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

Optimal. Leaf size=115 \[ -\frac{\left (4 a c+3 b^2\right ) \tan ^{-1}\left (\frac{2 a-b x^2}{2 \sqrt{a} \sqrt{-a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{3 b \sqrt{-a+b x^2+c x^4}}{8 a^2 x^2}+\frac{\sqrt{-a+b x^2+c x^4}}{4 a x^4} \]

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

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Rubi [A]  time = 0.123475, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {1114, 744, 806, 724, 204} \[ -\frac{\left (4 a c+3 b^2\right ) \tan ^{-1}\left (\frac{2 a-b x^2}{2 \sqrt{a} \sqrt{-a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{3 b \sqrt{-a+b x^2+c x^4}}{8 a^2 x^2}+\frac{\sqrt{-a+b x^2+c x^4}}{4 a x^4} \]

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

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]

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 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 204

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

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 ) \tan ^{-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*}

Mathematica [A]  time = 0.0462942, size = 95, normalized size = 0.83 \[ \frac{\left (4 a c+3 b^2\right ) \tan ^{-1}\left (\frac{b x^2-2 a}{2 \sqrt{a} \sqrt{-a+b x^2+c x^4}}\right )}{16 a^{5/2}}+\frac{\left (2 a+3 b x^2\right ) \sqrt{-a+b x^2+c x^4}}{8 a^2 x^4} \]

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)*ArcTan[(-2*a + b*x^2)/(2*Sqrt[a]*Sqr
t[-a + b*x^2 + c*x^4])])/(16*a^(5/2))

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Maple [A]  time = 0.17, size = 149, normalized size = 1.3 \begin{align*}{\frac{1}{4\,a{x}^{4}}\sqrt{c{x}^{4}+b{x}^{2}-a}}+{\frac{3\,b}{8\,{a}^{2}{x}^{2}}\sqrt{c{x}^{4}+b{x}^{2}-a}}-{\frac{3\,{b}^{2}}{16\,{a}^{2}}\ln \left ({\frac{1}{{x}^{2}} \left ( -2\,a+b{x}^{2}+2\,\sqrt{-a}\sqrt{c{x}^{4}+b{x}^{2}-a} \right ) } \right ){\frac{1}{\sqrt{-a}}}}-{\frac{c}{4\,a}\ln \left ({\frac{1}{{x}^{2}} \left ( -2\,a+b{x}^{2}+2\,\sqrt{-a}\sqrt{c{x}^{4}+b{x}^{2}-a} \right ) } \right ){\frac{1}{\sqrt{-a}}}} \end{align*}

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

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

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

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Fricas [A]  time = 1.71857, size = 510, normalized size = 4.43 \begin{align*} \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{align*}

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)*s
qrt(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{5} \sqrt{- a + b x^{2} + c x^{4}}}\, dx \end{align*}

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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Giac [A]  time = 1.84369, size = 122, normalized size = 1.06 \begin{align*} \frac{1}{8} \, \sqrt{c + \frac{b}{x^{2}} - \frac{a}{x^{4}}}{\left (\frac{3 \, b}{a^{2}} + \frac{2}{a x^{2}}\right )} + \frac{{\left (3 \, b^{2} + 4 \, a c\right )} \log \left ({\left | -2 \, \sqrt{-a}{\left (\sqrt{c + \frac{b}{x^{2}} - \frac{a}{x^{4}}} - \frac{\sqrt{-a}}{x^{2}}\right )} + b \right |}\right )}{16 \, \sqrt{-a} a^{2}} \end{align*}

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*sqrt(c + b/x^2 - a/x^4)*(3*b/a^2 + 2/(a*x^2)) + 1/16*(3*b^2 + 4*a*c)*log(abs(-2*sqrt(-a)*(sqrt(c + b/x^2 -
a/x^4) - sqrt(-a)/x^2) + b))/(sqrt(-a)*a^2)