3.272 \(\int (e \sec (c+d x))^{4/3} \sqrt{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=325 \[ \frac{4\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 e \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right ),-7-4 \sqrt{3}\right )}{5 d (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac{6 a e \tan (c+d x) \sqrt [3]{e \sec (c+d x)}}{5 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

(6*a*e*(e*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (4*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*
e*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])/(5*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*Se
c[c + d*x])^(1/3))^2])

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Rubi [A]  time = 0.383354, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {3806, 50, 63, 218} \[ \frac{4\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 e \tan (c+d x) \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right ) \sqrt{\frac{\sqrt [3]{e} \sqrt [3]{e \sec (c+d x)}+(e \sec (c+d x))^{2/3}+e^{2/3}}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}{\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}}\right )|-7-4 \sqrt{3}\right )}{5 d (a-a \sec (c+d x)) \sqrt{a \sec (c+d x)+a} \sqrt{\frac{\sqrt [3]{e} \left (\sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{e}-\sqrt [3]{e \sec (c+d x)}\right )^2}}}+\frac{6 a e \tan (c+d x) \sqrt [3]{e \sec (c+d x)}}{5 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a*e*(e*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(5*d*Sqrt[a + a*Sec[c + d*x]]) + (4*3^(3/4)*Sqrt[2 + Sqrt[3]]*a^2*
e*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])/(5*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*Se
c[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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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]

Rubi steps

\begin{align*} \int (e \sec (c+d x))^{4/3} \sqrt{a+a \sec (c+d x)} \, dx &=-\frac{\left (a^2 e \tan (c+d x)\right ) \operatorname{Subst}\left (\int \frac{\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{6 a e \sqrt [3]{e \sec (c+d x)} \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (2 a^2 e^2 \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 )}{5 d \sqrt{a-a \sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=\frac{6 a e \sqrt [3]{e \sec (c+d x)} \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (6 a^2 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 )}{5 d \sqrt{a-a \sec (c+d x)} \sqrt{a+a \sec (c+d x)}}\\ &=\frac{6 a e \sqrt [3]{e \sec (c+d x)} \tan (c+d x)}{5 d \sqrt{a+a \sec (c+d x)}}+\frac{4\ 3^{3/4} \sqrt{2+\sqrt{3}} a^2 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)}{5 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 = 0.230085, size = 71, normalized size = 0.22 \[ \frac{2 \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\sec (c+d x)+1)} (e \sec (c+d x))^{4/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{3}{2},1-\sec (c+d x)\right )}{d \sec ^{\frac{4}{3}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(e*Sec[c + d*x])^(4/3)*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])^(4/3)*Sqrt[a*(1 + Sec[c + d*x])]*Tan[(
c + d*x)/2])/(d*Sec[c + d*x]^(4/3))

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Maple [F]  time = 0.194, size = 0, normalized size = 0. \begin{align*} \int \left ( e\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{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))^(4/3)*(a+a*sec(d*x+c))^(1/2),x)

[Out]

int((e*sec(d*x+c))^(4/3)*(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{4}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(4/3)*(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))^(4/3), 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{1}{3}} e \sec \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sec(d*x+c))^(4/3)*(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)*e*sec(d*x + c), 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((e*sec(d*x+c))**(4/3)*(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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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{4}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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