\(\int \frac {(12-3 e^2 x^2)^{3/2}}{(2+e x)^{13/2}} \, dx\) [222]

Optimal result

Integrand size = 24, antiderivative size = 144 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (2+e x)^4}+\frac {3 \sqrt {3} \sqrt {2-e x}}{8 e (2+e x)^3}-\frac {3 \sqrt {3} \sqrt {2-e x}}{128 e (2+e x)^2}-\frac {9 \sqrt {3} \sqrt {2-e x}}{1024 e (2+e x)}-\frac {9 \sqrt {3} \text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{2048 e} \] Output:

-3/4*3^(1/2)*(-e*x+2)^(3/2)/e/(e*x+2)^4+3/8*3^(1/2)*(-e*x+2)^(1/2)/e/(e*x+ 
2)^3-3/128*3^(1/2)*(-e*x+2)^(1/2)/e/(e*x+2)^2-9/1024*3^(1/2)*(-e*x+2)^(1/2 
)/e/(e*x+2)-9/2048*3^(1/2)*arctanh(1/2*(-e*x+2)^(1/2))/e
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=\frac {3 \sqrt {3} \left (-\frac {2 \sqrt {4-e^2 x^2} \left (312-316 e x+26 e^2 x^2+3 e^3 x^3\right )}{(2+e x)^{9/2}}-3 \text {arctanh}\left (\frac {2 \sqrt {2+e x}}{\sqrt {4-e^2 x^2}}\right )\right )}{2048 e} \] Input:

Integrate[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(13/2),x]
 

Output:

(3*Sqrt[3]*((-2*Sqrt[4 - e^2*x^2]*(312 - 316*e*x + 26*e^2*x^2 + 3*e^3*x^3) 
)/(2 + e*x)^(9/2) - 3*ArcTanh[(2*Sqrt[2 + e*x])/Sqrt[4 - e^2*x^2]]))/(2048 
*e)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {456, 51, 27, 51, 52, 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.

\(\displaystyle \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(e x+2)^{13/2}} \, dx\)

\(\Big \downarrow \) 456

\(\displaystyle \int \frac {(6-3 e x)^{3/2}}{(e x+2)^5}dx\)

\(\Big \downarrow \) 51

\(\displaystyle -\frac {9}{8} \int \frac {\sqrt {3} \sqrt {2-e x}}{(e x+2)^4}dx-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (e x+2)^4}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {9}{8} \sqrt {3} \int \frac {\sqrt {2-e x}}{(e x+2)^4}dx-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (e x+2)^4}\)

\(\Big \downarrow \) 51

\(\displaystyle -\frac {9}{8} \sqrt {3} \left (-\frac {1}{6} \int \frac {1}{\sqrt {2-e x} (e x+2)^3}dx-\frac {\sqrt {2-e x}}{3 e (e x+2)^3}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (e x+2)^4}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {9}{8} \sqrt {3} \left (\frac {1}{6} \left (\frac {\sqrt {2-e x}}{8 e (e x+2)^2}-\frac {3}{16} \int \frac {1}{\sqrt {2-e x} (e x+2)^2}dx\right )-\frac {\sqrt {2-e x}}{3 e (e x+2)^3}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (e x+2)^4}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {9}{8} \sqrt {3} \left (\frac {1}{6} \left (\frac {\sqrt {2-e x}}{8 e (e x+2)^2}-\frac {3}{16} \left (\frac {1}{8} \int \frac {1}{\sqrt {2-e x} (e x+2)}dx-\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )\right )-\frac {\sqrt {2-e x}}{3 e (e x+2)^3}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (e x+2)^4}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {9}{8} \sqrt {3} \left (\frac {1}{6} \left (\frac {\sqrt {2-e x}}{8 e (e x+2)^2}-\frac {3}{16} \left (-\frac {\int \frac {1}{e x+2}d\sqrt {2-e x}}{4 e}-\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )\right )-\frac {\sqrt {2-e x}}{3 e (e x+2)^3}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (e x+2)^4}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {9}{8} \sqrt {3} \left (\frac {1}{6} \left (\frac {\sqrt {2-e x}}{8 e (e x+2)^2}-\frac {3}{16} \left (-\frac {\text {arctanh}\left (\frac {1}{2} \sqrt {2-e x}\right )}{8 e}-\frac {\sqrt {2-e x}}{4 e (e x+2)}\right )\right )-\frac {\sqrt {2-e x}}{3 e (e x+2)^3}\right )-\frac {3 \sqrt {3} (2-e x)^{3/2}}{4 e (e x+2)^4}\)

Input:

Int[(12 - 3*e^2*x^2)^(3/2)/(2 + e*x)^(13/2),x]
 

Output:

(-3*Sqrt[3]*(2 - e*x)^(3/2))/(4*e*(2 + e*x)^4) - (9*Sqrt[3]*(-1/3*Sqrt[2 - 
 e*x]/(e*(2 + e*x)^3) + (Sqrt[2 - e*x]/(8*e*(2 + e*x)^2) - (3*(-1/4*Sqrt[2 
 - e*x]/(e*(2 + e*x)) - ArcTanh[Sqrt[2 - e*x]/2]/(8*e)))/16)/6))/8
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 51

