3.940 \(\int \frac{1}{\sqrt{2+e x} \sqrt [4]{12-3 e^2 x^2}} \, dx\)

Optimal. Leaf size=241 \[ -\frac{\log \left (\frac{\sqrt{6-3 e x}+\sqrt{3} \sqrt{e x+2}-\sqrt{6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt{e x+2}}\right )}{\sqrt{2} \sqrt [4]{3} e}+\frac{\log \left (\frac{\sqrt{6-3 e x}+\sqrt{3} \sqrt{e x+2}+\sqrt{6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt{e x+2}}\right )}{\sqrt{2} \sqrt [4]{3} e}+\frac{\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt [4]{3} e}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt [4]{3} e} \]

[Out]

(Sqrt[2]*ArcTan[1 - (Sqrt[2]*(2 - e*x)^(1/4))/(2 + e*x)^(1/4)])/(3^(1/4)*e) - (Sqrt[2]*ArcTan[1 + (Sqrt[2]*(2
- e*x)^(1/4))/(2 + e*x)^(1/4)])/(3^(1/4)*e) - Log[(Sqrt[6 - 3*e*x] - Sqrt[6]*(2 - e*x)^(1/4)*(2 + e*x)^(1/4) +
Sqrt[3]*Sqrt[2 + e*x])/Sqrt[2 + e*x]]/(Sqrt[2]*3^(1/4)*e) + Log[(Sqrt[6 - 3*e*x] + Sqrt[6]*(2 - e*x)^(1/4)*(2
+ e*x)^(1/4) + Sqrt[3]*Sqrt[2 + e*x])/Sqrt[2 + e*x]]/(Sqrt[2]*3^(1/4)*e)

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Rubi [A]  time = 0.244801, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {675, 63, 331, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\log \left (\frac{\sqrt{6-3 e x}+\sqrt{3} \sqrt{e x+2}-\sqrt{6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt{e x+2}}\right )}{\sqrt{2} \sqrt [4]{3} e}+\frac{\log \left (\frac{\sqrt{6-3 e x}+\sqrt{3} \sqrt{e x+2}+\sqrt{6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt{e x+2}}\right )}{\sqrt{2} \sqrt [4]{3} e}+\frac{\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt [4]{3} e}-\frac{\sqrt{2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt [4]{3} e} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 + e*x]*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

(Sqrt[2]*ArcTan[1 - (Sqrt[2]*(2 - e*x)^(1/4))/(2 + e*x)^(1/4)])/(3^(1/4)*e) - (Sqrt[2]*ArcTan[1 + (Sqrt[2]*(2
- e*x)^(1/4))/(2 + e*x)^(1/4)])/(3^(1/4)*e) - Log[(Sqrt[6 - 3*e*x] - Sqrt[6]*(2 - e*x)^(1/4)*(2 + e*x)^(1/4) +
Sqrt[3]*Sqrt[2 + e*x])/Sqrt[2 + e*x]]/(Sqrt[2]*3^(1/4)*e) + Log[(Sqrt[6 - 3*e*x] + Sqrt[6]*(2 - e*x)^(1/4)*(2
+ e*x)^(1/4) + Sqrt[3]*Sqrt[2 + e*x])/Sqrt[2 + e*x]]/(Sqrt[2]*3^(1/4)*e)

Rule 675

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^p,
x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && GtQ[a, 0] && GtQ[d, 0] &&  !I
GtQ[m, 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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{2+e x} \sqrt [4]{12-3 e^2 x^2}} \, dx &=\int \frac{1}{\sqrt [4]{6-3 e x} (2+e x)^{3/4}} \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{x^2}{\left (4-\frac{x^4}{3}\right )^{3/4}} \, dx,x,\sqrt [4]{6-3 e x}\right )}{3 e}\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{x^2}{1+\frac{x^4}{3}} \, dx,x,\frac{\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{3 e}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{\sqrt{3}-x^2}{1+\frac{x^4}{3}} \, dx,x,\frac{\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{3 e}-\frac{2 \operatorname{Subst}\left (\int \frac{\sqrt{3}+x^2}{1+\frac{x^4}{3}} \, dx,x,\frac{\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{3 e}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{3}-\sqrt{2} \sqrt [4]{3} x+x^2} \, dx,x,\frac{\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{e}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{3}+\sqrt{2} \sqrt [4]{3} x+x^2} \, dx,x,\frac{\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{e}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt [4]{3}+2 x}{-\sqrt{3}-\sqrt{2} \sqrt [4]{3} x-x^2} \, dx,x,\frac{\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt{2} \sqrt [4]{3} e}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2} \sqrt [4]{3}-2 x}{-\sqrt{3}+\sqrt{2} \sqrt [4]{3} x-x^2} \, dx,x,\frac{\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt{2} \sqrt [4]{3} e}\\ &=-\frac{\log \left (\frac{\sqrt{2-e x}-\sqrt{2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt{2+e x}}{\sqrt{2+e x}}\right )}{\sqrt{2} \sqrt [4]{3} e}+\frac{\log \left (\frac{\sqrt{2-e x}+\sqrt{2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt{2+e x}}{\sqrt{2+e x}}\right )}{\sqrt{2} \sqrt [4]{3} e}-\frac{\sqrt{2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt [4]{3} e}+\frac{\sqrt{2} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt [4]{3} e}\\ &=\frac{\sqrt{2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt [4]{3} e}-\frac{\sqrt{2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt [4]{3} e}-\frac{\log \left (\frac{\sqrt{2-e x}-\sqrt{2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt{2+e x}}{\sqrt{2+e x}}\right )}{\sqrt{2} \sqrt [4]{3} e}+\frac{\log \left (\frac{\sqrt{2-e x}+\sqrt{2} \sqrt [4]{2-e x} \sqrt [4]{2+e x}+\sqrt{2+e x}}{\sqrt{2+e x}}\right )}{\sqrt{2} \sqrt [4]{3} e}\\ \end{align*}

