\(\int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx\) [75]

Optimal result

Integrand size = 40, antiderivative size = 294 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {((b c+4 a d) e+2 d (b c+2 a d) x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 c d \left (a+b x^2\right )}-\frac {a \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{c x \left (a+b x^2\right )}+\frac {\left (4 b c d+8 a d^2-b e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{8 d^{3/2} \left (a+b x^2\right )}-\frac {a e \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{2 \sqrt {c} \left (a+b x^2\right )} \] Output:

1/4*((4*a*d+b*c)*e+2*d*(2*a*d+b*c)*x)*(d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1 
/2)/c/d/(b*x^2+a)-a*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/2)/c/x/(b*x^2+a)+ 
1/8*(8*a*d^2+4*b*c*d-b*e^2)*((b*x^2+a)^2)^(1/2)*arctanh(1/2*(2*d*x+e)/d^(1 
/2)/(d*x^2+e*x+c)^(1/2))/d^(3/2)/(b*x^2+a)-1/2*a*e*((b*x^2+a)^2)^(1/2)*arc 
tanh(1/2*(e*x+2*c)/c^(1/2)/(d*x^2+e*x+c)^(1/2))/c^(1/2)/(b*x^2+a)
 

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (\sqrt {c} \left (4 b c d+8 a d^2-b e^2\right ) x \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+x (e+d x)}}\right )+2 \sqrt {d} \left (\sqrt {c} \sqrt {c+x (e+d x)} (-4 a d+b x (e+2 d x))+4 a d e x \text {arctanh}\left (\frac {\sqrt {d} x-\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )\right )\right )}{8 \sqrt {c} d^{3/2} x \left (a+b x^2\right )} \] Input:

Integrate[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^2,x]
 

Output:

(Sqrt[(a + b*x^2)^2]*(Sqrt[c]*(4*b*c*d + 8*a*d^2 - b*e^2)*x*ArcTanh[(e + 2 
*d*x)/(2*Sqrt[d]*Sqrt[c + x*(e + d*x)])] + 2*Sqrt[d]*(Sqrt[c]*Sqrt[c + x*( 
e + d*x)]*(-4*a*d + b*x*(e + 2*d*x)) + 4*a*d*e*x*ArcTanh[(Sqrt[d]*x - Sqrt 
[c + x*(e + d*x)])/Sqrt[c]])))/(8*Sqrt[c]*d^(3/2)*x*(a + b*x^2))
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.72, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1384, 27, 2181, 27, 1231, 25, 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.

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

\(\Big \downarrow \) 1384

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2181

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {\int -\frac {(a e+2 (b c+2 a d) x) \sqrt {d x^2+e x+c}}{2 x}dx}{c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\int \frac {(a e+2 (b c+2 a d) x) \sqrt {d x^2+e x+c}}{x}dx}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1231

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}-\frac {\int -\frac {c \left (4 a d e+\left (8 a d^2+4 b c d-b e^2\right ) x\right )}{x \sqrt {d x^2+e x+c}}dx}{4 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {\int \frac {c \left (4 a d e+\left (8 a d^2+4 b c d-b e^2\right ) x\right )}{x \sqrt {d x^2+e x+c}}dx}{4 d}+\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {c \int \frac {4 a d e+\left (8 a d^2+4 b c d-b e^2\right ) x}{x \sqrt {d x^2+e x+c}}dx}{4 d}+\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {c \left (\left (8 a d^2+4 b c d-b e^2\right ) \int \frac {1}{\sqrt {d x^2+e x+c}}dx+4 a d e \int \frac {1}{x \sqrt {d x^2+e x+c}}dx\right )}{4 d}+\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {c \left (2 \left (8 a d^2+4 b c d-b e^2\right ) \int \frac {1}{4 d-\frac {(e+2 d x)^2}{d x^2+e x+c}}d\frac {e+2 d x}{\sqrt {d x^2+e x+c}}+4 a d e \int \frac {1}{x \sqrt {d x^2+e x+c}}dx\right )}{4 d}+\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {c \left (4 a d e \int \frac {1}{x \sqrt {d x^2+e x+c}}dx+\frac {\left (8 a d^2+4 b c d-b e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{\sqrt {d}}\right )}{4 d}+\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {c \left (\frac {\left (8 a d^2+4 b c d-b e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{\sqrt {d}}-8 a d e \int \frac {1}{4 c-\frac {(2 c+e x)^2}{d x^2+e x+c}}d\frac {2 c+e x}{\sqrt {d x^2+e x+c}}\right )}{4 d}+\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {\frac {c \left (\frac {\left (8 a d^2+4 b c d-b e^2\right ) \text {arctanh}\left (\frac {2 d x+e}{2 \sqrt {d} \sqrt {c+d x^2+e x}}\right )}{\sqrt {d}}-\frac {4 a d e \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{\sqrt {c}}\right )}{4 d}+\frac {\sqrt {c+d x^2+e x} (e (4 a d+b c)+2 d x (2 a d+b c))}{2 d}}{2 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{c x}\right )}{a+b x^2}\)