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]
 

rule 52

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]
 

rule 73

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]
 

rule 219

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])
 

rule 456

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]))
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.43

method	result	size
default	\(-\frac {3 \sqrt {-e^{2} x^{2}+4}\, \left (3 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{4} x^{4}+24 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{3} x^{3}+6 e^{3} x^{3} \sqrt {-3 e x +6}+72 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e^{2} x^{2}+52 e^{2} x^{2} \sqrt {-3 e x +6}+96 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right ) e x -632 e x \sqrt {-3 e x +6}+48 \sqrt {3}\, \operatorname {arctanh}\left (\frac {\sqrt {-3 e x +6}\, \sqrt {3}}{6}\right )+624 \sqrt {-3 e x +6}\right ) \sqrt {3}}{2048 \left (e x +2\right )^{\frac {9}{2}} \sqrt {-3 e x +6}\, e}\)	\(206\)

Input:

int((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(13/2),x,method=_RETURNVERBOSE)
 

Output:

-3/2048*(-e^2*x^2+4)^(1/2)*(3*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2) 
)*e^4*x^4+24*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*e^3*x^3+6*e^3*x 
^3*(-3*e*x+6)^(1/2)+72*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^(1/2))*e^2*x 
^2+52*e^2*x^2*(-3*e*x+6)^(1/2)+96*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2)*3^( 
1/2))*e*x-632*e*x*(-3*e*x+6)^(1/2)+48*3^(1/2)*arctanh(1/6*(-3*e*x+6)^(1/2) 
*3^(1/2))+624*(-3*e*x+6)^(1/2))*3^(1/2)/(e*x+2)^(9/2)/(-3*e*x+6)^(1/2)/e
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.30 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=\frac {3 \, {\left (3 \, \sqrt {3} {\left (e^{5} x^{5} + 10 \, e^{4} x^{4} + 40 \, e^{3} x^{3} + 80 \, e^{2} x^{2} + 80 \, e x + 32\right )} \log \left (-\frac {3 \, e^{2} x^{2} - 12 \, e x + 4 \, \sqrt {3} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2} - 36}{e^{2} x^{2} + 4 \, e x + 4}\right ) - 4 \, {\left (3 \, e^{3} x^{3} + 26 \, e^{2} x^{2} - 316 \, e x + 312\right )} \sqrt {-3 \, e^{2} x^{2} + 12} \sqrt {e x + 2}\right )}}{4096 \, {\left (e^{6} x^{5} + 10 \, e^{5} x^{4} + 40 \, e^{4} x^{3} + 80 \, e^{3} x^{2} + 80 \, e^{2} x + 32 \, e\right )}} \] Input:

integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(13/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

3/4096*(3*sqrt(3)*(e^5*x^5 + 10*e^4*x^4 + 40*e^3*x^3 + 80*e^2*x^2 + 80*e*x 
 + 32)*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^3*x^3 + 26*e^2*x^2 - 316*e*x + 
 312)*sqrt(-3*e^2*x^2 + 12)*sqrt(e*x + 2))/(e^6*x^5 + 10*e^5*x^4 + 40*e^4* 
x^3 + 80*e^3*x^2 + 80*e^2*x + 32*e)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((-3*e**2*x**2+12)**(3/2)/(e*x+2)**(13/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=\int { \frac {{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{2}}}{{\left (e x + 2\right )}^{\frac {13}{2}}} \,d x } \] Input:

integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(13/2),x, algorithm="maxima")
 

Output:

integrate((-3*e^2*x^2 + 12)^(3/2)/(e*x + 2)^(13/2), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.70 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=-\frac {3 \, \sqrt {3} {\left (\frac {4 \, {\left (3 \, {\left (e x - 2\right )}^{3} \sqrt {-e x + 2} + 44 \, {\left (e x - 2\right )}^{2} \sqrt {-e x + 2} + 176 \, {\left (-e x + 2\right )}^{\frac {3}{2}} - 192 \, \sqrt {-e x + 2}\right )}}{{\left (e x + 2\right )}^{4}} + 3 \, \log \left (\sqrt {-e x + 2} + 2\right ) - 3 \, \log \left (-\sqrt {-e x + 2} + 2\right )\right )}}{4096 \, e} \] Input:

integrate((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(13/2),x, algorithm="giac")
 

Output:

-3/4096*sqrt(3)*(4*(3*(e*x - 2)^3*sqrt(-e*x + 2) + 44*(e*x - 2)^2*sqrt(-e* 
x + 2) + 176*(-e*x + 2)^(3/2) - 192*sqrt(-e*x + 2))/(e*x + 2)^4 + 3*log(sq 
rt(-e*x + 2) + 2) - 3*log(-sqrt(-e*x + 2) + 2))/e
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=\int \frac {{\left (12-3\,e^2\,x^2\right )}^{3/2}}{{\left (e\,x+2\right )}^{13/2}} \,d x \] Input:

int((12 - 3*e^2*x^2)^(3/2)/(e*x + 2)^(13/2),x)
 

Output:

int((12 - 3*e^2*x^2)^(3/2)/(e*x + 2)^(13/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.68 \[ \int \frac {\left (12-3 e^2 x^2\right )^{3/2}}{(2+e x)^{13/2}} \, dx=\frac {3 \sqrt {3}\, \left (48 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )^{8}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )^{16}-8 \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )^{12}+8 \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )^{4}-1\right )}{32768 \tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +2}}{2}\right )}{2}\right )^{8} e} \] Input:

int((-3*e^2*x^2+12)^(3/2)/(e*x+2)^(13/2),x)
 

Output:

(3*sqrt(3)*(48*log(tan(asin(sqrt(e*x + 2)/2)/2))*tan(asin(sqrt(e*x + 2)/2) 
/2)**8 + tan(asin(sqrt(e*x + 2)/2)/2)**16 - 8*tan(asin(sqrt(e*x + 2)/2)/2) 
**12 + 8*tan(asin(sqrt(e*x + 2)/2)/2)**4 - 1))/(32768*tan(asin(sqrt(e*x + 
2)/2)/2)**8*e)