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

Optimal. Leaf size=309 \[ -\frac {(2-e x)^{3/4} (e x+2)^{5/4}}{2 \sqrt [4]{3} e}-\frac {5 (2-e x)^{3/4} \sqrt [4]{e x+2}}{2 \sqrt [4]{3} e}-\frac {5 \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}-\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} \sqrt [4]{3} e}+\frac {5 \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}+\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} \sqrt [4]{3} e}+\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} \sqrt [4]{3} e}-\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} \sqrt [4]{3} e} \]

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {675, 50, 63, 331, 297, 1162, 617, 204, 1165, 628} \begin {gather*} -\frac {(2-e x)^{3/4} (e x+2)^{5/4}}{2 \sqrt [4]{3} e}-\frac {5 (2-e x)^{3/4} \sqrt [4]{e x+2}}{2 \sqrt [4]{3} e}-\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}+\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}+\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} \sqrt [4]{3} e}-\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} \sqrt [4]{3} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(2 - e*x)^(3/4)*(2 + e*x)^(1/4))/(2*3^(1/4)*e) - ((2 - e*x)^(3/4)*(2 + e*x)^(5/4))/(2*3^(1/4)*e) + (5*ArcT
an[1 - (Sqrt[2]*(2 - e*x)^(1/4))/(2 + e*x)^(1/4)])/(Sqrt[2]*3^(1/4)*e) - (5*ArcTan[1 + (Sqrt[2]*(2 - e*x)^(1/4
))/(2 + e*x)^(1/4)])/(Sqrt[2]*3^(1/4)*e) - (5*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]])/(2*Sqrt[2]*3^(1/4)*e) + (5*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]])/(2*Sqrt[2]*3^(1/4)*e)

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 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 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 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 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 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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {(2+e x)^{3/2}}{\sqrt [4]{12-3 e^2 x^2}} \, dx &=\int \frac {(2+e x)^{5/4}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}+\frac {5}{2} \int \frac {\sqrt [4]{2+e x}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {5 (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 \sqrt [4]{3} e}-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}+\frac {5}{2} \int \frac {1}{\sqrt [4]{6-3 e x} (2+e x)^{3/4}} \, dx\\ &=-\frac {5 (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 \sqrt [4]{3} e}-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}-\frac {10 \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 {5 (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 \sqrt [4]{3} e}-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}-\frac {10 \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 {5 (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 \sqrt [4]{3} e}-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}+\frac {5 \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 {5 \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 {5 (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 \sqrt [4]{3} e}-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}-\frac {5 \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 )}{2 e}-\frac {5 \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 )}{2 e}-\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}-\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}\\ &=-\frac {5 (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 \sqrt [4]{3} e}-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}-\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}+\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}-\frac {5 \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 {2} \sqrt [4]{3} e}+\frac {5 \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 {2} \sqrt [4]{3} e}\\ &=-\frac {5 (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 \sqrt [4]{3} e}-\frac {(2-e x)^{3/4} (2+e x)^{5/4}}{2 \sqrt [4]{3} e}+\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} \sqrt [4]{3} e}-\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} \sqrt [4]{3} e}-\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}+\frac {5 \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 )}{2 \sqrt {2} \sqrt [4]{3} e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 60, normalized size = 0.19 \begin {gather*} \frac {16 \sqrt {2} (e x-2) \sqrt [4]{e x+2} \, _2F_1\left (-\frac {5}{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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(16*Sqrt[2]*(-2 + e*x)*(2 + e*x)^(1/4)*Hypergeometric2F1[-5/4, 3/4, 7/4, 1/2 - (e*x)/4])/(3*e*(12 - 3*e^2*x^2)
^(1/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.73, size = 215, normalized size = 0.70 \begin {gather*} -\frac {\left (4 (e x+2)-(e x+2)^2\right )^{3/4} (e x+7)}{2 \sqrt [4]{3} e \sqrt {e x+2}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {e x+2} \sqrt [4]{4 (e x+2)-(e x+2)^2}}{-e x+\sqrt {4 (e x+2)-(e x+2)^2}-2}\right )}{\sqrt {2} \sqrt [4]{3} e}+\frac {5 \tanh ^{-1}\left (\frac {\frac {e x+2}{\sqrt {2}}+\frac {\sqrt {4 (e x+2)-(e x+2)^2}}{\sqrt {2}}}{\sqrt {e x+2} \sqrt [4]{4 (e x+2)-(e x+2)^2}}\right )}{\sqrt {2} \sqrt [4]{3} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((7 + e*x)*(4*(2 + e*x) - (2 + e*x)^2)^(3/4))/(3^(1/4)*e*Sqrt[2 + e*x]) + (5*ArcTan[(Sqrt[2]*Sqrt[2 + e*x
]*(4*(2 + e*x) - (2 + e*x)^2)^(1/4))/(-2 - e*x + Sqrt[4*(2 + e*x) - (2 + e*x)^2])])/(Sqrt[2]*3^(1/4)*e) + (5*A
rcTanh[((2 + e*x)/Sqrt[2] + Sqrt[4*(2 + e*x) - (2 + e*x)^2]/Sqrt[2])/(Sqrt[2 + e*x]*(4*(2 + e*x) - (2 + e*x)^2
)^(1/4))])/(Sqrt[2]*3^(1/4)*e)

