3.164 \(\int \frac{\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx\)
Optimal. Leaf size=90 \[ \frac{3 \sqrt{2} \tan (c+d x) F_1\left (-\frac{7}{6};\frac{1}{2},2;-\frac{1}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 a d \sqrt{1-\sec (c+d x)} (\sec (c+d x)+1) (a \sec (c+d x)+a)^{2/3}} \]
[Out]
(3*Sqrt[2]*AppellF1[-7/6, 1/2, 2, -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))
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Rubi [A] time = 0.110847, antiderivative size = 90, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used =
{3828, 3827, 136} \[ \frac{3 \sqrt{2} \tan (c+d x) F_1\left (-\frac{7}{6};\frac{1}{2},2;-\frac{1}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{7 a d \sqrt{1-\sec (c+d x)} (\sec (c+d x)+1) (a \sec (c+d x)+a)^{2/3}} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]/(a + a*Sec[c + d*x])^(5/3),x]
[Out]
(3*Sqrt[2]*AppellF1[-7/6, 1/2, 2, -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 3828
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
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]
Rule 3827
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !In
tegerQ[m] && GtQ[a, 0]
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])
Rubi steps
\begin{align*} \int \frac{\cos (c+d x)}{(a+a \sec (c+d x))^{5/3}} \, dx &=\frac{(1+\sec (c+d x))^{2/3} \int \frac{\cos (c+d x)}{(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^2 (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},2;-\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*}
Mathematica [B] time = 16.028, size = 3011, normalized size = 33.46 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]/(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)*((-48*Sin[c + d*x])/7 + (51*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)) - (5*2^(1/3)*(1 + Sec[c +
d*x])^(5/3)*((-30*(1 + Sec[c + d*x])^(1/3))/7 + (55*Cos[c + d*x]*(1 + Sec[c + d*x])^(1/3))/7)*Tan[(c + d*x)/2]
*((-11*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]*S
ec[(c + d*x)/2]^2)^(2/3) + 9*Cos[(c + d*x)/2]^2*(-11 - 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)))))/(63*d*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)*
(a*(1 + Sec[c + d*x]))^(5/3)*((-5*Sec[(c + d*x)/2]^2*((-11*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) + 9*Cos[(c + d*x)/2]^2*(-11 - App
ellF1[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, -Ta
n[(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
)))))/(63*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2/3)) - (5*2^(1/3)*Tan[(c + d*x)/2]*((-11*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) - (11*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) + (22*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]))/(3*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(5/3)) - 9*Cos[(c + d*
x)/2]*Sin[(c + d*x)/2]*(-11 - 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))) + 9*Cos[(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]*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)) - (
-(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])/3 +
(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)/
((-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*Appel
lF1[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)) + (AppellF1[1/2, 1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*
x)/2]^2]*(-(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])/3 + (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, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/9)^2))))/(63*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(2
/3)) + (10*2^(1/3)*Tan[(c + d*x)/2]*((-11*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) + 9*Cos[(c + d*x)/2]^2*(-11 - 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))))*(-(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]))/(189*(Cos[(c + d*x)/2]
^2*Sec[c + d*x])^(5/3))))
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Maple [F] time = 0.102, size = 0, normalized size = 0. \begin{align*} \int{\cos \left ( dx+c \right ) \left ( a+a\sec \left ( dx+c \right ) \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)/(a+a*sec(d*x+c))^(5/3),x)
[Out]
int(cos(d*x+c)/(a+a*sec(d*x+c))^(5/3),x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)/(a+a*sec(d*x+c))^(5/3),x, algorithm="maxima")
[Out]
integrate(cos(d*x + c)/(a*sec(d*x + c) + a)^(5/3), x)
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)/(a+a*sec(d*x+c))^(5/3),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)/(a+a*sec(d*x+c))**(5/3),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)/(a+a*sec(d*x+c))^(5/3),x, algorithm="giac")
[Out]
integrate(cos(d*x + c)/(a*sec(d*x + c) + a)^(5/3), x)