\(\int \frac {1}{(27 a^3+27 a^2 b x-4 b^3 x^3)^{3/2}} \, dx\) [13]

Optimal result

Integrand size = 25, antiderivative size = 244 \[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=-\frac {(3 a-b x) (3 a+2 b x)}{18 a b \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}-\frac {5 (3 a-b x) (3 a+2 b x)^2}{324 a^2 b \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}+\frac {5 (3 a-b x) (3 a+2 b x)^3}{972 a^3 b \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}-\frac {5 (3 a+2 b x)^3 \left (3-\frac {b x}{a}\right )^{3/2} \text {arctanh}\left (\frac {1}{3} \sqrt {2} \sqrt {3-\frac {b x}{a}}\right )}{1458 \sqrt {2} a^2 b \left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \] Output:

-1/18*(-b*x+3*a)*(2*b*x+3*a)/a/b/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2)-5/32 
4*(-b*x+3*a)*(2*b*x+3*a)^2/a^2/b/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2)+5/97 
2*(-b*x+3*a)*(2*b*x+3*a)^3/a^3/b/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2)-5/29 
16*(2*b*x+3*a)^3*(3-b*x/a)^(3/2)*arctanh(1/3*2^(1/2)*(3-b*x/a)^(1/2))*2^(1 
/2)/a^2/b/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=\frac {6 \sqrt {a} \left (-27 a^2+15 a b x+10 b^2 x^2\right )-5 \sqrt {6 a-2 b x} (3 a+2 b x)^2 \text {arctanh}\left (\frac {\sqrt {6 a-2 b x}}{3 \sqrt {a}}\right )}{2916 a^{7/2} b (3 a+2 b x) \sqrt {(3 a-b x) (3 a+2 b x)^2}} \] Input:

Integrate[(27*a^3 + 27*a^2*b*x - 4*b^3*x^3)^(-3/2),x]
 

Output:

(6*Sqrt[a]*(-27*a^2 + 15*a*b*x + 10*b^2*x^2) - 5*Sqrt[6*a - 2*b*x]*(3*a + 
2*b*x)^2*ArcTanh[Sqrt[6*a - 2*b*x]/(3*Sqrt[a])])/(2916*a^(7/2)*b*(3*a + 2* 
b*x)*Sqrt[(3*a - b*x)*(3*a + 2*b*x)^2])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2473, 27, 52, 52, 61, 73, 221}

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.

\(\displaystyle \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2473

\(\displaystyle \frac {1594323 \sqrt {3} a^6 (3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2} \int \frac {1}{1594323 \sqrt {3} a^6 (3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2}}dx}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2} \int \frac {1}{(3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2}}dx}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2} \left (\frac {5 \int \frac {1}{(3 a+2 b x)^2 \left (3 a^3-a^2 b x\right )^{3/2}}dx}{36 a}-\frac {1}{18 a^3 b (3 a+2 b x)^2 \sqrt {3 a^3-a^2 b x}}\right )}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2} \left (\frac {5 \left (\frac {\int \frac {1}{(3 a+2 b x) \left (3 a^3-a^2 b x\right )^{3/2}}dx}{6 a}-\frac {1}{9 a^3 b (3 a+2 b x) \sqrt {3 a^3-a^2 b x}}\right )}{36 a}-\frac {1}{18 a^3 b (3 a+2 b x)^2 \sqrt {3 a^3-a^2 b x}}\right )}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {(3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2} \left (\frac {5 \left (\frac {\frac {2 \int \frac {1}{(3 a+2 b x) \sqrt {3 a^3-a^2 b x}}dx}{9 a^3}+\frac {2}{9 a^3 b \sqrt {3 a^3-a^2 b x}}}{6 a}-\frac {1}{9 a^3 b (3 a+2 b x) \sqrt {3 a^3-a^2 b x}}\right )}{36 a}-\frac {1}{18 a^3 b (3 a+2 b x)^2 \sqrt {3 a^3-a^2 b x}}\right )}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2} \left (\frac {5 \left (\frac {\frac {2}{9 a^3 b \sqrt {3 a^3-a^2 b x}}-\frac {4 \int \frac {1}{9 a-\frac {2 \left (3 a^3-a^2 b x\right )}{a^2}}d\sqrt {3 a^3-a^2 b x}}{9 a^5 b}}{6 a}-\frac {1}{9 a^3 b (3 a+2 b x) \sqrt {3 a^3-a^2 b x}}\right )}{36 a}-\frac {1}{18 a^3 b (3 a+2 b x)^2 \sqrt {3 a^3-a^2 b x}}\right )}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(3 a+2 b x)^3 \left (3 a^3-a^2 b x\right )^{3/2} \left (\frac {5 \left (\frac {\frac {2}{9 a^3 b \sqrt {3 a^3-a^2 b x}}-\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {3 a^3-a^2 b x}}{3 a^{3/2}}\right )}{27 a^{9/2} b}}{6 a}-\frac {1}{9 a^3 b (3 a+2 b x) \sqrt {3 a^3-a^2 b x}}\right )}{36 a}-\frac {1}{18 a^3 b (3 a+2 b x)^2 \sqrt {3 a^3-a^2 b x}}\right )}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}}\)

