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

Optimal. Leaf size=78 \[ \frac{2^{5/6} \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac{1}{2};-\frac{1}{3},\frac{1}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}} \]

[Out]

(2^(5/6)*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])^(1/3)*Tan[
c + d*x])/(d*(1 + Sec[c + d*x])^(5/6))

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Rubi [A]  time = 0.109453, antiderivative size = 78, 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} \[ \frac{2^{5/6} \tan (c+d x) \sqrt [3]{a \sec (c+d x)+a} F_1\left (\frac{1}{2};-\frac{1}{3},\frac{1}{6};\frac{3}{2};1-\sec (c+d x),\frac{1}{2} (1-\sec (c+d x))\right )}{d (\sec (c+d x)+1)^{5/6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 14.654, size = 1982, normalized size = 25.41 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*Sec[c + d*x]^(1/3)*((1 + Cos[c + d*x])*Sec[c + d*x])^(1/3)*(a*(1 + Sec[c + d*x]))^(1/3)*Sin[c + d*x])/(2*d*
(1 + Sec[c + d*x])^(1/3)) + (3*(a*(1 + Sec[c + d*x]))^(1/3)*(-((1 + Sec[c + d*x])^(1/3)/Sec[c + d*x]^(2/3)) +
(Sec[c + d*x]^(1/3)*(1 + Sec[c + d*x])^(1/3))/2)*Tan[(c + d*x)/2]*(-1 + (3*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(
c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
- 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4},
{7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2)))/(2^(2/3)*d*(Sec[(c + d*x)/2]^2)^(2/3)*(Cos[(c + d*x)/2]^2*S
ec[c + d*x])^(1/3)*(1 + Sec[c + d*x])^(1/3)*((3*(Sec[(c + d*x)/2]^2)^(1/3)*(-1 + (3*AppellF1[1/2, -1/3, 2/3, 3
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d
*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2
/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2)))/(2*2^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)
) - (2^(1/3)*Tan[(c + d*x)/2]^2*(-1 + (3*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2
])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(
c + d*x)/2]^2)))/((Sec[(c + d*x)/2]^2)^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)) + (3*Tan[(c + d*x)/2]*((
3*((-2*AppellF1[3/2, -1/3, 5/3, 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 - (HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/9))/(9
*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, T
an[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d
*x)/2]^2) - (3*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(-2*(2*AppellF1[3/2, -1/
3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^
4])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] + 9*((-2*AppellF1[3/2, -1/3, 5/3, 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 - (HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4]*S
ec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/9) - 2*Tan[(c + d*x)/2]^2*(2*(-(AppellF1[5/2, -1/3, 8/3, 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]) - (AppellF1[5/2, 2/3, 5/3, 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) + (3*Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(-
HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4] + (1 - Tan[(c + d*x)/2]^4)^(-2/3)))/2)))/(9*AppellF1[
1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*
x)/2]^2, -Tan[(c + d*x)/2]^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2)^
2))/(2^(2/3)*(Sec[(c + d*x)/2]^2)^(2/3)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(1/3)) - (Tan[(c + d*x)/2]*(-1 + (3*
AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])/(9*AppellF1[1/2, -1/3, 2/3, 3/2, Tan[(
c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*(2*AppellF1[3/2, -1/3, 5/3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2] + HypergeometricPFQ[{2/3, 3/4}, {7/4}, Tan[(c + d*x)/2]^4])*Tan[(c + d*x)/2]^2))*(-(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]))/(2^(2/3)*(Sec[(c + d*x)/2]^2)^(2/3)
*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(4/3))))

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Maple [F]  time = 0.112, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}}\sqrt [3]{a+a\sec \left ( dx+c \right ) }\, 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))^(1/3),x)

[Out]

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

[Out]

integrate((a*sec(d*x + c) + a)^(1/3)*sec(d*x + c)^(4/3), 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{1}{3}} \sec \left (d x + c\right )^{\frac{4}{3}}, x\right ) \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))^(1/3),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^(1/3)*sec(d*x + c)^(4/3), 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)**(4/3)*(a+a*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{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{1}{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))^(1/3),x, algorithm="giac")

[Out]

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