3.10.30 \(\int \sqrt {2+e x} \sqrt [4]{12-3 e^2 x^2} \, dx\) [930]

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

[Out]

3/2*3^(1/4)*(-e*x+2)^(1/4)*(e*x+2)^(3/4)/e-1/2*3^(1/4)*(-e*x+2)^(5/4)*(e*x+2)^(3/4)/e-3/2*3^(1/4)*arctan(-1+(-
e*x+2)^(1/4)*2^(1/2)/(e*x+2)^(1/4))/e*2^(1/2)-3/2*3^(1/4)*arctan(1+(-e*x+2)^(1/4)*2^(1/2)/(e*x+2)^(1/4))/e*2^(
1/2)+3/4*3^(1/4)*ln(3^(1/2)-(-e*x+2)^(1/4)*6^(1/2)/(e*x+2)^(1/4)+3^(1/2)*(-e*x+2)^(1/2)/(e*x+2)^(1/2))/e*2^(1/
2)-3/4*3^(1/4)*ln(3^(1/2)+(-e*x+2)^(1/4)*6^(1/2)/(e*x+2)^(1/4)+3^(1/2)*(-e*x+2)^(1/2)/(e*x+2)^(1/2))/e*2^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.21, 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 = {689, 52, 65, 246, 217, 1179, 642, 1176, 631, 210} \begin {gather*} \frac {3 \sqrt [4]{3} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} e}-\frac {3 \sqrt [4]{3} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} e}-\frac {\sqrt [4]{3} (e x+2)^{3/4} (2-e x)^{5/4}}{2 e}+\frac {3 \sqrt [4]{3} (e x+2)^{3/4} \sqrt [4]{2-e x}}{2 e}+\frac {3 \sqrt [4]{3} \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} e}-\frac {3 \sqrt [4]{3} \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} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 52

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 65

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 210

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

Rule 217

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

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p
+ 1/n]

Rule 631

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 642

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 689

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^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 1176

