Integrand size = 25, antiderivative size = 175 \[ \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx=\frac {\left (16 c^2 d-8 b c e+5 b^2 f-4 a c f\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {(8 c e-5 b f) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (b^2-4 a c\right ) \left (16 c^2 d+5 b^2 f-4 c (2 b e+a f)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}} \]
1/24*(-5*b*f+8*c*e)*(c*x^2+b*x+a)^(3/2)/c^2+1/4*f*x*(c*x^2+b*x+a)^(3/2)/c- 1/128*(-4*a*c+b^2)*(16*c^2*d+5*b^2*f-4*c*(a*f+2*b*e))*arctanh(1/2*(2*c*x+b )/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)+1/64*(-4*a*c*f+5*b^2*f-8*b*c*e+16*c ^2*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^3
Time = 0.07 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.99 \[ \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx=\frac {\sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^3 f-2 b^2 c (12 e+5 f x)+4 b c \left (-13 a f+2 c \left (6 d+2 e x+f x^2\right )\right )+8 c^2 \left (a (8 e+3 f x)+2 c x \left (6 d+4 e x+3 f x^2\right )\right )\right )-3 \left (b^2-4 a c\right ) \left (16 c^2 d+5 b^2 f-4 c (2 b e+a f)\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{192 c^{7/2}} \]
(Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^3*f - 2*b^2*c*(12*e + 5*f*x) + 4*b*c* (-13*a*f + 2*c*(6*d + 2*e*x + f*x^2)) + 8*c^2*(a*(8*e + 3*f*x) + 2*c*x*(6* d + 4*e*x + 3*f*x^2))) - 3*(b^2 - 4*a*c)*(16*c^2*d + 5*b^2*f - 4*c*(2*b*e + a*f))*ArcTanh[(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/(192*c^(7 /2))
Time = 0.33 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2192, 27, 1160, 1087, 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.
\(\displaystyle \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx\) |
\(\Big \downarrow \) 2192 |
\(\displaystyle \frac {\int \frac {1}{2} (8 c d-2 a f+(8 c e-5 b f) x) \sqrt {c x^2+b x+a}dx}{4 c}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (2 (4 c d-a f)+(8 c e-5 b f) x) \sqrt {c x^2+b x+a}dx}{8 c}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\frac {\left (-4 a c f+5 b^2 f-8 b c e+16 c^2 d\right ) \int \sqrt {c x^2+b x+a}dx}{2 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 c e-5 b f)}{3 c}}{8 c}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {\frac {\left (-4 a c f+5 b^2 f-8 b c e+16 c^2 d\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c}\right )}{2 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 c e-5 b f)}{3 c}}{8 c}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {\frac {\left (-4 a c f+5 b^2 f-8 b c e+16 c^2 d\right ) \left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c}\right )}{2 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 c e-5 b f)}{3 c}}{8 c}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\left (\frac {(b+2 c x) \sqrt {a+b x+c x^2}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2}}\right ) \left (-4 a c f+5 b^2 f-8 b c e+16 c^2 d\right )}{2 c}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 c e-5 b f)}{3 c}}{8 c}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}\) |
(f*x*(a + b*x + c*x^2)^(3/2))/(4*c) + (((8*c*e - 5*b*f)*(a + b*x + c*x^2)^ (3/2))/(3*c) + ((16*c^2*d - 8*b*c*e + 5*b^2*f - 4*a*c*f)*(((b + 2*c*x)*Sqr t[a + b*x + c*x^2])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]* Sqrt[a + b*x + c*x^2])])/(8*c^(3/2))))/(2*c))/(8*c)
