3.2.57 \(\int \frac {1}{\sqrt [3]{a+a \sec (c+d x)}} \, dx\) [157]

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

[Out]

3*AppellF1(1/6,1,1/2,7/6,1+sec(d*x+c),1/2+1/2*sec(d*x+c))*2^(1/2)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/3)/(1-sec(d
*x+c))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 = {3864, 3863, 141} \begin {gather*} \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}} \end {gather*}

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 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 3863

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 3864

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

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) \text {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*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(718\) vs. \(2(75)=150\).
time = 4.81, size = 718, normalized size = 9.57 \begin {gather*} \frac {45 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 ) \cos (c+d x) (1+\sec (c+d x))^2 \tan \left (\frac {1}{2} (c+d x)\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 \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 ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt [3]{a (1+\sec (c+d x))} \left (40 \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 \sec (c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+6 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 ) \sec ^2(c+d x) \sin ^2\left (\frac {1}{2} (c+d x)\right ) \left (-15 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 ) (1-10 \cos (c+d x)+3 \cos (2 (c+d x)))-5 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 ) (1-10 \cos (c+d x)+3 \cos (2 (c+d x)))-24 \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 ) \cos (c+d x) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+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 (3+3 \sec (c+d x)-3 \sin (c+d x) \tan (c+d x)-\tan ^2(c+d x)\right )\right )} \end {gather*}

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 = 0.05, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [3]{a \sec {\left (c + d x \right )} + a}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a/cos(c + d*x))^(1/3),x)

[Out]

int(1/(a + a/cos(c + d*x))^(1/3), x)

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