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 1179

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 \sqrt {2+e x} \sqrt [4]{12-3 e^2 x^2} \, dx &=\int \sqrt [4]{6-3 e x} (2+e x)^{3/4} \, dx\\ &=-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {3}{2} \int \frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}} \, dx\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {9}{2} \int \frac {1}{(6-3 e x)^{3/4} \sqrt [4]{2+e x}} \, dx\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}-\frac {6 \text {Subst}\left (\int \frac {1}{\sqrt [4]{4-\frac {x^4}{3}}} \, dx,x,\sqrt [4]{6-3 e x}\right )}{e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}-\frac {6 \text {Subst}\left (\int \frac {1}{1+\frac {x^4}{3}} \, dx,x,\frac {\sqrt [4]{6-3 e x}}{\sqrt [4]{2+e x}}\right )}{e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}-\frac {\sqrt {3} \text {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 )}{e}-\frac {\sqrt {3} \text {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 )}{e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {\left (3 \sqrt [4]{3}\right ) \text {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} e}+\frac {\left (3 \sqrt [4]{3}\right ) \text {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} e}-\frac {\left (3 \sqrt {3}\right ) \text {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 {\left (3 \sqrt {3}\right ) \text {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 {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {3 \sqrt [4]{3} \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} e}-\frac {3 \sqrt [4]{3} \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} e}-\frac {\left (3 \sqrt [4]{3}\right ) \text {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} e}+\frac {\left (3 \sqrt [4]{3}\right ) \text {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} e}\\ &=\frac {3 \sqrt [4]{3} \sqrt [4]{2-e x} (2+e x)^{3/4}}{2 e}-\frac {\sqrt [4]{3} (2-e x)^{5/4} (2+e x)^{3/4}}{2 e}+\frac {3 \sqrt [4]{3} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {3 \sqrt [4]{3} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}+\frac {3 \sqrt [4]{3} \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} e}-\frac {3 \sqrt [4]{3} \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} e}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.51, size = 150, normalized size = 0.49 \begin {gather*} \frac {\sqrt [4]{3} \left ((1+e x) \sqrt {2+e x} \sqrt [4]{4-e^2 x^2}-3 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {4+2 e x} \sqrt [4]{4-e^2 x^2}}{2+e x-\sqrt {4-e^2 x^2}}\right )-3 \sqrt {2} \tanh ^{-1}\left (\frac {2+e x+\sqrt {4-e^2 x^2}}{\sqrt {4+2 e x} \sqrt [4]{4-e^2 x^2}}\right )\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \sqrt {e x +2}\, \left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 3.40, size = 446, normalized size = 1.44 \begin {gather*} \frac {1}{4} \, {\left (2 \, {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} {\left (x e + 1\right )} + 12 \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {3^{\frac {3}{4}} \sqrt {2} {\left (x e^{4} + 2 \, e^{3}\right )} \sqrt {\frac {\sqrt {3} {\left (x e^{3} + 2 \, e^{2}\right )} e^{\left (-2\right )} + 3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} + \sqrt {-3 \, x^{2} e^{2} + 12}}{x e + 2}} e^{\left (-3\right )} - 3^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} - 3 \, x e - 6}{3 \, {\left (x e + 2\right )}}\right ) + 12 \cdot 3^{\frac {1}{4}} \sqrt {2} \arctan \left (\frac {3^{\frac {3}{4}} \sqrt {2} {\left (x e^{4} + 2 \, e^{3}\right )} \sqrt {\frac {\sqrt {3} {\left (x e^{3} + 2 \, e^{2}\right )} e^{\left (-2\right )} - 3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} + \sqrt {-3 \, x^{2} e^{2} + 12}}{x e + 2}} e^{\left (-3\right )} - 3^{\frac {3}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} + 3 \, x e + 6}{3 \, {\left (x e + 2\right )}}\right ) - 3 \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (\frac {\sqrt {3} {\left (x e^{3} + 2 \, e^{2}\right )} e^{\left (-2\right )} + 3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} + \sqrt {-3 \, x^{2} e^{2} + 12}}{x e + 2}\right ) + 3 \cdot 3^{\frac {1}{4}} \sqrt {2} \log \left (\frac {\sqrt {3} {\left (x e^{3} + 2 \, e^{2}\right )} e^{\left (-2\right )} - 3^{\frac {1}{4}} \sqrt {2} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {1}{4}} \sqrt {x e + 2} + \sqrt {-3 \, x^{2} e^{2} + 12}}{x e + 2}\right )\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(-3*x^2*e^2 + 12)^(1/4)*sqrt(x*e + 2)*(x*e + 1) + 12*3^(1/4)*sqrt(2)*arctan(1/3*(3^(3/4)*sqrt(2)*(x*e^4
+ 2*e^3)*sqrt((sqrt(3)*(x*e^3 + 2*e^2)*e^(-2) + 3^(1/4)*sqrt(2)*(-3*x^2*e^2 + 12)^(1/4)*sqrt(x*e + 2) + sqrt(
-3*x^2*e^2 + 12))/(x*e + 2))*e^(-3) - 3^(3/4)*sqrt(2)*(-3*x^2*e^2 + 12)^(1/4)*sqrt(x*e + 2) - 3*x*e - 6)/(x*e
+ 2)) + 12*3^(1/4)*sqrt(2)*arctan(1/3*(3^(3/4)*sqrt(2)*(x*e^4 + 2*e^3)*sqrt((sqrt(3)*(x*e^3 + 2*e^2)*e^(-2) -
3^(1/4)*sqrt(2)*(-3*x^2*e^2 + 12)^(1/4)*sqrt(x*e + 2) + sqrt(-3*x^2*e^2 + 12))/(x*e + 2))*e^(-3) - 3^(3/4)*sqr
t(2)*(-3*x^2*e^2 + 12)^(1/4)*sqrt(x*e + 2) + 3*x*e + 6)/(x*e + 2)) - 3*3^(1/4)*sqrt(2)*log((sqrt(3)*(x*e^3 + 2
*e^2)*e^(-2) + 3^(1/4)*sqrt(2)*(-3*x^2*e^2 + 12)^(1/4)*sqrt(x*e + 2) + sqrt(-3*x^2*e^2 + 12))/(x*e + 2)) + 3*3
^(1/4)*sqrt(2)*log((sqrt(3)*(x*e^3 + 2*e^2)*e^(-2) - 3^(1/4)*sqrt(2)*(-3*x^2*e^2 + 12)^(1/4)*sqrt(x*e + 2) + s
qrt(-3*x^2*e^2 + 12))/(x*e + 2)))*e^(-1)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument Value

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________