Integrand size = 26, antiderivative size = 132 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2}{\left (b^2-4 a c\right ) d^3 (b+2 c x)^2 \sqrt {a+b x+c x^2}}-\frac {12 c \sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right )^2 d^3 (b+2 c x)^2}-\frac {6 \sqrt {c} \arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{5/2} d^3} \] Output:
-2/(-4*a*c+b^2)/d^3/(2*c*x+b)^2/(c*x^2+b*x+a)^(1/2)-12*c*(c*x^2+b*x+a)^(1/ 2)/(-4*a*c+b^2)^2/d^3/(2*c*x+b)^2-6*c^(1/2)*arctan(2*c^(1/2)*(c*x^2+b*x+a) ^(1/2)/(-4*a*c+b^2)^(1/2))/(-4*a*c+b^2)^(5/2)/d^3
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.04 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.45 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\frac {4 c (a+x (b+c x))}{-b^2+4 a c}\right )}{\left (b^2-4 a c\right )^2 d^3 \sqrt {a+x (b+c x)}} \] Input:
Integrate[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)),x]
Output:
(-2*Hypergeometric2F1[-1/2, 2, 1/2, (4*c*(a + x*(b + c*x)))/(-b^2 + 4*a*c) ])/((b^2 - 4*a*c)^2*d^3*Sqrt[a + x*(b + c*x)])
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1111, 27, 1117, 1112, 218}
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.
| \(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^{3/2} (b d+2 c d x)^3} \, dx\) |
| \(\Big \downarrow \) 1111 |
| \(\displaystyle -\frac {12 c \int \frac {1}{d^3 (b+2 c x)^3 \sqrt {c x^2+b x+a}}dx}{b^2-4 a c}-\frac {2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {12 c \int \frac {1}{(b+2 c x)^3 \sqrt {c x^2+b x+a}}dx}{d^3 \left (b^2-4 a c\right )}-\frac {2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 1117 |
| \(\displaystyle -\frac {12 c \left (\frac {\int \frac {1}{(b+2 c x) \sqrt {c x^2+b x+a}}dx}{2 \left (b^2-4 a c\right )}+\frac {\sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) (b+2 c x)^2}\right )}{d^3 \left (b^2-4 a c\right )}-\frac {2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 1112 |
| \(\displaystyle -\frac {12 c \left (\frac {2 c \int \frac {1}{8 \left (c x^2+b x+a\right ) c^2+2 \left (b^2-4 a c\right ) c}d\sqrt {c x^2+b x+a}}{b^2-4 a c}+\frac {\sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) (b+2 c x)^2}\right )}{d^3 \left (b^2-4 a c\right )}-\frac {2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt {a+b x+c x^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {12 c \left (\frac {\arctan \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}{2 \sqrt {c} \left (b^2-4 a c\right )^{3/2}}+\frac {\sqrt {a+b x+c x^2}}{\left (b^2-4 a c\right ) (b+2 c x)^2}\right )}{d^3 \left (b^2-4 a c\right )}-\frac {2}{d^3 \left (b^2-4 a c\right ) (b+2 c x)^2 \sqrt {a+b x+c x^2}}\) |
Input:
Int[1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)),x]
Output:
-2/((b^2 - 4*a*c)*d^3*(b + 2*c*x)^2*Sqrt[a + b*x + c*x^2]) - (12*c*(Sqrt[a + b*x + c*x^2]/((b^2 - 4*a*c)*(b + 2*c*x)^2) + ArcTan[(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c]]/(2*Sqrt[c]*(b^2 - 4*a*c)^(3/2))))/((b^2 - 4*a*c)*d^3)
