3.925 $$\int \frac{1}{(2+e x)^{3/2} (12-3 e^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=108 $\frac{5}{256 \sqrt{3} e \sqrt{2-e x}}-\frac{5}{192 \sqrt{3} e \sqrt{2-e x} (e x+2)}-\frac{1}{24 \sqrt{3} e \sqrt{2-e x} (e x+2)^2}-\frac{5 \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{512 \sqrt{3} e}$

[Out]

5/(256*Sqrt[3]*e*Sqrt[2 - e*x]) - 1/(24*Sqrt[3]*e*Sqrt[2 - e*x]*(2 + e*x)^2) - 5/(192*Sqrt[3]*e*Sqrt[2 - e*x]*
(2 + e*x)) - (5*ArcTanh[Sqrt[2 - e*x]/2])/(512*Sqrt[3]*e)

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Rubi [A]  time = 0.0414894, antiderivative size = 115, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {627, 51, 63, 206} $-\frac{5 \sqrt{2-e x}}{256 \sqrt{3} e (e x+2)}-\frac{5 \sqrt{2-e x}}{96 \sqrt{3} e (e x+2)^2}+\frac{1}{6 \sqrt{3} e \sqrt{2-e x} (e x+2)^2}-\frac{5 \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{512 \sqrt{3} e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

1/(6*Sqrt[3]*e*Sqrt[2 - e*x]*(2 + e*x)^2) - (5*Sqrt[2 - e*x])/(96*Sqrt[3]*e*(2 + e*x)^2) - (5*Sqrt[2 - e*x])/(
256*Sqrt[3]*e*(2 + e*x)) - (5*ArcTanh[Sqrt[2 - e*x]/2])/(512*Sqrt[3]*e)

Rule 627

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
ntegerQ[m + p]))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 206

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

Rubi steps

\begin{align*} \int \frac{1}{(2+e x)^{3/2} \left (12-3 e^2 x^2\right )^{3/2}} \, dx &=\int \frac{1}{(6-3 e x)^{3/2} (2+e x)^3} \, dx\\ &=\frac{1}{6 \sqrt{3} e \sqrt{2-e x} (2+e x)^2}+\frac{5}{12} \int \frac{1}{\sqrt{6-3 e x} (2+e x)^3} \, dx\\ &=\frac{1}{6 \sqrt{3} e \sqrt{2-e x} (2+e x)^2}-\frac{5 \sqrt{2-e x}}{96 \sqrt{3} e (2+e x)^2}+\frac{5}{64} \int \frac{1}{\sqrt{6-3 e x} (2+e x)^2} \, dx\\ &=\frac{1}{6 \sqrt{3} e \sqrt{2-e x} (2+e x)^2}-\frac{5 \sqrt{2-e x}}{96 \sqrt{3} e (2+e x)^2}-\frac{5 \sqrt{2-e x}}{256 \sqrt{3} e (2+e x)}+\frac{5}{512} \int \frac{1}{\sqrt{6-3 e x} (2+e x)} \, dx\\ &=\frac{1}{6 \sqrt{3} e \sqrt{2-e x} (2+e x)^2}-\frac{5 \sqrt{2-e x}}{96 \sqrt{3} e (2+e x)^2}-\frac{5 \sqrt{2-e x}}{256 \sqrt{3} e (2+e x)}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{4-\frac{x^2}{3}} \, dx,x,\sqrt{6-3 e x}\right )}{768 e}\\ &=\frac{1}{6 \sqrt{3} e \sqrt{2-e x} (2+e x)^2}-\frac{5 \sqrt{2-e x}}{96 \sqrt{3} e (2+e x)^2}-\frac{5 \sqrt{2-e x}}{256 \sqrt{3} e (2+e x)}-\frac{5 \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{512 \sqrt{3} e}\\ \end{align*}

Mathematica [C]  time = 0.0501724, size = 48, normalized size = 0.44 $\frac{\sqrt{e x+2} \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{1}{2}-\frac{e x}{4}\right )}{96 e \sqrt{12-3 e^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Sqrt[2 + e*x]*Hypergeometric2F1[-1/2, 3, 1/2, 1/2 - (e*x)/4])/(96*e*Sqrt[12 - 3*e^2*x^2])

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Maple [A]  time = 0.059, size = 135, normalized size = 1.3 \begin{align*}{\frac{1}{ \left ( 4608\,ex-9216 \right ) e}\sqrt{-3\,{e}^{2}{x}^{2}+12} \left ( 5\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) \sqrt{-3\,ex+6}{x}^{2}{e}^{2}+20\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) \sqrt{-3\,ex+6}xe-30\,{e}^{2}{x}^{2}+20\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) \sqrt{-3\,ex+6}-80\,ex+24 \right ) \left ( ex+2 \right ) ^{-{\frac{5}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4608/(e*x+2)^(5/2)*(-3*e^2*x^2+12)^(1/2)*(5*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*(-3*e*x+6)^(1/2)*x
^2*e^2+20*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*(-3*e*x+6)^(1/2)*x*e-30*e^2*x^2+20*3^(1/2)*arctanh(1/6
*3^(1/2)*(-3*e*x+6)^(1/2))*(-3*e*x+6)^(1/2)-80*e*x+24)/(e*x-2)/e

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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{3}{2}}{\left (e x + 2\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.83962, size = 359, normalized size = 3.32 \begin{align*} \frac{15 \, \sqrt{3}{\left (e^{4} x^{4} + 4 \, e^{3} x^{3} - 16 \, e x - 16\right )} \log \left (-\frac{3 \, e^{2} x^{2} - 12 \, e x + 4 \, \sqrt{3} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 4 \,{\left (15 \, e^{2} x^{2} + 40 \, e x - 12\right )} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}}{9216 \,{\left (e^{5} x^{4} + 4 \, e^{4} x^{3} - 16 \, e^{2} x - 16 \, e\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9216*(15*sqrt(3)*(e^4*x^4 + 4*e^3*x^3 - 16*e*x - 16)*log(-(3*e^2*x^2 - 12*e*x + 4*sqrt(3)*sqrt(-3*e^2*x^2 +
12)*sqrt(e*x + 2) - 36)/(e^2*x^2 + 4*e*x + 4)) - 4*(15*e^2*x^2 + 40*e*x - 12)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x +
2))/(e^5*x^4 + 4*e^4*x^3 - 16*e^2*x - 16*e)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, \mathit{undef}, -2\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[undef, undef, undef, undef, undef, undef, undef, -2]