3.285 \(\int \sec ^{\frac{4}{3}}(c+d x) (a+a \sec (c+d x))^{2/3} \, dx\)

Optimal. Leaf size=79 \[ \frac{2 \sqrt [6]{2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} F_1\left (\frac{1}{2};-\frac{1}{3},-\frac{1}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{7/6}} \]

[Out]

(2*2^(1/6)*AppellF1[1/2, -1/3, -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))

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Rubi [A]  time = 0.123868, antiderivative size = 79, 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} \[ \frac{2 \sqrt [6]{2} \tan (c+d x) (a \sec (c+d x)+a)^{2/3} F_1\left (\frac{1}{2};-\frac{1}{3},-\frac{1}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{7/6}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^(4/3)*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

(2*2^(1/6)*AppellF1[1/2, -1/3, -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{4}{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{4}{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{\sqrt [3]{1-x} \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{1}{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 [C]  time = 19.2487, size = 2618, normalized size = 33.14 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^(4/3)*(a + a*Sec[c + d*x])^(2/3),x]

[Out]

(Sec[c + d*x]^(1/3)*((1 + Cos[c + d*x])*Sec[c + d*x])^(2/3)*(a*(1 + Sec[c + d*x]))^(2/3)*Tan[(c + d*x)/2])/(d*
(1 + Sec[c + d*x])^(2/3)) + (15*AppellF1[1/2, 2/3, 1/3, 3/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*(a*(1 +
Sec[c + d*x]))^(2/3)*(9*AppellF1[1/2, 2/3, 1/3, 3/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 2*(AppellF1[3/
2, 2/3, 4/3, 5/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 2*AppellF1[3/2, 5/3, 1/3, 5/2, Tan[(c + d*x)/4]^2
, -Tan[(c + d*x)/4]^2])*Tan[(c + d*x)/4]^2)*Tan[(c + d*x)/2])/(d*(Sec[c + d*x]/(1 + Sec[c + d*x]))^(2/3)*(1 +
Sec[c + d*x])^(2/3)*((135*AppellF1[1/2, 2/3, 1/3, 3/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]^2*(5 + Cos[c +
d*x]))/2 + 20*(AppellF1[3/2, 2/3, 4/3, 5/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 2*AppellF1[3/2, 5/3, 1
/3, 5/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2])^2*Cos[(c + d*x)/2]*Tan[(c + d*x)/4]^4 - 3*AppellF1[1/2, 2/3
, 1/3, 3/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*Tan[(c + d*x)/4]^2*(5*AppellF1[3/2, 2/3, 4/3, 5/2, Tan[(c
+ d*x)/4]^2, -Tan[(c + d*x)/4]^2]*(5 - 12*Cos[(c + d*x)/2] + Cos[c + d*x]) - 10*AppellF1[3/2, 5/3, 1/3, 5/2,
Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2]*(5 - 12*Cos[(c + d*x)/2] + Cos[c + d*x]) + 24*(2*AppellF1[5/2, 2/3, 7
/3, 7/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2] - 2*AppellF1[5/2, 5/3, 4/3, 7/2, Tan[(c + d*x)/4]^2, -Tan[(c
+ d*x)/4]^2] + 5*AppellF1[5/2, 8/3, 1/3, 7/2, Tan[(c + d*x)/4]^2, -Tan[(c + d*x)/4]^2])*Cos[(c + d*x)/2]*Tan[
(c + d*x)/4]^2))) - ((Sec[(c + d*x)/2]^2)^(4/3)*Sec[c + d*x]^(1/3)*(a*(1 + Sec[c + d*x]))^(2/3)*(AppellF1[-2/3
, -1/3, -1/3, 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)) - AppellF1[-2/3, -1/
3, -1/3, 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 + T
an[(c + d*x)/2]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3))))/(2^(1/3)*d*(-(((Sec[(c + d*x)
/2]^2)^(1/3)*Tan[(c + d*x)/2]*(AppellF1[-2/3, -1/3, -1/3, 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 + Ta
n[(c + d*x)/2]))^(1/3)) - AppellF1[-2/3, -1/3, -1/3, 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))))/2^(1/3)) - (3*(Sec[(c + d*x)/2]^2)^(1/3)*((((1/3 - I/3)*AppellF1[1/3, -1/3, 2/3, 4/3, (-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 + ((1
/3 + I/3)*AppellF1[1/3, 2/3, -1/3, 4/3, (-1 - I)/(-1 + Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*Se
c[(c + d*x)/2]^2)/(-1 + Tan[(c + d*x)/2])^2)/(((-I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(1/3)*((I + Ta
n[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(1/3)) - (AppellF1[-2/3, -1/3, -1/3, 1/3, (-1 - I)/(-1 + Tan[(c + d*x
)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*(Sec[(c + d*x)/2]^2/(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]))
^(4/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(1/3)) - (AppellF1[-2/3, -1/3, -1/3, 1/3, (-1 - I)/(-1
+ Tan[(c + d*x)/2]), (-1 + I)/(-1 + Tan[(c + d*x)/2])]*(Sec[(c + d*x)/2]^2/(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]))^(1/3)*((I + Tan[(c + d*x)/2])/(-1 + Tan[(c + d*x)/2]))^(4/3)) - (((-1/3 - I/3)*AppellF1[1/3, -1
/3, 2/3, 4/3, (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 - ((1/3 - I/3)*AppellF1[1/3, 2/3, -1/3, 4/3, (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]))^(1
/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)) + (AppellF1[-2/3, -1/3, -1/3, 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]))/(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]))^(4/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(1/3)) + (AppellF1[-2/3, -1/3, -1/3, 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]))/(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]))^(1/3)*((I + Tan[(c + d*x)/2])/(1 + Tan[(c + d*x)/2]))^(4/3))))/2^(1/3)))

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Maple [F]  time = 0.111, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{{\frac{4}{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)^(4/3)*(a+a*sec(d*x+c))^(2/3),x)

[Out]

int(sec(d*x+c)^(4/3)*(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{4}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^(4/3)*(a+a*sec(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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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)**(4/3)*(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{4}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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