Input:

Int[(27*a^3 + 27*a^2*b*x - 4*b^3*x^3)^(-3/2),x]
 

Output:

((3*a + 2*b*x)^3*(3*a^3 - a^2*b*x)^(3/2)*(-1/18*1/(a^3*b*(3*a + 2*b*x)^2*S 
qrt[3*a^3 - a^2*b*x]) + (5*(-1/9*1/(a^3*b*(3*a + 2*b*x)*Sqrt[3*a^3 - a^2*b 
*x]) + (2/(9*a^3*b*Sqrt[3*a^3 - a^2*b*x]) - (2*Sqrt[2]*ArcTanh[(Sqrt[2]*Sq 
rt[3*a^3 - a^2*b*x])/(3*a^(3/2))])/(27*a^(9/2)*b))/(6*a)))/(36*a)))/(27*a^ 
3 + 27*a^2*b*x - 4*b^3*x^3)^(3/2)
 

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 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 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 2473

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Simp[(a + b*x + 
d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*x)^(2*p))   Int[(3*a - b*x)^p*(3*a + 2*b 
*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0] && 
 !IntegerQ[p]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {\left (20 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-b x +3 a}\, \sqrt {2}}{3 \sqrt {a}}\right ) \sqrt {-b x +3 a}\, b^{2} x^{2}+60 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-b x +3 a}\, \sqrt {2}}{3 \sqrt {a}}\right ) \sqrt {-b x +3 a}\, a b x +45 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-b x +3 a}\, \sqrt {2}}{3 \sqrt {a}}\right ) \sqrt {-b x +3 a}\, a^{2}+162 a^{\frac {5}{2}}-90 a^{\frac {3}{2}} b x -60 \sqrt {a}\, b^{2} x^{2}\right ) \left (-b x +3 a \right ) \left (2 b x +3 a \right )}{2916 a^{\frac {7}{2}} b \left (-4 b^{3} x^{3}+27 b \,a^{2} x +27 a^{3}\right )^{\frac {3}{2}}}\)	\(186\)

Input:

int(1/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2916*(20*2^(1/2)*arctanh(1/3*(-b*x+3*a)^(1/2)*2^(1/2)/a^(1/2))*(-b*x+3* 
a)^(1/2)*b^2*x^2+60*2^(1/2)*arctanh(1/3*(-b*x+3*a)^(1/2)*2^(1/2)/a^(1/2))* 
(-b*x+3*a)^(1/2)*a*b*x+45*2^(1/2)*arctanh(1/3*(-b*x+3*a)^(1/2)*2^(1/2)/a^( 
1/2))*(-b*x+3*a)^(1/2)*a^2+162*a^(5/2)-90*a^(3/2)*b*x-60*a^(1/2)*b^2*x^2)* 
(-b*x+3*a)*(2*b*x+3*a)/a^(7/2)/b/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.77 \[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=\left [\frac {5 \, \sqrt {2} {\left (8 \, b^{4} x^{4} + 12 \, a b^{3} x^{3} - 54 \, a^{2} b^{2} x^{2} - 135 \, a^{3} b x - 81 \, a^{4}\right )} \sqrt {a} \log \left (\frac {4 \, b^{2} x^{2} - 24 \, a b x - 45 \, a^{2} + 6 \, \sqrt {2} \sqrt {-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}} \sqrt {a}}{4 \, b^{2} x^{2} + 12 \, a b x + 9 \, a^{2}}\right ) - 12 \, \sqrt {-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}} {\left (10 \, a b^{2} x^{2} + 15 \, a^{2} b x - 27 \, a^{3}\right )}}{5832 \, {\left (8 \, a^{4} b^{5} x^{4} + 12 \, a^{5} b^{4} x^{3} - 54 \, a^{6} b^{3} x^{2} - 135 \, a^{7} b^{2} x - 81 \, a^{8} b\right )}}, -\frac {5 \, \sqrt {2} {\left (8 \, b^{4} x^{4} + 12 \, a b^{3} x^{3} - 54 \, a^{2} b^{2} x^{2} - 135 \, a^{3} b x - 81 \, a^{4}\right )} \sqrt {-a} \arctan \left (\frac {3 \, \sqrt {2} \sqrt {-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}} \sqrt {-a}}{2 \, {\left (2 \, b^{2} x^{2} - 3 \, a b x - 9 \, a^{2}\right )}}\right ) + 6 \, \sqrt {-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}} {\left (10 \, a b^{2} x^{2} + 15 \, a^{2} b x - 27 \, a^{3}\right )}}{2916 \, {\left (8 \, a^{4} b^{5} x^{4} + 12 \, a^{5} b^{4} x^{3} - 54 \, a^{6} b^{3} x^{2} - 135 \, a^{7} b^{2} x - 81 \, a^{8} b\right )}}\right ] \] Input:

