3.10.25 \(\int \frac {1}{(2+e x)^{3/2} (12-3 e^2 x^2)^{3/2}} \, dx\) [925]

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

[Out]

-5/1536*arctanh(1/2*(-e*x+2)^(1/2))*3^(1/2)/e+5/768/e*3^(1/2)/(-e*x+2)^(1/2)-1/72/e/(e*x+2)^2*3^(1/2)/(-e*x+2)
^(1/2)-5/576/e/(e*x+2)*3^(1/2)/(-e*x+2)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 108, normalized size of antiderivative = 1.00, 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 = {641, 44, 53, 65, 212} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[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)

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 53

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 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 {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 \text {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*}

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Mathematica [A]
time = 0.36, size = 87, normalized size = 0.81 \begin {gather*} \frac {-\frac {2 \sqrt {4-e^2 x^2} \left (-12+40 e x+15 e^2 x^2\right )}{(-2+e x) (2+e x)^{5/2}}-15 \tanh ^{-1}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )}{1536 \sqrt {3} e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[4 - e^2*x^2]*(-12 + 40*e*x + 15*e^2*x^2))/((-2 + e*x)*(2 + e*x)^(5/2)) - 15*ArcTanh[(2*Sqrt[2 + e*x]
)/Sqrt[4 - e^2*x^2]])/(1536*Sqrt[3]*e)

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Maple [A]
time = 0.54, size = 135, normalized size = 1.25

method	result	size
default	\(\frac {\sqrt {-3 e^{2} x^{2}+12}\, \left (5 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) \sqrt {-3 e x +6}\, e^{2} x^{2}+20 \sqrt {3}\, \sqrt {-3 e x +6}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x -30 e^{2} x^{2}+20 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) \sqrt {-3 e x +6}-80 e x +24\right )}{4608 \left (e x +2\right )^{\frac {5}{2}} \left (e x -2\right ) e}\)	\(135\)

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,method=_RETURNVERBOSE)

[Out]

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

[Out]

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

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Fricas [A]
time = 2.20, size = 144, normalized size = 1.33 \begin {gather*} \frac {15 \, \sqrt {3} {\left (x^{4} e^{4} + 4 \, x^{3} e^{3} - 16 \, x e - 16\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 \, {\left (15 \, x^{2} e^{2} + 40 \, x e - 12\right )} \sqrt {-3 \, x^{2} e^{2} + 12} \sqrt {x e + 2}}{9216 \, {\left (x^{4} e^{5} + 4 \, x^{3} e^{4} - 16 \, x e^{2} - 16 \, e\right )}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\sqrt {3} \int \frac {1}{- e^{3} x^{3} \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4} - 2 e^{2} x^{2} \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4} + 4 e x \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4} + 8 \sqrt {e x + 2} \sqrt {- e^{2} x^{2} + 4}}\, dx}{9} \end {gather*}

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]

sqrt(3)*Integral(1/(-e**3*x**3*sqrt(e*x + 2)*sqrt(-e**2*x**2 + 4) - 2*e**2*x**2*sqrt(e*x + 2)*sqrt(-e**2*x**2
+ 4) + 4*e*x*sqrt(e*x + 2)*sqrt(-e**2*x**2 + 4) + 8*sqrt(e*x + 2)*sqrt(-e**2*x**2 + 4)), x)/9

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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(1/(e*x+2)^(3/2)/(-3*e^2*x^2+12)^(3/2),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 {1}{{\left (12-3\,e^2\,x^2\right )}^{3/2}\,{\left (e\,x+2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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