3.10.38 \(\int \frac {(2+e x)^{3/2}}{\sqrt [4]{12-3 e^2 x^2}} \, dx\) [938]

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

[Out]

-5/6*3^(3/4)*(-e*x+2)^(3/4)*(e*x+2)^(1/4)/e-1/6*3^(3/4)*(-e*x+2)^(3/4)*(e*x+2)^(5/4)/e-5/6*3^(3/4)*arctan(-1+(
-e*x+2)^(1/4)*2^(1/2)/(e*x+2)^(1/4))/e*2^(1/2)-5/6*3^(3/4)*arctan(1+(-e*x+2)^(1/4)*2^(1/2)/(e*x+2)^(1/4))/e*2^
(1/2)-5/12*3^(3/4)*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)+5/12*3^(3/4)*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)

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Rubi [A]
time = 0.19, 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, 338, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {5 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} \sqrt [4]{3} e}-\frac {5 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} \sqrt [4]{3} e}-\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} \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 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 303

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 338

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 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 \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 \text {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 \text {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 \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 )}{3 e}-\frac {5 \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 )}{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 \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 {5 \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 {5 \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} \sqrt [4]{3} e}-\frac {5 \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} \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 \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} \sqrt [4]{3} e}+\frac {5 \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} \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*}

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Mathematica [A]
time = 0.51, size = 161, normalized size = 0.52 \begin {gather*} \frac {-\left ((7+e x) \left (4-e^2 x^2\right )^{3/4}\right )-5 \sqrt {4+2 e x} \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 )+5 \sqrt {4+2 e x} \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 )}{2 \sqrt [4]{3} e \sqrt {2+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+2)^(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)

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 568 vs. \(2 (232) = 464\).
time = 3.16, size = 568, normalized size = 1.84 \begin {gather*} \frac {60 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (x e^{2} + 2 \, e\right )} \arctan \left (\frac {\sqrt {3} \sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (x^{2} e^{5} - 4 \, e^{3}\right )} \sqrt {\frac {3 \, \sqrt {\frac {1}{3}} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} + \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}}{x^{2} e^{2} - 4}} e^{\left (-3\right )} - x^{2} e^{2} - \sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} + 4}{x^{2} e^{2} - 4}\right ) e^{\left (-1\right )} + 60 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (x e^{2} + 2 \, e\right )} \arctan \left (\frac {\sqrt {3} \sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (x^{2} e^{5} - 4 \, e^{3}\right )} \sqrt {\frac {3 \, \sqrt {\frac {1}{3}} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} - \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}}{x^{2} e^{2} - 4}} e^{\left (-3\right )} + x^{2} e^{2} - \sqrt {2} \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - 4}{x^{2} e^{2} - 4}\right ) e^{\left (-1\right )} - 15 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (x e^{2} + 2 \, e\right )} e^{\left (-1\right )} \log \left (\frac {3 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} + \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}\right )}}{x^{2} e^{2} - 4}\right ) + 15 \, \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (x e^{2} + 2 \, e\right )} e^{\left (-1\right )} \log \left (\frac {3 \, {\left (3 \, \sqrt {\frac {1}{3}} {\left (x^{2} e^{4} - 4 \, e^{2}\right )} e^{\left (-2\right )} - \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} \sqrt {x e + 2} - \sqrt {-3 \, x^{2} e^{2} + 12} {\left (x e + 2\right )}\right )}}{x^{2} e^{2} - 4}\right ) - 2 \, {\left (-3 \, x^{2} e^{2} + 12\right )}^{\frac {3}{4}} {\left (x e + 7\right )} \sqrt {x e + 2}}{12 \, {\left (x e^{2} + 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)*(x*e^2 + 2*e)*arctan((sqrt(3)*sqrt(2)*(1/3)^(3/4)*(x^2*e^5 - 4*e^3)*sqrt((3*sqrt(
1/3)*(x^2*e^4 - 4*e^2)*e^(-2) + sqrt(2)*(1/3)^(1/4)*(-3*x^2*e^2 + 12)^(3/4)*sqrt(x*e + 2) - sqrt(-3*x^2*e^2 +
12)*(x*e + 2))/(x^2*e^2 - 4))*e^(-3) - x^2*e^2 - sqrt(2)*(1/3)^(3/4)*(-3*x^2*e^2 + 12)^(3/4)*sqrt(x*e + 2) + 4
)/(x^2*e^2 - 4))*e^(-1) + 60*sqrt(2)*(1/3)^(1/4)*(x*e^2 + 2*e)*arctan((sqrt(3)*sqrt(2)*(1/3)^(3/4)*(x^2*e^5 -
4*e^3)*sqrt((3*sqrt(1/3)*(x^2*e^4 - 4*e^2)*e^(-2) - sqrt(2)*(1/3)^(1/4)*(-3*x^2*e^2 + 12)^(3/4)*sqrt(x*e + 2)
- sqrt(-3*x^2*e^2 + 12)*(x*e + 2))/(x^2*e^2 - 4))*e^(-3) + x^2*e^2 - sqrt(2)*(1/3)^(3/4)*(-3*x^2*e^2 + 12)^(3/
4)*sqrt(x*e + 2) - 4)/(x^2*e^2 - 4))*e^(-1) - 15*sqrt(2)*(1/3)^(1/4)*(x*e^2 + 2*e)*e^(-1)*log(3*(3*sqrt(1/3)*(
x^2*e^4 - 4*e^2)*e^(-2) + sqrt(2)*(1/3)^(1/4)*(-3*x^2*e^2 + 12)^(3/4)*sqrt(x*e + 2) - sqrt(-3*x^2*e^2 + 12)*(x
*e + 2))/(x^2*e^2 - 4)) + 15*sqrt(2)*(1/3)^(1/4)*(x*e^2 + 2*e)*e^(-1)*log(3*(3*sqrt(1/3)*(x^2*e^4 - 4*e^2)*e^(
-2) - sqrt(2)*(1/3)^(1/4)*(-3*x^2*e^2 + 12)^(3/4)*sqrt(x*e + 2) - sqrt(-3*x^2*e^2 + 12)*(x*e + 2))/(x^2*e^2 -
4)) - 2*(-3*x^2*e^2 + 12)^(3/4)*(x*e + 7)*sqrt(x*e + 2))/(x*e^2 + 2*e)