integrate(1/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2),x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

[1/5832*(5*sqrt(2)*(8*b^4*x^4 + 12*a*b^3*x^3 - 54*a^2*b^2*x^2 - 135*a^3*b* 
x - 81*a^4)*sqrt(a)*log((4*b^2*x^2 - 24*a*b*x - 45*a^2 + 6*sqrt(2)*sqrt(-4 
*b^3*x^3 + 27*a^2*b*x + 27*a^3)*sqrt(a))/(4*b^2*x^2 + 12*a*b*x + 9*a^2)) - 
 12*sqrt(-4*b^3*x^3 + 27*a^2*b*x + 27*a^3)*(10*a*b^2*x^2 + 15*a^2*b*x - 27 
*a^3))/(8*a^4*b^5*x^4 + 12*a^5*b^4*x^3 - 54*a^6*b^3*x^2 - 135*a^7*b^2*x - 
81*a^8*b), -1/2916*(5*sqrt(2)*(8*b^4*x^4 + 12*a*b^3*x^3 - 54*a^2*b^2*x^2 - 
 135*a^3*b*x - 81*a^4)*sqrt(-a)*arctan(3/2*sqrt(2)*sqrt(-4*b^3*x^3 + 27*a^ 
2*b*x + 27*a^3)*sqrt(-a)/(2*b^2*x^2 - 3*a*b*x - 9*a^2)) + 6*sqrt(-4*b^3*x^ 
3 + 27*a^2*b*x + 27*a^3)*(10*a*b^2*x^2 + 15*a^2*b*x - 27*a^3))/(8*a^4*b^5* 
x^4 + 12*a^5*b^4*x^3 - 54*a^6*b^3*x^2 - 135*a^7*b^2*x - 81*a^8*b)]
 

Sympy [F]

\[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (27 a^{3} + 27 a^{2} b x - 4 b^{3} x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(-4*b**3*x**3+27*a**2*b*x+27*a**3)**(3/2),x)
 

Output:

Integral((27*a**3 + 27*a**2*b*x - 4*b**3*x**3)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-4 \, b^{3} x^{3} + 27 \, a^{2} b x + 27 \, a^{3}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2),x, algorithm="maxima")
 

Output:

integrate((-4*b^3*x^3 + 27*a^2*b*x + 27*a^3)^(-3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.55 \[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=-\frac {5 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-b x + 3 \, a}}{3 \, \sqrt {-a}}\right )}{2916 \, \sqrt {-a} a^{3} b \mathrm {sgn}\left (-2 \, b x - 3 \, a\right )} - \frac {2}{729 \, \sqrt {-b x + 3 \, a} a^{3} b \mathrm {sgn}\left (-2 \, b x - 3 \, a\right )} - \frac {14 \, {\left (-b x + 3 \, a\right )}^{\frac {3}{2}} - 81 \, \sqrt {-b x + 3 \, a} a}{1458 \, {\left (2 \, b x + 3 \, a\right )}^{2} a^{3} b \mathrm {sgn}\left (-2 \, b x - 3 \, a\right )} \] Input:

integrate(1/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2),x, algorithm="giac")
 

