Integrand size = 24, antiderivative size = 113 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=-\frac {\sqrt {3} (2-e x)^{3/2}}{e (2+e x)^3}+\frac {3 \sqrt {3} \sqrt {2-e x}}{4 e (2+e x)^2}-\frac {3 \sqrt {3} \sqrt {2-e x}}{32 e (2+e x)}-\frac {3 \sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{64 e} \]
-(-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)
Time = 0.56 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.71 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\frac {\sqrt {3} \left (-\frac {2 \sqrt {4-e^2 x^2} \left (28-44 e x+3 e^2 x^2\right )}{(2+e x)^{7/2}}-3 \text {arctanh}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )\right )}{64 e} \]
(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)
Time = 0.21 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {456, 51, 27, 51, 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.
\(\displaystyle \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(e x+2)^{11/2}} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {(6-3 e x)^{3/2}}{(e x+2)^4}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {3}{2} \int \frac {\sqrt {3} \sqrt {2-e x}}{(e x+2)^3}dx-\frac {\sqrt {3} (2-e x)^{3/2}}{e (e x+2)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3}{2} \sqrt {3} \int \frac {\sqrt {2-e x}}{(e x+2)^3}dx-\frac {\sqrt {3} (2-e x)^{3/2}}{e (e x+2)^3}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {3}{2} \sqrt {3} \left (-\frac {1}{4} \int \frac {1}{\sqrt {2-e x} (e x+2)^2}dx-\frac {\sqrt {2-e x}}{2 e (e x+2)^2}\right )-\frac {\sqrt {3} (2-e x)^{3/2}}{e (e x+2)^3}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {3}{2} \sqrt {3} \left (\frac {1}{4} \left (\frac {\sqrt {2-e x}}{4 e (e x+2)}-\frac {1}{8} \int \frac {1}{\sqrt {2-e x} (e x+2)}dx\right )-\frac {\sqrt {2-e x}}{2 e (e x+2)^2}\right )-\frac {\sqrt {3} (2-e x)^{3/2}}{e (e x+2)^3}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {3}{2} \sqrt {3} \left (\frac {1}{4} \left (\frac {\int \frac {1}{e x+2}d\sqrt {2-e x}}{4 e}+\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )-\frac {\sqrt {2-e x}}{2 e (e x+2)^2}\right )-\frac {\sqrt {3} (2-e x)^{3/2}}{e (e x+2)^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {3}{2} \sqrt {3} \left (\frac {1}{4} \left (\frac {\text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{8 e}+\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )-\frac {\sqrt {2-e x}}{2 e (e x+2)^2}\right )-\frac {\sqrt {3} (2-e x)^{3/2}}{e (e x+2)^3}\) |
-((Sqrt[3]*(2 - e*x)^(3/2))/(e*(2 + e*x)^3)) - (3*Sqrt[3]*(-1/2*Sqrt[2 - e *x]/(e*(2 + e*x)^2) + (Sqrt[2 - e*x]/(4*e*(2 + e*x)) + ArcTanh[Sqrt[2 - e* x]/2]/(8*e))/4))/2
3.10.9.3.1 Defintions of rubi rules used
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 = 2.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.46
method | result | size |
default | \(-\frac {\sqrt {-x^{2} e^{2}+4}\, \left (3 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{3} x^{3}+18 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{2} x^{2}+6 e^{2} x^{2} \sqrt {-3 e x +6}+36 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x -88 e x \sqrt {-3 e x +6}+24 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )+56 \sqrt {-3 e x +6}\right ) \sqrt {3}}{64 \left (e x +2\right )^{\frac {7}{2}} \sqrt {-3 e x +6}\, e}\) | \(165\) |
-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-88* 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
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.44 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\frac {3 \, \sqrt {3} {\left (e^{4} x^{4} + 8 \, e^{3} x^{3} + 24 \, e^{2} x^{2} + 32 \, e x + 16\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x + 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 4 \, {\left (3 \, e^{2} x^{2} - 44 \, e x + 28\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}}{128 \, {\left (e^{5} x^{4} + 8 \, e^{4} x^{3} + 24 \, e^{3} x^{2} + 32 \, e^{2} x + 16 \, e\right )}} \]
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)*sq rt(e*x + 2))/(e^5*x^4 + 8*e^4*x^3 + 24*e^3*x^2 + 32*e^2*x + 16*e)
Timed out. \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\int { \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{2}}}{{\left (e x + 2\right )}^{\frac {11}{2}}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.74 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=-\frac {\sqrt {3} {\left (\frac {4 \, {\left (3 \, {\left (e x - 2\right )}^{2} \sqrt {-e x + 2} + 32 \, {\left (-e x + 2\right )}^{\frac {3}{2}} - 48 \, \sqrt {-e x + 2}\right )}}{{\left (e x + 2\right )}^{3}} + 3 \, \log \left (\sqrt {-e x + 2} + 2\right ) - 3 \, \log \left (-\sqrt {-e x + 2} + 2\right )\right )}}{128 \, e} \]
-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
Timed out. \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{11/2}} \, dx=\int \frac {{\left (12-3\,e^2\,x^2\right )}^{3/2}}{{\left (e\,x+2\right )}^{11/2}} \,d x \]