3.163 \(\int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx\)
Optimal. Leaf size=90 \[ -\frac {3 \sqrt {2} \tan (c+d x) F_1\left (-\frac {7}{6};\frac {1}{2},1;-\frac {1}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 a d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1) (a \sec (c+d x)+a)^{2/3}} \]
[Out]
-3/7*AppellF1(-7/6,1,1/2,-1/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan(d*x+c)/a/d/(1+sec(d*x+c))/(a+a*sec(
d*x+c))^(2/3)/(1-sec(d*x+c))^(1/2)
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Rubi [A] time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used =
{3779, 3778, 136} \[ -\frac {3 \sqrt {2} \tan (c+d x) F_1\left (-\frac {7}{6};\frac {1}{2},1;-\frac {1}{6};\frac {1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 a d \sqrt {1-\sec (c+d x)} (\sec (c+d x)+1) (a \sec (c+d x)+a)^{2/3}} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])^(-5/3),x]
[Out]
(-3*Sqrt[2]*AppellF1[-7/6, 1/2, 1, -1/6, (1 + Sec[c + d*x])/2, 1 + Sec[c + d*x]]*Tan[c + d*x])/(7*a*d*Sqrt[1 -
Sec[c + d*x]]*(1 + Sec[c + d*x])*(a + a*Sec[c + d*x])^(2/3))
Rule 136
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)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), 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 3778
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 3779
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Csc[c + d*x])^FracPart
[n])/(1 + (b*Csc[c + d*x])/a)^FracPart[n], Int[(1 + (b*Csc[c + d*x])/a)^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*} \int \frac {1}{(a+a \sec (c+d x))^{5/3}} \, dx &=\frac {(1+\sec (c+d x))^{2/3} \int \frac {1}{(1+\sec (c+d x))^{5/3}} \, dx}{a (a+a \sec (c+d x))^{2/3}}\\ &=-\frac {\left (\sqrt [6]{1+\sec (c+d x)} \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x (1+x)^{13/6}} \, dx,x,\sec (c+d x)\right )}{a d \sqrt {1-\sec (c+d x)} (a+a \sec (c+d x))^{2/3}}\\ &=-\frac {3 \sqrt {2} F_1\left (-\frac {7}{6};\frac {1}{2},1;-\frac {1}{6};\frac {1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{7 a d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x)) (a+a \sec (c+d x))^{2/3}}\\ \end {align*}
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Mathematica [B] time = 16.72, size = 3007, normalized size = 33.41 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[c + d*x])^(-5/3),x]
[Out]
(((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(1 + Sec[c + d*x])^(5/3)*((27*Sin[c + d*x])/7 - (30*Tan[(c + d*x)/2])
/7 + (3*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/14))/(d*(a*(1 + Sec[c + d*x]))^(5/3)) + (2^(1/3)*(1 + Sec[c + d*x
])^(5/3)*((16*(1 + Sec[c + d*x])^(1/3))/7 - (27*Cos[c + d*x]*(1 + Sec[c + d*x])^(1/3))/7)*Tan[(c + d*x)/2]*((-
3*AppellF1[3/2, 1/3, 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]*Sec[(c
+ d*x)/2]^2)^(2/3) + Cos[(c + d*x)/2]^2*(-27 - (5*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*
x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2
*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 4/3, 1, 5/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9)))))/(7*d*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*(a*(
1 + Sec[c + d*x]))^(5/3)*((Sec[(c + d*x)/2]^2*((-3*AppellF1[3/2, 1/3, 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]*Sec[(c + d*x)/2]^2)^(2/3) + Cos[(c + d*x)/2]^2*(-27 - (5*AppellF1[1
/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(AppellF1[1/2, 1/3, 1, 3
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c +
d*x)/2]^2] + AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9)))))/
(7*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)) + (2^(1/3)*Tan[(c + d*x)/2]*((-3*AppellF1[3/2, 1/3, 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])/(Cos[c + d*x]*Sec[(c + d*x)/2]
^2)^(2/3) - (3*Tan[(c + d*x)/2]^2*((-3*AppellF1[5/2, 1/3, 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, 4/3, 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))/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(2/3) + (2*AppellF1[3/2, 1/3, 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]))/(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/3) - Cos[(c + d*x)/2]*Sin[(c + d*x
)/2]*(-27 - (5*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)
*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[
(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Ta
n[(c + d*x)/2]^2)/9))) + Cos[(c + d*x)/2]^2*((5*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/
2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/((-1 + Tan[(c + d*x)/2]^2)^2*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d
*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] +
AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9)) - (5*(-1/3*(Appel
lF1[3/2, 1/3, 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]) + (AppellF
1[3/2, 4/3, 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 + T
an[(c + d*x)/2]^2)*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2,
1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(
c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9)) + (5*AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2]*(-1/3*(AppellF1[3/2, 1/3, 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]) + (AppellF1[3/2, 4/3, 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*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 4/3, 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
*AppellF1[5/2, 4/3, 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 +
(4*AppellF1[5/2, 7/3, 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
- 3*((-6*AppellF1[5/2, 1/3, 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 + (AppellF1[5/2, 4/3, 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)))/9))/((-1 + Tan[(c + d*x)/2]^2)*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] +
(2*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 4/3, 1, 5/2, Tan[(
c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9)^2))))/(7*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3))
- (2*2^(1/3)*Tan[(c + d*x)/2]*((-3*AppellF1[3/2, 1/3, 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]*Sec[(c + d*x)/2]^2)^(2/3) + Cos[(c + d*x)/2]^2*(-27 - (5*AppellF1[1/2, 1/3, 1, 3/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/((-1 + Tan[(c + d*x)/2]^2)*(AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x
)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2, 1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + Ap
pellF1[3/2, 4/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9))))*(-(Cos[(c + d*x)/
2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x]))/(21*(Cos[(c + d*x)/2]^2*Sec
[c + d*x])^(5/3))))
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sec(d*x+c))^(5/3),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sec(d*x+c))^(5/3),x, algorithm="giac")
[Out]
integrate((a*sec(d*x + c) + a)^(-5/3), x)
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maple [F] time = 0.70, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {5}{3}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a+a*sec(d*x+c))^(5/3),x)
[Out]
int(1/(a+a*sec(d*x+c))^(5/3),x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{3}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sec(d*x+c))^(5/3),x, algorithm="maxima")
[Out]
integrate((a*sec(d*x + c) + a)^(-5/3), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/3}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(a + a/cos(c + d*x))^(5/3),x)
[Out]
int(1/(a + a/cos(c + d*x))^(5/3), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \sec {\left (c + d x \right )} + a\right )^{\frac {5}{3}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(a+a*sec(d*x+c))**(5/3),x)
[Out]
Integral((a*sec(c + d*x) + a)**(-5/3), x)
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