\(\int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx\) [78]

Optimal result

Integrand size = 40, antiderivative size = 293 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=-\frac {\left (16 b c^2-4 a c d+5 a e^2\right ) (2 c+e x) \sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{64 c^3 x^2 \left (a+b x^2\right )}-\frac {a \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 c x^4 \left (a+b x^2\right )}+\frac {5 a e \left (c+e x+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 c^2 x^3 \left (a+b x^2\right )}-\frac {\left (4 c d-e^2\right ) \left (16 b c^2-4 a c d+5 a e^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+e x+d x^2}}\right )}{128 c^{7/2} \left (a+b x^2\right )} \] Output:

-1/64*(-4*a*c*d+5*a*e^2+16*b*c^2)*(e*x+2*c)*(d*x^2+e*x+c)^(1/2)*((b*x^2+a) 
^2)^(1/2)/c^3/x^2/(b*x^2+a)-1/4*a*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/2)/ 
c/x^4/(b*x^2+a)+5/24*a*e*(d*x^2+e*x+c)^(3/2)*((b*x^2+a)^2)^(1/2)/c^2/x^3/( 
b*x^2+a)-1/128*(4*c*d-e^2)*(-4*a*c*d+5*a*e^2+16*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^(7/2)/(b*x^2+a)
 

Mathematica [A] (verified)

Time = 1.46 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (-\sqrt {c} \sqrt {c+x (e+d x)} \left (48 b c^2 x^2 (2 c+e x)+a \left (48 c^3+15 e^3 x^3+8 c^2 x (e+3 d x)-2 c e x^2 (5 e+26 d x)\right )\right )+3 \left (64 b c^3 d+24 a c d e^2-5 a e^4\right ) x^4 \text {arctanh}\left (\frac {\sqrt {d} x-\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )+48 c^2 \left (a d^2+b e^2\right ) x^4 \text {arctanh}\left (\frac {-\sqrt {d} x+\sqrt {c+x (e+d x)}}{\sqrt {c}}\right )\right )}{192 c^{7/2} x^4 \left (a+b x^2\right )} \] Input:

Integrate[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^5,x]
 

Output:

(Sqrt[(a + b*x^2)^2]*(-(Sqrt[c]*Sqrt[c + x*(e + d*x)]*(48*b*c^2*x^2*(2*c + 
 e*x) + a*(48*c^3 + 15*e^3*x^3 + 8*c^2*x*(e + 3*d*x) - 2*c*e*x^2*(5*e + 26 
*d*x)))) + 3*(64*b*c^3*d + 24*a*c*d*e^2 - 5*a*e^4)*x^4*ArcTanh[(Sqrt[d]*x 
- Sqrt[c + x*(e + d*x)])/Sqrt[c]] + 48*c^2*(a*d^2 + b*e^2)*x^4*ArcTanh[(-( 
Sqrt[d]*x) + Sqrt[c + x*(e + d*x)])/Sqrt[c]]))/(192*c^(7/2)*x^4*(a + b*x^2 
))
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.68, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1384, 27, 2181, 27, 1228, 1152, 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.

\(\displaystyle \int \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2+e x}}{x^5} \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b \left (b x^2+a\right ) \sqrt {d x^2+e x+c}}{x^5}dx}{b \left (a+b x^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right ) \sqrt {d x^2+e x+c}}{x^5}dx}{a+b x^2}\)

\(\Big \downarrow \) 2181

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {\int \frac {(5 a e-2 (4 b c-a d) x) \sqrt {d x^2+e x+c}}{2 x^4}dx}{4 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{4 c x^4}\right )}{a+b x^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {\int \frac {(5 a e-2 (4 b c-a d) x) \sqrt {d x^2+e x+c}}{x^4}dx}{8 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{4 c x^4}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1228

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {-\frac {\left (-4 a c d+5 a e^2+16 b c^2\right ) \int \frac {\sqrt {d x^2+e x+c}}{x^3}dx}{2 c}-\frac {5 a e \left (c+d x^2+e x\right )^{3/2}}{3 c x^3}}{8 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{4 c x^4}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1152

