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

Optimal. Leaf size=327 \[ -\frac{5 \tan ^3(c+d x) (a (\sec (c+d x)+1))^{2/3} \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right )} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{3}{4},\frac{7}{4},\tan ^4\left (\frac{1}{2} (c+d x)\right )\right )}{8 d \sqrt [3]{\frac{1}{\cos (c+d x)+1}} (\sec (c+d x)+1)^{10/3}}+\frac{\tan (c+d x) (a (\sec (c+d x)+1))^{2/3} \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right )} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{3},\frac{5}{4},\tan ^4\left (\frac{1}{2} (c+d x)\right )\right )}{8 d \sqrt [3]{\frac{1}{\cos (c+d x)+1}} (\sec (c+d x)+1)^{4/3}}+\frac{9 \sin (c+d x) \sec ^{\frac{2}{3}}(c+d x) (a (\sec (c+d x)+1))^{2/3}}{4 d}-\frac{3 a \sin (c+d x) \sec ^{\frac{5}{3}}(c+d x)}{2 d \sqrt [3]{a (\sec (c+d x)+1)}}-\frac{9 \tan (c+d x) (a (\sec (c+d x)+1))^{2/3}}{4 d \sqrt [3]{\frac{1}{\cos (c+d x)+1}} (\sec (c+d x)+1)^{7/3}} \]

[Out]

(-3*a*Sec[c + d*x]^(5/3)*Sin[c + d*x])/(2*d*(a*(1 + Sec[c + d*x]))^(1/3)) + (9*Sec[c + d*x]^(2/3)*(a*(1 + Sec[
c + d*x]))^(2/3)*Sin[c + d*x])/(4*d) - (9*(a*(1 + Sec[c + d*x]))^(2/3)*Tan[c + d*x])/(4*d*((1 + Cos[c + d*x])^
(-1))^(1/3)*(1 + Sec[c + d*x])^(7/3)) + (Hypergeometric2F1[1/4, 1/3, 5/4, Tan[(c + d*x)/2]^4]*(Cos[c + d*x]*Se
c[(c + d*x)/2]^4)^(1/3)*(a*(1 + Sec[c + d*x]))^(2/3)*Tan[c + d*x])/(8*d*((1 + Cos[c + d*x])^(-1))^(1/3)*(1 + S
ec[c + d*x])^(4/3)) - (5*Hypergeometric2F1[1/3, 3/4, 7/4, Tan[(c + d*x)/2]^4]*(Cos[c + d*x]*Sec[(c + d*x)/2]^4
)^(1/3)*(a*(1 + Sec[c + d*x]))^(2/3)*Tan[c + d*x]^3)/(8*d*((1 + Cos[c + d*x])^(-1))^(1/3)*(1 + Sec[c + d*x])^(
10/3))

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Rubi [C]  time = 0.122051, antiderivative size = 79, normalized size of antiderivative = 0.24, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3828, 3825, 133} \[ \frac{2 \sqrt [6]{2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} F_1\left (\frac{1}{2};-\frac{2}{3},-\frac{1}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{7/6}} \]

Warning: Unable to verify antiderivative.

[In]

Int[Sec[c + d*x]^(5/3)*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

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

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 3825

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[(((a*
d)/b)^n*Cot[e + f*x])/(a^(n - 2)*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((a - x)^(n -
1)*(2*a - x)^(m - 1/2))/Sqrt[x], x], x, a - b*Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2
- b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] &&  !IntegerQ[n] && GtQ[(a*d)/b, 0]

Rule 133

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

Rubi steps

\begin{align*} \int \sec ^{\frac{5}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx &=\frac{(a+a \sec (c+d x))^{2/3} \int \sec ^{\frac{5}{3}}(c+d x) (1+\sec (c+d x))^{2/3} \, dx}{(1+\sec (c+d x))^{2/3}}\\ &=\frac{\left ((a+a \sec (c+d x))^{2/3} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(1-x)^{2/3} \sqrt [6]{2-x}}{\sqrt{x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{7/6}}\\ &=\frac{2 \sqrt [6]{2} F_1\left (\frac{1}{2};-\frac{2}{3},-\frac{1}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (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}}\\ \end{align*}

Mathematica [A]  time = 7.39605, size = 274, normalized size = 0.84 \[ \frac{(a (\sec (c+d x)+1))^{2/3} \left (-5 \sqrt [3]{2} \tan ^3\left (\frac{1}{2} (c+d x)\right ) \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right )} \sqrt [3]{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{3}{4},\frac{7}{4},\tan ^4\left (\frac{1}{2} (c+d x)\right )\right )+\sqrt [3]{2} \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt [3]{\cos (c+d x) \sec ^4\left (\frac{1}{2} (c+d x)\right )} \sqrt [3]{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{3},\frac{5}{4},\tan ^4\left (\frac{1}{2} (c+d x)\right )\right )-3 \left (\sin \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{3}{2} (c+d x)\right )\right ) \sec (c+d x) \sqrt [3]{\sec (c+d x)+1} \sec ^3\left (\frac{1}{2} (c+d x)\right )\right )}{8 d \sqrt [3]{\frac{1}{\cos (c+d x)+1}} (\sec (c+d x)+1)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^(5/3)*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

((a*(1 + Sec[c + d*x]))^(2/3)*(-3*Sec[(c + d*x)/2]^3*Sec[c + d*x]*(1 + Sec[c + d*x])^(1/3)*(Sin[(c + d*x)/2] -
2*Sin[(3*(c + d*x))/2]) + 2^(1/3)*Hypergeometric2F1[1/4, 1/3, 5/4, Tan[(c + d*x)/2]^4]*(Cos[c + d*x]*Sec[(c +
d*x)/2]^4)^(1/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)*Tan[(c + d*x)/2] - 5*2^(1/3)*Hypergeometric2F1[1/3,
3/4, 7/4, Tan[(c + d*x)/2]^4]*(Cos[c + d*x]*Sec[(c + d*x)/2]^4)^(1/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)*
Tan[(c + d*x)/2]^3))/(8*d*((1 + Cos[c + d*x])^(-1))^(1/3)*(1 + Sec[c + d*x])^(2/3))

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Maple [F]  time = 0.121, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{{\frac{5}{3}}} \left ( a+a\sec \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x)

[Out]

int(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{\frac{5}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^(5/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{\frac{5}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^(5/3), x)

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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(sec(d*x+c)**(5/3)*(a+a*sec(d*x+c))**(2/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{2}{3}} \sec \left (d x + c\right )^{\frac{5}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(5/3)*(a+a*sec(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(2/3)*sec(d*x + c)^(5/3), x)