\(\int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx\) [926]

Optimal result

Integrand size = 20, antiderivative size = 88 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx=-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4}+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{3/2}} \]

[Out]

1/16*(-4*a*c+b^2)*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/a^(3/2)-1/8*(b*x^2+2*a)*(c*x^4+b*x^2+
a)^(1/2)/a/x^4

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1128, 734, 738, 212} \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx=\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{3/2}}-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4} \]

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^5,x]

[Out]

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> 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] + Dist[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] /; Free
Q[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m
+ 2*p + 2, 0] && GtQ[p, 0]

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 1128

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4}-\frac {\left (b^2-4 a c\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 a} \\ & = -\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4}+\frac {\left (b^2-4 a c\right ) \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )}{8 a} \\ & = -\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4}+\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx=-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{8 a^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {\left (b \,x^{2}+2 a \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 a \,x^{4}}-\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}\)	\(81\)
pseudoelliptic	\(\frac {-4 \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) a c \,x^{4}+\ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right ) b^{2} x^{4}-2 b \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2} \sqrt {a}-4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{\frac {3}{2}}}{16 a^{\frac {3}{2}} x^{4}}\)	\(132\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}+\frac {b \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{8 a^{2} x^{2}}-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 a^{2}}+\frac {b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}-\frac {b c \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{8 a^{2}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 a}-\frac {c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 \sqrt {a}}\)	\(193\)
elliptic	\(-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}+\frac {b \left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{8 a^{2} x^{2}}-\frac {b^{2} \sqrt {c \,x^{4}+b \,x^{2}+a}}{8 a^{2}}+\frac {b^{2} \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}-\frac {b c \sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{8 a^{2}}+\frac {c \sqrt {c \,x^{4}+b \,x^{2}+a}}{4 a}-\frac {c \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 \sqrt {a}}\)	\(193\)

[In]

int((c*x^4+b*x^2+a)^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/8*(b*x^2+2*a)*(c*x^4+b*x^2+a)^(1/2)/a/x^4-1/16/a^(3/2)*(4*a*c-b^2)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^
(1/2))/x^2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.44 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {a} x^{4} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (a b x^{2} + 2 \, a^{2}\right )}}{32 \, a^{2} x^{4}}, -\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} + a} {\left (a b x^{2} + 2 \, a^{2}\right )}}{16 \, a^{2} x^{4}}\right ] \]

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^5,x, algorithm="fricas")

[Out]

[-1/32*((b^2 - 4*a*c)*sqrt(a)*x^4*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*b*x^2 + 2*a^2))/(a^2*x^4), -1/16*((b^2 - 4*a*c)*sqrt(-a
)*x^4*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*sqrt(c*x^4 + b*
x^2 + a)*(a*b*x^2 + 2*a^2))/(a^2*x^4)]

Sympy [F]

\[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx=\int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{5}}\, dx \]

[In]

integrate((c*x**4+b*x**2+a)**(1/2)/x**5,x)

[Out]

Integral(sqrt(a + b*x**2 + c*x**4)/x**5, x)

Maxima [F(-2)]

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

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more deta

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (74) = 148\).

Time = 0.30 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.74 \[ \int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx=-\frac {{\left (b^{2} - 4 \, a c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a c + 8 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a b \sqrt {c} + {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} c}{8 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{2} a} \]

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^5,x, algorithm="giac")

[Out]

-1/8*(b^2 - 4*a*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a) + 1/8*((sqrt(c)*x^2
- sqrt(c*x^4 + b*x^2 + a))^3*b^2 + 4*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*a*c + 8*(sqrt(c)*x^2 - sqrt(c*x
^4 + b*x^2 + a))^2*a*b*sqrt(c) + (sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a*b^2 + 4*(sqrt(c)*x^2 - sqrt(c*x^4 +
b*x^2 + a))*a^2*c)/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^2*a)

Mupad [F(-1)]

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

[In]

int((a + b*x^2 + c*x^4)^(1/2)/x^5,x)

[Out]

int((a + b*x^2 + c*x^4)^(1/2)/x^5, x)