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {-\frac {\left (-4 a c d+5 a e^2+16 b c^2\right ) \left (\frac {\left (4 c d-e^2\right ) \int \frac {1}{x \sqrt {d x^2+e x+c}}dx}{8 c}-\frac {(2 c+e x) \sqrt {c+d x^2+e x}}{4 c x^2}\right )}{2 c}-\frac {5 a e \left (c+d x^2+e x\right )^{3/2}}{3 c x^3}}{8 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{4 c x^4}\right )}{a+b x^2}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {-\frac {\left (-4 a c d+5 a e^2+16 b c^2\right ) \left (-\frac {\left (4 c d-e^2\right ) \int \frac {1}{4 c-\frac {(2 c+e x)^2}{d x^2+e x+c}}d\frac {2 c+e x}{\sqrt {d x^2+e x+c}}}{4 c}-\frac {(2 c+e x) \sqrt {c+d x^2+e x}}{4 c x^2}\right )}{2 c}-\frac {5 a e \left (c+d x^2+e x\right )^{3/2}}{3 c x^3}}{8 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{4 c x^4}\right )}{a+b x^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {-\frac {\left (-4 a c d+5 a e^2+16 b c^2\right ) \left (-\frac {\left (4 c d-e^2\right ) \text {arctanh}\left (\frac {2 c+e x}{2 \sqrt {c} \sqrt {c+d x^2+e x}}\right )}{8 c^{3/2}}-\frac {(2 c+e x) \sqrt {c+d x^2+e x}}{4 c x^2}\right )}{2 c}-\frac {5 a e \left (c+d x^2+e x\right )^{3/2}}{3 c x^3}}{8 c}-\frac {a \left (c+d x^2+e x\right )^{3/2}}{4 c x^4}\right )}{a+b x^2}\)

Input:

Int[(Sqrt[c + e*x + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^5,x]
 

Output:

(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/4*(a*(c + e*x + d*x^2)^(3/2))/(c*x^4) 
 - ((-5*a*e*(c + e*x + d*x^2)^(3/2))/(3*c*x^3) - ((16*b*c^2 - 4*a*c*d + 5* 
a*e^2)*(-1/4*((2*c + e*x)*Sqrt[c + e*x + d*x^2])/(c*x^2) - ((4*c*d - e^2)* 
ArcTanh[(2*c + e*x)/(2*Sqrt[c]*Sqrt[c + e*x + d*x^2])])/(8*c^(3/2))))/(2*c 
))/(8*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 1152

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-(d + e*x)^(m + 1))*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b 
*x + c*x^2)^p/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a 
*c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)))   Int[(d + e*x)^(m + 2)*(a + b*x + 
 c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] 
 && GtQ[p, 0]
 

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 1228

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + 
 b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e 
*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^ 
(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
] && EqQ[Simplify[m + 2*p + 3], 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 = 208, normalized size of antiderivative = 0.71

method	result	size
risch	\(-\frac {\sqrt {d \,x^{2}+e x +c}\, \left (-52 a c d e \,x^{3}+15 e^{3} a \,x^{3}+48 x^{3} b \,c^{2} e +24 a \,c^{2} d \,x^{2}-10 a c \,e^{2} x^{2}+96 b \,c^{3} x^{2}+8 a \,c^{2} e x +48 a \,c^{3}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{192 x^{4} c^{3} \left (b \,x^{2}+a \right )}+\frac {\left (16 a \,c^{2} d^{2}-24 a c d \,e^{2}+5 e^{4} a -64 b \,c^{3} d +16 b \,c^{2} e^{2}\right ) \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{128 c^{\frac {7}{2}} \left (b \,x^{2}+a \right )}\)	\(208\)
default	\(-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (192 d \,c^{\frac {7}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b \,x^{4}-48 d^{2} c^{\frac {5}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,x^{4}-48 c^{\frac {5}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) b \,e^{2} x^{4}+72 d \,c^{\frac {3}{2}} \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,e^{2} x^{4}-15 \sqrt {c}\, \ln \left (\frac {2 c +e x +2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}}{x}\right ) a \,e^{4} x^{4}-24 d^{2} \sqrt {d \,x^{2}+e x +c}\, a c e \,x^{5}+30 d \sqrt {d \,x^{2}+e x +c}\, a \,e^{3} x^{5}+96 d \sqrt {d \,x^{2}+e x +c}\, b \,c^{2} e \,x^{5}+24 d \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a c e \,x^{3}-30 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,e^{3} x^{3}-96 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,c^{2} e \,x^{3}+48 d^{2} \sqrt {d \,x^{2}+e x +c}\, a \,c^{2} x^{4}-84 d \sqrt {d \,x^{2}+e x +c}\, a c \,e^{2} x^{4}+30 \sqrt {d \,x^{2}+e x +c}\, a \,e^{4} x^{4}-192 d \sqrt {d \,x^{2}+e x +c}\, b \,c^{3} x^{4}+96 \sqrt {d \,x^{2}+e x +c}\, b \,c^{2} e^{2} x^{4}-48 d \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,c^{2} x^{2}+60 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a c \,e^{2} x^{2}+192 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} b \,c^{3} x^{2}-80 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,c^{2} e x +96 \left (d \,x^{2}+e x +c \right )^{\frac {3}{2}} a \,c^{3}\right )}{384 x^{4} c^{4} \left (b \,x^{2}+a \right )}\)	\(583\)

