3.10.9 \(\int \frac {(12-3 e^2 x^2)^{3/2}}{(2+e x)^{11/2}} \, dx\) [909]

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

[Out]

-(-e*x+2)^(3/2)*3^(1/2)/e/(e*x+2)^3-3/64*arctanh(1/2*(-e*x+2)^(1/2))*3^(1/2)/e+3/4*3^(1/2)*(-e*x+2)^(1/2)/e/(e
*x+2)^2-3/32*3^(1/2)*(-e*x+2)^(1/2)/e/(e*x+2)

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Rubi [A]
time = 0.03, antiderivative size = 113, 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, 43, 44, 65, 212} \begin {gather*} -\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} \end {gather*}

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 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 {\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 \text {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*}

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Mathematica [A]
time = 0.32, size = 80, normalized size = 0.71 \begin {gather*} \frac {\sqrt {3} \left (-\frac {2 \sqrt {4-e^2 x^2} \left (28-44 e x+3 e^2 x^2\right )}{(2+e x)^{7/2}}-3 \tanh ^{-1}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )\right )}{64 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*((-2*Sqrt[4 - e^2*x^2]*(28 - 44*e*x + 3*e^2*x^2))/(2 + e*x)^(7/2) - 3*ArcTanh[(2*Sqrt[2 + e*x])/Sqrt[
4 - e^2*x^2]]))/(64*e)

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Maple [A]
time = 0.48, size = 167, normalized size = 1.48

method	result	size
default	\(-\frac {\sqrt {-e^{2} x^{2}+4}\, \left (3 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{3} x^{3}+18 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{2} x^{2}+6 e^{2} x^{2} \sqrt {-3 e x +6}+36 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x -88 e x \sqrt {-3 e x +6}+24 \sqrt {3}\, \arctanh \left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )+56 \sqrt {-3 e x +6}\right ) \sqrt {3}}{64 \sqrt {\left (e x +2\right )^{7}}\, \sqrt {-3 e x +6}\, e}\)	\(167\)

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,method=_RETURNVERBOSE)

[Out]

-1/64*(-e^2*x^2+4)^(1/2)*(3*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*e^3*x^3+18*3^(1/2)*arctanh(1/6*(-3*e
*x+6)^(1/2)*3^(1/2))*e^2*x^2+6*e^2*x^2*(-3*e*x+6)^(1/2)+36*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*e*x-8
8*e*x*(-3*e*x+6)^(1/2)+24*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(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 = 0.00, 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)^(3/2)/(e*x+2)^(11/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 4.92, size = 158, normalized size = 1.40 \begin {gather*} \frac {3 \, \sqrt {3} {\left (x^{4} e^{4} + 8 \, x^{3} e^{3} + 24 \, x^{2} e^{2} + 32 \, 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 (3 \, x^{2} e^{2} - 44 \, x e + 28\right )} \sqrt {-3 \, x^{2} e^{2} + 12} \sqrt {x e + 2}}{128 \, {\left (x^{4} e^{5} + 8 \, x^{3} e^{4} + 24 \, x^{2} e^{3} + 32 \, x e^{2} + 16 \, e\right )}} \end {gather*}

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)*(x^4*e^4 + 8*x^3*e^3 + 24*x^2*e^2 + 32*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*(3*x^2*e^2 - 44*x*e + 28)*sqrt(-3*x^2*e^2 + 12)*
sqrt(x*e + 2))/(x^4*e^5 + 8*x^3*e^4 + 24*x^2*e^3 + 32*x*e^2 + 16*e)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 8856 deep

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Giac [A]
time = 1.81, size = 90, normalized size = 0.80 \begin {gather*} -\frac {1}{128} \, \sqrt {3} {\left (\frac {4 \, {\left (3 \, {\left (x e - 2\right )}^{2} \sqrt {-x e + 2} + 32 \, {\left (-x e + 2\right )}^{\frac {3}{2}} - 48 \, \sqrt {-x e + 2}\right )}}{{\left (x e + 2\right )}^{3}} + 3 \, \log \left (\sqrt {-x e + 2} + 2\right ) - 3 \, \log \left (-\sqrt {-x e + 2} + 2\right )\right )} e^{\left (-1\right )} \end {gather*}

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]

-1/128*sqrt(3)*(4*(3*(x*e - 2)^2*sqrt(-x*e + 2) + 32*(-x*e + 2)^(3/2) - 48*sqrt(-x*e + 2))/(x*e + 2)^3 + 3*log
(sqrt(-x*e + 2) + 2) - 3*log(-sqrt(-x*e + 2) + 2))*e^(-1)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (12-3\,e^2\,x^2\right )}^{3/2}}{{\left (e\,x+2\right )}^{11/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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