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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*c*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(e*(p + 1)* (b^2 - 4*a*c))), x] - Simp[2*c*e*((m + 2*p + 3)/(e*(p + 1)*(b^2 - 4*a*c))) Int[(d + e*x)^m*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e , m}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[p, -1] && !G tQ[m, 1] && RationalQ[m] && IntegerQ[2*p]
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symb ol] :> Simp[4*c Subst[Int[1/(b^2*e - 4*a*c*e + 4*c*e*x^2), x], x, Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[-2*b*d*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(d^2*(m + 1)*(b^2 - 4*a*c))), x] + Simp[b^2*((m + 2*p + 3)/(d^2*(m + 1)*(b^2 - 4*a* c))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[2*c*d - b*e, 0] && NeQ[m + 2*p + 3, 0] && LtQ[m, -1] & & (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || IntegerQ[(m + 2*p + 3) /2])
Time = 1.17 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.87
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {2}{\sqrt {c \,x^{2}+b x +a}}-4 c \left (\frac {\sqrt {c \,x^{2}+b x +a}}{4 c^{2} x^{2}+4 c b x +b^{2}}-\frac {3 \,\operatorname {arctanh}\left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, c}{\sqrt {\left (4 a c -b^{2}\right ) c}}\right )}{2 \sqrt {\left (4 a c -b^{2}\right ) c}}\right )}{\left (4 a c -b^{2}\right )^{2} d^{3}}\) | \(115\) |
| default | \(\frac {-\frac {2 c}{\left (4 a c -b^{2}\right ) \left (x +\frac {b}{2 c}\right )^{2} \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}-\frac {6 c^{2} \left (\frac {4 c}{\left (4 a c -b^{2}\right ) \sqrt {c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{4 c}}}-\frac {8 c \ln \left (\frac {\frac {4 a c -b^{2}}{2 c}+\frac {\sqrt {\frac {4 a c -b^{2}}{c}}\, \sqrt {4 c \left (x +\frac {b}{2 c}\right )^{2}+\frac {4 a c -b^{2}}{c}}}{2}}{x +\frac {b}{2 c}}\right )}{\left (4 a c -b^{2}\right ) \sqrt {\frac {4 a c -b^{2}}{c}}}\right )}{4 a c -b^{2}}}{8 d^{3} c^{3}}\) | \(235\) |
Input:
int(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/(4*a*c-b^2)^2*(-1/(c*x^2+b*x+a)^(1/2)-2*c*((c*x^2+b*x+a)^(1/2)/(4*c^2*x^ 2+4*b*c*x+b^2)-3/2/((4*a*c-b^2)*c)^(1/2)*arctanh(2*(c*x^2+b*x+a)^(1/2)*c/( (4*a*c-b^2)*c)^(1/2))))/d^3
Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (118) = 236\).
Time = 0.59 (sec) , antiderivative size = 702, normalized size of antiderivative = 5.32 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (4 \, c^{3} x^{4} + 8 \, b c^{2} x^{3} + a b^{2} + {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} + 4 \, a b c\right )} x\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}} \log \left (-\frac {4 \, c^{2} x^{2} + 4 \, b c x - b^{2} + 8 \, a c - 4 \, \sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {-\frac {c}{b^{2} - 4 \, a c}}}{4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}\right ) - 2 \, {\left (6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \, {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} + {\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} + {\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x + {\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}}, -\frac {2 \, {\left (3 \, {\left (4 \, c^{3} x^{4} + 8 \, b