Input:

int((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)
 

Output:

-1/192*(d*x^2+e*x+c)^(1/2)*(-52*a*c*d*e*x^3+15*a*e^3*x^3+48*b*c^2*e*x^3+24 
*a*c^2*d*x^2-10*a*c*e^2*x^2+96*b*c^3*x^2+8*a*c^2*e*x+48*a*c^3)/x^4/c^3*((b 
*x^2+a)^2)^(1/2)/(b*x^2+a)+1/128*(16*a*c^2*d^2-24*a*c*d*e^2+5*a*e^4-64*b*c 
^3*d+16*b*c^2*e^2)/c^(7/2)*ln((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/2))/x)*( 
(b*x^2+a)^2)^(1/2)/(b*x^2+a)
 

Fricas [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=\left [-\frac {3 \, {\left (64 \, b c^{3} d - 16 \, a c^{2} d^{2} - 5 \, a e^{4} - 8 \, {\left (2 \, b c^{2} - 3 \, a c d\right )} e^{2}\right )} \sqrt {c} x^{4} \log \left (\frac {8 \, c e x + {\left (4 \, c d + e^{2}\right )} x^{2} + 4 \, \sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {c} + 8 \, c^{2}}{x^{2}}\right ) + 4 \, {\left (8 \, a c^{3} e x + 48 \, a c^{4} + {\left (15 \, a c e^{3} + 4 \, {\left (12 \, b c^{3} - 13 \, a c^{2} d\right )} e\right )} x^{3} + 2 \, {\left (48 \, b c^{4} + 12 \, a c^{3} d - 5 \, a c^{2} e^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + e x + c}}{768 \, c^{4} x^{4}}, \frac {3 \, {\left (64 \, b c^{3} d - 16 \, a c^{2} d^{2} - 5 \, a e^{4} - 8 \, {\left (2 \, b c^{2} - 3 \, a c d\right )} e^{2}\right )} \sqrt {-c} x^{4} \arctan \left (\frac {\sqrt {d x^{2} + e x + c} {\left (e x + 2 \, c\right )} \sqrt {-c}}{2 \, {\left (c d x^{2} + c e x + c^{2}\right )}}\right ) - 2 \, {\left (8 \, a c^{3} e x + 48 \, a c^{4} + {\left (15 \, a c e^{3} + 4 \, {\left (12 \, b c^{3} - 13 \, a c^{2} d\right )} e\right )} x^{3} + 2 \, {\left (48 \, b c^{4} + 12 \, a c^{3} d - 5 \, a c^{2} e^{2}\right )} x^{2}\right )} \sqrt {d x^{2} + e x + c}}{384 \, c^{4} x^{4}}\right ] \] Input:

integrate((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^5,x, algorithm="fricas 
")
 

Output:

[-1/768*(3*(64*b*c^3*d - 16*a*c^2*d^2 - 5*a*e^4 - 8*(2*b*c^2 - 3*a*c*d)*e^ 
2)*sqrt(c)*x^4*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*(8*a*c^3*e*x + 48*a*c^4 + (15*a*c*e^ 
3 + 4*(12*b*c^3 - 13*a*c^2*d)*e)*x^3 + 2*(48*b*c^4 + 12*a*c^3*d - 5*a*c^2* 
e^2)*x^2)*sqrt(d*x^2 + e*x + c))/(c^4*x^4), 1/384*(3*(64*b*c^3*d - 16*a*c^ 
2*d^2 - 5*a*e^4 - 8*(2*b*c^2 - 3*a*c*d)*e^2)*sqrt(-c)*x^4*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*(8*a*c^ 
3*e*x + 48*a*c^4 + (15*a*c*e^3 + 4*(12*b*c^3 - 13*a*c^2*d)*e)*x^3 + 2*(48* 
b*c^4 + 12*a*c^3*d - 5*a*c^2*e^2)*x^2)*sqrt(d*x^2 + e*x + c))/(c^4*x^4)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=\text {Timed out} \] Input:

integrate((d*x**2+e*x+c)**(1/2)*((b*x**2+a)**2)**(1/2)/x**5,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^5,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1172 vs. \(2 (227) = 454\).

Time = 0.17 (sec) , antiderivative size = 1172, normalized size of antiderivative = 4.00 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=\text {Too large to display} \] Input:

integrate((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^5,x, algorithm="giac")
 

Output:

1/64*(64*b*c^3*d*sgn(b*x^2 + a) - 16*a*c^2*d^2*sgn(b*x^2 + a) - 16*b*c^2*e 
^2*sgn(b*x^2 + a) + 24*a*c*d*e^2*sgn(b*x^2 + a) - 5*a*e^4*sgn(b*x^2 + a))* 
arctan(-(sqrt(d)*x - sqrt(d*x^2 + e*x + c))/sqrt(-c))/(sqrt(-c)*c^3) + 1/1 
92*(192*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^7*b*c^3*d*sgn(b*x^2 + a) + 48* 
(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^7*a*c^2*d^2*sgn(b*x^2 + a) + 48*(sqrt( 
d)*x - sqrt(d*x^2 + e*x + c))^7*b*c^2*e^2*sgn(b*x^2 + a) - 72*(sqrt(d)*x - 
 sqrt(d*x^2 + e*x + c))^7*a*c*d*e^2*sgn(b*x^2 + a) + 15*(sqrt(d)*x - sqrt( 
d*x^2 + e*x + c))^7*a*e^4*sgn(b*x^2 + a) + 384*(sqrt(d)*x - sqrt(d*x^2 + e 
*x + c))^6*b*c^3*sqrt(d)*e*sgn(b*x^2 + a) - 192*(sqrt(d)*x - sqrt(d*x^2 + 
e*x + c))^5*b*c^4*d*sgn(b*x^2 + a) + 336*(sqrt(d)*x - sqrt(d*x^2 + e*x + c 
))^5*a*c^3*d^2*sgn(b*x^2 + a) - 48*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^5*b 
*c^3*e^2*sgn(b*x^2 + a) + 264*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^5*a*c^2* 
d*e^2*sgn(b*x^2 + a) - 55*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^5*a*c*e^4*sg 
n(b*x^2 + a) - 768*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^4*b*c^4*sqrt(d)*e*s 
gn(b*x^2 + a) + 1152*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^4*a*c^3*d^(3/2)*e 
*sgn(b*x^2 + a) - 192*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^3*b*c^5*d*sgn(b* 
x^2 + a) + 336*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^3*a*c^4*d^2*sgn(b*x^2 + 
 a) - 48*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^3*b*c^4*e^2*sgn(b*x^2 + a) + 
648*(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^3*a*c^3*d*e^2*sgn(b*x^2 + a) + 73* 
(sqrt(d)*x - sqrt(d*x^2 + e*x + c))^3*a*c^2*e^4*sgn(b*x^2 + a) + 384*(s...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=\int \frac {\sqrt {{\left (b\,x^2+a\right )}^2}\,\sqrt {d\,x^2+e\,x+c}}{x^5} \,d x \] Input:

int((((a + b*x^2)^2)^(1/2)*(c + e*x + d*x^2)^(1/2))/x^5,x)
 

