Integrand size = 24, antiderivative size = 73 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=-\frac {9 \sqrt {3} \sqrt {2-e x}}{e}-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (2+e x)}+\frac {18 \sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e} \] Output:
-9*3^(1/2)*(-e*x+2)^(1/2)/e-3*3^(1/2)*(-e*x+2)^(3/2)/e/(e*x+2)+18*3^(1/2)* arctanh(1/2*(-e*x+2)^(1/2))/e
Time = 0.36 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=\frac {6 \sqrt {3} \left (-\frac {(4+e x) \sqrt {4-e^2 x^2}}{(2+e x)^{3/2}}+3 \text {arctanh}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )\right )}{e} \] Input:
Integrate[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(7/2),x]
Output:
(6*Sqrt[3]*(-(((4 + e*x)*Sqrt[4 - e^2*x^2])/(2 + e*x)^(3/2)) + 3*ArcTanh[( 2*Sqrt[2 + e*x])/Sqrt[4 - e^2*x^2]]))/e
Time = 0.30 (sec) , antiderivative size = 73, 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, 60, 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.
\(\displaystyle \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(e x+2)^{7/2}} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {(6-3 e x)^{3/2}}{(e x+2)^2}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {9}{2} \int \frac {\sqrt {3} \sqrt {2-e x}}{e x+2}dx-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (e x+2)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {9}{2} \sqrt {3} \int \frac {\sqrt {2-e x}}{e x+2}dx-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (e x+2)}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {9}{2} \sqrt {3} \left (4 \int \frac {1}{\sqrt {2-e x} (e x+2)}dx+\frac {2 \sqrt {2-e x}}{e}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (e x+2)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {9}{2} \sqrt {3} \left (\frac {2 \sqrt {2-e x}}{e}-\frac {8 \int \frac {1}{e x+2}d\sqrt {2-e x}}{e}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (e x+2)}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {9}{2} \sqrt {3} \left (\frac {2 \sqrt {2-e x}}{e}-\frac {4 \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{e}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{e (e x+2)}\) |
Input:
Int[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(7/2),x]
Output:
(-3*Sqrt[3]*(2 - e*x)^(3/2))/(e*(2 + e*x)) - (9*Sqrt[3]*((2*Sqrt[2 - e*x]) /e - (4*ArcTanh[Sqrt[2 - e*x]/2])/e))/2
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/(b*(m + n + 1))), x] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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 = 0.22 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.36
method | result | size |
default | \(\frac {6 \sqrt {-e^{2} x^{2}+4}\, \left (3 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x -e x \sqrt {-3 e x +6}+6 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )-4 \sqrt {-3 e x +6}\right ) \sqrt {3}}{\left (e x +2\right )^{\frac {3}{2}} \sqrt {-3 e x +6}\, e}\) | \(99\) |
risch | \(\frac {18 \left (e x -2\right ) \sqrt {\frac {-3 e^{2} x^{2}+12}{e x +2}}\, \sqrt {e x +2}}{e \sqrt {-3 e x +6}\, \sqrt {-3 e^{2} x^{2}+12}}-\frac {72 \left (-\frac {\sqrt {-3 e x +6}}{2 \left (-3 e x -6\right )}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )}{4}\right ) \sqrt {\frac {-3 e^{2} x^{2}+12}{e x +2}}\, \sqrt {e x +2}}{e \sqrt {-3 e^{2} x^{2}+12}}\) | \(141\) |
Input:
int((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(7/2),x,method=_RETURNVERBOSE)
Output:
6*(-e^2*x^2+4)^(1/2)*(3*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*e*x- e*x*(-3*e*x+6)^(1/2)+6*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))-4*(-3 *e*x+6)^(1/2))/(e*x+2)^(3/2)/(-3*e*x+6)^(1/2)*3^(1/2)/e
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (59) = 118\).
Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.67 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=\frac {3 \, {\left (3 \, \sqrt {3} {\left (e^{2} x^{2} + 4 \, e x + 4\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x - 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 2 \, \sqrt {-3 \, e^{2} x^{2} + 12} {\left (e x + 4\right )} \sqrt {e x + 2}\right )}}{e^{3} x^{2} + 4 \, e^{2} x + 4 \, e} \] Input:
integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(7/2),x, algorithm="fricas")
Output:
3*(3*sqrt(3)*(e^2*x^2 + 4*e*x + 4)*log(-(3*e^2*x^2 - 12*e*x - 4*sqrt(3)*sq rt(-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)
\[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=3 \sqrt {3} \left (\int \frac {4 \sqrt {- e^{2} x^{2} + 4}}{e^{3} x^{3} \sqrt {e x + 2} + 6 e^{2} x^{2} \sqrt {e x + 2} + 12 e x \sqrt {e x + 2} + 8 \sqrt {e x + 2}}\, dx + \int \left (- \frac {e^{2} x^{2} \sqrt {- e^{2} x^{2} + 4}}{e^{3} x^{3} \sqrt {e x + 2} + 6 e^{2} x^{2} \sqrt {e x + 2} + 12 e x \sqrt {e x + 2} + 8 \sqrt {e x + 2}}\right )\, dx\right ) \] Input:
integrate((-3*e**2*x**2+12)**(3/2)/(e*x+2)**(7/2),x)
Output:
3*sqrt(3)*(Integral(4*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) + Inte gral(-e**2*x**2*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))
\[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=\int { \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{2}}}{{\left (e x + 2\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(7/2),x, algorithm="maxima")
Output:
integrate((-3*e^2*x^2 + 12)^(3/2)/(e*x + 2)^(7/2), x)
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=-\frac {3 \, \sqrt {3} {\left (2 \, \sqrt {-e x + 2} + \frac {4 \, \sqrt {-e x + 2}}{e x + 2} - 3 \, \log \left (\sqrt {-e x + 2} + 2\right ) + 3 \, \log \left (-\sqrt {-e x + 2} + 2\right )\right )}}{e} \] Input:
integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(7/2),x, algorithm="giac")
Output:
-3*sqrt(3)*(2*sqrt(-e*x + 2) + 4*sqrt(-e*x + 2)/(e*x + 2) - 3*log(sqrt(-e* x + 2) + 2) + 3*log(-sqrt(-e*x + 2) + 2))/e
Timed out. \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=\int \frac {{\left (12-3\,e^2\,x^2\right )}^{3/2}}{{\left (e\,x+2\right )}^{7/2}} \,d x \] Input:
int((12 - 3*e^2*x^2)^(3/2)/(e*x + 2)^(7/2),x)
Output:
int((12 - 3*e^2*x^2)^(3/2)/(e*x + 2)^(7/2), x)
Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{7/2}} \, dx=\frac {3 \sqrt {3}\, \left (-4 \sqrt {-e x +2}\, e x -16 \sqrt {-e x +2}-12 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )\right ) e x -24 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )\right )+9 e x +18\right )}{2 e \left (e x +2\right )} \] Input:
int((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(7/2),x)
Output:
(3*sqrt(3)*( - 4*sqrt( - e*x + 2)*e*x - 16*sqrt( - e*x + 2) - 12*log(tan(a sin(sqrt(e*x + 2)/2)/2))*e*x - 24*log(tan(asin(sqrt(e*x + 2)/2)/2)) + 9*e* x + 18))/(2*e*(e*x + 2))