3.2.89.3.1 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[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(q + 2*p + 1))), x] + Simp[1/(c*(q + 2*p + 1)) Int[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b *e*(q + p)*x^(q - 1) - c*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c , p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Time = 0.67 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.13
method | result | size |
risch | \(-\frac {\left (-48 f \,c^{3} x^{3}-8 b \,c^{2} f \,x^{2}-64 c^{3} e \,x^{2}-24 a \,c^{2} f x +10 b^{2} c f x -16 b \,c^{2} e x -96 c^{3} d x +52 a b c f -64 a \,c^{2} e -15 b^{3} f +24 b^{2} c e -48 b d \,c^{2}\right ) \sqrt {c \,x^{2}+b x +a}}{192 c^{3}}-\frac {\left (16 a^{2} c^{2} f -24 a \,b^{2} c f +32 a b \,c^{2} e -64 a \,c^{3} d +5 b^{4} f -8 b^{3} c e +16 b^{2} d \,c^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}\) | \(197\) |
default | \(d \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )+f \left (\frac {x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{4 c}-\frac {5 b \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )}{8 c}-\frac {a \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{4 c}\right )+e \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 c}-\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2 c}\right )\) | \(343\) |
-1/192*(-48*c^3*f*x^3-8*b*c^2*f*x^2-64*c^3*e*x^2-24*a*c^2*f*x+10*b^2*c*f*x -16*b*c^2*e*x-96*c^3*d*x+52*a*b*c*f-64*a*c^2*e-15*b^3*f+24*b^2*c*e-48*b*c^ 2*d)*(c*x^2+b*x+a)^(1/2)/c^3-1/128*(16*a^2*c^2*f-24*a*b^2*c*f+32*a*b*c^2*e -64*a*c^3*d+5*b^4*f-8*b^3*c*e+16*b^2*c^2*d)/c^(7/2)*ln((1/2*b+c*x)/c^(1/2) +(c*x^2+b*x+a)^(1/2))
Time = 0.30 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.66 \[ \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx=\left [\frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} f\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} f x^{3} + 48 \, b c^{3} d + 8 \, {\left (8 \, c^{4} e + b c^{3} f\right )} x^{2} - 8 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} f + 2 \, {\left (48 \, c^{4} d + 8 \, b c^{3} e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} f\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{4}}, \frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} f\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} f x^{3} + 48 \, b c^{3} d + 8 \, {\left (8 \, c^{4} e + b c^{3} f\right )} x^{2} - 8 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} f + 2 \, {\left (48 \, c^{4} d + 8 \, b c^{3} e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} f\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{4}}\right ] \]
[1/768*(3*(16*(b^2*c^2 - 4*a*c^3)*d - 8*(b^3*c - 4*a*b*c^2)*e + (5*b^4 - 2 4*a*b^2*c + 16*a^2*c^2)*f)*sqrt(c)*log(-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*(48*c^4*f*x^3 + 48*b*c^ 3*d + 8*(8*c^4*e + b*c^3*f)*x^2 - 8*(3*b^2*c^2 - 8*a*c^3)*e + (15*b^3*c - 52*a*b*c^2)*f + 2*(48*c^4*d + 8*b*c^3*e - (5*b^2*c^2 - 12*a*c^3)*f)*x)*sqr t(c*x^2 + b*x + a))/c^4, 1/384*(3*(16*(b^2*c^2 - 4*a*c^3)*d - 8*(b^3*c - 4 *a*b*c^2)*e + (5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*f)*sqrt(-c)*arctan(1/2*sqr t(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(48*c ^4*f*x^3 + 48*b*c^3*d + 8*(8*c^4*e + b*c^3*f)*x^2 - 8*(3*b^2*c^2 - 8*a*c^3 )*e + (15*b^3*c - 52*a*b*c^2)*f + 2*(48*c^4*d + 8*b*c^3*e - (5*b^2*c^2 - 1 2*a*c^3)*f)*x)*sqrt(c*x^2 + b*x + a))/c^4]
Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (168) = 336\).
