\(\int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx\) [119]

Optimal result

Integrand size = 33, antiderivative size = 139 \[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx=\frac {g \sqrt {a+b x+c x^2}}{c}-\frac {d \sqrt {a+b x+c x^2}}{a x}+\frac {(b d-2 a e) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{3/2}}+\frac {(2 c f-b g) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}} \] Output:

g*(c*x^2+b*x+a)^(1/2)/c-d*(c*x^2+b*x+a)^(1/2)/a/x+1/2*(-2*a*e+b*d)*arctanh 
(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(3/2)+1/2*(-b*g+2*c*f)*arcta 
nh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2)
 

Mathematica [A] (verified)

Time = 1.24 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.05 \[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx=\frac {(-c d+a g x) \sqrt {a+x (b+c x)}}{a c x}+\frac {(2 c f-b g) \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{c^{3/2}}+\frac {(b d-2 a e) \log (x)}{2 a^{3/2}}+\frac {(-b d+2 a e) \log \left (a \left (2 a+b x-2 \sqrt {a} \sqrt {a+x (b+c x)}\right )\right )}{2 a^{3/2}} \] Input:

Integrate[(d + e*x + f*x^2 + g*x^3)/(x^2*Sqrt[a + b*x + c*x^2]),x]
 

Output:

((-(c*d) + a*g*x)*Sqrt[a + x*(b + c*x)])/(a*c*x) + ((2*c*f - b*g)*ArcTanh[ 
(Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/c^(3/2) + ((b*d - 2*a*e)* 
Log[x])/(2*a^(3/2)) + ((-(b*d) + 2*a*e)*Log[a*(2*a + b*x - 2*Sqrt[a]*Sqrt[ 
a + x*(b + c*x)])])/(2*a^(3/2))
 

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2181, 27, 2184, 1269, 1092, 219, 1154, 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.

\(\displaystyle \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx\)

\(\Big \downarrow \) 2181

\(\displaystyle -\frac {\int \frac {-2 a g x^2-2 a f x+b d-2 a e}{2 x \sqrt {c x^2+b x+a}}dx}{a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\int \frac {-2 a g x^2-2 a f x+b d-2 a e}{x \sqrt {c x^2+b x+a}}dx}{2 a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

\(\Big \downarrow \) 2184

\(\displaystyle -\frac {\frac {\int \frac {c (b d-2 a e)-a (2 c f-b g) x}{x \sqrt {c x^2+b x+a}}dx}{c}-\frac {2 a g \sqrt {a+b x+c x^2}}{c}}{2 a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

\(\Big \downarrow \) 1269

\(\displaystyle -\frac {\frac {c (b d-2 a e) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-a (2 c f-b g) \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{c}-\frac {2 a g \sqrt {a+b x+c x^2}}{c}}{2 a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

\(\Big \downarrow \) 1092

\(\displaystyle -\frac {\frac {c (b d-2 a e) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-2 a (2 c f-b g) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c}-\frac {2 a g \sqrt {a+b x+c x^2}}{c}}{2 a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {c (b d-2 a e) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-\frac {a (2 c f-b g) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}}{c}-\frac {2 a g \sqrt {a+b x+c x^2}}{c}}{2 a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

\(\Big \downarrow \) 1154

\(\displaystyle -\frac {\frac {-2 c (b d-2 a e) \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}-\frac {a (2 c f-b g) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}}{c}-\frac {2 a g \sqrt {a+b x+c x^2}}{c}}{2 a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {-\frac {c (b d-2 a e) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a}}-\frac {a (2 c f-b g) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}}{c}-\frac {2 a g \sqrt {a+b x+c x^2}}{c}}{2 a}-\frac {d \sqrt {a+b x+c x^2}}{a x}\)

Input:

Int[(d + e*x + f*x^2 + g*x^3)/(x^2*Sqrt[a + b*x + c*x^2]),x]
 

Output:

-((d*Sqrt[a + b*x + c*x^2])/(a*x)) - ((-2*a*g*Sqrt[a + b*x + c*x^2])/c + ( 
-((c*(b*d - 2*a*e)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])]) 
/Sqrt[a]) - (a*(2*c*f - b*g)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + 
 c*x^2])])/Sqrt[c])/c)/(2*a)
 

