\(\int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx\) [76]

Optimal result

Integrand size = 40, antiderivative size = 288 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {(a e+2 (2 b c+a d) x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 c x \left (a+b x^2\right )}-\frac {a \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 c x^2 \left (a+b x^2\right )}+\frac {b e \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {e+2 d x}{2 \sqrt {d} \sqrt {c+e x+d x^2}}\right )}{2 \sqrt {d} \left (a+b x^2\right )}-\frac {\left (8 b c^2+4 a c d-a e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{8 c^{3/2} \left (a+b x^2\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (\sqrt {d} \left (8 b c^2+4 a c d-a e^2\right ) x^2 \text {arctanh}\left (\frac {-\sqrt {d} x+\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )+\sqrt {c} \left (\sqrt {d} \left (2 a c+a e x-4 b c x^2\right ) \sqrt {c+x (e+d x)}+2 b c e x^2 \log \left (e+2 d x-2 \sqrt {d} \sqrt {c+x (e+d x)}\right )\right )\right )}{4 c^{3/2} \sqrt {d} x^2 \left (a+b x^2\right )} \] Input:

Integrate[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^3,x]
 

Output:

-1/4*(Sqrt[(a + b*x^2)^2]*(Sqrt[d]*(8*b*c^2 + 4*a*c*d - a*e^2)*x^2*ArcTanh 
[(-(Sqrt[d]*x) + Sqrt[c + x*(e + d*x)])/Sqrt[c]] + Sqrt[c]*(Sqrt[d]*(2*a*c 
 + a*e*x - 4*b*c*x^2)*Sqrt[c + x*(e + d*x)] + 2*b*c*e*x^2*Log[e + 2*d*x - 
2*Sqrt[d]*Sqrt[c + x*(e + d*x)]])))/(c^(3/2)*Sqrt[d]*x^2*(a + b*x^2))
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.69, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1384, 27, 2181, 27, 1230, 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.

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

\(\Big \downarrow \) 1384

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2181

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1230

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

Int[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^3,x]
 

Output:

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/2*(a*(c + e*x + d*x^2)^(3/2))/(c*x^2) 
 - (-(((a*e + 2*(2*b*c + a*d)*x)*Sqrt[c + e*x + d*x^2])/x) + ((-4*b*c*e*Ar 
cTanh[(e + 2*d*x)/(2*Sqrt[d]*Sqrt[c + e*x + d*x^2])])/Sqrt[d] + ((8*b*c^2 
+ 4*a*c*d - a*e^2)*ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + e*x + d*x^2])]) 
/Sqrt[c])/2)/(4*c)))/(a + b*x^2)
 

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 1230

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - 
 d*g*(2*p + 1) + e*g*(m + 1)*x)*((a + b*x + c*x^2)^p/(e^2*(m + 1)*(m + 2*p 
+ 2))), x] + Simp[p/(e^2*(m + 1)*(m + 2*p + 2))   Int[(d + e*x)^(m + 1)*(a 
+ b*x + c*x^2)^(p - 1)*Simp[g*(b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m 
+ 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, 
 x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && GtQ[p, 0] && (LtQ[m, - 
1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&  !ILtQ 
[m + 2*p + 1, 0] && (IntegerQ[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 = 0.32 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {a \sqrt {d \,x^{2}+e x +c}\, \left (e x +2 c \right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 x^{2} c \left (b \,x^{2}+a \right )}+\frac {\left (-\frac {\left (4 a c d -a \,e^{2}+8 b \,c^{2}\right ) \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right )}{\sqrt {c}}+\frac {8 b c e \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )}{\sqrt {d}}+8 c b d \left (\frac {\sqrt {d \,x^{2}+e x +c}}{d}-\frac {e \ln \left (\frac {\frac {e}{2}+d x}{\sqrt {d}}+\sqrt {d \,x^{2}+e x +c}\right )}{2 d^{\frac {3}{2}}}\right )\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{8 c \left (b \,x^{2}+a \right )}\)	\(214\)
default	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-4 d^{\frac {5}{2}} c^{\frac {3}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,x^{2}-8 d^{\frac {3}{2}} c^{\frac {5}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b \,x^{2}-2 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, a e \,x^{3}+4 d^{\frac {5}{2}} \sqrt {d \,x^{2}+e x +c}\, a c \,x^{2}+d^{\frac {3}{2}} \sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,e^{2} x^{2}+2 d^{\frac {3}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a e x -2 d^{\frac {3}{2}} \sqrt {d \,x^{2}+e x +c}\, a \,e^{2} x^{2}+8 d^{\frac {3}{2}} \sqrt {d \,x^{2}+e x +c}\, b \,c^{2} x^{2}+4 \ln \left (\frac {2 \sqrt {d \,x^{2}+e x +c}\, \sqrt {d}+2 d x +e}{2 \sqrt {d}}\right ) d b \,c^{2} e \,x^{2}-4 d^{\frac {3}{2}} \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a c \right )}{8 d^{\frac {3}{2}} x^{2} c^{2} \left (b \,x^{2}+a \right )}\)	\(329\)