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 656, normalized size = 2.12 \begin {gather*} \frac {60 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e^{3} \frac {1}{e^{4}}^{\frac {3}{4}} + e^{2} x^{2} - \sqrt {3} \sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (e^{5} x^{2} - 4 \, e^{3}\right )} \sqrt {\frac {\sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} + 3 \, \sqrt {\frac {1}{3}} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} - \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}}{e^{2} x^{2} - 4}} \frac {1}{e^{4}}^{\frac {3}{4}} - 4}{e^{2} x^{2} - 4}\right ) + 60 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e^{3} \frac {1}{e^{4}}^{\frac {3}{4}} - e^{2} x^{2} - \sqrt {3} \sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (e^{5} x^{2} - 4 \, e^{3}\right )} \sqrt {-\frac {\sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} - 3 \, \sqrt {\frac {1}{3}} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} + \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}}{e^{2} x^{2} - 4}} \frac {1}{e^{4}}^{\frac {3}{4}} + 4}{e^{2} x^{2} - 4}\right ) - 15 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \log \left (\frac {3 \, {\left (\sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} + 3 \, \sqrt {\frac {1}{3}} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} - \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}\right )}}{e^{2} x^{2} - 4}\right ) + 15 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \log \left (-\frac {3 \, {\left (\sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e \frac {1}{e^{4}}^{\frac {1}{4}} - 3 \, \sqrt {\frac {1}{3}} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} + \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}\right )}}{e^{2} x^{2} - 4}\right ) - 2 \, {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} {\left (e x + 7\right )} \sqrt {e x + 2}}{12 \, {\left (e^{2} x + 2 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(60*sqrt(2)*(1/3)^(1/4)*(e^2*x + 2*e)*(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)) + 60*sqrt(2)*(1/3)^(1/4)*(e
^2*x + 2*e)*(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)) - 15*sqrt(2)*(1/3)^(1/4)*(e^2*x + 2*e)*(e^(-4))^(1/4)*log
(3*(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)) + 15*sqrt(2)*(1/3)^(1/4)*(e^2*x + 2*e)*(e^(-4)
)^(1/4)*log(-3*(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)) - 2*(-3*e^2*x^2 + 12)^(3/4)*(e*x +
7)*sqrt(e*x + 2))/(e^2*x + 2*e)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (e x + 2\right )}^{\frac {3}{2}}}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.12, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e x +2\right )^{\frac {3}{2}}}{\left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (e x + 2\right )}^{\frac {3}{2}}}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,x+2\right )}^{3/2}}{{\left (12-3\,e^2\,x^2\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {3^{\frac {3}{4}} \left (\int \frac {2 \sqrt {e x + 2}}{\sqrt [4]{- e^{2} x^{2} + 4}}\, dx + \int \frac {e x \sqrt {e x + 2}}{\sqrt [4]{- e^{2} x^{2} + 4}}\, dx\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________