\(\int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx\) [953]

Optimal result

Integrand size = 20, antiderivative size = 112 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx=-\frac {\sqrt {a+b x^2+c x^4}}{2 x^2}-\frac {b \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 \sqrt {a}}+\frac {1}{2} \sqrt {c} \text {arctanh}\left (\frac {b+2 c x^2}{2 \sqrt {c} \sqrt {a+b x^2+c x^4}}\right ) \] Output:

-1/2*(c*x^4+b*x^2+a)^(1/2)/x^2-1/4*b*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^ 
4+b*x^2+a)^(1/2))/a^(1/2)+1/2*c^(1/2)*arctanh(1/2*(2*c*x^2+b)/c^(1/2)/(c*x 
^4+b*x^2+a)^(1/2))
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx=\frac {1}{2} \left (-\frac {\sqrt {a+b x^2+c x^4}}{x^2}+\frac {b \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{\sqrt {a}}-\sqrt {c} \log \left (b+2 c x^2-2 \sqrt {c} \sqrt {a+b x^2+c x^4}\right )\right ) \] Input:

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

Output:

(-(Sqrt[a + b*x^2 + c*x^4]/x^2) + (b*ArcTanh[(Sqrt[c]*x^2 - Sqrt[a + b*x^2 
 + c*x^4])/Sqrt[a]])/Sqrt[a] - Sqrt[c]*Log[b + 2*c*x^2 - 2*Sqrt[c]*Sqrt[a 
+ b*x^2 + c*x^4]])/2
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1434, 1161, 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.

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

\(\Big \downarrow \) 1434

\(\displaystyle \frac {1}{2} \int \frac {\sqrt {c x^4+b x^2+a}}{x^4}dx^2\)

\(\Big \downarrow \) 1161

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(-(Sqrt[a + b*x^2 + c*x^4]/x^2) + (-((b*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*S 
qrt[a + b*x^2 + c*x^4])])/Sqrt[a]) + 2*Sqrt[c]*ArcTanh[(b + 2*c*x^2)/(2*Sq 
rt[c]*Sqrt[a + b*x^2 + c*x^4])])/2)/2
 

Defintions of rubi rules used

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 1161

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 1))), x] - Si 
mp[p/(e*(m + 1))   Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^(p - 
 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && GtQ[p, 0] && (IntegerQ[p] || 
 LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, 
 c, d, e, m, p, 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 1434

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp 
[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x + c*x^2)^p, x], x, x^2], x] /; Free 
Q[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.84

method	result	size
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 x^{2}}-\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {\sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2}\)	\(94\)
pseudoelliptic	\(-\frac {-\sqrt {c}\, x^{2} \sqrt {a}\, \ln \left (\frac {2 c \,x^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {c}+b}{\sqrt {c}}\right )+\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) x^{2}}{2}+\sqrt {a}\, \left (\ln \left (2\right ) \sqrt {c}\, x^{2}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2 \sqrt {a}\, x^{2}}\)	\(120\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a}-\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{2 a}+\frac {\sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2}\)	\(140\)
elliptic	\(-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a}-\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{2 a}+\frac {\sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c \,x^{2}}{\sqrt {c}}+\sqrt {c \,x^{4}+b \,x^{2}+a}\right )}{2}\)	\(140\)

Input:

int((c*x^4+b*x^2+a)^(1/2)/x^3,x,method=_RETURNVERBOSE)
 

Output:

-1/2*(c*x^4+b*x^2+a)^(1/2)/x^2-1/4*b/a^(1/2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^ 
4+b*x^2+a)^(1/2))/x^2)+1/2*c^(1/2)*ln((1/2*b+c*x^2)/c^(1/2)+(c*x^4+b*x^2+a 
)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 601, normalized size of antiderivative = 5.37 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx=\left [\frac {2 \, a \sqrt {c} x^{2} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) + \sqrt {a} b x^{2} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, \sqrt {c x^{4} + b x^{2} + a} a}{8 \, a x^{2}}, -\frac {4 \, a \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - \sqrt {a} b x^{2} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, \sqrt {c x^{4} + b x^{2} + a} a}{8 \, a x^{2}}, \frac {\sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + a \sqrt {c} x^{2} \log \left (-8 \, c^{2} x^{4} - 8 \, b c x^{2} - b^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {c} - 4 \, a c\right ) - 2 \, \sqrt {c x^{4} + b x^{2} + a} a}{4 \, a x^{2}}, \frac {\sqrt {-a} b x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) - 2 \, a \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (2 \, c x^{2} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )}}\right ) - 2 \, \sqrt {c x^{4} + b x^{2} + a} a}{4 \, a x^{2}}\right ] \] Input:

integrate((c*x^4+b*x^2+a)^(1/2)/x^3,x, algorithm="fricas")
 

