Integrand size = 24, antiderivative size = 86 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (2+e x)^2}+\frac {\sqrt {3} \sqrt {2-e x}}{16 e (2+e x)}+\frac {\sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{32 e} \] Output:
-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)+1/32*3^(1/2)*arctanh(1/2*(-e*x+2)^(1/2))/e
Time = 0.32 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=\frac {\sqrt {3} \left (\frac {4 (-6+e x) \sqrt {4-e^2 x^2}}{(2+e x)^{5/2}}-\log \left (e \left (-2 \sqrt {2+e x}+\sqrt {4-e^2 x^2}\right )\right )+\log \left (2 \sqrt {2+e x}+\sqrt {4-e^2 x^2}\right )\right )}{64 e} \] Input:
Integrate[Sqrt[12 - 3*e^2*x^2]/(2 + e*x)^(7/2),x]
Output:
(Sqrt[3]*((4*(-6 + e*x)*Sqrt[4 - e^2*x^2])/(2 + e*x)^(5/2) - Log[e*(-2*Sqr t[2 + e*x] + Sqrt[4 - e^2*x^2])] + Log[2*Sqrt[2 + e*x] + Sqrt[4 - e^2*x^2] ]))/(64*e)
Time = 0.31 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {456, 51, 27, 52, 73, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {12-3 e^2 x^2}}{(e x+2)^{7/2}} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {\sqrt {6-3 e x}}{(e x+2)^3}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {3}{4} \int \frac {1}{\sqrt {3} \sqrt {2-e x} (e x+2)^2}dx-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (e x+2)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} \sqrt {3} \int \frac {1}{\sqrt {2-e x} (e x+2)^2}dx-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (e x+2)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {1}{4} \sqrt {3} \left (\frac {1}{8} \int \frac {1}{\sqrt {2-e x} (e x+2)}dx-\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (e x+2)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {1}{4} \sqrt {3} \left (-\frac {\int \frac {1}{e x+2}d\sqrt {2-e x}}{4 e}-\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (e x+2)^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {1}{4} \sqrt {3} \left (-\frac {\text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{8 e}-\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )-\frac {\sqrt {3} \sqrt {2-e x}}{2 e (e x+2)^2}\) |
Input:
Int[Sqrt[12 - 3*e^2*x^2]/(2 + e*x)^(7/2),x]
Output:
-1/2*(Sqrt[3]*Sqrt[2 - e*x])/(e*(2 + e*x)^2) - (Sqrt[3]*(-1/4*Sqrt[2 - e*x ]/(e*(2 + e*x)) - ArcTanh[Sqrt[2 - e*x]/2]/(8*e)))/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] - Simp[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 ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !Integ erQ[n]))
Time = 0.17 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.43
method | result | size |
default | \(\frac {\sqrt {-e^{2} x^{2}+4}\, \left (\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{2} x^{2}+4 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x +2 e x \sqrt {-3 e x +6}+4 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )-12 \sqrt {-3 e x +6}\right ) \sqrt {3}}{32 \left (e x +2\right )^{\frac {5}{2}} \sqrt {-3 e x +6}\, e}\) | \(123\) |
Input:
int((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x,method=_RETURNVERBOSE)
Output:
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)^(5/2)/(-3*e*x+6)^(1/2)/e
Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (66) = 132\).
Time = 0.12 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.59 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=\frac {\sqrt {3} {\left (e^{3} x^{3} + 6 \, e^{2} x^{2} + 12 \, e x + 8\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x - 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) + 4 \, \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} {\left (e x - 6\right )}}{64 \, {\left (e^{4} x^{3} + 6 \, e^{3} x^{2} + 12 \, e^{2} x + 8 \, e\right )}} \] Input:
integrate((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x, algorithm="fricas")
Output:
1/64*(sqrt(3)*(e^3*x^3 + 6*e^2*x^2 + 12*e*x + 8)*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*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2)*(e*x - 6))/(e^4*x^3 + 6*e^3*x^2 + 12*e^2*x + 8*e)
\[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=\sqrt {3} \int \frac {\sqrt {- e^{2} x^{2} + 4}}{e^{3} x^{3} \sqrt {e x + 2} + 6 e^{2} x^{2} \sqrt {e x + 2} + 12 e x \sqrt {e x + 2} + 8 \sqrt {e x + 2}}\, dx \] Input:
integrate((-3*e**2*x**2+12)**(1/2)/(e*x+2)**(7/2),x)
Output:
sqrt(3)*Integral(sqrt(-e**2*x**2 + 4)/(e**3*x**3*sqrt(e*x + 2) + 6*e**2*x* *2*sqrt(e*x + 2) + 12*e*x*sqrt(e*x + 2) + 8*sqrt(e*x + 2)), x)
\[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=\int { \frac {\sqrt {-3 \, e^{2} x^{2} + 12}}{{\left (e x + 2\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x, algorithm="maxima")
Output:
integrate(sqrt(-3*e^2*x^2 + 12)/(e*x + 2)^(7/2), x)
Time = 0.12 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=-\frac {\sqrt {3} {\left (\frac {4 \, {\left ({\left (-e x + 2\right )}^{\frac {3}{2}} + 4 \, \sqrt {-e x + 2}\right )}}{{\left (e x + 2\right )}^{2}} - \log \left (\sqrt {-e x + 2} + 2\right ) + \log \left (-\sqrt {-e x + 2} + 2\right )\right )}}{64 \, e} \] Input:
integrate((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x, algorithm="giac")
Output:
-1/64*sqrt(3)*(4*((-e*x + 2)^(3/2) + 4*sqrt(-e*x + 2))/(e*x + 2)^2 - log(s qrt(-e*x + 2) + 2) + log(-sqrt(-e*x + 2) + 2))/e
Timed out. \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=\int \frac {\sqrt {12-3\,e^2\,x^2}}{{\left (e\,x+2\right )}^{7/2}} \,d x \] Input:
int((12 - 3*e^2*x^2)^(1/2)/(e*x + 2)^(7/2),x)
Output:
int((12 - 3*e^2*x^2)^(1/2)/(e*x + 2)^(7/2), x)
Time = 0.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {12-3 e^2 x^2}}{(2+e x)^{7/2}} \, dx=\frac {\sqrt {3}\, \left (2 \sqrt {-e x +2}\, e x -12 \sqrt {-e x +2}-\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )\right ) e^{2} x^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )\right ) e x -4 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )\right )\right )}{32 e \left (e^{2} x^{2}+4 e x +4\right )} \] Input:
int((-3*e^2*x^2+12)^(1/2)/(e*x+2)^(7/2),x)
Output:
(sqrt(3)*(2*sqrt( - e*x + 2)*e*x - 12*sqrt( - e*x + 2) - log(tan(asin(sqrt (e*x + 2)/2)/2))*e**2*x**2 - 4*log(tan(asin(sqrt(e*x + 2)/2)/2))*e*x - 4*l og(tan(asin(sqrt(e*x + 2)/2)/2))))/(32*e*(e**2*x**2 + 4*e*x + 4))