Input:

Int[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^2,x]
 

Output:

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-((a*(c + e*x + d*x^2)^(3/2))/(c*x)) + ( 
(((b*c + 4*a*d)*e + 2*d*(b*c + 2*a*d)*x)*Sqrt[c + e*x + d*x^2])/(2*d) + (c 
*(((4*b*c*d + 8*a*d^2 - b*e^2)*ArcTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x 
 + d*x^2])])/Sqrt[d] - (4*a*d*e*ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + e* 
x + d*x^2])])/Sqrt[c]))/(4*d))/(2*c)))/(a + b*x^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 1384

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac 
Part[p]))   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n 
- 1)] && NeQ[u, x^(2*n - 1)] &&  !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
 

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]
 

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {a \sqrt {d \,x^{2}+e x +c}\, \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x \left (b \,x^{2}+a \right )}+\frac {\left (-\frac {a e \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right )}{2 \sqrt {c}}+a \sqrt {d}\, \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )+\frac {b c \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )}{2 \sqrt {d}}+\frac {b x \sqrt {d \,x^{2}+e x +c}}{2}+\frac {b e \sqrt {d \,x^{2}+e x +c}}{4 d}-\frac {b \,e^{2} \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )}{8 d^{\frac {3}{2}}}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\)	\(229\)
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (8 d^{\frac {7}{2}} \sqrt {d \,x^{2}+e x +c}\, a \,x^{2}-4 d^{\frac {5}{2}} \sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a e x +4 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, b c \,x^{2}-8 d^{\frac {5}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a +8 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, a e x +2 d^{\frac {3}{2}} \sqrt {d \,x^{2}+e x +c}\, b c e x +8 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) a c \,d^{3} x +4 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b \,c^{2} d^{2} x -\ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) b c d \,e^{2} x \right )}{8 \left (b \,x^{2}+a \right ) c x \,d^{\frac {5}{2}}}\)	\(289\)

Input:

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

Output:

-a/x*(d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+(-1/2*a*e/c^(1/2)*l 
n((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/2))/x)+a*d^(1/2)*ln((1/2*e+d*x)/d^(1 
/2)+(d*x^2+e*x+c)^(1/2))+1/2*b*c*ln((1/2*e+d*x)/d^(1/2)+(d*x^2+e*x+c)^(1/2 
))/d^(1/2)+1/2*b*x*(d*x^2+e*x+c)^(1/2)+1/4*b/d*e*(d*x^2+e*x+c)^(1/2)-1/8*b 
/d^(3/2)*e^2*ln((1/2*e+d*x)/d^(1/2)+(d*x^2+e*x+c)^(1/2)))*((b*x^2+a)^2)^(1 
/2)/(b*x^2+a)
 

Fricas [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 731, normalized size of antiderivative = 2.49 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*(4*a*sqrt(c)*d^2*e*x*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^2*d + 8*a*c*d^2 - b 
*c*e^2)*sqrt(d)*x*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*c*d^2*x^2 + b*c*d*e*x - 4*a*c*d^2)*s 
qrt(d*x^2 + e*x + c))/(c*d^2*x), 1/8*(2*a*sqrt(c)*d^2*e*x*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^2*d + 8*a*c*d^2 - b*c*e^2)*sqrt(-d)*x*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*c*d^2*x^2 
+ b*c*d*e*x - 4*a*c*d^2)*sqrt(d*x^2 + e*x + c))/(c*d^2*x), 1/16*(8*a*sqrt( 
-c)*d^2*e*x*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^2*d + 8*a*c*d^2 - b*c*e^2)*sqrt(d)*x*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*c*d^2*x^2 + b*c*d*e*x - 4*a*c*d^2)*sqrt(d*x^2 + e*x + c))/(c*d^2* 
x), 1/8*(4*a*sqrt(-c)*d^2*e*x*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^2*d + 8*a*c*d^2 - b*c*e^2)*sqr 
t(-d)*x*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*c*d^2*x^2 + b*c*d*e*x - 4*a*c*d^2)*sqrt(d*x^2 + e*x 
+ c))/(c*d^2*x)]
 

