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

Optimal. Leaf size=88 \[ \frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{3/2}}-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4} \]

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Rubi [A]  time = 0.07, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1114, 720, 724, 206} \begin {gather*} \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}}-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^5,x]

[Out]

-((2*a + b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 720

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] /; FreeQ[
{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 724

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 1114

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*} \int \frac {\sqrt {a+b x^2+c x^4}}{x^5} \, dx &=\frac {1}{2} \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [A]  time = 0.04, size = 88, normalized size = 1.00 \begin {gather*} \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}}-\frac {\left (2 a+b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4} \end {gather*}

Antiderivative was successfully verified.

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

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IntegrateAlgebraic [A]  time = 0.33, size = 91, normalized size = 1.03 \begin {gather*} \frac {\left (4 a c-b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{8 a^{3/2}}+\frac {\left (-2 a-b x^2\right ) \sqrt {a+b x^2+c x^4}}{8 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.17, size = 215, normalized size = 2.44 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [B]  time = 0.22, size = 241, normalized size = 2.74 \begin {gather*} -\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.01, size = 193, normalized size = 2.19 \begin {gather*} -\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b c \,x^{2}}{8 a^{2}}-\frac {c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{4 \sqrt {a}}+\frac {b^{2} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{16 a^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, c}{4 a}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}}{8 a^{2}}+\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}} b}{8 a^{2} x^{2}}-\frac {\left (c \,x^{4}+b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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 details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {c\,x^4+b\,x^2+a}}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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