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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

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Giac [A]
time = 1.20, size = 193, normalized size = 0.62 \begin {gather*} -\frac {1}{24} \cdot 3^{\frac {3}{4}} {\left ({\left (x e + 2\right )}^{2} {\left (5 \, {\left (\frac {4}{x e + 2} - 1\right )}^{\frac {7}{4}} + 9 \, {\left (\frac {4}{x e + 2} - 1\right )}^{\frac {3}{4}}\right )} + 20 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, {\left (\frac {4}{x e + 2} - 1\right )}^{\frac {1}{4}}\right )}\right ) + 20 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, {\left (\frac {4}{x e + 2} - 1\right )}^{\frac {1}{4}}\right )}\right ) - 10 \, \sqrt {2} \log \left (\sqrt {2} {\left (\frac {4}{x e + 2} - 1\right )}^{\frac {1}{4}} + \sqrt {\frac {4}{x e + 2} - 1} + 1\right ) + 10 \, \sqrt {2} \log \left (-\sqrt {2} {\left (\frac {4}{x e + 2} - 1\right )}^{\frac {1}{4}} + \sqrt {\frac {4}{x e + 2} - 1} + 1\right )\right )} e^{\left (-1\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="giac")

[Out]

-1/24*3^(3/4)*((x*e + 2)^2*(5*(4/(x*e + 2) - 1)^(7/4) + 9*(4/(x*e + 2) - 1)^(3/4)) + 20*sqrt(2)*arctan(1/2*sqr
t(2)*(sqrt(2) + 2*(4/(x*e + 2) - 1)^(1/4))) + 20*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*(4/(x*e + 2) - 1)^(1
/4))) - 10*sqrt(2)*log(sqrt(2)*(4/(x*e + 2) - 1)^(1/4) + sqrt(4/(x*e + 2) - 1) + 1) + 10*sqrt(2)*log(-sqrt(2)*
(4/(x*e + 2) - 1)^(1/4) + sqrt(4/(x*e + 2) - 1) + 1))*e^(-1)

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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)

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