + d*x])^(5/3),x]
 

Output:

(3*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(5/3)*Ta 
n[(c + d*x)/2])/(2*d*(1 + Sec[c + d*x])^(5/3)) + ((Cos[(c + d*x)/2]^2*Sec[ 
c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(5/3)*((3*Sec[(c + d*x)/2]^2*(1 + S 
ec[c + d*x])^(2/3))/4 + (Cos[c + d*x]*Sec[(c + d*x)/2]^2*(1 + Sec[c + d*x] 
)^(2/3))/2)*Tan[(c + d*x)/2]*(AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^ 
2, -Tan[(c + d*x)/2]^2]*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d 
*x)/2]^2 + (135*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d 
*x)/2]^2]*Cos[(c + d*x)/2]^2)/(9*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/ 
2]^2, -Tan[(c + d*x)/2]^2] + 2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x 
)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/ 
2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)))/(3*2^(1/3)*d*(1 + Sec[c 
+ d*x])^(5/3)*((Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3) 
*(AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(Cos 
[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d*x)/2]^2 + (135*AppellF1[1/2 
, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2 
)/(9*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 
 2*(-3*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] 
 + 2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])* 
Tan[(c + d*x)/2]^2)))/(6*2^(1/3)) + ((Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/ 
3)*Tan[(c + d*x)/2]*(AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Ta...
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4266, 3042, 4265, 149, 25, 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.

\(\displaystyle \int (a \sec (c+d x)+a)^{5/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/3}dx\)

\(\Big \downarrow \) 4266

\(\displaystyle \frac {a (a \sec (c+d x)+a)^{2/3} \int (\sec (c+d x)+1)^{5/3}dx}{(\sec (c+d x)+1)^{2/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (a \sec (c+d x)+a)^{2/3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )^{5/3}dx}{(\sec (c+d x)+1)^{2/3}}\)

\(\Big \downarrow \) 4265

\(\displaystyle -\frac {a \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos (c+d x) (\sec (c+d x)+1)^{7/6}}{\sqrt {1-\sec (c+d x)}}d\sec (c+d x)}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\)

\(\Big \downarrow \) 149

\(\displaystyle -\frac {6 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int \frac {\cos (c+d x) (\sec (c+d x)+1)^2}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {6 a \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \int -\frac {\cos (c+d x) (\sec (c+d x)+1)^2}{\sqrt {1-\sec (c+d x)}}d\sqrt [6]{\sec (c+d x)+1}}{d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1)^{7/6}}\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {3 \sqrt {2} a \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {13}{6},1,\frac {1}{2},\frac {19}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{13 d \sqrt {1-\sec (c+d x)}}\)

Input:

Int[(a + a*Sec[c + d*x])^(5/3),x]
 

Output:

(3*Sqrt[2]*a*AppellF1[13/6, 1, 1/2, 19/6, 1 + Sec[c + d*x], (1 + Sec[c + d 
*x])/2]*(1 + Sec[c + d*x])*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(13*d* 
Sqrt[1 - Sec[c + d*x]])
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 149

Int[((a_) + (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 - a*(d/b) + d*(x^k/b))^n*(e - a*(f/b) + f*(x^k/b))^p, x], x, (a + 
 b*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && FractionQ[m] && 
IntegerQ[2*n] && 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 4265

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^n*(Cot 
[c + d*x]/(d*Sqrt[1 + Csc[c + d*x]]*Sqrt[1 - Csc[c + d*x]]))   Subst[Int[(1 
 + b*(x/a))^(n - 1/2)/(x*Sqrt[1 - b*(x/a)]), x], x, Csc[c + d*x]], x] /; Fr 
eeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0 
]
 

rule 4266

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[a^IntPar 
t[n]*((a + b*Csc[c + d*x])^FracPart[n]/(1 + (b/a)*Csc[c + d*x])^FracPart[n] 
)   Int[(1 + (b/a)*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && 
EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]
 

Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}d x\]

Input:

int((a+a*sec(d*x+c))^(5/3),x)
 

Output:

int((a+a*sec(d*x+c))^(5/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/3} \, dx=\text {Timed out} \] Input:

integrate((a+a*sec(d*x+c))^(5/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int (a+a \sec (c+d x))^{5/3} \, dx=\int \left (a \sec {\left (c + d x \right )} + a\right )^{\frac {5}{3}}\, dx \] Input:

integrate((a+a*sec(d*x+c))**(5/3),x)
 

Output:

Integral((a*sec(c + d*x) + a)**(5/3), x)
 

Maxima [F]

\[ \int (a+a \sec (c+d x))^{5/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^(5/3),x, algorithm="maxima")
 

Output:

integrate((a*sec(d*x + c) + a)^(5/3), x)
 

Giac [F]

\[ \int (a+a \sec (c+d x))^{5/3} \, dx=\int { {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}} \,d x } \] Input:

integrate((a+a*sec(d*x+c))^(5/3),x, algorithm="giac")
 

Output:

integrate((a*sec(d*x + c) + a)^(5/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{5/3} \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3} \,d x \] Input:

int((a + a/cos(c + d*x))^(5/3),x)
 

Output:

int((a + a/cos(c + d*x))^(5/3), x)
 

Reduce [F]

\[ \int (a+a \sec (c+d x))^{5/3} \, dx=a^{\frac {5}{3}} \left (\int \left (\sec \left (d x +c \right )+1\right )^{\frac {2}{3}}d x +\int \left (\sec \left (d x +c \right )+1\right )^{\frac {2}{3}} \sec \left (d x +c \right )d x \right ) \] Input:

int((a+a*sec(d*x+c))^(5/3),x)
                                                                                    
                                                                                    
 

Output:

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