Sympy [F]

\[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\int \frac {\sqrt {c + d x^{2} + e x} \sqrt {\left (a + b x^{2}\right )^{2}}}{x^{2}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^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(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {a e \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + e x + c}}{\sqrt {-c}}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {-c}} + \frac {1}{4} \, \sqrt {d x^{2} + e x + c} {\left (2 \, b x \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {b e \mathrm {sgn}\left (b x^{2} + a\right )}{d}\right )} - \frac {{\left (4 \, b c d \mathrm {sgn}\left (b x^{2} + a\right ) + 8 \, a d^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - b e^{2} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} + e \right |}\right )}{8 \, d^{\frac {3}{2}}} + \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} a e \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a c \sqrt {d} \mathrm {sgn}\left (b x^{2} + a\right )}{{\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )}^{2} - c} \] Input:

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

Output:

a*e*arctan(-(sqrt(d)*x - sqrt(d*x^2 + e*x + c))/sqrt(-c))*sgn(b*x^2 + a)/s 
qrt(-c) + 1/4*sqrt(d*x^2 + e*x + c)*(2*b*x*sgn(b*x^2 + a) + b*e*sgn(b*x^2 
+ a)/d) - 1/8*(4*b*c*d*sgn(b*x^2 + a) + 8*a*d^2*sgn(b*x^2 + a) - b*e^2*sgn 
(b*x^2 + a))*log(abs(2*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))*sqrt(d) + e))/d 
^(3/2) + ((sqrt(d)*x - sqrt(d*x^2 + e*x + c))*a*e*sgn(b*x^2 + a) + 2*a*c*s 
qrt(d)*sgn(b*x^2 + a))/((sqrt(d)*x - sqrt(d*x^2 + e*x + c))^2 - c)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\int \frac {\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {d\,x^2+e\,x+c}}{x^2} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^2} \, dx=\frac {-8 \sqrt {d \,x^{2}+e x +c}\, a c \,d^{2}+4 \sqrt {d \,x^{2}+e x +c}\, b c \,d^{2} x^{2}+2 \sqrt {d \,x^{2}+e x +c}\, b c d e x +4 \sqrt {c}\, \mathrm {log}\left (2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) a \,d^{2} e x -4 \sqrt {c}\, \mathrm {log}\left (x \right ) a \,d^{2} e x +8 \sqrt {d}\, \mathrm {log}\left (-2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}-2 d x -e \right ) a c \,d^{2} x +4 \sqrt {d}\, \mathrm {log}\left (-2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}-2 d x -e \right ) b \,c^{2} d x -\sqrt {d}\, \mathrm {log}\left (-2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}-2 d x -e \right ) b c \,e^{2} x}{8 c \,d^{2} x} \] Input:

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

Output:

( - 8*sqrt(c + d*x**2 + e*x)*a*c*d**2 + 4*sqrt(c + d*x**2 + e*x)*b*c*d**2* 
x**2 + 2*sqrt(c + d*x**2 + e*x)*b*c*d*e*x + 4*sqrt(c)*log(2*sqrt(c)*sqrt(c 
 + d*x**2 + e*x) - 2*c - e*x)*a*d**2*e*x - 4*sqrt(c)*log(x)*a*d**2*e*x + 8 
*sqrt(d)*log( - 2*sqrt(d)*sqrt(c + d*x**2 + e*x) - 2*d*x - e)*a*c*d**2*x + 
 4*sqrt(d)*log( - 2*sqrt(d)*sqrt(c + d*x**2 + e*x) - 2*d*x - e)*b*c**2*d*x 
 - sqrt(d)*log( - 2*sqrt(d)*sqrt(c + d*x**2 + e*x) - 2*d*x - e)*b*c*e**2*x 
)/(8*c*d**2*x)