Integrand size = 26, antiderivative size = 165 \[ \int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx=-\frac {\left (b^2-4 a c\right )^2 d^4 (b+2 c x) \sqrt {a+b x+c x^2}}{32 c}-\frac {\left (b^2-4 a c\right ) d^4 (b+2 c x)^3 \sqrt {a+b x+c x^2}}{48 c}+\frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (b^2-4 a c\right )^3 d^4 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{64 c^{3/2}} \]
-1/64*(-4*a*c+b^2)^3*d^4*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2) )/c^(3/2)-1/32*(-4*a*c+b^2)^2*d^4*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c-1/48*(-4 *a*c+b^2)*d^4*(2*c*x+b)^3*(c*x^2+b*x+a)^(1/2)/c+1/12*d^4*(2*c*x+b)^5*(c*x^ 2+b*x+a)^(1/2)/c
Time = 2.07 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.92 \[ \int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx=\frac {d^4 \left (\sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)} \left (3 b^4+56 b^3 c x+32 b c^2 x \left (a+8 c x^2\right )+8 b^2 c \left (4 a+23 c x^2\right )+16 c^2 \left (-3 a^2+2 a c x^2+8 c^2 x^4\right )\right )-3 \left (b^2-4 a c\right )^3 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )\right )}{96 c^{3/2}} \]
(d^4*(Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(3*b^4 + 56*b^3*c*x + 32*b *c^2*x*(a + 8*c*x^2) + 8*b^2*c*(4*a + 23*c*x^2) + 16*c^2*(-3*a^2 + 2*a*c*x ^2 + 8*c^2*x^4)) - 3*(b^2 - 4*a*c)^3*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[ a + x*(b + c*x)])]))/(96*c^(3/2))
Time = 0.32 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1109, 27, 1116, 1116, 1092, 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.
\(\displaystyle \int \sqrt {a+b x+c x^2} (b d+2 c d x)^4 \, dx\) |
\(\Big \downarrow \) 1109 |
\(\displaystyle \frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {\left (b^2-4 a c\right ) \int \frac {d^4 (b+2 c x)^4}{\sqrt {c x^2+b x+a}}dx}{24 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {d^4 \left (b^2-4 a c\right ) \int \frac {(b+2 c x)^4}{\sqrt {c x^2+b x+a}}dx}{24 c}\) |
\(\Big \downarrow \) 1116 |
\(\displaystyle \frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {d^4 \left (b^2-4 a c\right ) \left (\frac {3}{4} \left (b^2-4 a c\right ) \int \frac {(b+2 c x)^2}{\sqrt {c x^2+b x+a}}dx+\frac {1}{2} (b+2 c x)^3 \sqrt {a+b x+c x^2}\right )}{24 c}\) |
\(\Big \downarrow \) 1116 |
\(\displaystyle \frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {d^4 \left (b^2-4 a c\right ) \left (\frac {3}{4} \left (b^2-4 a c\right ) \left (\frac {1}{2} \left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx+(b+2 c x) \sqrt {a+b x+c x^2}\right )+\frac {1}{2} (b+2 c x)^3 \sqrt {a+b x+c x^2}\right )}{24 c}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {d^4 \left (b^2-4 a c\right ) \left (\frac {3}{4} \left (b^2-4 a c\right ) \left (\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}+(b+2 c x) \sqrt {a+b x+c x^2}\right )+\frac {1}{2} (b+2 c x)^3 \sqrt {a+b x+c x^2}\right )}{24 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {d^4 (b+2 c x)^5 \sqrt {a+b x+c x^2}}{12 c}-\frac {d^4 \left (b^2-4 a c\right ) \left (\frac {3}{4} \left (b^2-4 a c\right ) \left (\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c}}+(b+2 c x) \sqrt {a+b x+c x^2}\right )+\frac {1}{2} (b+2 c x)^3 \sqrt {a+b x+c x^2}\right )}{24 c}\) |
