### 3.906 $$\int \frac{(12-3 e^2 x^2)^{3/2}}{(2+e x)^{5/2}} \, dx$$

Optimal. Leaf size=66 $\frac{2 \sqrt{3} (2-e x)^{3/2}}{e}+\frac{24 \sqrt{3} \sqrt{2-e x}}{e}-\frac{48 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{e}$

[Out]

(24*Sqrt[3]*Sqrt[2 - e*x])/e + (2*Sqrt[3]*(2 - e*x)^(3/2))/e - (48*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/e

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Rubi [A]  time = 0.028687, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {627, 50, 63, 206} $\frac{2 \sqrt{3} (2-e x)^{3/2}}{e}+\frac{24 \sqrt{3} \sqrt{2-e x}}{e}-\frac{48 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{e}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(24*Sqrt[3]*Sqrt[2 - e*x])/e + (2*Sqrt[3]*(2 - e*x)^(3/2))/e - (48*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/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 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 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{\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{5/2}} \, dx &=\int \frac{(6-3 e x)^{3/2}}{2+e x} \, dx\\ &=\frac{2 \sqrt{3} (2-e x)^{3/2}}{e}+12 \int \frac{\sqrt{6-3 e x}}{2+e x} \, dx\\ &=\frac{24 \sqrt{3} \sqrt{2-e x}}{e}+\frac{2 \sqrt{3} (2-e x)^{3/2}}{e}+144 \int \frac{1}{\sqrt{6-3 e x} (2+e x)} \, dx\\ &=\frac{24 \sqrt{3} \sqrt{2-e x}}{e}+\frac{2 \sqrt{3} (2-e x)^{3/2}}{e}-\frac{96 \operatorname{Subst}\left (\int \frac{1}{4-\frac{x^2}{3}} \, dx,x,\sqrt{6-3 e x}\right )}{e}\\ &=\frac{24 \sqrt{3} \sqrt{2-e x}}{e}+\frac{2 \sqrt{3} (2-e x)^{3/2}}{e}-\frac{48 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{e}\\ \end{align*}

Mathematica [A]  time = 0.0691722, size = 69, normalized size = 1.05 $-\frac{2 \sqrt{12-3 e^2 x^2} \left (\sqrt{e x-2} (e x-14)+24 \tan ^{-1}\left (\frac{1}{2} \sqrt{e x-2}\right )\right )}{e \sqrt{e x-2} \sqrt{e x+2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*Sqrt[12 - 3*e^2*x^2]*((-14 + e*x)*Sqrt[-2 + e*x] + 24*ArcTan[Sqrt[-2 + e*x]/2]))/(e*Sqrt[-2 + e*x]*Sqrt[2
+ e*x])

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Maple [A]  time = 0.124, size = 77, normalized size = 1.2 \begin{align*} -2\,{\frac{\sqrt{-{e}^{2}{x}^{2}+4} \left ( xe\sqrt{-3\,ex+6}+24\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) -14\,\sqrt{-3\,ex+6} \right ) \sqrt{3}}{\sqrt{ex+2}\sqrt{-3\,ex+6}e}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(-e^2*x^2+4)^(1/2)*(x*e*(-3*e*x+6)^(1/2)+24*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))-14*(-3*e*x+6)^(1/
2))*3^(1/2)/(e*x+2)^(1/2)/(-3*e*x+6)^(1/2)/e

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac{3}{2}}}{{\left (e x + 2\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.89295, size = 258, normalized size = 3.91 \begin{align*} \frac{2 \,{\left (12 \, \sqrt{3}{\left (e x + 2\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 ) - \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}{\left (e x - 14\right )}\right )}}{e^{2} x + 2 \, e} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(12*sqrt(3)*(e*x + 2)*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)) - sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)*(e*x - 14))/(e^2*x + 2*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((-3*e**2*x**2+12)**(3/2)/(e*x+2)**(5/2),x)

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError