3.287 \(\int \frac{(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx\)

Optimal. Leaf size=80 \[ \frac{2\ 2^{5/6} a \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac{1}{2};\frac{4}{3},-\frac{5}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}} \]

[Out]

(2*2^(5/6)*a*AppellF1[1/2, 4/3, -5/6, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(1/3)*
Tan[c + d*x])/(d*(1 + Sec[c + d*x])^(5/6))

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Rubi [A]  time = 0.13262, antiderivative size = 80, normalized size of antiderivative = 1., 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} \[ \frac{2\ 2^{5/6} a \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac{1}{2};\frac{4}{3},-\frac{5}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(4/3)/Sec[c + d*x]^(1/3),x]

[Out]

(2*2^(5/6)*a*AppellF1[1/2, 4/3, -5/6, 3/2, 1 - Sec[c + d*x], (1 - Sec[c + d*x])/2]*(a + a*Sec[c + d*x])^(1/3)*
Tan[c + d*x])/(d*(1 + Sec[c + d*x])^(5/6))

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 \frac{(a+a \sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx &=\frac{\left (a \sqrt [3]{a+a \sec (c+d x)}\right ) \int \frac{(1+\sec (c+d x))^{4/3}}{\sqrt [3]{\sec (c+d x)}} \, dx}{\sqrt [3]{1+\sec (c+d x)}}\\ &=\frac{\left (a \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(2-x)^{5/6}}{(1-x)^{4/3} \sqrt{x}} \, dx,x,1-\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} (1+\sec (c+d x))^{5/6}}\\ &=\frac{2\ 2^{5/6} a F_1\left (\frac{1}{2};\frac{4}{3},-\frac{5}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right ) \sqrt [3]{a+a \sec (c+d x)} \tan (c+d x)}{d (1+\sec (c+d x))^{5/6}}\\ \end{align*}

Mathematica [C]  time = 14.7493, size = 2325, normalized size = 29.06 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^(4/3)/Sec[c + d*x]^(1/3),x]

[Out]

(-3*(a*(1 + Sec[c + d*x]))^(4/3)*((1 + Sec[c + d*x])^(1/3)/Sec[c + d*x]^(1/3) + Sec[c + d*x]^(2/3)*(1 + Sec[c
+ d*x])^(1/3))*(-8*Tan[(c + d*x)/2] + (AppellF1[-4/3, -2/3, -2/3, -1/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1
+ I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(((-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)*((I
+ Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) - AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c +
d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*((I + Tan[(c
+ d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^2))/(4*2^(2/3)*d*(Sec[(c + d*x)/2]^2)^(1/3)*(1
+ Sec[c + d*x])^(4/3)*((Tan[(c + d*x)/2]*(-8*Tan[(c + d*x)/2] + (AppellF1[-4/3, -2/3, -2/3, -1/3, (-1 - I)/(-
1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(((-I + Tan[(c + d*x)/2])/(-1 + T
an[(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) - AppellF1[-4/3, -2/3, -2/3, -
1/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*
x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^2))/(4*2^(2/3)*(Sec
[(c + d*x)/2]^2)^(1/3)) - (3*(-4*Sec[(c + d*x)/2]^2 + (Sec[(c + d*x)/2]^2*(((-4/3 + (4*I)/3)*AppellF1[-1/3, -2
/3, 1/3, 2/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(-1 + Ta
n[(c + d*x)/2])^2 - ((4/3 + (4*I)/3)*AppellF1[-1/3, 1/3, -2/3, 2/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)
/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(-1 + Tan[(c + d*x)/2])^2))/(((-I + Tan[(c + d*x)/2])/(-1 + Tan[
(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) + (AppellF1[-4/3, -2/3, -2/3, -1/
3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/((
(-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3))
- AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*Sec[(c + d
*x)/2]^2*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2])
)^(1/3)*(1 + Tan[(c + d*x)/2]) - (2*AppellF1[-4/3, -2/3, -2/3, -1/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I
)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2*(Sec[(c + d*x)/2]^2/(2*(-1 + Tan[(c + d*x)/2])) - (Sec[(c + d*x)
/2]^2*(-I + Tan[(c + d*x)/2]))/(2*(-1 + Tan[(c + d*x)/2])^2)))/(3*((-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)
/2]))^(5/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)) - (2*AppellF1[-4/3, -2/3, -2/3, -1/3, (-1
- I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2*(Sec[(c + d*x)/2]^2/(2*(-1
+ Tan[(c + d*x)/2])) - (Sec[(c + d*x)/2]^2*(I + Tan[(c + d*x)/2]))/(2*(-1 + Tan[(c + d*x)/2])^2)))/(3*((-I + T
an[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(2/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(5/3)) - ((-I
+ Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 +
Tan[(c + d*x)/2])^2*(((4/3 + (4*I)/3)*AppellF1[-1/3, -2/3, 1/3, 2/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(
1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2])^2 + ((4/3 - (4*I)/3)*AppellF1[-1/3, 1/3, -2/
3, 2/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*Sec[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)
/2])^2) - (AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x)/2]), (1 + I)/(1 + Tan[(c + d*x)/2])]*((
I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^2*(-(Sec[(c + d*x)/2]^2*(-I + Tan[(
c + d*x)/2]))/(2*(1 + Tan[(c + d*x)/2])^2) + Sec[(c + d*x)/2]^2/(2*(1 + Tan[(c + d*x)/2]))))/(3*((-I + Tan[(c
+ d*x)/2])/(1 + Tan[(c + d*x)/2]))^(2/3)) - (AppellF1[-4/3, -2/3, -2/3, -1/3, (1 - I)/(1 + Tan[(c + d*x)/2]),
(1 + I)/(1 + Tan[(c + d*x)/2])]*((-I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)*(1 + Tan[(c + d*x)/2])^
2*(-(Sec[(c + d*x)/2]^2*(I + Tan[(c + d*x)/2]))/(2*(1 + Tan[(c + d*x)/2])^2) + Sec[(c + d*x)/2]^2/(2*(1 + Tan[
(c + d*x)/2]))))/(3*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(2/3))))/(4*2^(2/3)*(Sec[(c + d*x)/2]^2)^(
1/3))))

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Maple [F]  time = 0.111, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+a\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}}{\frac{1}{\sqrt [3]{\sec \left ( dx+c \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x)

[Out]

int((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{4}{3}}}{\sec \left (d x + c\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^(4/3)/sec(d*x + c)^(1/3), 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((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),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((a+a*sec(d*x+c))**(4/3)/sec(d*x+c)**(1/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{4}{3}}}{\sec \left (d x + c\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^(4/3)/sec(d*x+c)^(1/3),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^(4/3)/sec(d*x + c)^(1/3), x)