(d^4*(b + 2*c*x)^5*Sqrt[a + b*x + c*x^2])/(12*c) - ((b^2 - 4*a*c)*d^4*(((b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/2 + (3*(b^2 - 4*a*c)*((b + 2*c*x)*Sqrt[ a + b*x + c*x^2] + ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(2*Sqrt[c])))/4))/(24*c)
3.12.91.3.1 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_)^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[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[d*p*((b^2 - 4*a*c)/(b*e*(m + 2*p + 1))) 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] && GtQ[p, 0] && !LtQ[m, -1] && !(IGtQ[(m - 1 )/2, 0] && ( !IntegerQ[p] || LtQ[m, 2*p])) && RationalQ[m] && IntegerQ[2*p]
Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S ymbol] :> Simp[2*d*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] + Simp[d^2*(m - 1)*((b^2 - 4*a*c)/(b^2*(m + 2*p + 1))) 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] && GtQ[m, 1] && NeQ[m + 2*p + 1, 0] && (IntegerQ[2*p] || (IntegerQ[m] && RationalQ[p]) || OddQ[m])
Time = 2.79 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.17
method | result | size |
risch | \(-\frac {\left (-256 c^{5} x^{5}-640 b \,x^{4} c^{4}-64 a \,c^{4} x^{3}-624 b^{2} c^{3} x^{3}-96 a b \,c^{3} x^{2}-296 x^{2} b^{3} c^{2}+96 a^{2} c^{3} x -96 a \,c^{2} b^{2} x -62 c x \,b^{4}+48 a^{2} b \,c^{2}-32 a \,b^{3} c -3 b^{5}\right ) \sqrt {c \,x^{2}+b x +a}\, d^{4}}{96 c}+\frac {\left (64 c^{3} a^{3}-48 a^{2} b^{2} c^{2}+12 a \,b^{4} c -b^{6}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) d^{4}}{64 c^{\frac {3}{2}}}\) | \(193\) |
default | \(\text {Expression too large to display}\) | \(1194\) |
-1/96/c*(-256*c^5*x^5-640*b*c^4*x^4-64*a*c^4*x^3-624*b^2*c^3*x^3-96*a*b*c^ 3*x^2-296*b^3*c^2*x^2+96*a^2*c^3*x-96*a*b^2*c^2*x-62*b^4*c*x+48*a^2*b*c^2- 32*a*b^3*c-3*b^5)*(c*x^2+b*x+a)^(1/2)*d^4+1/64*(64*a^3*c^3-48*a^2*b^2*c^2+ 12*a*b^4*c-b^6)/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d^4
Time = 0.31 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.87 \[ \int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx=\left [-\frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} d^{4} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (256 \, c^{6} d^{4} x^{5} + 640 \, b c^{5} d^{4} x^{4} + 16 \, {\left (39 \, b^{2} c^{4} + 4 \, a c^{5}\right )} d^{4} x^{3} + 8 \, {\left (37 \, b^{3} c^{3} + 12 \, a b c^{4}\right )} d^{4} x^{2} + 2 \, {\left (31 \, b^{4} c^{2} + 48 \, a b^{2} c^{3} - 48 \, a^{2} c^{4}\right )} d^{4} x + {\left (3 \, b^{5} c + 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{2}}, \frac {3 \, {\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} d^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (256 \, c^{6} d^{4} x^{5} + 640 \, b c^{5} d^{4} x^{4} + 16 \, {\left (39 \, b^{2} c^{4} + 4 \, a c^{5}\right )} d^{4} x^{3} + 8 \, {\left (37 \, b^{3} c^{3} + 12 \, a b c^{4}\right )} d^{4} x^{2} + 2 \, {\left (31 \, b^{4} c^{2} + 48 \, a b^{2} c^{3} - 48 \, a^{2} c^{4}\right )} d^{4} x + {\left (3 \, b^{5} c + 32 \, a b^{3} c^{2} - 48 \, a^{2} b c^{3}\right )} d^{4}\right )} \sqrt {c x^{2} + b x + a}}{192 \, c^{2}}\right ] \]
