3.277 \(\int (e \sec (c+d x))^{2/3} \sqrt{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=624 \[ -\frac{2 \sqrt{2} 3^{3/4} a^2 \sqrt [3]{e} \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right ),-7-4 \sqrt{3}\right )}{d (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac{3 \sqrt [4]{3} \sqrt{2-\sqrt{3}} a^2 \sqrt [3]{e} \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} E\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt{3}\right )}{d (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}-\frac{6 a e \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )} \]

[Out]

(-6*a*e*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]*((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))) + (3*3^(1/
4)*Sqrt[2 - Sqrt[3]]*a^2*e^(1/3)*EllipticE[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]) - (2*Sqrt[2]*3^(3/4)*a^2*e^(1/3)*EllipticF[ArcSin[((1 - Sq
rt[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])

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Rubi [A]  time = 0.435255, antiderivative size = 624, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3806, 63, 303, 218, 1877} \[ -\frac{2 \sqrt{2} 3^{3/4} a^2 \sqrt [3]{e} \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt{3}\right )}{d (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac{3 \sqrt [4]{3} \sqrt{2-\sqrt{3}} a^2 \sqrt [3]{e} \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} E\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt{3}\right )}{d (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}-\frac{6 a e \tan (c+d x)}{d \sqrt{a \sec (c+d x)+a} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*e*Tan[c + d*x])/(d*Sqrt[a + a*Sec[c + d*x]]*((1 + Sqrt[3])*e^(1/3) - (e*Sec[c + d*x])^(1/3))) + (3*3^(1/
4)*Sqrt[2 - Sqrt[3]]*a^2*e^(1/3)*EllipticE[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]) - (2*Sqrt[2]*3^(3/4)*a^2*e^(1/3)*EllipticF[ArcSin[((1 - Sq
rt[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 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]

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 303

Int[(x_)/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Dist[(Sq
rt[2]*s)/(Sqrt[2 + Sqrt[3]]*r), Int[1/Sqrt[a + b*x^3], x], x] + Dist[1/r, Int[((1 - Sqrt[3])*s + r*x)/Sqrt[a +
b*x^3], x], x]] /; FreeQ[{a, b}, x] && PosQ[a]

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 1877

Int[((c_) + (d_.)*(x_))/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Simplify[((1 - Sqrt[3])*d)/c]]
, s = Denom[Simplify[((1 - Sqrt[3])*d)/c]]}, Simp[(2*d*s^3*Sqrt[a + b*x^3])/(a*r^2*((1 + Sqrt[3])*s + r*x)), x
] - Simp[(3^(1/4)*Sqrt[2 - Sqrt[3]]*d*s*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*Elli
pticE[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(r^2*Sqrt[a + b*x^3]*Sqrt[(s*(
s + r*x))/((1 + Sqrt[3])*s + r*x)^2]), x]] /; FreeQ[{a, b, c, d}, x] && PosQ[a] && EqQ[b*c^3 - 2*(5 - 3*Sqrt[3
])*a*d^3, 0]

Rubi steps

\begin{align*} \int (e \sec (c+d x))^{2/3} \sqrt{a+a \sec (c+d x)} \, dx &=-\frac{\left (a^2 e \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{e x} \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{x}{\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{\left (3 a^2 \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-x}{\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{\left (3 \sqrt{2 \left (2-\sqrt{3}\right )} a^2 \sqrt [3]{e} \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{6 a e \tan (c+d x)}{d \sqrt{a+a \sec (c+d x)} \left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}+\frac{3 \sqrt [4]{3} \sqrt{2-\sqrt{3}} a^2 \sqrt [3]{e} E\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}}}-\frac{2 \sqrt{2} 3^{3/4} a^2 \sqrt [3]{e} 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*}

Mathematica [C]  time = 0.138841, size = 71, normalized size = 0.11 \[ \frac{2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (e \sec (c+d x))^{2/3} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{1}{2},\frac{3}{2},1-\sec (c+d x)\right )}{d \sec ^{\frac{2}{3}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.188, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}\sqrt{a+a\sec \left ( dx+c \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac{2}{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )} \left (e \sec{\left (c + d x \right )}\right )^{\frac{2}{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \sec \left (d x + c\right ) + a} \left (e \sec \left (d x + c\right )\right )^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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