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 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 1092

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]
 

rule 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 2181

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = Polynomi 
alRemainder[Pq, d + e*x, x]}, Simp[(e*R*(d + e*x)^(m + 1)*(a + b*x + c*x^2) 
^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Simp[1/((m + 1)*(c*d^2 - 
b*d*e + a*e^2))   Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*ExpandToSum[(m 
+ 1)*(c*d^2 - b*d*e + a*e^2)*Qx + c*d*R*(m + 1) - b*e*R*(m + p + 2) - c*e*R 
*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, p}, x] && PolyQ[Pq, 
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1]
 

rule 2184

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S 
imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q 
+ 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + 
b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 
1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q - 1) - c 
*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[ 
q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && Pol 
yQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !(IGt 
Q[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {d \sqrt {c \,x^{2}+b x +a}}{a x}+\frac {-\frac {\left (2 a e -b d \right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\sqrt {a}}+\frac {2 a f \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+2 a g \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )}{2 a}\)	\(156\)
default	\(\frac {f \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{\sqrt {c}}+d \left (-\frac {\sqrt {c \,x^{2}+b x +a}}{a x}+\frac {b \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )-\frac {e \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\sqrt {a}}+g \left (\frac {\sqrt {c \,x^{2}+b x +a}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {3}{2}}}\right )\)	\(175\)

Input:

int((g*x^3+f*x^2+e*x+d)/x^2/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-d*(c*x^2+b*x+a)^(1/2)/a/x+1/2/a*(-(2*a*e-b*d)/a^(1/2)*ln((2*a+b*x+2*a^(1/ 
2)*(c*x^2+b*x+a)^(1/2))/x)+2*a*f*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2 
))/c^(1/2)+2*a*g*(1/c*(c*x^2+b*x+a)^(1/2)-1/2*b/c^(3/2)*ln((1/2*b+c*x)/c^( 
1/2)+(c*x^2+b*x+a)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 1.15 (sec) , antiderivative size = 703, normalized size of antiderivative = 5.06 \[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx =\text {Too large to display} \] Input:

integrate((g*x^3+f*x^2+e*x+d)/x^2/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas 
")
 

Output:

[-1/4*((2*a^2*c*f - a^2*b*g)*sqrt(c)*x*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) + (b*c^2*d - 2*a*c^2*e) 
*sqrt(a)*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 + b*x + a)*(b* 
x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 4*(a^2*c*g*x - a*c^2*d)*sqrt(c*x^2 + b*x 
+ a))/(a^2*c^2*x), -1/4*(2*(2*a^2*c*f - a^2*b*g)*sqrt(-c)*x*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)) + (b*c^2* 
d - 2*a*c^2*e)*sqrt(a)*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqrt(c*x^2 
+ b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 4*(a^2*c*g*x - a*c^2*d)*sqr 
t(c*x^2 + b*x + a))/(a^2*c^2*x), -1/4*(2*(b*c^2*d - 2*a*c^2*e)*sqrt(-a)*x* 
arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a 
^2)) + (2*a^2*c*f - a^2*b*g)*sqrt(c)*x*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*(a^2*c*g*x - a*c^2* 
d)*sqrt(c*x^2 + b*x + a))/(a^2*c^2*x), -1/2*((b*c^2*d - 2*a*c^2*e)*sqrt(-a 
)*x*arctan(1/2*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x 
 + a^2)) + (2*a^2*c*f - a^2*b*g)*sqrt(-c)*x*arctan(1/2*sqrt(c*x^2 + b*x + 
a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 + b*c*x + a*c)) - 2*(a^2*c*g*x - a*c^2*d) 
*sqrt(c*x^2 + b*x + a))/(a^2*c^2*x)]
 

Sympy [F]

\[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx=\int \frac {d + e x + f x^{2} + g x^{3}}{x^{2} \sqrt {a + b x + c x^{2}}}\, dx \] Input:

integrate((g*x**3+f*x**2+e*x+d)/x**2/(c*x**2+b*x+a)**(1/2),x)
 