Time = 0.54 (sec) , antiderivative size = 384, normalized size of antiderivative = 2.19 \[ \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx=\begin {cases} \sqrt {a + b x + c x^{2}} \left (\frac {f x^{3}}{4} + \frac {x^{2} \left (\frac {b f}{8} + c e\right )}{3 c} + \frac {x \left (\frac {a f}{4} + b e - \frac {5 b \left (\frac {b f}{8} + c e\right )}{6 c} + c d\right )}{2 c} + \frac {a e - \frac {2 a \left (\frac {b f}{8} + c e\right )}{3 c} + b d - \frac {3 b \left (\frac {a f}{4} + b e - \frac {5 b \left (\frac {b f}{8} + c e\right )}{6 c} + c d\right )}{4 c}}{c}\right ) + \left (a d - \frac {a \left (\frac {a f}{4} + b e - \frac {5 b \left (\frac {b f}{8} + c e\right )}{6 c} + c d\right )}{2 c} - \frac {b \left (a e - \frac {2 a \left (\frac {b f}{8} + c e\right )}{3 c} + b d - \frac {3 b \left (\frac {a f}{4} + b e - \frac {5 b \left (\frac {b f}{8} + c e\right )}{6 c} + c d\right )}{4 c}\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {a + b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: a - \frac {b^{2}}{4 c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (\frac {f \left (a + b x\right )^{\frac {7}{2}}}{7 b^{2}} + \frac {\left (a + b x\right )^{\frac {5}{2}} \left (- 2 a f + b e\right )}{5 b^{2}} + \frac {\left (a + b x\right )^{\frac {3}{2}} \left (a^{2} f - a b e + b^{2} d\right )}{3 b^{2}}\right )}{b} & \text {for}\: b \neq 0 \\\sqrt {a} \left (d x + \frac {e x^{2}}{2} + \frac {f x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x + c*x**2)*(f*x**3/4 + x**2*(b*f/8 + c*e)/(3*c) + x *(a*f/4 + b*e - 5*b*(b*f/8 + c*e)/(6*c) + c*d)/(2*c) + (a*e - 2*a*(b*f/8 + c*e)/(3*c) + b*d - 3*b*(a*f/4 + b*e - 5*b*(b*f/8 + c*e)/(6*c) + c*d)/(4*c ))/c) + (a*d - a*(a*f/4 + b*e - 5*b*(b*f/8 + c*e)/(6*c) + c*d)/(2*c) - b*( a*e - 2*a*(b*f/8 + c*e)/(3*c) + b*d - 3*b*(a*f/4 + b*e - 5*b*(b*f/8 + c*e) /(6*c) + c*d)/(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*c) + x)**2), True)), Ne(c, 0)), (2*(f*(a + b*x)**(7/2)/ (7*b**2) + (a + b*x)**(5/2)*(-2*a*f + b*e)/(5*b**2) + (a + b*x)**(3/2)*(a* *2*f - a*b*e + b**2*d)/(3*b**2))/b, Ne(b, 0)), (sqrt(a)*(d*x + e*x**2/2 + f*x**3/3), True))
Exception generated. \[ \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx=\text {Exception raised: ValueError} \]
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 = 204, normalized size of antiderivative = 1.17 \[ \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, f x + \frac {8 \, c^{3} e + b c^{2} f}{c^{3}}\right )} x + \frac {48 \, c^{3} d + 8 \, b c^{2} e - 5 \, b^{2} c f + 12 \, a c^{2} f}{c^{3}}\right )} x + \frac {48 \, b c^{2} d - 24 \, b^{2} c e + 64 \, a c^{2} e + 15 \, b^{3} f - 52 \, a b c f}{c^{3}}\right )} + \frac {{\left (16 \, b^{2} c^{2} d - 64 \, a c^{3} d - 8 \, b^{3} c e + 32 \, a b c^{2} e + 5 \, b^{4} f - 24 \, a b^{2} c f + 16 \, a^{2} c^{2} f\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]
1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*f*x + (8*c^3*e + b*c^2*f)/c^3)*x + (4 8*c^3*d + 8*b*c^2*e - 5*b^2*c*f + 12*a*c^2*f)/c^3)*x + (48*b*c^2*d - 24*b^ 2*c*e + 64*a*c^2*e + 15*b^3*f - 52*a*b*c*f)/c^3) + 1/128*(16*b^2*c^2*d - 6 4*a*c^3*d - 8*b^3*c*e + 32*a*b*c^2*e + 5*b^4*f - 24*a*b^2*c*f + 16*a^2*c^2 *f)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)
Time = 13.54 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.83 \[ \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx=d\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}-\frac {a\,f\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}+\frac {d\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}+\frac {e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}-\frac {5\,b\,f\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}+\frac {e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}+\frac {f\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \]
d*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) - (a*f*((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) + (d*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)) + (e*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)) - (5*b*f*((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) + (e*(8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/( 24*c^2) + (f*x*(a + b*x + c*x^2)^(3/2))/(4*c)