Output:

[1/8*(2*a*sqrt(c)*x^2*log(-8*c^2*x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b* 
x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 4*a*c) + sqrt(a)*b*x^2*log(-((b^2 + 4*a*c 
)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^ 
2)/x^4) - 4*sqrt(c*x^4 + b*x^2 + a)*a)/(a*x^2), -1/8*(4*a*sqrt(-c)*x^2*arc 
tan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(-c)/(c^2*x^4 + b*c*x^2 
+ a*c)) - sqrt(a)*b*x^2*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 - 4*sqrt(c*x^4 
 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) + 4*sqrt(c*x^4 + b*x^2 + 
 a)*a)/(a*x^2), 1/4*(sqrt(-a)*b*x^2*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(b* 
x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + a*sqrt(c)*x^2*log(-8*c^2* 
x^4 - 8*b*c*x^2 - b^2 - 4*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sqrt(c) - 
4*a*c) - 2*sqrt(c*x^4 + b*x^2 + a)*a)/(a*x^2), 1/4*(sqrt(-a)*b*x^2*arctan( 
1/2*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^ 
2)) - 2*a*sqrt(-c)*x^2*arctan(1/2*sqrt(c*x^4 + b*x^2 + a)*(2*c*x^2 + b)*sq 
rt(-c)/(c^2*x^4 + b*c*x^2 + a*c)) - 2*sqrt(c*x^4 + b*x^2 + a)*a)/(a*x^2)]
 

Sympy [F]

\[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx=\int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{3}}\, dx \] Input:

integrate((c*x**4+b*x**2+a)**(1/2)/x**3,x)
 

Output:

Integral(sqrt(a + b*x**2 + c*x**4)/x**3, x)
 

Maxima [F(-2)]

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

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx=\frac {b \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a}} - \frac {1}{2} \, \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} \sqrt {c} - b \right |}\right ) + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} b + 2 \, a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}} \] Input:

integrate((c*x^4+b*x^2+a)^(1/2)/x^3,x, algorithm="giac")
 

Output:

1/2*b*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/sqrt(-a) - 
 1/2*sqrt(c)*log(abs(-2*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*sqrt(c) - 
b)) + 1/2*((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*b + 2*a*sqrt(c))/((sqrt 
(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)
 

Mupad [B] (verification not implemented)

Time = 18.38 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx=\frac {\sqrt {c}\,\ln \left (\sqrt {c\,x^4+b\,x^2+a}+\frac {c\,x^2+\frac {b}{2}}{\sqrt {c}}\right )}{2}-\frac {\sqrt {c\,x^4+b\,x^2+a}}{2\,x^2}-\frac {b\,\ln \left (\frac {b}{2}+\frac {a}{x^2}+\frac {\sqrt {a}\,\sqrt {c\,x^4+b\,x^2+a}}{x^2}\right )}{4\,\sqrt {a}} \] Input:

int((a + b*x^2 + c*x^4)^(1/2)/x^3,x)
 

Output:

(c^(1/2)*log((a + b*x^2 + c*x^4)^(1/2) + (b/2 + c*x^2)/c^(1/2)))/2 - (a + 
b*x^2 + c*x^4)^(1/2)/(2*x^2) - (b*log(b/2 + a/x^2 + (a^(1/2)*(a + b*x^2 + 
c*x^4)^(1/2))/x^2))/(4*a^(1/2))
 

Reduce [F]

\[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^3} \, dx=\frac {-\sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+\sqrt {c}\, \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{5}+b \,x^{3}+a x}d x \right ) b \,x^{2}+\mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{4}+b \,x^{2}+a}+b +2 c \,x^{2}}{\sqrt {4 a c -b^{2}}}\right ) c \,x^{2}}{2 \sqrt {c}\, x^{2}} \] Input:

int((c*x^4+b*x^2+a)^(1/2)/x^3,x)
 

Output:

( - sqrt(c)*sqrt(a + b*x**2 + c*x**4) + sqrt(c)*int(sqrt(a + b*x**2 + c*x* 
*4)/(a*x + b*x**3 + c*x**5),x)*b*x**2 + log((2*sqrt(c)*sqrt(a + b*x**2 + c 
*x**4) + b + 2*c*x**2)/sqrt(4*a*c - b**2))*c*x**2)/(2*sqrt(c)*x**2)