3.8.78 \(\int \frac {(2+e x)^{5/2}}{\sqrt [4]{12-3 e^2 x^2}} \, dx\)

Optimal. Leaf size=340 \[ -\frac {(2-e x)^{3/4} (e x+2)^{9/4}}{3 \sqrt [4]{3} e}-\frac {3^{3/4} (2-e x)^{3/4} (e x+2)^{5/4}}{2 e}-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{e x+2}}{2 e}-\frac {5\ 3^{3/4} \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}-\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} e}+\frac {5\ 3^{3/4} \log \left (\frac {\sqrt {6-3 e x}+\sqrt {3} \sqrt {e x+2}+\sqrt {6} \sqrt [4]{2-e x} \sqrt [4]{e x+2}}{\sqrt {e x+2}}\right )}{2 \sqrt {2} e}+\frac {5\ 3^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} e} \]

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Rubi [A]  time = 0.30, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 15, 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)^{9/4}}{3 \sqrt [4]{3} e}-\frac {3^{3/4} (2-e x)^{3/4} (e x+2)^{5/4}}{2 e}-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{e x+2}}{2 e}-\frac {5\ 3^{3/4} \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 {5\ 3^{3/4} \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 {5\ 3^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{e x+2}}+1\right )}{\sqrt {2} e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + e*x)^(5/2)/(12 - 3*e^2*x^2)^(1/4),x]

[Out]

(-5*3^(3/4)*(2 - e*x)^(3/4)*(2 + e*x)^(1/4))/(2*e) - (3^(3/4)*(2 - e*x)^(3/4)*(2 + e*x)^(5/4))/(2*e) - ((2 - e
*x)^(3/4)*(2 + e*x)^(9/4))/(3*3^(1/4)*e) + (5*3^(3/4)*ArcTan[1 - (Sqrt[2]*(2 - e*x)^(1/4))/(2 + e*x)^(1/4)])/(
Sqrt[2]*e) - (5*3^(3/4)*ArcTan[1 + (Sqrt[2]*(2 - e*x)^(1/4))/(2 + e*x)^(1/4)])/(Sqrt[2]*e) - (5*3^(3/4)*Log[(S
qrt[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
) + (5*3^(3/4)*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 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)^{5/2}}{\sqrt [4]{12-3 e^2 x^2}} \, dx &=\int \frac {(2+e x)^{9/4}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+3 \int \frac {(2+e x)^{5/4}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {15}{2} \int \frac {\sqrt [4]{2+e x}}{\sqrt [4]{6-3 e x}} \, dx\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {15}{2} \int \frac {1}{\sqrt [4]{6-3 e x} (2+e x)^{3/4}} \, dx\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \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 )}{e}\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \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 )}{e}\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \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 )}{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 )}{e}\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}-\frac {15 \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 {15 \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 {\left (5\ 3^{3/4}\right ) \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} e}-\frac {\left (5\ 3^{3/4}\right ) \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} e}\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}-\frac {5\ 3^{3/4} \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 {5\ 3^{3/4} \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 (5\ 3^{3/4}\right ) \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} e}+\frac {\left (5\ 3^{3/4}\right ) \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} e}\\ &=-\frac {5\ 3^{3/4} (2-e x)^{3/4} \sqrt [4]{2+e x}}{2 e}-\frac {3^{3/4} (2-e x)^{3/4} (2+e x)^{5/4}}{2 e}-\frac {(2-e x)^{3/4} (2+e x)^{9/4}}{3 \sqrt [4]{3} e}+\frac {5\ 3^{3/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{2-e x}}{\sqrt [4]{2+e x}}\right )}{\sqrt {2} e}-\frac {5\ 3^{3/4} \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 {5\ 3^{3/4} \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*}

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Mathematica [C]  time = 0.07, size = 60, normalized size = 0.18 \begin {gather*} \frac {64 \sqrt {2} (e x-2) \sqrt [4]{e x+2} \, _2F_1\left (-\frac {9}{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)^(5/2)/(12 - 3*e^2*x^2)^(1/4),x]

[Out]

(64*Sqrt[2]*(-2 + e*x)*(2 + e*x)^(1/4)*Hypergeometric2F1[-9/4, 3/4, 7/4, 1/2 - (e*x)/4])/(3*e*(12 - 3*e^2*x^2)
^(1/4))

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IntegrateAlgebraic [A]  time = 0.77, size = 228, normalized size = 0.67 \begin {gather*} -\frac {\left (4 (e x+2)-(e x+2)^2\right )^{3/4} \left (2 (e x+2)^2+9 (e x+2)+45\right )}{6 \sqrt [4]{3} e \sqrt {e x+2}}+\frac {5\ 3^{3/4} \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} e}+\frac {5\ 3^{3/4} \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} e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 + e*x)^(5/2)/(12 - 3*e^2*x^2)^(1/4),x]

[Out]

-1/6*((4*(2 + e*x) - (2 + e*x)^2)^(3/4)*(45 + 9*(2 + e*x) + 2*(2 + e*x)^2))/(3^(1/4)*e*Sqrt[2 + e*x]) + (5*3^(
3/4)*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]*e) + (5*3^(3/4)*ArcTanh[((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]*e)

