3.9.100 \(\int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx\) [900]

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

[Out]

1/32*arctanh(1/2*(-e*x+2)^(1/2))*3^(1/2)/e-1/2*3^(1/2)*(-e*x+2)^(1/2)/e/(e*x+2)^2+1/16*3^(1/2)*(-e*x+2)^(1/2)/
e/(e*x+2)

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Rubi [A]
time = 0.02, 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 = {641, 43, 44, 65, 212} \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]

-1/2*(Sqrt[3]*Sqrt[2 - e*x])/(e*(2 + e*x)^2) + (Sqrt[3]*Sqrt[2 - e*x])/(16*e*(2 + e*x)) + (Sqrt[3]*ArcTanh[Sqr
t[2 - e*x]/2])/(32*e)

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 212

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

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 {\text {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 [A]
time = 0.29, size = 101, normalized size = 1.17 \begin {gather*} \frac {\sqrt {3} \left (\frac {4 (-6+e x) \sqrt {4-e^2 x^2}}{(2+e x)^{5/2}}-\log \left (e \left (-2 \sqrt {2+e x}+\sqrt {4-e^2 x^2}\right )\right )+\log \left (2 \sqrt {2+e x}+\sqrt {4-e^2 x^2}\right )\right )}{64 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*((4*(-6 + e*x)*Sqrt[4 - e^2*x^2])/(2 + e*x)^(5/2) - Log[e*(-2*Sqrt[2 + e*x] + Sqrt[4 - e^2*x^2])] + L
og[2*Sqrt[2 + e*x] + Sqrt[4 - e^2*x^2]]))/(64*e)

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Maple [A]
time = 0.49, size = 125, normalized size = 1.45

method	result	size
default	\(\frac {\sqrt {-e^{2} x^{2}+4}\, \left (\sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{2} x^{2}+4 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x +2 e x \sqrt {-3 e x +6}+4 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )-12 \sqrt {-3 e x +6}\right ) \sqrt {3}}{32 \sqrt {\left (e x +2\right )^{5}}\, \sqrt {-3 e x +6}\, e}\)	\(125\)

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,method=_RETURNVERBOSE)

[Out]

1/32*(-e^2*x^2+4)^(1/2)*(3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*e^2*x^2+4*3^(1/2)*arctanh(1/6*(-3*e*x+6
)^(1/2)*3^(1/2))*e*x+2*e*x*(-3*e*x+6)^(1/2)+4*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(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*} \text {Failed to integrate} \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*x^2*e^2 + 12)/(x*e + 2)^(7/2), x)

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Fricas [A]
time = 2.57, size = 135, normalized size = 1.57 \begin {gather*} \frac {\sqrt {3} {\left (x^{3} e^{3} + 6 \, x^{2} e^{2} + 12 \, x e + 8\right )} \log \left (-\frac {3 \, x^{2} e^{2} - 12 \, x e - 4 \, \sqrt {3} \sqrt {-3 \, x^{2} e^{2} + 12} \sqrt {x e + 2} - 36}{x^{2} e^{2} + 4 \, x e + 4}\right ) + 4 \, \sqrt {-3 \, x^{2} e^{2} + 12} \sqrt {x e + 2} {\left (x e - 6\right )}}{64 \, {\left (x^{3} e^{4} + 6 \, x^{2} e^{3} + 12 \, x e^{2} + 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)*(x^3*e^3 + 6*x^2*e^2 + 12*x*e + 8)*log(-(3*x^2*e^2 - 12*x*e - 4*sqrt(3)*sqrt(-3*x^2*e^2 + 12)*sq
rt(x*e + 2) - 36)/(x^2*e^2 + 4*x*e + 4)) + 4*sqrt(-3*x^2*e^2 + 12)*sqrt(x*e + 2)*(x*e - 6))/(x^3*e^4 + 6*x^2*e
^3 + 12*x*e^2 + 8*e)

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Sympy [F(-1)] Timed out
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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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,sageVARx):;OUTP
UT:Error index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument Value

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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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