3.1.49 \(\int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x} \, dx\) [49]

3.1.49.1 Optimal result
3.1.49.2 Mathematica [A] (verified)
3.1.49.3 Rubi [A] (verified)
3.1.49.4 Maple [C] (warning: unable to verify)
3.1.49.5 Fricas [A] (verification not implemented)
3.1.49.6 Sympy [F]
3.1.49.7 Maxima [F]
3.1.49.8 Giac [F(-2)]
3.1.49.9 Mupad [F(-1)]

3.1.49.1 Optimal result

Integrand size = 38, antiderivative size = 211 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x} \, dx=\frac {(4 a d+b e+2 b d x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{4 d (a+b x)}+\frac {\left (4 b c d+4 a d e-b e^2\right ) \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{8 d^{3/2} (a+b x)}-\frac {a \sqrt {c} \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{a+b x} \]

output
1/8*(4*a*d*e+4*b*c*d-b*e^2)*arctanh(1/2*(2*d*x+e)/d^(1/2)/(d*x^2+e*x+c)^(1 
/2))*((b*x+a)^2)^(1/2)/d^(3/2)/(b*x+a)-a*arctanh(1/2*(e*x+2*c)/c^(1/2)/(d* 
x^2+e*x+c)^(1/2))*c^(1/2)*((b*x+a)^2)^(1/2)/(b*x+a)+1/4*(2*b*d*x+4*a*d+b*e 
)*((b*x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/d/(b*x+a)
 
3.1.49.2 Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x} \, dx=\frac {\sqrt {(a+b x)^2} \left (\left (4 b c d+4 a d e-b e^2\right ) \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+x (e+d x)}}\right )+2 \sqrt {d} \left (\sqrt {c+x (e+d x)} (4 a d+b (e+2 d x))+8 a \sqrt {c} d \text {arctanh}\left (\frac {\sqrt {d} x-\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )\right )\right )}{8 d^{3/2} (a+b x)} \]

input
Integrate[(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2])/x,x]
 
output
(Sqrt[(a + b*x)^2]*((4*b*c*d + 4*a*d*e - b*e^2)*ArcTanh[(e + 2*d*x)/(2*Sqr 
t[d]*Sqrt[c + x*(e + d*x)])] + 2*Sqrt[d]*(Sqrt[c + x*(e + d*x)]*(4*a*d + b 
*(e + 2*d*x)) + 8*a*Sqrt[c]*d*ArcTanh[(Sqrt[d]*x - Sqrt[c + x*(e + d*x)])/ 
Sqrt[c]])))/(8*d^(3/2)*(a + b*x))
 
3.1.49.3 Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.78, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {1333, 27, 1231, 27, 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.

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

\(\Big \downarrow \) 1333

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1231

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d}-\frac {\int -\frac {8 a c d+\left (-b e^2+4 a d e+4 b c d\right ) x}{2 x \sqrt {d x^2+e x+c}}dx}{4 d}\right )}{a+b x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\int \frac {8 a c d+\left (-b e^2+4 a d e+4 b c d\right ) x}{x \sqrt {d x^2+e x+c}}dx}{8 d}+\frac {\sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d}\right )}{a+b x}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\left (4 a d e+4 b c d-b e^2\right ) \int \frac {1}{\sqrt {d x^2+e x+c}}dx+8 a c d \int \frac {1}{x \sqrt {d x^2+e x+c}}dx}{8 d}+\frac {\sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d}\right )}{a+b x}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {2 \left (4 a d e+4 b c d-b e^2\right ) \int \frac {1}{4 d-\frac {(e+2 d x)^2}{d x^2+e x+c}}d\frac {e+2 d x}{\sqrt {d x^2+e x+c}}+8 a c d \int \frac {1}{x \sqrt {d x^2+e x+c}}dx}{8 d}+\frac {\sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d}\right )}{a+b x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {8 a c d \int \frac {1}{x \sqrt {d x^2+e x+c}}dx+\frac {\left (4 a d e+4 b c d-b e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{\sqrt {d}}}{8 d}+\frac {\sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d}\right )}{a+b x}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\frac {\left (4 a d e+4 b c d-b e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{\sqrt {d}}-16 a c d \int \frac {1}{4 c-\frac {(2 c+e x)^2}{d x^2+e x+c}}d\frac {2 c+e x}{\sqrt {d x^2+e x+c}}}{8 d}+\frac {\sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d}\right )}{a+b x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {\frac {\left (4 a d e+4 b c d-b e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{\sqrt {d}}-8 a \sqrt {c} d \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{8 d}+\frac {\sqrt {c+d x^2+e x} (4 a d+2 b d x+b e)}{4 d}\right )}{a+b x}\)

input
Int[(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + e*x + d*x^2])/x,x]
 
output
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(((4*a*d + b*e + 2*b*d*x)*Sqrt[c + e*x + d* 
x^2])/(4*d) + (((4*b*c*d + 4*a*d*e - b*e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d] 
*Sqrt[c + e*x + d*x^2])])/Sqrt[d] - 8*a*Sqrt[c]*d*ArcTanh[(2*c + e*x)/(2*S 
qrt[c]*Sqrt[c + e*x + d*x^2])])/(8*d)))/(a + b*x)
 

3.1.49.3.1 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 1231
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) 
 - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^2)^p/ 
(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Simp[p/(c*e^2*(m + 2*p + 1)*(m + 
 2*p + 2))   Int[(d + e*x)^m*(a + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2* 
a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p - c*d - 2* 
c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c 
^2*d^2*(1 + 2*p) - c*e*(b*d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x 
] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (IntegerQ[p] ||  !R 
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (Integer 
Q[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