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fricas [B]  time = 0.46, size = 661, normalized size = 1.94 \begin {gather*} \frac {180 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \arctan \left (-\frac {27^{\frac {3}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e^{3} \frac {1}{e^{4}}^{\frac {3}{4}} + 27 \, e^{2} x^{2} - 27^{\frac {3}{4}} \sqrt {2} {\left (e^{5} x^{2} - 4 \, e^{3}\right )} \sqrt {\frac {27^{\frac {1}{4}} \sqrt {2} {\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 {3} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} - 3 \, \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}}{e^{2} x^{2} - 4}} \frac {1}{e^{4}}^{\frac {3}{4}} - 108}{27 \, {\left (e^{2} x^{2} - 4\right )}}\right ) + 180 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \arctan \left (-\frac {27^{\frac {3}{4}} \sqrt {2} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2} e^{3} \frac {1}{e^{4}}^{\frac {3}{4}} - 27 \, e^{2} x^{2} - 27^{\frac {3}{4}} \sqrt {2} {\left (e^{5} x^{2} - 4 \, e^{3}\right )} \sqrt {-\frac {27^{\frac {1}{4}} \sqrt {2} {\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 {3} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} + 3 \, \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}}{e^{2} x^{2} - 4}} \frac {1}{e^{4}}^{\frac {3}{4}} + 108}{27 \, {\left (e^{2} x^{2} - 4\right )}}\right ) - 45 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \log \left (\frac {27^{\frac {1}{4}} \sqrt {2} {\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 {3} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} - 3 \, \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}}{e^{2} x^{2} - 4}\right ) + 45 \cdot 27^{\frac {1}{4}} \sqrt {2} {\left (e^{2} x + 2 \, e\right )} \frac {1}{e^{4}}^{\frac {1}{4}} \log \left (-\frac {27^{\frac {1}{4}} \sqrt {2} {\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 {3} {\left (e^{4} x^{2} - 4 \, e^{2}\right )} \sqrt {\frac {1}{e^{4}}} + 3 \, \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 2\right )}}{e^{2} x^{2} - 4}\right ) - 2 \, {\left (2 \, e^{2} x^{2} + 17 \, e x + 71\right )} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2}}{36 \, {\left (e^{2} x + 2 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+2)^(5/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="fricas")

[Out]

1/36*(180*27^(1/4)*sqrt(2)*(e^2*x + 2*e)*(e^(-4))^(1/4)*arctan(-1/27*(27^(3/4)*sqrt(2)*(-3*e^2*x^2 + 12)^(3/4)
*sqrt(e*x + 2)*e^3*(e^(-4))^(3/4) + 27*e^2*x^2 - 27^(3/4)*sqrt(2)*(e^5*x^2 - 4*e^3)*sqrt((27^(1/4)*sqrt(2)*(-3
*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e*(e^(-4))^(1/4) + 3*sqrt(3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) - 3*sqrt(-3*e^2
*x^2 + 12)*(e*x + 2))/(e^2*x^2 - 4))*(e^(-4))^(3/4) - 108)/(e^2*x^2 - 4)) + 180*27^(1/4)*sqrt(2)*(e^2*x + 2*e)
*(e^(-4))^(1/4)*arctan(-1/27*(27^(3/4)*sqrt(2)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e^3*(e^(-4))^(3/4) - 27*e
^2*x^2 - 27^(3/4)*sqrt(2)*(e^5*x^2 - 4*e^3)*sqrt(-(27^(1/4)*sqrt(2)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e*(e
^(-4))^(1/4) - 3*sqrt(3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) + 3*sqrt(-3*e^2*x^2 + 12)*(e*x + 2))/(e^2*x^2 - 4))*(e
^(-4))^(3/4) + 108)/(e^2*x^2 - 4)) - 45*27^(1/4)*sqrt(2)*(e^2*x + 2*e)*(e^(-4))^(1/4)*log((27^(1/4)*sqrt(2)*(-
3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e*(e^(-4))^(1/4) + 3*sqrt(3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) - 3*sqrt(-3*e^
2*x^2 + 12)*(e*x + 2))/(e^2*x^2 - 4)) + 45*27^(1/4)*sqrt(2)*(e^2*x + 2*e)*(e^(-4))^(1/4)*log(-(27^(1/4)*sqrt(2
)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2)*e*(e^(-4))^(1/4) - 3*sqrt(3)*(e^4*x^2 - 4*e^2)*sqrt(e^(-4)) + 3*sqrt(-
3*e^2*x^2 + 12)*(e*x + 2))/(e^2*x^2 - 4)) - 2*(2*e^2*x^2 + 17*e*x + 71)*(-3*e^2*x^2 + 12)^(3/4)*sqrt(e*x + 2))
/(e^2*x + 2*e)

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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)^(5/2)/(-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,x):;OUTPUT:Erro
r index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument Value

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e x +2\right )^{\frac {5}{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)^(5/2)/(-3*e^2*x^2+12)^(1/4),x)

[Out]

int((e*x+2)^(5/2)/(-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*} \int \frac {{\left (e x + 2\right )}^{\frac {5}{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)^(5/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="maxima")

[Out]

integrate((e*x + 2)^(5/2)/(-3*e^2*x^2 + 12)^(1/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (e\,x+2\right )}^{5/2}}{{\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)^(5/2)/(12 - 3*e^2*x^2)^(1/4),x)

[Out]

int((e*x + 2)^(5/2)/(12 - 3*e^2*x^2)^(1/4), x)

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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 {4 \sqrt {e x + 2}}{\sqrt [4]{- e^{2} x^{2} + 4}}\, dx + \int \frac {4 e x \sqrt {e x + 2}}{\sqrt [4]{- e^{2} x^{2} + 4}}\, dx + \int \frac {e^{2} x^{2} \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)**(5/2)/(-3*e**2*x**2+12)**(1/4),x)

[Out]

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

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