3.909 \(\int \frac{(12-3 e^2 x^2)^{3/2}}{(2+e x)^{11/2}} \, dx\)

Optimal. Leaf size=113 \[ -\frac{\sqrt{3} (2-e x)^{3/2}}{e (e x+2)^3}-\frac{3 \sqrt{3} \sqrt{2-e x}}{32 e (e x+2)}+\frac{3 \sqrt{3} \sqrt{2-e x}}{4 e (e x+2)^2}-\frac{3 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{64 e} \]

[Out]

-((Sqrt[3]*(2 - e*x)^(3/2))/(e*(2 + e*x)^3)) + (3*Sqrt[3]*Sqrt[2 - e*x])/(4*e*(2 + e*x)^2) - (3*Sqrt[3]*Sqrt[2
- e*x])/(32*e*(2 + e*x)) - (3*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/(64*e)

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Rubi [A]  time = 0.0389677, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {627, 47, 51, 63, 206} \[ -\frac{\sqrt{3} (2-e x)^{3/2}}{e (e x+2)^3}-\frac{3 \sqrt{3} \sqrt{2-e x}}{32 e (e x+2)}+\frac{3 \sqrt{3} \sqrt{2-e x}}{4 e (e x+2)^2}-\frac{3 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{64 e} \]

Antiderivative was successfully verified.

[In]

Int[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(11/2),x]

[Out]

-((Sqrt[3]*(2 - e*x)^(3/2))/(e*(2 + e*x)^3)) + (3*Sqrt[3]*Sqrt[2 - e*x])/(4*e*(2 + e*x)^2) - (3*Sqrt[3]*Sqrt[2
- e*x])/(32*e*(2 + e*x)) - (3*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2])/(64*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 47

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

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{\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx &=\int \frac{(6-3 e x)^{3/2}}{(2+e x)^4} \, dx\\ &=-\frac{\sqrt{3} (2-e x)^{3/2}}{e (2+e x)^3}-\frac{3}{2} \int \frac{\sqrt{6-3 e x}}{(2+e x)^3} \, dx\\ &=-\frac{\sqrt{3} (2-e x)^{3/2}}{e (2+e x)^3}+\frac{3 \sqrt{3} \sqrt{2-e x}}{4 e (2+e x)^2}+\frac{9}{8} \int \frac{1}{\sqrt{6-3 e x} (2+e x)^2} \, dx\\ &=-\frac{\sqrt{3} (2-e x)^{3/2}}{e (2+e x)^3}+\frac{3 \sqrt{3} \sqrt{2-e x}}{4 e (2+e x)^2}-\frac{3 \sqrt{3} \sqrt{2-e x}}{32 e (2+e x)}+\frac{9}{64} \int \frac{1}{\sqrt{6-3 e x} (2+e x)} \, dx\\ &=-\frac{\sqrt{3} (2-e x)^{3/2}}{e (2+e x)^3}+\frac{3 \sqrt{3} \sqrt{2-e x}}{4 e (2+e x)^2}-\frac{3 \sqrt{3} \sqrt{2-e x}}{32 e (2+e x)}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{4-\frac{x^2}{3}} \, dx,x,\sqrt{6-3 e x}\right )}{32 e}\\ &=-\frac{\sqrt{3} (2-e x)^{3/2}}{e (2+e x)^3}+\frac{3 \sqrt{3} \sqrt{2-e x}}{4 e (2+e x)^2}-\frac{3 \sqrt{3} \sqrt{2-e x}}{32 e (2+e x)}-\frac{3 \sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{2-e x}\right )}{64 e}\\ \end{align*}

Mathematica [C]  time = 0.0846051, size = 55, normalized size = 0.49 \[ -\frac{3 (e x-2)^2 \sqrt{12-3 e^2 x^2} \, _2F_1\left (\frac{5}{2},4;\frac{7}{2};\frac{1}{2}-\frac{e x}{4}\right )}{640 e \sqrt{e x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(11/2),x]

[Out]

(-3*(-2 + e*x)^2*Sqrt[12 - 3*e^2*x^2]*Hypergeometric2F1[5/2, 4, 7/2, 1/2 - (e*x)/4])/(640*e*Sqrt[2 + e*x])

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Maple [A]  time = 0.154, size = 167, normalized size = 1.5 \begin{align*} -{\frac{\sqrt{3}}{64\,e}\sqrt{-{e}^{2}{x}^{2}+4} \left ( 3\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ){x}^{3}{e}^{3}+18\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ){x}^{2}{e}^{2}+6\,{x}^{2}{e}^{2}\sqrt{-3\,ex+6}+36\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) xe-88\,xe\sqrt{-3\,ex+6}+24\,\sqrt{3}{\it Artanh} \left ( 1/6\,\sqrt{3}\sqrt{-3\,ex+6} \right ) +56\,\sqrt{-3\,ex+6} \right ){\frac{1}{\sqrt{ \left ( ex+2 \right ) ^{7}}}}{\frac{1}{\sqrt{-3\,ex+6}}}} \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)^(11/2),x)

[Out]

-1/64*(-e^2*x^2+4)^(1/2)*(3*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*x^3*e^3+18*3^(1/2)*arctanh(1/6*3^(1/
2)*(-3*e*x+6)^(1/2))*x^2*e^2+6*x^2*e^2*(-3*e*x+6)^(1/2)+36*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*x*e-8
8*x*e*(-3*e*x+6)^(1/2)+24*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))+56*(-3*e*x+6)^(1/2))*3^(1/2)/((e*x+2)^
7)^(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{11}{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)^(11/2),x, algorithm="maxima")

[Out]

integrate((-3*e^2*x^2 + 12)^(3/2)/(e*x + 2)^(11/2), x)

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Fricas [A]  time = 1.85877, size = 390, normalized size = 3.45 \begin{align*} \frac{3 \, \sqrt{3}{\left (e^{4} x^{4} + 8 \, e^{3} x^{3} + 24 \, e^{2} x^{2} + 32 \, 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 (3 \, e^{2} x^{2} - 44 \, e x + 28\right )} \sqrt{-3 \, e^{2} x^{2} + 12} \sqrt{e x + 2}}{128 \,{\left (e^{5} x^{4} + 8 \, e^{4} x^{3} + 24 \, e^{3} x^{2} + 32 \, e^{2} x + 16 \, e\right )}} \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)^(11/2),x, algorithm="fricas")

[Out]

1/128*(3*sqrt(3)*(e^4*x^4 + 8*e^3*x^3 + 24*e^2*x^2 + 32*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*(3*e^2*x^2 - 44*e*x + 28)*sqrt(-3*e^2*x^2 + 12)*
sqrt(e*x + 2))/(e^5*x^4 + 8*e^4*x^3 + 24*e^3*x^2 + 32*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((-3*e**2*x**2+12)**(3/2)/(e*x+2)**(11/2),x)

[Out]

Timed out

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Giac [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{11}{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)^(11/2),x, algorithm="giac")

[Out]

integrate((-3*e^2*x^2 + 12)^(3/2)/(e*x + 2)^(11/2), x)