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

Optimal. Leaf size=89 \[ \frac{-2 a c+b^2+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0808448, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1114, 740, 12, 724, 206} \[ \frac{-2 a c+b^2+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.107392, size = 89, normalized size = 1. \[ \frac{-2 a c+b^2+b c x^2}{a \left (b^2-4 a c\right ) \sqrt{a+b x^2+c x^4}}-\frac{\tanh ^{-1}\left (\frac{2 a+b x^2}{2 \sqrt{a} \sqrt{a+b x^2+c x^4}}\right )}{2 a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.166, size = 99, normalized size = 1.1 \begin{align*}{\frac{1}{2\,a}{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}}-{\frac{b \left ( 2\,c{x}^{2}+b \right ) }{2\,a \left ( 4\,ac-{b}^{2} \right ) }{\frac{1}{\sqrt{c{x}^{4}+b{x}^{2}+a}}}}-{\frac{1}{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 ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.94186, size = 837, normalized size = 9.4 \begin{align*} \left [\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a 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 (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt{c x^{4} + b x^{2} + a}}{4 \,{\left (a^{3} b^{2} - 4 \, a^{4} c +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}}, \frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{4} + a b^{2} - 4 \, a^{2} c +{\left (b^{3} - 4 \, a 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 (a b c x^{2} + a b^{2} - 2 \, a^{2} c\right )} \sqrt{c x^{4} + b x^{2} + a}}{2 \,{\left (a^{3} b^{2} - 4 \, a^{4} c +{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{4} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (a + b x^{2} + c x^{4}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{4} + b x^{2} + a\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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