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

Optimal. Leaf size=73 \[ -\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (e x+2)}-\frac {9 \sqrt {3} \sqrt {2-e x}}{e}+\frac {18 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e} \]

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Rubi [A]  time = 0.03, antiderivative size = 73, 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, 50, 63, 206} \begin {gather*} -\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (e x+2)}-\frac {9 \sqrt {3} \sqrt {2-e x}}{e}+\frac {18 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*Sqrt[3]*Sqrt[2 - e*x])/e - (3*Sqrt[3]*(2 - e*x)^(3/2))/(e*(2 + e*x)) + (18*Sqrt[3]*ArcTanh[Sqrt[2 - e*x]/2
])/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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx &=\int \frac {(6-3 e x)^{3/2}}{(2+e x)^2} \, dx\\ &=-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (2+e x)}-\frac {9}{2} \int \frac {\sqrt {6-3 e x}}{2+e x} \, dx\\ &=-\frac {9 \sqrt {3} \sqrt {2-e x}}{e}-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (2+e x)}-54 \int \frac {1}{\sqrt {6-3 e x} (2+e x)} \, dx\\ &=-\frac {9 \sqrt {3} \sqrt {2-e x}}{e}-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (2+e x)}+\frac {36 \operatorname {Subst}\left (\int \frac {1}{4-\frac {x^2}{3}} \, dx,x,\sqrt {6-3 e x}\right )}{e}\\ &=-\frac {9 \sqrt {3} \sqrt {2-e x}}{e}-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (2+e x)}+\frac {18 \sqrt {3} \tanh ^{-1}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 55, normalized size = 0.75 \begin {gather*} -\frac {3 (e x-2)^2 \sqrt {12-3 e^2 x^2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {1}{2}-\frac {e x}{4}\right )}{40 e \sqrt {e x+2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(-2 + e*x)^2*Sqrt[12 - 3*e^2*x^2]*Hypergeometric2F1[2, 5/2, 7/2, 1/2 - (e*x)/4])/(40*e*Sqrt[2 + e*x])

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IntegrateAlgebraic [B]  time = 0.39, size = 177, normalized size = 2.42 \begin {gather*} \frac {\frac {\left (\frac {6 \sqrt {3} (e x+2)}{e}+\frac {12 \sqrt {3}}{e}\right ) \sqrt {4 (e x+2)-(e x+2)^2}}{\sqrt {e x+2}}-\frac {18 \sqrt {3} (e x+2) \tanh ^{-1}\left (\frac {2 \sqrt {e x+2}}{\sqrt {4 (e x+2)-(e x+2)^2}}\right )}{e}}{\left (\frac {\sqrt {4 (e x+2)-(e x+2)^2}}{\sqrt {e x+2}}-2\right ) \left (\frac {\sqrt {4 (e x+2)-(e x+2)^2}}{\sqrt {e x+2}}+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((((12*Sqrt[3])/e + (6*Sqrt[3]*(2 + e*x))/e)*Sqrt[4*(2 + e*x) - (2 + e*x)^2])/Sqrt[2 + e*x] - (18*Sqrt[3]*(2 +
e*x)*ArcTanh[(2*Sqrt[2 + e*x])/Sqrt[4*(2 + e*x) - (2 + e*x)^2]])/e)/((-2 + Sqrt[4*(2 + e*x) - (2 + e*x)^2]/Sq
rt[2 + e*x])*(2 + Sqrt[4*(2 + e*x) - (2 + e*x)^2]/Sqrt[2 + e*x]))

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fricas [B]  time = 0.41, size = 122, normalized size = 1.67 \begin {gather*} \frac {3 \, {\left (3 \, \sqrt {3} {\left (e^{2} x^{2} + 4 \, e x + 4\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 ) - 2 \, \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 4\right )} \sqrt {e x + 2}\right )}}{e^{3} x^{2} + 4 \, e^{2} x + 4 \, e} \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)^(7/2),x, algorithm="fricas")

[Out]

3*(3*sqrt(3)*(e^2*x^2 + 4*e*x + 4)*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)) - 2*sqrt(-3*e^2*x^2 + 12)*(e*x + 4)*sqrt(e*x + 2))/(e^3*x^2 + 4*e^2*x + 4*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)^(3/2)/(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 = 0.06, size = 101, normalized size = 1.38 \begin {gather*} \frac {6 \sqrt {-e^{2} x^{2}+4}\, \left (3 \sqrt {3}\, e x \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )-\sqrt {-3 e x +6}\, e x +6 \sqrt {3}\, \arctanh \left (\frac {\sqrt {3}\, \sqrt {-3 e x +6}}{6}\right )-4 \sqrt {-3 e x +6}\right ) \sqrt {3}}{\sqrt {\left (e x +2\right )^{3}}\, \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)^(3/2)/(e*x+2)^(7/2),x)

[Out]

6*(-e^2*x^2+4)^(1/2)*(3*arctanh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))*3^(1/2)*x*e-(-3*e*x+6)^(1/2)*e*x+6*3^(1/2)*arcta
nh(1/6*3^(1/2)*(-3*e*x+6)^(1/2))-4*(-3*e*x+6)^(1/2))*3^(1/2)/((e*x+2)^3)^(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 {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{2}}}{{\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)^(3/2)/(e*x+2)^(7/2),x, algorithm="maxima")

[Out]

integrate((-3*e^2*x^2 + 12)^(3/2)/(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 {{\left (12-3\,e^2\,x^2\right )}^{3/2}}{{\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)^(3/2)/(e*x + 2)^(7/2),x)

[Out]

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

[Out]

Timed out

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