Output:

-5/2916*sqrt(2)*arctan(1/3*sqrt(2)*sqrt(-b*x + 3*a)/sqrt(-a))/(sqrt(-a)*a^ 
3*b*sgn(-2*b*x - 3*a)) - 2/729/(sqrt(-b*x + 3*a)*a^3*b*sgn(-2*b*x - 3*a)) 
- 1/1458*(14*(-b*x + 3*a)^(3/2) - 81*sqrt(-b*x + 3*a)*a)/((2*b*x + 3*a)^2* 
a^3*b*sgn(-2*b*x - 3*a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=\int \frac {1}{{\left (27\,a^3+27\,a^2\,b\,x-4\,b^3\,x^3\right )}^{3/2}} \,d x \] Input:

int(1/(27*a^3 - 4*b^3*x^3 + 27*a^2*b*x)^(3/2),x)
 

Output:

int(1/(27*a^3 - 4*b^3*x^3 + 27*a^2*b*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (27 a^3+27 a^2 b x-4 b^3 x^3\right )^{3/2}} \, dx=\frac {45 \sqrt {a}\, \sqrt {-b x +3 a}\, \sqrt {2}\, \mathrm {log}\left (2 \sqrt {-b x +3 a}-3 \sqrt {a}\, \sqrt {2}\right ) a^{2}+60 \sqrt {a}\, \sqrt {-b x +3 a}\, \sqrt {2}\, \mathrm {log}\left (2 \sqrt {-b x +3 a}-3 \sqrt {a}\, \sqrt {2}\right ) a b x +20 \sqrt {a}\, \sqrt {-b x +3 a}\, \sqrt {2}\, \mathrm {log}\left (2 \sqrt {-b x +3 a}-3 \sqrt {a}\, \sqrt {2}\right ) b^{2} x^{2}-45 \sqrt {a}\, \sqrt {-b x +3 a}\, \sqrt {2}\, \mathrm {log}\left (2 \sqrt {-b x +3 a}+3 \sqrt {a}\, \sqrt {2}\right ) a^{2}-60 \sqrt {a}\, \sqrt {-b x +3 a}\, \sqrt {2}\, \mathrm {log}\left (2 \sqrt {-b x +3 a}+3 \sqrt {a}\, \sqrt {2}\right ) a b x -20 \sqrt {a}\, \sqrt {-b x +3 a}\, \sqrt {2}\, \mathrm {log}\left (2 \sqrt {-b x +3 a}+3 \sqrt {a}\, \sqrt {2}\right ) b^{2} x^{2}-324 a^{3}+180 a^{2} b x +120 a \,b^{2} x^{2}}{5832 \sqrt {-b x +3 a}\, a^{4} b \left (4 b^{2} x^{2}+12 a b x +9 a^{2}\right )} \] Input:

int(1/(-4*b^3*x^3+27*a^2*b*x+27*a^3)^(3/2),x)
 

Output:

(45*sqrt(a)*sqrt(3*a - b*x)*sqrt(2)*log(2*sqrt(3*a - b*x) - 3*sqrt(a)*sqrt 
(2))*a**2 + 60*sqrt(a)*sqrt(3*a - b*x)*sqrt(2)*log(2*sqrt(3*a - b*x) - 3*s 
qrt(a)*sqrt(2))*a*b*x + 20*sqrt(a)*sqrt(3*a - b*x)*sqrt(2)*log(2*sqrt(3*a 
- b*x) - 3*sqrt(a)*sqrt(2))*b**2*x**2 - 45*sqrt(a)*sqrt(3*a - b*x)*sqrt(2) 
*log(2*sqrt(3*a - b*x) + 3*sqrt(a)*sqrt(2))*a**2 - 60*sqrt(a)*sqrt(3*a - b 
*x)*sqrt(2)*log(2*sqrt(3*a - b*x) + 3*sqrt(a)*sqrt(2))*a*b*x - 20*sqrt(a)* 
sqrt(3*a - b*x)*sqrt(2)*log(2*sqrt(3*a - b*x) + 3*sqrt(a)*sqrt(2))*b**2*x* 
*2 - 324*a**3 + 180*a**2*b*x + 120*a*b**2*x**2)/(5832*sqrt(3*a - b*x)*a**4 
*b*(9*a**2 + 12*a*b*x + 4*b**2*x**2))