c^{2} x^{3} + a b^{2} + {\left (5 \, b^{2} c + 4 \, a c^{2}\right )} x^{2} + {\left (b^{3} + 4 \, a b c\right )} x\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}} \arctan \left (-\frac {\sqrt {c x^{2} + b x + a} {\left (b^{2} - 4 \, a c\right )} \sqrt {\frac {c}{b^{2} - 4 \, a c}}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + {\left (6 \, c^{2} x^{2} + 6 \, b c x + b^{2} + 2 \, a c\right )} \sqrt {c x^{2} + b x + a}\right )}}{4 \, {\left (b^{4} c^{3} - 8 \, a b^{2} c^{4} + 16 \, a^{2} c^{5}\right )} d^{3} x^{4} + 8 \, {\left (b^{5} c^{2} - 8 \, a b^{3} c^{3} + 16 \, a^{2} b c^{4}\right )} d^{3} x^{3} + {\left (5 \, b^{6} c - 36 \, a b^{4} c^{2} + 48 \, a^{2} b^{2} c^{3} + 64 \, a^{3} c^{4}\right )} d^{3} x^{2} + {\left (b^{7} - 4 \, a b^{5} c - 16 \, a^{2} b^{3} c^{2} + 64 \, a^{3} b c^{3}\right )} d^{3} x + {\left (a b^{6} - 8 \, a^{2} b^{4} c + 16 \, a^{3} b^{2} c^{2}\right )} d^{3}}\right ] \] Input:
integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
Output:
[(3*(4*c^3*x^4 + 8*b*c^2*x^3 + a*b^2 + (5*b^2*c + 4*a*c^2)*x^2 + (b^3 + 4* a*b*c)*x)*sqrt(-c/(b^2 - 4*a*c))*log(-(4*c^2*x^2 + 4*b*c*x - b^2 + 8*a*c - 4*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(-c/(b^2 - 4*a*c)))/(4*c^2*x^2 + 4*b*c*x + b^2)) - 2*(6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)*sqrt(c*x^2 + b*x + a))/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8*(b^5*c^2 - 8*a* b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3) *d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3), -2*(3*(4*c^3*x^4 + 8 *b*c^2*x^3 + a*b^2 + (5*b^2*c + 4*a*c^2)*x^2 + (b^3 + 4*a*b*c)*x)*sqrt(c/( b^2 - 4*a*c))*arctan(-1/2*sqrt(c*x^2 + b*x + a)*(b^2 - 4*a*c)*sqrt(c/(b^2 - 4*a*c))/(c^2*x^2 + b*c*x + a*c)) + (6*c^2*x^2 + 6*b*c*x + b^2 + 2*a*c)*s qrt(c*x^2 + b*x + a))/(4*(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*d^3*x^4 + 8* (b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4)*d^3*x^3 + (5*b^6*c - 36*a*b^4*c^2 + 48*a^2*b^2*c^3 + 64*a^3*c^4)*d^3*x^2 + (b^7 - 4*a*b^5*c - 16*a^2*b^3*c^2 + 64*a^3*b*c^3)*d^3*x + (a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*d^3)]
\[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {\int \frac {1}{a b^{3} \sqrt {a + b x + c x^{2}} + 6 a b^{2} c x \sqrt {a + b x + c x^{2}} + 12 a b c^{2} x^{2} \sqrt {a + b x + c x^{2}} + 8 a c^{3} x^{3} \sqrt {a + b x + c x^{2}} + b^{4} x \sqrt {a + b x + c x^{2}} + 7 b^{3} c x^{2} \sqrt {a + b x + c x^{2}} + 18 b^{2} c^{2} x^{3} \sqrt {a + b x + c x^{2}} + 20 b c^{3} x^{4} \sqrt {a + b x + c x^{2}} + 8 c^{4} x^{5} \sqrt {a + b x + c x^{2}}}\, dx}{d^{3}} \] Input:
integrate(1/(2*c*d*x+b*d)**3/(c*x**2+b*x+a)**(3/2),x)
Output:
Integral(1/(a*b**3*sqrt(a + b*x + c*x**2) + 6*a*b**2*c*x*sqrt(a + b*x + c* x**2) + 12*a*b*c**2*x**2*sqrt(a + b*x + c*x**2) + 8*a*c**3*x**3*sqrt(a + b *x + c*x**2) + b**4*x*sqrt(a + b*x + c*x**2) + 7*b**3*c*x**2*sqrt(a + b*x + c*x**2) + 18*b**2*c**2*x**3*sqrt(a + b*x + c*x**2) + 20*b*c**3*x**4*sqrt (a + b*x + c*x**2) + 8*c**4*x**5*sqrt(a + b*x + c*x**2)), x)/d**3
Exception generated. \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (118) = 236\).
