3.8.50 \(\int \frac {1}{x^5 \sqrt {a+b x^2+c x^4}} \, dx\)

Optimal. Leaf size=108 \[ -\frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac {3 b \sqrt {a+b x^2+c x^4}}{8 a^2 x^2}-\frac {\sqrt {a+b x^2+c x^4}}{4 a x^4} \]

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Rubi [A]  time = 0.11, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1114, 744, 806, 724, 206} \begin {gather*} -\frac {\left (3 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^{5/2}}+\frac {3 b \sqrt {a+b x^2+c x^4}}{8 a^2 x^2}-\frac {\sqrt {a+b x^2+c x^4}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

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

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 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 744

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(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] + Dist[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(a + b*x + c*x^2)^p, x], x]
/; FreeQ[{a, b, c, d, e, m, p}, 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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((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] - Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 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 {1}{x^5 \sqrt {a+b x^2+c x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{4 a x^4}-\frac {\operatorname {Subst}\left (\int \frac {\frac {3 b}{2}+c x}{x^2 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{4 a x^4}+\frac {3 b \sqrt {a+b x^2+c x^4}}{8 a^2 x^2}+\frac {\left (3 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^2}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{4 a x^4}+\frac {3 b \sqrt {a+b x^2+c x^4}}{8 a^2 x^2}-\frac {\left (3 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^2}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{4 a x^4}+\frac {3 b \sqrt {a+b x^2+c x^4}}{8 a^2 x^2}-\frac {\left (3 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^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 91, normalized size = 0.84 \begin {gather*} \frac {\left (4 a c-3 b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac {\left (3 b x^2-2 a\right ) \sqrt {a+b x^2+c x^4}}{8 a^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

((-2*a + 3*b*x^2)*Sqrt[a + b*x^2 + c*x^4])/(8*a^2*x^4) + ((-3*b^2 + 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sq
rt[a + b*x^2 + c*x^4])])/(16*a^(5/2))

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IntegrateAlgebraic [A]  time = 0.31, size = 91, normalized size = 0.84 \begin {gather*} \frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {\left (3 b x^2-2 a\right ) \sqrt {a+b x^2+c x^4}}{8 a^2 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^5*Sqrt[a + b*x^2 + c*x^4]),x]

[Out]

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

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fricas [A]  time = 1.20, size = 221, normalized size = 2.05 \begin {gather*} \left [-\frac {{\left (3 \, 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 (3 \, a b x^{2} - 2 \, a^{2}\right )}}{32 \, a^{3} x^{4}}, \frac {{\left (3 \, 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 (3 \, a b x^{2} - 2 \, a^{2}\right )}}{16 \, a^{3} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*((3*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)*(3*a*b*x^2 - 2*a^2))/(a^3*x^4), 1/16*((3*b^2 - 4*a*c)*sq
rt(-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)*(3*a*b*x^2 - 2*a^2))/(a^3*x^4)]

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giac [B]  time = 0.30, size = 221, normalized size = 2.05 \begin {gather*} \frac {{\left (3 \, 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^{2}} - \frac {3 \, {\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 - 5 \, {\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 \, a^{2} b \sqrt {c}}{8 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*b^2 - 4*a*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) - 1/8*(3*(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 - 5*(sqrt(c)*x^2 - sqr
t(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 - 8*a^2*b*sqrt(c))/(((sqrt(c)*x^
2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)^2*a^2)

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maple [A]  time = 0.01, size = 127, normalized size = 1.18 \begin {gather*} \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 a^{\frac {3}{2}}}-\frac {3 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 {5}{2}}}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b}{8 a^{2} x^{2}}-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{4 a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(c*x^4+b*x^2+a)^(1/2)/a/x^4+3/8*b*(c*x^4+b*x^2+a)^(1/2)/a^2/x^2-3/16*b^2/a^(5/2)*ln((b*x^2+2*a+2*(c*x^4+b
*x^2+a)^(1/2)*a^(1/2))/x^2)+1/4*c/a^(3/2)*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(1/x^5/(c*x^4+b*x^2+a)^(1/2),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 {1}{x^5\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \sqrt {a + b x^{2} + c x^{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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