3.987 \(\int \frac {1}{x^3 (a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=139 \[ \frac {3 b \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{5/2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{2 a^2 x^2 \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{a x^2 \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 139, 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, 740, 806, 724, 206} \[ -\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{2 a^2 x^2 \left (b^2-4 a c\right )}+\frac {3 b \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{5/2}}+\frac {-2 a c+b^2+b c x^2}{a x^2 \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

(b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*x^2*Sqrt[a + b*x^2 + c*x^4]) - ((3*b^2 - 8*a*c)*Sqrt[a + b*x^2 + c*x^
4])/(2*a^2*(b^2 - 4*a*c)*x^2) + (3*b*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(4*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 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, 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] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

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^3 \left (a+b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x^2+c x^4}}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (-3 b^2+8 a c\right )-b c x}{x^2 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{a \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x^2+c x^4}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{2 a^2 \left (b^2-4 a c\right ) x^2}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{4 a^2}\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x^2+c x^4}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {(3 b) \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 )}{2 a^2}\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^2 \sqrt {a+b x^2+c x^4}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a+b x^2+c x^4}}{2 a^2 \left (b^2-4 a c\right ) x^2}+\frac {3 b \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 137, normalized size = 0.99 \[ \frac {\frac {2 \sqrt {a} \left (-4 a^2 c+a \left (b^2-10 b c x^2-8 c^2 x^4\right )+3 b^2 x^2 \left (b+c x^2\right )\right )}{x^2 \sqrt {a+b x^2+c x^4}}-3 b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{5/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

((2*Sqrt[a]*(-4*a^2*c + 3*b^2*x^2*(b + c*x^2) + a*(b^2 - 10*b*c*x^2 - 8*c^2*x^4)))/(x^2*Sqrt[a + b*x^2 + c*x^4
]) - 3*b*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(4*a^(5/2)*(-b^2 + 4*a*c))

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fricas [A]  time = 0.99, size = 485, normalized size = 3.49 \[ \left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{6} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{4} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{8 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{6} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{4} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}, -\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{6} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{4} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{4} + a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{4 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{6} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{4} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*((b^3*c - 4*a*b*c^2)*x^6 + (b^4 - 4*a*b^2*c)*x^4 + (a*b^3 - 4*a^2*b*c)*x^2)*sqrt(a)*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*((3*a*b^2*c - 8*a^2*c^2)
*x^4 + a^2*b^2 - 4*a^3*c + (3*a*b^3 - 10*a^2*b*c)*x^2)*sqrt(c*x^4 + b*x^2 + a))/((a^3*b^2*c - 4*a^4*c^2)*x^6 +
(a^3*b^3 - 4*a^4*b*c)*x^4 + (a^4*b^2 - 4*a^5*c)*x^2), -1/4*(3*((b^3*c - 4*a*b*c^2)*x^6 + (b^4 - 4*a*b^2*c)*x^
4 + (a*b^3 - 4*a^2*b*c)*x^2)*sqrt(-a)*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*((3*a*b^2*c - 8*a^2*c^2)*x^4 + a^2*b^2 - 4*a^3*c + (3*a*b^3 - 10*a^2*b*c)*x^2)*sqrt(c*x^4 + b
*x^2 + a))/((a^3*b^2*c - 4*a^4*c^2)*x^6 + (a^3*b^3 - 4*a^4*b*c)*x^4 + (a^4*b^2 - 4*a^5*c)*x^2)]

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giac [A]  time = 0.28, size = 200, normalized size = 1.44 \[ -\frac {\frac {{\left (a^{2} b^{2} c - 2 \, a^{3} c^{2}\right )} x^{2}}{a^{4} b^{2} - 4 \, a^{5} c} + \frac {a^{2} b^{3} - 3 \, a^{3} b c}{a^{4} b^{2} - 4 \, a^{5} c}}{\sqrt {c x^{4} + b x^{2} + a}} - \frac {3 \, b \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a^{2}} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} b + 2 \, a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 195, normalized size = 1.40 \[ \frac {3 b^{2} c \,x^{2}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2}}+\frac {3 b^{3}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2}}-\frac {2 \left (2 c \,x^{2}+b \right ) c}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, a}+\frac {3 b \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}-\frac {3 b}{4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2}}-\frac {1}{2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(c*x^4+b*x^2+a)^(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(c*x^4+b*x^2+a)^(3/2),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 \[ \int \frac {1}{x^3\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*x^2 + c*x^4)^(3/2)),x)

[Out]

int(1/(x^3*(a + b*x^2 + c*x^4)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(1/(x**3*(a + b*x**2 + c*x**4)**(3/2)), x)

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