Mathematica [C]  time = 0.050139, size = 60, normalized size = 0.25 \[ \frac{\sqrt{2} (e x-2) \sqrt [4]{e x+2} \, _2F_1\left (\frac{3}{4},\frac{3}{4};\frac{7}{4};\frac{1}{2}-\frac{e x}{4}\right )}{3 e \sqrt [4]{12-3 e^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[2 + e*x]*(12 - 3*e^2*x^2)^(1/4)),x]

[Out]

(Sqrt[2]*(-2 + e*x)*(2 + e*x)^(1/4)*Hypergeometric2F1[3/4, 3/4, 7/4, 1/2 - (e*x)/4])/(3*e*(12 - 3*e^2*x^2)^(1/
4))

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Maple [F]  time = 0.465, size = 0, normalized size = 0. \begin{align*} \int{{\frac{1}{\sqrt{ex+2}}}{\frac{1}{\sqrt [4]{-3\,{e}^{2}{x}^{2}+12}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+2)^(1/2)/(-3*e^2*x^2+12)^(1/4),x)

[Out]

int(1/(e*x+2)^(1/2)/(-3*e^2*x^2+12)^(1/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{1}{4}} \sqrt{e x + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(1/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((-3*e^2*x^2 + 12)^(1/4)*sqrt(e*x + 2)), x)

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Fricas [B]  time = 2.15836, size = 1654, normalized size = 6.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(1/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="fricas")

[Out]

2*sqrt(2)*(1/3)^(1/4)*(e^(-4))^(1/4)*arctan(-(sqrt(2)*(1/3)^(3/4)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e^3*(e
^(-4))^(3/4) + e^2*x^2 - sqrt(3)*sqrt(2)*(1/3)^(3/4)*(e^5*x^2 - 4*e^3)*sqrt((sqrt(2)*(1/3)^(1/4)*(-3*e^2*x^2 +
12)^(3/4)*sqrt(e*x + 2)*e*(e^(-4))^(1/4) + 3*sqrt(1/3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) - sqrt(-3*e^2*x^2 + 12)
*(e*x + 2))/(e^2*x^2 - 4))*(e^(-4))^(3/4) - 4)/(e^2*x^2 - 4)) + 2*sqrt(2)*(1/3)^(1/4)*(e^(-4))^(1/4)*arctan(-(
sqrt(2)*(1/3)^(3/4)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e^3*(e^(-4))^(3/4) - e^2*x^2 - sqrt(3)*sqrt(2)*(1/3)
^(3/4)*(e^5*x^2 - 4*e^3)*sqrt(-(sqrt(2)*(1/3)^(1/4)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e*(e^(-4))^(1/4) - 3
*sqrt(1/3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) + sqrt(-3*e^2*x^2 + 12)*(e*x + 2))/(e^2*x^2 - 4))*(e^(-4))^(3/4) + 4
)/(e^2*x^2 - 4)) - 1/2*sqrt(2)*(1/3)^(1/4)*(e^(-4))^(1/4)*log(3*(sqrt(2)*(1/3)^(1/4)*(-3*e^2*x^2 + 12)^(3/4)*s
qrt(e*x + 2)*e*(e^(-4))^(1/4) + 3*sqrt(1/3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) - sqrt(-3*e^2*x^2 + 12)*(e*x + 2))/
(e^2*x^2 - 4)) + 1/2*sqrt(2)*(1/3)^(1/4)*(e^(-4))^(1/4)*log(-3*(sqrt(2)*(1/3)^(1/4)*(-3*e^2*x^2 + 12)^(3/4)*sq
rt(e*x + 2)*e*(e^(-4))^(1/4) - 3*sqrt(1/3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) + sqrt(-3*e^2*x^2 + 12)*(e*x + 2))/(
e^2*x^2 - 4))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{3^{\frac{3}{4}} \int \frac{1}{\sqrt{e x + 2} \sqrt [4]{- e^{2} x^{2} + 4}}\, dx}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)**(1/2)/(-3*e**2*x**2+12)**(1/4),x)

[Out]

3**(3/4)*Integral(1/(sqrt(e*x + 2)*(-e**2*x**2 + 4)**(1/4)), x)/3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{1}{4}} \sqrt{e x + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+2)^(1/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((-3*e^2*x^2 + 12)^(1/4)*sqrt(e*x + 2)), x)