Output:

Integral((d + e*x + f*x**2 + g*x**3)/(x**2*sqrt(a + b*x + c*x**2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((g*x^3+f*x^2+e*x+d)/x^2/(c*x^2+b*x+a)^(1/2),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
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.22 \[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x + a} g}{c} - \frac {{\left (b d - 2 \, a e\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} - \frac {{\left (2 \, c f - b g\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {3}{2}}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} b d + 2 \, a \sqrt {c} d}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )} a} \] Input:

integrate((g*x^3+f*x^2+e*x+d)/x^2/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

sqrt(c*x^2 + b*x + a)*g/c - (b*d - 2*a*e)*arctan(-(sqrt(c)*x - sqrt(c*x^2 
+ b*x + a))/sqrt(-a))/(sqrt(-a)*a) - 1/2*(2*c*f - b*g)*log(abs(2*(sqrt(c)* 
x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(3/2) + ((sqrt(c)*x - sqrt(c*x^ 
2 + b*x + a))*b*d + 2*a*sqrt(c)*d)/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 
 - a)*a)
 

Mupad [B] (verification not implemented)

Time = 20.22 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.19 \[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx=\frac {g\,\sqrt {c\,x^2+b\,x+a}}{c}-\frac {e\,\ln \left (\frac {b}{2}+\frac {a}{x}+\frac {\sqrt {a}\,\sqrt {c\,x^2+b\,x+a}}{x}\right )}{\sqrt {a}}+\frac {f\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{\sqrt {c}}-\frac {b\,g\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{2\,c^{3/2}}-\frac {d\,\sqrt {c\,x^2+b\,x+a}}{a\,x}+\frac {b\,d\,\mathrm {atanh}\left (\frac {a+\frac {b\,x}{2}}{\sqrt {a}\,\sqrt {c\,x^2+b\,x+a}}\right )}{2\,a^{3/2}} \] Input:

int((d + e*x + f*x^2 + g*x^3)/(x^2*(a + b*x + c*x^2)^(1/2)),x)
 

Output:

(g*(a + b*x + c*x^2)^(1/2))/c - (e*log(b/2 + a/x + (a^(1/2)*(a + b*x + c*x 
^2)^(1/2))/x))/a^(1/2) + (f*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1 
/2)))/c^(1/2) - (b*g*log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2)))/( 
2*c^(3/2)) - (d*(a + b*x + c*x^2)^(1/2))/(a*x) + (b*d*atanh((a + (b*x)/2)/ 
(a^(1/2)*(a + b*x + c*x^2)^(1/2))))/(2*a^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.50 \[ \int \frac {d+e x+f x^2+g x^3}{x^2 \sqrt {a+b x+c x^2}} \, dx=\frac {2 \sqrt {c \,x^{2}+b x +a}\, a^{2} c g x -2 \sqrt {c \,x^{2}+b x +a}\, a \,c^{2} d +2 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a \,c^{2} e x -\sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) b \,c^{2} d x -2 \sqrt {a}\, \mathrm {log}\left (x \right ) a \,c^{2} e x +\sqrt {a}\, \mathrm {log}\left (x \right ) b \,c^{2} d x -\sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a^{2} b g x +2 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a^{2} c f x}{2 a^{2} c^{2} x} \] Input:

int((g*x^3+f*x^2+e*x+d)/x^2/(c*x^2+b*x+a)^(1/2),x)
 

Output:

(2*sqrt(a + b*x + c*x**2)*a**2*c*g*x - 2*sqrt(a + b*x + c*x**2)*a*c**2*d + 
 2*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a*c**2*e*x - 
sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*b*c**2*d*x - 2*s 
qrt(a)*log(x)*a*c**2*e*x + sqrt(a)*log(x)*b*c**2*d*x - sqrt(c)*log( - 2*sq 
rt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)*a**2*b*g*x + 2*sqrt(c)*log( - 2* 
sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)*a**2*c*f*x)/(2*a**2*c**2*x)