3.8.41 \(\int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx\)

Optimal. Leaf size=86 \[ \frac {\sqrt {3} \sqrt {2-e x}}{16 e (e x+2)}-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (e x+2)^2}+\frac {\sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{32 e} \]

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Rubi [A]  time = 0.03, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \begin {gather*} \frac {\sqrt {3} \sqrt {2-e x}}{16 e (e x+2)}-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (e x+2)^2}+\frac {\sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{32 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[12 - 3*e^2*x^2]/(2 + e*x)^(7/2),x]

[Out]

-(Sqrt[3]*Sqrt[2 - e*x])/(2*e*(2 + e*x)^2) + (Sqrt[3]*Sqrt[2 - e*x])/(16*e*(2 + e*x)) + (Sqrt[3]*ArcTanh[Sqrt[
2 - e*x]/2])/(32*e)

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])

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]))

Rubi steps

\begin {align*} \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx &=\int \frac {\sqrt {6-3 e x}}{(2+e x)^3} \, dx\\ &=-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (2+e x)^2}-\frac {3}{4} \int \frac {1}{\sqrt {6-3 e x} (2+e x)^2} \, dx\\ &=-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (2+e x)^2}+\frac {\sqrt {3} \sqrt {2-e x}}{16 e (2+e x)}-\frac {3}{32} \int \frac {1}{\sqrt {6-3 e x} (2+e x)} \, dx\\ &=-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (2+e x)^2}+\frac {\sqrt {3} \sqrt {2-e x}}{16 e (2+e x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{4-\frac {x^2}{3}} \, dx,x,\sqrt {6-3 e x}\right )}{16 e}\\ &=-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (2+e x)^2}+\frac {\sqrt {3} \sqrt {2-e x}}{16 e (2+e x)}+\frac {\sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{32 e}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 54, normalized size = 0.63 \begin {gather*} \frac {(e x-2) \sqrt {4-e^2 x^2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {1}{2}-\frac {e x}{4}\right )}{32 e \sqrt {3 e x+6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[12 - 3*e^2*x^2]/(2 + e*x)^(7/2),x]

[Out]

((-2 + e*x)*Sqrt[4 - e^2*x^2]*Hypergeometric2F1[3/2, 3, 5/2, 1/2 - (e*x)/4])/(32*e*Sqrt[6 + 3*e*x])

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IntegrateAlgebraic [A]  time = 0.37, size = 143, normalized size = 1.66 \begin {gather*} \frac {\sqrt {3} \sqrt {4 (e x+2)-(e x+2)^2} (e x-6)}{16 e (e x+2)^{5/2}}+\frac {\sqrt {3} \log \left (2 \sqrt {e x+2}+\sqrt {4 (e x+2)-(e x+2)^2}\right )}{64 e}-\frac {\sqrt {3} \log \left (e \sqrt {4 (e x+2)-(e x+2)^2}-2 e \sqrt {e x+2}\right )}{64 e} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[12 - 3*e^2*x^2]/(2 + e*x)^(7/2),x]

[Out]

(Sqrt[3]*(-6 + e*x)*Sqrt[4*(2 + e*x) - (2 + e*x)^2])/(16*e*(2 + e*x)^(5/2)) + (Sqrt[3]*Log[2*Sqrt[2 + e*x] + S
qrt[4*(2 + e*x) - (2 + e*x)^2]])/(64*e) - (Sqrt[3]*Log[-2*e*Sqrt[2 + e*x] + e*Sqrt[4*(2 + e*x) - (2 + e*x)^2]]
)/(64*e)

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fricas [B]  time = 0.41, size = 137, normalized size = 1.59 \begin {gather*} \frac {\sqrt {3} {\left (e^{3} x^{3} + 6 \, e^{2} x^{2} + 12 \, e x + 8\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 \, \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} {\left (e x - 6\right )}}{64 \, {\left (e^{4} x^{3} + 6 \, e^{3} x^{2} + 12 \, e^{2} x + 8 \, e\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x, algorithm="fricas")

[Out]

1/64*(sqrt(3)*(e^3*x^3 + 6*e^2*x^2 + 12*e*x + 8)*log(-(3*e^2*x^2 - 12*e*x - 4*sqrt(3)*sqrt(-3*e^2*x^2 + 12)*sq
rt(e*x + 2) - 36)/(e^2*x^2 + 4*e*x + 4)) + 4*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)*(e*x - 6))/(e^4*x^3 + 6*e^3*x
^2 + 12*e^2*x + 8*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((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),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 [A]  time = 0.10, size = 125, normalized size = 1.45 \begin {gather*} \frac {\sqrt {-e^{2} x^{2}+4}\, \left (\sqrt {3}\, e^{2} x^{2} \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )+4 \sqrt {3}\, e x \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )+2 \sqrt {-3 e x +6}\, e x +4 \sqrt {3}\, \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )-12 \sqrt {-3 e x +6}\right ) \sqrt {3}}{32 \sqrt {\left (e x +2\right )^{5}}\, \sqrt {-3 e x +6}\, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x)

[Out]

1/32*(-e^2*x^2+4)^(1/2)*(3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*x^2*e^2+4*arctanh(1/6*3^(1/2)*(-3*e*x+6
)^(1/2))*3^(1/2)*x*e+2*x*e*(-3*e*x+6)^(1/2)+4*3^(1/2)*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))-12*(-3*e*x+6)^(1/2
))*3^(1/2)/((e*x+2)^5)^(1/2)/(-3*e*x+6)^(1/2)/e

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {-3 \, e^{2} x^{2} + 12}}{{\left (e x + 2\right )}^{\frac {7}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-3*e^2*x^2 + 12)/(e*x + 2)^(7/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {12-3\,e^2\,x^2}}{{\left (e\,x+2\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12 - 3*e^2*x^2)^(1/2)/(e*x + 2)^(7/2),x)

[Out]

int((12 - 3*e^2*x^2)^(1/2)/(e*x + 2)^(7/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*e**2*x**2+12)**(1/2)/(e*x+2)**(7/2),x)

[Out]

Timed out

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