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

Optimal result

Integrand size = 24, antiderivative size = 113 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=-\frac {\sqrt {3} (2-e x)^{3/2}}{e (2+e x)^3}+\frac {3 \sqrt {3} \sqrt {2-e x}}{4 e (2+e x)^2}-\frac {3 \sqrt {3} \sqrt {2-e x}}{32 e (2+e x)}-\frac {3 \sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{64 e} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {641, 43, 44, 65, 212} \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=-\frac {3 \sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{64 e}-\frac {\sqrt {3} (2-e x)^{3/2}}{e (e x+2)^3}-\frac {3 \sqrt {3} \sqrt {2-e x}}{32 e (e x+2)}+\frac {3 \sqrt {3} \sqrt {2-e x}}{4 e (e x+2)^2} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\frac {\sqrt {3} \left (-\frac {2 \sqrt {4-e^2 x^2} \left (28-44 e x+3 e^2 x^2\right )}{(2+e x)^{7/2}}-3 \text {arctanh}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )\right )}{64 e} \]

[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)

Maple [A] (verified)

Time = 2.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.46

method	result	size
default	\(-\frac {\sqrt {-x^{2} e^{2}+4}\, \left (3 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{3} x^{3}+18 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{2} x^{2}+6 e^{2} x^{2} \sqrt {-3 e x +6}+36 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x -88 e x \sqrt {-3 e x +6}+24 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )+56 \sqrt {-3 e x +6}\right ) \sqrt {3}}{64 \left (e x +2\right )^{\frac {7}{2}} \sqrt {-3 e x +6}\, e}\)	\(165\)

[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)^(
7/2)/(-3*e*x+6)^(1/2)/e

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.44 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\frac {3 \, \sqrt {3} {\left (e^{4} x^{4} + 8 \, e^{3} x^{3} + 24 \, e^{2} x^{2} + 32 \, e x + 16\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x + 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 4 \, {\left (3 \, e^{2} x^{2} - 44 \, e x + 28\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{128 \, {\left (e^{5} x^{4} + 8 \, e^{4} x^{3} + 24 \, e^{3} x^{2} + 32 \, e^{2} x + 16 \, e\right )}} \]

[In]

integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(11/2),x, algorithm="fricas")

[Out]

1/128*(3*sqrt(3)*(e^4*x^4 + 8*e^3*x^3 + 24*e^2*x^2 + 32*e*x + 16)*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)) - 4*(3*e^2*x^2 - 44*e*x + 28)*sqrt(-3*e^2*x^2 + 12)*
sqrt(e*x + 2))/(e^5*x^4 + 8*e^4*x^3 + 24*e^3*x^2 + 32*e^2*x + 16*e)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\int { \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{2}}}{{\left (e x + 2\right )}^{\frac {11}{2}}} \,d x } \]

[In]

integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(11/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=-\frac {\sqrt {3} {\left (\frac {4 \, {\left (3 \, {\left (e x - 2\right )}^{2} \sqrt {-e x + 2} + 32 \, {\left (-e x + 2\right )}^{\frac {3}{2}} - 48 \, \sqrt {-e x + 2}\right )}}{{\left (e x + 2\right )}^{3}} + 3 \, \log \left (\sqrt {-e x + 2} + 2\right ) - 3 \, \log \left (-\sqrt {-e x + 2} + 2\right )\right )}}{128 \, e} \]

[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*(e*x - 2)^2*sqrt(-e*x + 2) + 32*(-e*x + 2)^(3/2) - 48*sqrt(-e*x + 2))/(e*x + 2)^3 + 3*log
(sqrt(-e*x + 2) + 2) - 3*log(-sqrt(-e*x + 2) + 2))/e

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\int \frac {{\left (12-3\,e^2\,x^2\right )}^{3/2}}{{\left (e\,x+2\right )}^{11/2}} \,d x \]

[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)