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

Optimal. Leaf size=70 \[ \frac{b \tanh ^{-1}\left (\frac{b-2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a \sqrt{b^2-4 a c}}-\frac{\log \left (a-b x^2+c x^4\right )}{4 a}+\frac{\log (x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.0837868, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {1114, 705, 29, 634, 618, 206, 628} \[ \frac{b \tanh ^{-1}\left (\frac{b-2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a \sqrt{b^2-4 a c}}-\frac{\log \left (a-b x^2+c x^4\right )}{4 a}+\frac{\log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a - b*x^2 + c*x^4)),x]

[Out]

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

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 705

Int[1/(((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 - b*d*e + a*e^2
), Int[1/(d + e*x), x], x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(c*d - b*e - c*e*x)/(a + b*x + c*x^2), x], x]
/; FreeQ[{a, b, c, d, e}, 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]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.0713104, size = 117, normalized size = 1.67 \[ \frac{\left (b-\sqrt{b^2-4 a c}\right ) \log \left (-\sqrt{b^2-4 a c}-b+2 c x^2\right )-\left (\sqrt{b^2-4 a c}+b\right ) \log \left (\sqrt{b^2-4 a c}-b+2 c x^2\right )+4 \log (x) \sqrt{b^2-4 a c}}{4 a \sqrt{b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.165, size = 69, normalized size = 1. \begin{align*}{\frac{\ln \left ( x \right ) }{a}}-{\frac{\ln \left ( c{x}^{4}-b{x}^{2}+a \right ) }{4\,a}}+{\frac{b}{2\,a}\arctan \left ({(2\,c{x}^{2}-b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.56106, size = 512, normalized size = 7.31 \begin{align*} \left [\frac{\sqrt{b^{2} - 4 \, a c} b \log \left (\frac{2 \, c^{2} x^{4} - 2 \, b c x^{2} + b^{2} - 2 \, a c -{\left (2 \, c x^{2} - b\right )} \sqrt{b^{2} - 4 \, a c}}{c x^{4} - b x^{2} + a}\right ) -{\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} - b x^{2} + a\right ) + 4 \,{\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{4 \,{\left (a b^{2} - 4 \, a^{2} c\right )}}, -\frac{2 \, \sqrt{-b^{2} + 4 \, a c} b \arctan \left (-\frac{{\left (2 \, c x^{2} - b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) +{\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} - b x^{2} + a\right ) - 4 \,{\left (b^{2} - 4 \, a c\right )} \log \left (x\right )}{4 \,{\left (a b^{2} - 4 \, a^{2} c\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),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 3.0117, size = 253, normalized size = 3.61 \begin{align*} \left (- \frac{b \sqrt{- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac{1}{4 a}\right ) \log{\left (x^{2} + \frac{8 a^{2} c \left (- \frac{b \sqrt{- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac{1}{4 a}\right ) - 2 a b^{2} \left (- \frac{b \sqrt{- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac{1}{4 a}\right ) + 2 a c - b^{2}}{b c} \right )} + \left (\frac{b \sqrt{- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac{1}{4 a}\right ) \log{\left (x^{2} + \frac{8 a^{2} c \left (\frac{b \sqrt{- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac{1}{4 a}\right ) - 2 a b^{2} \left (\frac{b \sqrt{- 4 a c + b^{2}}}{4 a \left (4 a c - b^{2}\right )} - \frac{1}{4 a}\right ) + 2 a c - b^{2}}{b c} \right )} + \frac{\log{\left (x \right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b*sqrt(-4*a*c + b**2)/(4*a*(4*a*c - b**2)) - 1/(4*a))*log(x**2 + (8*a**2*c*(-b*sqrt(-4*a*c + b**2)/(4*a*(4*a
*c - b**2)) - 1/(4*a)) - 2*a*b**2*(-b*sqrt(-4*a*c + b**2)/(4*a*(4*a*c - b**2)) - 1/(4*a)) + 2*a*c - b**2)/(b*c
)) + (b*sqrt(-4*a*c + b**2)/(4*a*(4*a*c - b**2)) - 1/(4*a))*log(x**2 + (8*a**2*c*(b*sqrt(-4*a*c + b**2)/(4*a*(
4*a*c - b**2)) - 1/(4*a)) - 2*a*b**2*(b*sqrt(-4*a*c + b**2)/(4*a*(4*a*c - b**2)) - 1/(4*a)) + 2*a*c - b**2)/(b
*c)) + log(x)/a

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Giac [A]  time = 1.33826, size = 96, normalized size = 1.37 \begin{align*} \frac{b \arctan \left (\frac{2 \, c x^{2} - b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt{-b^{2} + 4 \, a c} a} - \frac{\log \left (c x^{4} - b x^{2} + a\right )}{4 \, a} + \frac{\log \left (x^{2}\right )}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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