3.157 \(\int \frac{1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=75 \[ \frac{3 \sqrt{2} \tan (c+d x) F_1\left (\frac{1}{6};\frac{1}{2},1;\frac{7}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d \sqrt{1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.042687, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3779, 3778, 136} \[ \frac{3 \sqrt{2} \tan (c+d x) F_1\left (\frac{1}{6};\frac{1}{2},1;\frac{7}{6};\frac{1}{2} (\sec (c+d x)+1),\sec (c+d x)+1\right )}{d \sqrt{1-\sec (c+d x)} \sqrt [3]{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^(-1/3),x]

[Out]

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

Rule 3779

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Csc[c + d*x])^FracPart
[n])/(1 + (b*Csc[c + d*x])/a)^FracPart[n], Int[(1 + (b*Csc[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rule 3778

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^n*Cot[c + d*x])/(d*Sqrt[1 + Csc[c + d*x]
]*Sqrt[1 - Csc[c + d*x]]), Subst[Int[(1 + (b*x)/a)^(n - 1/2)/(x*Sqrt[1 - (b*x)/a]), x], x, Csc[c + d*x]], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && 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{1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx &=\frac{\sqrt [3]{1+\sec (c+d x)} \int \frac{1}{\sqrt [3]{1+\sec (c+d x)}} \, dx}{\sqrt [3]{a+a \sec (c+d x)}}\\ &=-\frac{\tan (c+d x) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} x (1+x)^{5/6}} \, dx,x,\sec (c+d x)\right )}{d \sqrt{1-\sec (c+d x)} \sqrt [6]{1+\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ &=\frac{3 \sqrt{2} F_1\left (\frac{1}{6};\frac{1}{2},1;\frac{7}{6};\frac{1}{2} (1+\sec (c+d x)),1+\sec (c+d x)\right ) \tan (c+d x)}{d \sqrt{1-\sec (c+d x)} \sqrt [3]{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [B]  time = 4.51008, size = 718, normalized size = 9.57 \[ \frac{45 \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 F_1\left (\frac{1}{2};-\frac{1}{3},1;\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \left (9 F_1\left (\frac{1}{2};-\frac{1}{3},1;\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-2 \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (3 F_1\left (\frac{3}{2};-\frac{1}{3},2;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+F_1\left (\frac{3}{2};\frac{2}{3},1;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{d \sqrt [3]{a (\sec (c+d x)+1)} \left (40 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \left (3 F_1\left (\frac{3}{2};-\frac{1}{3},2;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+F_1\left (\frac{3}{2};\frac{2}{3},1;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right ){}^2+135 F_1\left (\frac{1}{2};-\frac{1}{3},1;\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ){}^2 \left (-\tan ^2(c+d x)+3 \sec (c+d x)-3 \sin (c+d x) \tan (c+d x)+3\right )+6 \sin ^2\left (\frac{1}{2} (c+d x)\right ) \sec ^2(c+d x) F_1\left (\frac{1}{2};-\frac{1}{3},1;\frac{3}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right ) \left (-24 \cos (c+d x) \tan ^2\left (\frac{1}{2} (c+d x)\right ) \left (9 F_1\left (\frac{5}{2};-\frac{1}{3},3;\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )+3 F_1\left (\frac{5}{2};\frac{2}{3},2;\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-F_1\left (\frac{5}{2};\frac{5}{3},1;\frac{7}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )-15 (-10 \cos (c+d x)+3 \cos (2 (c+d x))+1) F_1\left (\frac{3}{2};-\frac{1}{3},2;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )-5 (-10 \cos (c+d x)+3 \cos (2 (c+d x))+1) F_1\left (\frac{3}{2};\frac{2}{3},1;\frac{5}{2};\tan ^2\left (\frac{1}{2} (c+d x)\right ),-\tan ^2\left (\frac{1}{2} (c+d x)\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(45*AppellF1[1/2, -1/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x]*(1 + Sec[c + d*x])^2*Tan
[(c + d*x)/2]*(9*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, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c +
d*x)/2]^2])*Tan[(c + d*x)/2]^2))/(d*(a*(1 + Sec[c + d*x]))^(1/3)*(40*(3*AppellF1[3/2, -1/3, 2, 5/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])^2*Sec[c
+ d*x]*Sin[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2 + 6*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*Sin[(c + d*x)/2]^2*(-15*AppellF1[3/2, -1/3, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)
/2]^2]*(1 - 10*Cos[c + d*x] + 3*Cos[2*(c + d*x)]) - 5*AppellF1[3/2, 2/3, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c +
d*x)/2]^2]*(1 - 10*Cos[c + d*x] + 3*Cos[2*(c + d*x)]) - 24*(9*AppellF1[5/2, -1/3, 3, 7/2, Tan[(c + d*x)/2]^2,
-Tan[(c + d*x)/2]^2] + 3*AppellF1[5/2, 2/3, 2, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[5/2,
5/3, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[c + d*x]*Tan[(c + d*x)/2]^2) + 135*AppellF1[1/2, -1
/3, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]^2*(3 + 3*Sec[c + d*x] - 3*Sin[c + d*x]*Tan[c + d*x] - Tan
[c + d*x]^2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(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 \frac{1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sec(c + d*x) + a)**(-1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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