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 1333
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ 
) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(a + b*x + c*x^2)^Fr 
acPart[p]/((4*c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p]))   Int[(g + h*x)^m* 
(b + 2*c*x)^(2*p)*(d + e*x + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g 
, h, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]
 
3.1.49.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.70 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.02

method result size
default \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (4 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} b x -8 \sqrt {c}\, d^{\frac {5}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a +8 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {5}{2}} a +2 \sqrt {d \,x^{2}+e x +c}\, d^{\frac {3}{2}} b e +4 d^{2} \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a e +4 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c \,d^{2}-\ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b d \,e^{2}\right )}{8 d^{\frac {5}{2}}}\) \(215\)

input
int(((b*x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/x,x,method=_RETURNVERBOSE)
 
output
1/8*csgn(b*x+a)*(4*(d*x^2+e*x+c)^(1/2)*d^(5/2)*b*x-8*c^(1/2)*d^(5/2)*ln((2 
*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/2))/x)*a+8*(d*x^2+e*x+c)^(1/2)*d^(5/2)*a 
+2*(d*x^2+e*x+c)^(1/2)*d^(3/2)*b*e+4*d^2*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^( 
1/2)+2*d*x+e)/d^(1/2))*a*e+4*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e 
)/d^(1/2))*b*c*d^2-ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2)) 
*b*d*e^2)/d^(5/2)
 
3.1.49.5 Fricas [A] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 651, normalized size of antiderivative = 3.09 \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x} \, dx=\left [\frac {8 \, a \sqrt {c} d^{2} \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) - {\left (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x - 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{16 \, d^{2}}, \frac {4 \, a \sqrt {c} d^{2} \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) - {\left (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{8 \, d^{2}}, \frac {16 \, a \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) - {\left (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {d} \log \left (8 \, d^{2} x^{2} + 8 \, d e x - 4 \, \sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {d} + 4 \, c d + e^{2}\right ) + 4 \, {\left (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{16 \, d^{2}}, \frac {8 \, a \sqrt {-c} d^{2} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) - {\left (4 \, b c d + 4 \, a d e - b e^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (2 \, d x + e\right )} \sqrt {-d}}{2 \, {\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \, {\left (2 \, b d^{2} x + 4 \, a d^{2} + b d e\right )} \sqrt {d x^{2} + e x + c}}{8 \, d^{2}}\right ] \]

input
integrate(((b*x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/x,x, algorithm="fricas")
 
output
[1/16*(8*a*sqrt(c)*d^2*log((8*c*e*x + (4*c*d + e^2)*x^2 - 4*sqrt(d*x^2 + e 
*x + c)*(e*x + 2*c)*sqrt(c) + 8*c^2)/x^2) - (4*b*c*d + 4*a*d*e - b*e^2)*sq 
rt(d)*log(8*d^2*x^2 + 8*d*e*x - 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d 
) + 4*c*d + e^2) + 4*(2*b*d^2*x + 4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/ 
d^2, 1/8*(4*a*sqrt(c)*d^2*log((8*c*e*x + (4*c*d + e^2)*x^2 - 4*sqrt(d*x^2 
+ e*x + c)*(e*x + 2*c)*sqrt(c) + 8*c^2)/x^2) - (4*b*c*d + 4*a*d*e - b*e^2) 
*sqrt(-d)*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + 
 d*e*x + c*d)) + 2*(2*b*d^2*x + 4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/d^ 
2, 1/16*(16*a*sqrt(-c)*d^2*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sq 
rt(-c)/(c*d*x^2 + c*e*x + c^2)) - (4*b*c*d + 4*a*d*e - b*e^2)*sqrt(d)*log( 
8*d^2*x^2 + 8*d*e*x - 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d 
+ e^2) + 4*(2*b*d^2*x + 4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/d^2, 1/8*( 
8*a*sqrt(-c)*d^2*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sqrt(-c)/(c* 
d*x^2 + c*e*x + c^2)) - (4*b*c*d + 4*a*d*e - b*e^2)*sqrt(-d)*arctan(1/2*sq 
rt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*(2*b 
*d^2*x + 4*a*d^2 + b*d*e)*sqrt(d*x^2 + e*x + c))/d^2]
 
3.1.49.6 Sympy [F]

\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x} \, dx=\int \frac {\sqrt {c + d x^{2} + e x} \sqrt {\left (a + b x\right )^{2}}}{x}\, dx \]

input
integrate(((b*x+a)**2)**(1/2)*(d*x**2+e*x+c)**(1/2)/x,x)
 
output
Integral(sqrt(c + d*x**2 + e*x)*sqrt((a + b*x)**2)/x, x)
 
3.1.49.7 Maxima [F]

\[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x} \, dx=\int { \frac {\sqrt {d x^{2} + e x + c} \sqrt {{\left (b x + a\right )}^{2}}}{x} \,d x } \]

input
integrate(((b*x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/x,x, algorithm="maxima")
 
output
integrate(sqrt(d*x^2 + e*x + c)*sqrt((b*x + a)^2)/x, x)
 
3.1.49.8 Giac [F(-2)]

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

input
integrate(((b*x+a)^2)^(1/2)*(d*x^2+e*x+c)^(1/2)/x,x, algorithm="giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 
3.1.49.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+e x+d x^2}}{x} \, dx=\int \frac {\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+e\,x+c}}{x} \,d x \]

input
int((((a + b*x)^2)^(1/2)*(c + e*x + d*x^2)^(1/2))/x,x)
 
output
int((((a + b*x)^2)^(1/2)*(c + e*x + d*x^2)^(1/2))/x, x)