Input:

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

Output:

-1/4*a*(d*x^2+e*x+c)^(1/2)*(e*x+2*c)/x^2/c*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1 
/8/c*(-(4*a*c*d-a*e^2+8*b*c^2)/c^(1/2)*ln((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c) 
^(1/2))/x)+8*b*c*e*ln((1/2*e+d*x)/d^(1/2)+(d*x^2+e*x+c)^(1/2))/d^(1/2)+8*c 
*b*d*(1/d*(d*x^2+e*x+c)^(1/2)-1/2*e/d^(3/2)*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 = 0.37 (sec) , antiderivative size = 749, normalized size of antiderivative = 2.60 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^3,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^2-4*c*d>0)', see `assume?` for 
 more deta
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=-\frac {b e \log \left ({\left | -2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} \sqrt {d} - e \right |}\right ) \mathrm {sgn}\left (b x^{2} + a\right )}{2 \, \sqrt {d}} + \sqrt {d x^{2} + e x + c} b \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {{\left (8 \, b c^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a c d \mathrm {sgn}\left (b x^{2} + a\right ) - a e^{2} \mathrm {sgn}\left (b x^{2} + a\right )\right )} \arctan \left (-\frac {\sqrt {d} x - \sqrt {d x^{2} + e x + c}}{\sqrt {-c}}\right )}{4 \, \sqrt {-c} c} + \frac {4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )}^{3} a c d \mathrm {sgn}\left (b x^{2} + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )}^{3} a e^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 8 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )}^{2} a c \sqrt {d} e \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} a c^{2} d \mathrm {sgn}\left (b x^{2} + a\right ) + {\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )} a c e^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{4 \, {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + e x + c}\right )}^{2} - c\right )}^{2} c} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\int \frac {\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {d\,x^2+e\,x+c}}{x^3} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx=\frac {-4 \sqrt {d \,x^{2}+e x +c}\, a \,c^{2} d -2 \sqrt {d \,x^{2}+e x +c}\, a c d e x +8 \sqrt {d \,x^{2}+e x +c}\, b \,c^{2} d \,x^{2}-4 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) a c \,d^{2} x^{2}+\sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) a d \,e^{2} x^{2}-8 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) b \,c^{2} d \,x^{2}+4 \sqrt {c}\, \mathrm {log}\left (x \right ) a c \,d^{2} x^{2}-\sqrt {c}\, \mathrm {log}\left (x \right ) a d \,e^{2} x^{2}+8 \sqrt {c}\, \mathrm {log}\left (x \right ) b \,c^{2} d \,x^{2}+4 \sqrt {d}\, \mathrm {log}\left (-2 \sqrt {d}\, \sqrt {d \,x^{2}+e x +c}-2 d x -e \right ) b \,c^{2} e \,x^{2}}{8 c^{2} d \,x^{2}} \] Input:

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

Output:

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