3.2.46 \(\int (a+a \sec (c+d x))^{2/3} \, dx\) [146]

3.2.46.1 Optimal result

Integrand size = 14, antiderivative size = 77 \[ \int (a+a \sec (c+d x))^{2/3} \, dx=\frac {3 \sqrt {2} \operatorname {AppellF1}\left (\frac {7}{6},\frac {1}{2},1,\frac {13}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{7 d \sqrt {1-\sec (c+d x)}} \]

output

3/7*AppellF1(7/6,1,1/2,13/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*(a+a*sec(d*x+ 
c))^(2/3)*2^(1/2)*tan(d*x+c)/d/(1-sec(d*x+c))^(1/2)
 

3.2.46.2 Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(691\) vs. \(2(77)=154\).

Time = 5.21 (sec) , antiderivative size = 691, normalized size of antiderivative = 8.97 \[ \int (a+a \sec (c+d x))^{2/3} \, dx=\frac {45 \operatorname {AppellF1}\left (\frac {1}{2},\frac {2}{3},1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (a (1+\sec (c+d x)))^{5/3} \sin (c+d x) \left (9 \operatorname {AppellF1}\left (\frac {1}{2},\frac {2}{3},1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+2 \left (-3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {2}{3},2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{3},1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{a d \left (40 \left (3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {2}{3},2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{3},1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )^2 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+6 \operatorname {AppellF1}\left (\frac {1}{2},\frac {2}{3},1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sec ^2(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (15 \operatorname {AppellF1}\left (\frac {3}{2},\frac {2}{3},2,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (-7+16 \cos (c+d x)-3 \cos (2 (c+d x)))+10 \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{3},1,\frac {5}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (7-16 \cos (c+d x)+3 \cos (2 (c+d x)))-24 \left (9 \operatorname {AppellF1}\left (\frac {5}{2},\frac {2}{3},3,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-6 \operatorname {AppellF1}\left (\frac {5}{2},\frac {5}{3},2,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+5 \operatorname {AppellF1}\left (\frac {5}{2},\frac {8}{3},1,\frac {7}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \cos (c+d x) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+135 \operatorname {AppellF1}\left (\frac {1}{2},\frac {2}{3},1,\frac {3}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^2 \left (3+3 \cos (c+d x)+2 \tan ^2(c+d x)\right )\right )} \]

input

Integrate[(a + a*Sec[c + d*x])^(2/3),x]
 

output

(45*AppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(a 
*(1 + Sec[c + d*x]))^(5/3)*Sin[c + d*x]*(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, Ta 
n[(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))/(a*d*(40*(3*App 
ellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*Appel 
lF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Sec[c + 
d*x]*Sin[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2 + 6*AppellF1[1/2, 2/3, 1, 3/2, 
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[c + d*x]^2*Sin[(c + d*x)/2]^2 
*(15*AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*( 
-7 + 16*Cos[c + d*x] - 3*Cos[2*(c + d*x)]) + 10*AppellF1[3/2, 5/3, 1, 5/2, 
 Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(7 - 16*Cos[c + d*x] + 3*Cos[2*( 
c + d*x)]) - 24*(9*AppellF1[5/2, 2/3, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c 
+ d*x)/2]^2] - 6*AppellF1[5/2, 5/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + 
d*x)/2]^2] + 5*AppellF1[5/2, 8/3, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d* 
x)/2]^2])*Cos[c + d*x]*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]^2*(3 + 3*Cos[c + d*x] + 2*Tan[c 
 + d*x]^2)))
 

3.2.46.3 Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4266

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4265

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

\(\Big \downarrow \) 149

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1012

\(\displaystyle \frac {3 \sqrt {2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {7}{6},1,\frac {1}{2},\frac {13}{6},\sec (c+d x)+1,\frac {1}{2} (\sec (c+d x)+1)\right )}{7 d \sqrt {1-\sec (c+d x)}}\)

input

Int[(a + a*Sec[c + d*x])^(2/3),x]
 

output

(3*Sqrt[2]*AppellF1[7/6, 1, 1/2, 13/6, 1 + Sec[c + d*x], (1 + Sec[c + d*x] 
)/2]*(a + a*Sec[c + d*x])^(2/3)*Tan[c + d*x])/(7*d*Sqrt[1 - Sec[c + d*x]])
 

3.2.46.3.1 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]
 

3.2.46.4 Maple [F]

\[\int \left (a +a \sec \left (d x +c \right )\right )^{\frac {2}{3}}d x\]

input

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

output

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

3.2.46.5 Fricas [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{2/3} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.2.46.6 Sympy [F]

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

input

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

output

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

3.2.46.7 Maxima [F]

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

input

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

output

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

3.2.46.8 Giac [F]

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

input

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

output

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

3.2.46.9 Mupad [F(-1)]

Timed out. \[ \int (a+a \sec (c+d x))^{2/3} \, dx=\int {\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{2/3} \,d x \]

input

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

output

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