Output:

int((((a + b*x^2)^2)^(1/2)*(c + e*x + d*x^2)^(1/2))/x^5, x)
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 431, normalized size of antiderivative = 1.47 \[ \int \frac {\sqrt {c+e x+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{x^5} \, dx=\frac {-96 \sqrt {d \,x^{2}+e x +c}\, a \,c^{4}-48 \sqrt {d \,x^{2}+e x +c}\, a \,c^{3} d \,x^{2}-16 \sqrt {d \,x^{2}+e x +c}\, a \,c^{3} e x +104 \sqrt {d \,x^{2}+e x +c}\, a \,c^{2} d e \,x^{3}+20 \sqrt {d \,x^{2}+e x +c}\, a \,c^{2} e^{2} x^{2}-30 \sqrt {d \,x^{2}+e x +c}\, a c \,e^{3} x^{3}-192 \sqrt {d \,x^{2}+e x +c}\, b \,c^{4} x^{2}-96 \sqrt {d \,x^{2}+e x +c}\, b \,c^{3} e \,x^{3}+48 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) a \,c^{2} d^{2} x^{4}-72 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) a c d \,e^{2} x^{4}+15 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) a \,e^{4} x^{4}-192 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) b \,c^{3} d \,x^{4}+48 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {d \,x^{2}+e x +c}-2 c -e x \right ) b \,c^{2} e^{2} x^{4}-48 \sqrt {c}\, \mathrm {log}\left (x \right ) a \,c^{2} d^{2} x^{4}+72 \sqrt {c}\, \mathrm {log}\left (x \right ) a c d \,e^{2} x^{4}-15 \sqrt {c}\, \mathrm {log}\left (x \right ) a \,e^{4} x^{4}+192 \sqrt {c}\, \mathrm {log}\left (x \right ) b \,c^{3} d \,x^{4}-48 \sqrt {c}\, \mathrm {log}\left (x \right ) b \,c^{2} e^{2} x^{4}}{384 c^{4} x^{4}} \] Input:

int((d*x^2+e*x+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^5,x)
 

Output:

( - 96*sqrt(c + d*x**2 + e*x)*a*c**4 - 48*sqrt(c + d*x**2 + e*x)*a*c**3*d* 
x**2 - 16*sqrt(c + d*x**2 + e*x)*a*c**3*e*x + 104*sqrt(c + d*x**2 + e*x)*a 
*c**2*d*e*x**3 + 20*sqrt(c + d*x**2 + e*x)*a*c**2*e**2*x**2 - 30*sqrt(c + 
d*x**2 + e*x)*a*c*e**3*x**3 - 192*sqrt(c + d*x**2 + e*x)*b*c**4*x**2 - 96* 
sqrt(c + d*x**2 + e*x)*b*c**3*e*x**3 + 48*sqrt(c)*log( - 2*sqrt(c)*sqrt(c 
+ d*x**2 + e*x) - 2*c - e*x)*a*c**2*d**2*x**4 - 72*sqrt(c)*log( - 2*sqrt(c 
)*sqrt(c + d*x**2 + e*x) - 2*c - e*x)*a*c*d*e**2*x**4 + 15*sqrt(c)*log( - 
2*sqrt(c)*sqrt(c + d*x**2 + e*x) - 2*c - e*x)*a*e**4*x**4 - 192*sqrt(c)*lo 
g( - 2*sqrt(c)*sqrt(c + d*x**2 + e*x) - 2*c - e*x)*b*c**3*d*x**4 + 48*sqrt 
(c)*log( - 2*sqrt(c)*sqrt(c + d*x**2 + e*x) - 2*c - e*x)*b*c**2*e**2*x**4 
- 48*sqrt(c)*log(x)*a*c**2*d**2*x**4 + 72*sqrt(c)*log(x)*a*c*d*e**2*x**4 - 
 15*sqrt(c)*log(x)*a*e**4*x**4 + 192*sqrt(c)*log(x)*b*c**3*d*x**4 - 48*sqr 
t(c)*log(x)*b*c**2*e**2*x**4)/(384*c**4*x**4)