[-1/384*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)*d^4*lo g(-8*c^2*x^2 - 8*b*c*x - b^2 - 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(256*c^6*d^4*x^5 + 640*b*c^5*d^4*x^4 + 16*(39*b^2*c^4 + 4*a* c^5)*d^4*x^3 + 8*(37*b^3*c^3 + 12*a*b*c^4)*d^4*x^2 + 2*(31*b^4*c^2 + 48*a* b^2*c^3 - 48*a^2*c^4)*d^4*x + (3*b^5*c + 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^4) *sqrt(c*x^2 + b*x + a))/c^2, 1/192*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(-c)*d^4*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqr t(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(256*c^6*d^4*x^5 + 640*b*c^5*d^4*x^4 + 16*(39*b^2*c^4 + 4*a*c^5)*d^4*x^3 + 8*(37*b^3*c^3 + 12*a*b*c^4)*d^4*x^2 + 2*(31*b^4*c^2 + 48*a*b^2*c^3 - 48*a^2*c^4)*d^4*x + (3*b^5*c + 32*a*b^3*c^2 - 48*a^2*b*c^3)*d^4)*sqrt(c*x^2 + b*x + a))/c^2]
Leaf count of result is larger than twice the leaf count of optimal. 1149 vs. \(2 (150) = 300\).
Time = 0.86 (sec) , antiderivative size = 1149, normalized size of antiderivative = 6.96 \[ \int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx=\text {Too large to display} \]
Piecewise((sqrt(a + b*x + c*x**2)*(20*b*c**3*d**4*x**4/3 + 8*c**4*d**4*x** 5/3 + x**3*(8*a*c**4*d**4/3 + 26*b**2*c**3*d**4)/(4*c) + x**2*(16*a*b*c**3 *d**4/3 + 32*b**3*c**2*d**4 - 7*b*(8*a*c**4*d**4/3 + 26*b**2*c**3*d**4)/(8 *c))/(3*c) + x*(24*a*b**2*c**2*d**4 - 3*a*(8*a*c**4*d**4/3 + 26*b**2*c**3* d**4)/(4*c) + 9*b**4*c*d**4 - 5*b*(16*a*b*c**3*d**4/3 + 32*b**3*c**2*d**4 - 7*b*(8*a*c**4*d**4/3 + 26*b**2*c**3*d**4)/(8*c))/(6*c))/(2*c) + (8*a*b** 3*c*d**4 - 2*a*(16*a*b*c**3*d**4/3 + 32*b**3*c**2*d**4 - 7*b*(8*a*c**4*d** 4/3 + 26*b**2*c**3*d**4)/(8*c))/(3*c) + b**5*d**4 - 3*b*(24*a*b**2*c**2*d* *4 - 3*a*(8*a*c**4*d**4/3 + 26*b**2*c**3*d**4)/(4*c) + 9*b**4*c*d**4 - 5*b *(16*a*b*c**3*d**4/3 + 32*b**3*c**2*d**4 - 7*b*(8*a*c**4*d**4/3 + 26*b**2* c**3*d**4)/(8*c))/(6*c))/(4*c))/c) + (a*b**4*d**4 - a*(24*a*b**2*c**2*d**4 - 3*a*(8*a*c**4*d**4/3 + 26*b**2*c**3*d**4)/(4*c) + 9*b**4*c*d**4 - 5*b*( 16*a*b*c**3*d**4/3 + 32*b**3*c**2*d**4 - 7*b*(8*a*c**4*d**4/3 + 26*b**2*c* *3*d**4)/(8*c))/(6*c))/(2*c) - b*(8*a*b**3*c*d**4 - 2*a*(16*a*b*c**3*d**4/ 3 + 32*b**3*c**2*d**4 - 7*b*(8*a*c**4*d**4/3 + 26*b**2*c**3*d**4)/(8*c))/( 3*c) + b**5*d**4 - 3*b*(24*a*b**2*c**2*d**4 - 3*a*(8*a*c**4*d**4/3 + 26*b* *2*c**3*d**4)/(4*c) + 9*b**4*c*d**4 - 5*b*(16*a*b*c**3*d**4/3 + 32*b**3*c* *2*d**4 - 7*b*(8*a*c**4*d**4/3 + 26*b**2*c**3*d**4)/(8*c))/(6*c))/(4*c))/( 2*c))*Piecewise((log(b + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + 2*c*x)/sqrt(c) , Ne(a - b**2/(4*c), 0)), ((b/(2*c) + x)*log(b/(2*c) + x)/sqrt(c*(b/(2*...
