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

Optimal result

Integrand size = 14, antiderivative size = 86 \[ \int (a+a \sec (c+d x))^{5/3} \, dx=\frac {3 \sqrt {2} a \operatorname {AppellF1}\left (\frac {13}{6},\frac {1}{2},1,\frac {19}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (1+\sec (c+d x)) (a+a \sec (c+d x))^{2/3} \tan (c+d x)}{13 d \sqrt {1-\sec (c+d x)}} \]

[Out]

3/13*a*AppellF1(13/6,1,1/2,19/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*(1+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)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3864, 3863, 141} \[ \int (a+a \sec (c+d x))^{5/3} \, dx=\frac {3 \sqrt {2} a \tan (c+d x) (\sec (c+d x)+1) (a \sec (c+d x)+a)^{2/3} \operatorname {AppellF1}\left (\frac {13}{6},\frac {1}{2},1,\frac {19}{6},\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{13 d \sqrt {1-\sec (c+d x)}} \]

[In]

Int[(a + a*Sec[c + d*x])^(5/3),x]

[Out]

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

Rule 141

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

Rule 3863

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

Rule 3864

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^IntPart[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a (a+a \sec (c+d x))^{2/3}\right ) \int (1+\sec (c+d x))^{5/3} \, dx}{(1+\sec (c+d x))^{2/3}} \\ & = -\frac {\left (a (a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \text {Subst}\left (\int \frac {(1+x)^{7/6}}{\sqrt {1-x} x} \, dx,x,\sec (c+d x)\right )}{d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}} \\ & = \frac {3 \sqrt {2} a \operatorname {AppellF1}\left (\frac {13}{6},\frac {1}{2},1,\frac {19}{6},\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) (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)}} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(2694\) vs. \(2(86)=172\).

Time = 15.93 (sec) , antiderivative size = 2694, normalized size of antiderivative = 31.33 \[ \int (a+a \sec (c+d x))^{5/3} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

(3*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(5/3)*Tan[(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 +
Sec[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]*(Appe
llF1[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])*Ta
n[(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]*(App
ellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*(Cos[c + d*x]*Sec[(c + d*x
)/2]^2)^(2/3)*Tan[(c + d*x)/2] + (Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3)*Tan[(c + d*x)/2]^2*((-3*AppellF1[5/2,
2/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + (2*AppellF1[5/
2, 5/3, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5) + (2*AppellF1
[3/2, 2/3, 1, 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*Sin[c + d
*x]) + Cos[c + d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(1/3)) - (135*A
ppellF1[1/2, 2/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]*Sin[(c + d*x)/2])/(9*Appel
lF1[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) + (135*Cos[(c + d*x)/2]^2*(-1/3*(AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]) + (2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/9))/(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) - (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*(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])*Sec[(c + d*x)/2]^2*Tan[(
c + d*x)/2] + 9*(-1/3*(AppellF1[3/2, 2/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*
Tan[(c + d*x)/2]) + (2*AppellF1[3/2, 5/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*
Tan[(c + d*x)/2])/9) + 2*Tan[(c + d*x)/2]^2*(-3*((-6*AppellF1[5/2, 2/3, 3, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c +
d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + (2*AppellF1[5/2, 5/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c
+ d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5) + 2*((-3*AppellF1[5/2, 5/3, 2, 7/2, Tan[(c + d*x)/2]^2, -
Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + AppellF1[5/2, 8/3, 1, 7/2, Tan[(c + d*x)/2]^2, -T
an[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/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)^2))/(3*2^(1/3)) + (2^(2/3)*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))*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*S
ec[c + d*x]*Tan[c + d*x]))/(9*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3))))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((a + a/cos(c + d*x))^(5/3),x)

[Out]

int((a + a/cos(c + d*x))^(5/3), x)