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]

-1/16*(4*a*c+3*b^2)*arctan(1/2*(-b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2-a)^(1/2))/a^(5/2)+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

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Rubi [A]  time = 0.12, antiderivative size = 115, normalized size of antiderivative = 1.00, 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 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])

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 ) \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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.75, size = 230, normalized size = 2.00 \[ \left [-\frac {{\left (3 \, b^{2} + 4 \, a c\right )} \sqrt {-a} x^{4} \log \left (\frac {{\left (b^{2} - 4 \, a c\right )} x^{4} - 8 \, a b x^{2} - 4 \, \sqrt {c x^{4} + b x^{2} - a} {\left (b x^{2} - 2 \, a\right )} \sqrt {-a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, \sqrt {c x^{4} + b x^{2} - a} {\left (3 \, a b x^{2} + 2 \, a^{2}\right )}}{32 \, a^{3} x^{4}}, \frac {{\left (3 \, b^{2} + 4 \, a c\right )} \sqrt {a} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} - a} {\left (b x^{2} - 2 \, a\right )} \sqrt {a}}{2 \, {\left (a c x^{4} + a b x^{2} - a^{2}\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{2} - a} {\left (3 \, a b x^{2} + 2 \, a^{2}\right )}}{16 \, a^{3} x^{4}}\right ] \]

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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giac [B]  time = 0.23, size = 224, normalized size = 1.95 \[ \frac {{\left (3 \, b^{2} + 4 \, a c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}}{\sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}\right )}^{3} b^{2} + 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}\right )}^{3} a c + 5 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}\right )} a b^{2} - 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}\right )} a^{2} c - 8 \, a^{2} b \sqrt {c}}{8 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} - a}\right )}^{2} + a\right )}^{2} a^{2}} \]

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))/a^(5/2) - 1/8*(3*(sqrt(c)*x^2 - s
qrt(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 - 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 - 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 = 149, normalized size = 1.30 \[ -\frac {c \ln \left (\frac {b \,x^{2}-2 a +2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{4 \sqrt {-a}\, a}-\frac {3 b^{2} \ln \left (\frac {b \,x^{2}-2 a +2 \sqrt {-a}\, \sqrt {c \,x^{4}+b \,x^{2}-a}}{x^{2}}\right )}{16 \sqrt {-a}\, a^{2}}+\frac {3 \sqrt {c \,x^{4}+b \,x^{2}-a}\, b}{8 a^{2} x^{2}}+\frac {\sqrt {c \,x^{4}+b \,x^{2}-a}}{4 a \,x^{4}} \]

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

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maxima [A]  time = 2.43, size = 126, normalized size = 1.10 \[ -\frac {3 \, b^{2} \arcsin \left (-\frac {b}{\sqrt {b^{2} + 4 \, a c}} + \frac {2 \, a}{\sqrt {b^{2} + 4 \, a c} x^{2}}\right )}{16 \, a^{\frac {5}{2}}} - \frac {c \arcsin \left (-\frac {b}{\sqrt {b^{2} + 4 \, a c}} + \frac {2 \, a}{\sqrt {b^{2} + 4 \, a c} x^{2}}\right )}{4 \, a^{\frac {3}{2}}} + \frac {3 \, \sqrt {c x^{4} + b x^{2} - a} b}{8 \, a^{2} x^{2}} + \frac {\sqrt {c x^{4} + b x^{2} - a}}{4 \, a x^{4}} \]

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]

-3/16*b^2*arcsin(-b/sqrt(b^2 + 4*a*c) + 2*a/(sqrt(b^2 + 4*a*c)*x^2))/a^(5/2) - 1/4*c*arcsin(-b/sqrt(b^2 + 4*a*
c) + 2*a/(sqrt(b^2 + 4*a*c)*x^2))/a^(3/2) + 3/8*sqrt(c*x^4 + b*x^2 - a)*b/(a^2*x^2) + 1/4*sqrt(c*x^4 + b*x^2 -
a)/(a*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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