3.273 \(\int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=280 \[ \frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right )}{d (a-a \sec (c+d x)) \sqrt {a \sec (c+d x)+a} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}} \]

[Out]

2*3^(3/4)*a^2*EllipticF((-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1-3^(1/2)))/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2))
),I*3^(1/2)+2*I)*(e^(1/3)-(e*sec(d*x+c))^(1/3))*(1/2*6^(1/2)+1/2*2^(1/2))*((e^(2/3)+e^(1/3)*(e*sec(d*x+c))^(1/
3)+(e*sec(d*x+c))^(2/3))/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2)))^2)^(1/2)*tan(d*x+c)/d/(a-a*sec(d*x+c))/(a
+a*sec(d*x+c))^(1/2)/(e^(1/3)*(e^(1/3)-(e*sec(d*x+c))^(1/3))/(-(e*sec(d*x+c))^(1/3)+e^(1/3)*(1+3^(1/2)))^2)^(1
/2)

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Rubi [A]  time = 0.20, antiderivative size = 280, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3806, 63, 218} \[ \frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right )}{d (a-a \sec (c+d x)) \sqrt {a \sec (c+d x)+a} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*EllipticF[ArcSin[((1 - Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))/((1 + Sqrt[
3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))], -7 - 4*Sqrt[3]]*(e^(1/3) - (e*Sec[c + d*x])^(1/3))*Sqrt[(e^(2/3) + e^(
1/3)*(e*Sec[c + d*x])^(1/3) + (e*Sec[c + d*x])^(2/3))/((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))^2]*Tan[
c + d*x])/(d*(a - a*Sec[c + d*x])*Sqrt[a + a*Sec[c + d*x]]*Sqrt[(e^(1/3)*(e^(1/3) - (e*Sec[c + d*x])^(1/3)))/(
(1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))^2])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 3806

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(a^2*d*
Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[(d*x)^(n - 1)/Sqrt[a - b*x], x]
, x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \sqrt [3]{e \sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx &=-\frac {\left (a^2 e \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{(e x)^{2/3} \sqrt {a-a x}} \, dx,x,\sec (c+d x)\right )}{d \sqrt {a-a \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=-\frac {\left (3 a^2 \tan (c+d x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-\frac {a x^3}{e}}} \, dx,x,\sqrt [3]{e \sec (c+d x)}\right )}{d \sqrt {a-a \sec (c+d x)} \sqrt {a+a \sec (c+d x)}}\\ &=\frac {2\ 3^{3/4} \sqrt {2+\sqrt {3}} a^2 F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt {3}\right ) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt {\frac {e^{2/3}+\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \tan (c+d x)}{d (a-a \sec (c+d x)) \sqrt {a+a \sec (c+d x)} \sqrt {\frac {\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 71, normalized size = 0.25 \[ \frac {2 \tan \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (\sec (c+d x)+1)} \sqrt [3]{e \sec (c+d x)} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {3}{2};1-\sec (c+d x)\right )}{d \sqrt [3]{\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Hypergeometric2F1[1/2, 2/3, 3/2, 1 - Sec[c + d*x]]*(e*Sec[c + d*x])^(1/3)*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(c
+ d*x)/2])/(d*Sec[c + d*x]^(1/3))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*sec(d*x + c) + a)*(e*sec(d*x + c))^(1/3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.60, size = 0, normalized size = 0.00 \[ \int \left (e \sec \left (d x +c \right )\right )^{\frac {1}{3}} \sqrt {a +a \sec \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

int((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac {1}{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(1/3)*(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}\,{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{1/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^(1/2)*(e/cos(c + d*x))^(1/3),x)

[Out]

int((a + a/cos(c + d*x))^(1/2)*(e/cos(c + d*x))^(1/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \sqrt [3]{e \sec {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))**(1/3)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

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

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