Time = 0.45 (sec) , antiderivative size = 432, normalized size of antiderivative = 3.27 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {12 \, c \arctan \left (-\frac {2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c + b \sqrt {c}}{\sqrt {b^{2} c - 4 \, a c^{2}}}\right )}{{\left (b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}\right )} \sqrt {b^{2} c - 4 \, a c^{2}}} - \frac {2 \, {\left (b^{6} d^{3} - 12 \, a b^{4} c d^{3} + 48 \, a^{2} b^{2} c^{2} d^{3} - 64 \, a^{3} c^{3} d^{3}\right )}}{{\left (b^{10} d^{6} - 20 \, a b^{8} c d^{6} + 160 \, a^{2} b^{6} c^{2} d^{6} - 640 \, a^{3} b^{4} c^{3} d^{6} + 1280 \, a^{4} b^{2} c^{4} d^{6} - 1024 \, a^{5} c^{5} d^{6}\right )} \sqrt {c x^{2} + b x + a}} + \frac {4 \, {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} c^{2} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} b c^{\frac {3}{2}} + {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b^{2} c + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} a c^{2} + a b c^{\frac {3}{2}}\right )}}{{\left (b^{4} d^{3} - 8 \, a b^{2} c d^{3} + 16 \, a^{2} c^{2} d^{3}\right )} {\left (2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} c + 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b \sqrt {c} + b^{2} - 2 \, a c\right )}^{2}} \] Input:
integrate(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
Output:
-12*c*arctan(-(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c + b*sqrt(c))/sqrt(b ^2*c - 4*a*c^2))/((b^4*d^3 - 8*a*b^2*c*d^3 + 16*a^2*c^2*d^3)*sqrt(b^2*c - 4*a*c^2)) - 2*(b^6*d^3 - 12*a*b^4*c*d^3 + 48*a^2*b^2*c^2*d^3 - 64*a^3*c^3* d^3)/((b^10*d^6 - 20*a*b^8*c*d^6 + 160*a^2*b^6*c^2*d^6 - 640*a^3*b^4*c^3*d ^6 + 1280*a^4*b^2*c^4*d^6 - 1024*a^5*c^5*d^6)*sqrt(c*x^2 + b*x + a)) + 4*( 2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*c^2 + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^(3/2) + (sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^2*c + 2*(sqr t(c)*x - sqrt(c*x^2 + b*x + a))*a*c^2 + a*b*c^(3/2))/((b^4*d^3 - 8*a*b^2*c *d^3 + 16*a^2*c^2*d^3)*(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c + 2*(sqr t(c)*x - sqrt(c*x^2 + b*x + a))*b*sqrt(c) + b^2 - 2*a*c)^2)
Timed out. \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (b\,d+2\,c\,d\,x\right )}^3\,{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \] Input:
int(1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)),x)
Output:
int(1/((b*d + 2*c*d*x)^3*(a + b*x + c*x^2)^(3/2)), x)
Time = 0.32 (sec) , antiderivative size = 1330, normalized size of antiderivative = 10.08 \[ \int \frac {1}{(b d+2 c d x)^3 \left (a+b x+c x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:
int(1/(2*c*d*x+b*d)^3/(c*x^2+b*x+a)^(3/2),x)
Output:
(2*( - 3*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c) *sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2 - 12*sqrt( c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c*x - 12*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*c**2*x**2 - 3*sqrt(c)*sqrt(4*a*c - b**2)*l og(( - sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/ sqrt(4*a*c - b**2))*b**3*x - 15*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4* a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b **2))*b**2*c*x**2 - 24*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b** 2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*b*c **2*x**3 - 12*sqrt(c)*sqrt(4*a*c - b**2)*log(( - sqrt(4*a*c - b**2) + 2*sq rt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*c**3*x**4 + 3*sqrt(c)*sqrt(4*a*c - b**2)*log((sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b**2 + 12*sqrt(c)*sqrt(4* a*c - b**2)*log((sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a*b*c*x + 12*sqrt(c)*sqrt(4*a*c - b**2)*log( (sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4 *a*c - b**2))*a*c**2*x**2 + 3*sqrt(c)*sqrt(4*a*c - b**2)*log((sqrt(4*a*c - b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**...