Exception generated. \[ \int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx=\text {Exception raised: ValueError} \]
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
Time = 0.30 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.56 \[ \int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx=\frac {1}{96} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, {\left (2 \, c^{4} d^{4} x + 5 \, b c^{3} d^{4}\right )} x + \frac {39 \, b^{2} c^{7} d^{4} + 4 \, a c^{8} d^{4}}{c^{5}}\right )} x + \frac {37 \, b^{3} c^{6} d^{4} + 12 \, a b c^{7} d^{4}}{c^{5}}\right )} x + \frac {31 \, b^{4} c^{5} d^{4} + 48 \, a b^{2} c^{6} d^{4} - 48 \, a^{2} c^{7} d^{4}}{c^{5}}\right )} x + \frac {3 \, b^{5} c^{4} d^{4} + 32 \, a b^{3} c^{5} d^{4} - 48 \, a^{2} b c^{6} d^{4}}{c^{5}}\right )} + \frac {{\left (b^{6} d^{4} - 12 \, a b^{4} c d^{4} + 48 \, a^{2} b^{2} c^{2} d^{4} - 64 \, a^{3} c^{3} d^{4}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{64 \, c^{\frac {3}{2}}} \]
1/96*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*c^4*d^4*x + 5*b*c^3*d^4)*x + (39 *b^2*c^7*d^4 + 4*a*c^8*d^4)/c^5)*x + (37*b^3*c^6*d^4 + 12*a*b*c^7*d^4)/c^5 )*x + (31*b^4*c^5*d^4 + 48*a*b^2*c^6*d^4 - 48*a^2*c^7*d^4)/c^5)*x + (3*b^5 *c^4*d^4 + 32*a*b^3*c^5*d^4 - 48*a^2*b*c^6*d^4)/c^5) + 1/64*(b^6*d^4 - 12* a*b^4*c*d^4 + 48*a^2*b^2*c^2*d^4 - 64*a^3*c^3*d^4)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(3/2)
Time = 11.34 (sec) , antiderivative size = 1144, normalized size of antiderivative = 6.93 \[ \int (b d+2 c d x)^4 \sqrt {a+b x+c x^2} \, dx=\text {Too large to display} \]
8*a*c^3*d^4*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*( b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b *x + c*x^2)^(1/2))/(24*c^2)))/(8*c) - (x*(a + b*x + c*x^2)^(3/2))/(4*c) + (a*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + ( a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c)) - 12*b*c^3*d^4 *((7*b*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) - (x*(a + b*x + c*x^2)^(3/2))/(4*c) + (a*(( x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + (a + b *x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c)))/(10*c) - (2*a*((lo g((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^ (5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24 *c^2)))/(5*c) + (x^2*(a + b*x + c*x^2)^(3/2))/(5*c)) + b^4*d^4*(x/2 + b/(4 *c))*(a + b*x + c*x^2)^(1/2) + (8*c^3*d^4*x^3*(a + b*x + c*x^2)^(3/2))/3 + (112*b^2*c^2*d^4*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1 /2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)* (a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) - (x*(a + b*x + c*x^2)^(3/2))/(4 *c) + (a*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/ 2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c)))/5 - 15* b^3*c*d^4*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 -...