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

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

[Out]

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

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Rubi [A]  time = 0.139838, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {1114, 709, 800, 634, 618, 206, 628} \[ \frac{\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac{b-2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^2 \sqrt{b^2-4 a c}}-\frac{b \log \left (a-b x^2+c x^4\right )}{4 a^2}+\frac{b \log (x)}{a^2}-\frac{1}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a - b*x^2 + c*x^4)),x]

[Out]

-1/(2*a*x^2) + ((b^2 - 2*a*c)*ArcTanh[(b - 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*a^2*Sqrt[b^2 - 4*a*c]) + (b*Log[x])
/a^2 - (b*Log[a - b*x^2 + c*x^4])/(4*a^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 709

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

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

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

Mathematica [A]  time = 0.142451, size = 139, normalized size = 1.56 \[ \frac{\frac{\left (-b \sqrt{b^2-4 a c}-2 a c+b^2\right ) \log \left (-\sqrt{b^2-4 a c}-b+2 c x^2\right )}{\sqrt{b^2-4 a c}}-\frac{\left (b \sqrt{b^2-4 a c}-2 a c+b^2\right ) \log \left (\sqrt{b^2-4 a c}-b+2 c x^2\right )}{\sqrt{b^2-4 a c}}-\frac{2 a}{x^2}+4 b \log (x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.166, size = 123, normalized size = 1.4 \begin{align*} -{\frac{1}{2\,a{x}^{2}}}+{\frac{b\ln \left ( x \right ) }{{a}^{2}}}-{\frac{b\ln \left ( c{x}^{4}-b{x}^{2}+a \right ) }{4\,{a}^{2}}}-{\frac{c}{a}\arctan \left ({(2\,c{x}^{2}-b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}}+{\frac{{b}^{2}}{2\,{a}^{2}}\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^3/(c*x^4-b*x^2+a),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59929, size = 664, normalized size = 7.46 \begin{align*} \left [-\frac{{\left (b^{2} - 2 \, a c\right )} \sqrt{b^{2} - 4 \, a c} x^{2} \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^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} - b x^{2} + a\right ) - 4 \,{\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}, -\frac{2 \,{\left (b^{2} - 2 \, a c\right )} \sqrt{-b^{2} + 4 \, a c} x^{2} \arctan \left (-\frac{{\left (2 \, c x^{2} - b\right )} \sqrt{-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) +{\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (c x^{4} - b x^{2} + a\right ) - 4 \,{\left (b^{3} - 4 \, a b c\right )} x^{2} \log \left (x\right ) + 2 \, a b^{2} - 8 \, a^{2} c}{4 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 7.50391, size = 350, normalized size = 3.93 \begin{align*} \left (- \frac{b}{4 a^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) \log{\left (x^{2} + \frac{- 8 a^{3} c \left (- \frac{b}{4 a^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} \left (- \frac{b}{4 a^{2}} - \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) - 3 a b c + b^{3}}{2 a c^{2} - b^{2} c} \right )} + \left (- \frac{b}{4 a^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) \log{\left (x^{2} + \frac{- 8 a^{3} c \left (- \frac{b}{4 a^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) + 2 a^{2} b^{2} \left (- \frac{b}{4 a^{2}} + \frac{\sqrt{- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 a^{2} \left (4 a c - b^{2}\right )}\right ) - 3 a b c + b^{3}}{2 a c^{2} - b^{2} c} \right )} - \frac{1}{2 a x^{2}} + \frac{b \log{\left (x \right )}}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b/(4*a**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2)))*log(x**2 + (-8*a**3*c*(-b/(4*a**2)
- sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2))) + 2*a**2*b**2*(-b/(4*a**2) - sqrt(-4*a*c + b**2)
*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2))) - 3*a*b*c + b**3)/(2*a*c**2 - b**2*c)) + (-b/(4*a**2) + sqrt(-4*a*c +
b**2)*(2*a*c - b**2)/(4*a**2*(4*a*c - b**2)))*log(x**2 + (-8*a**3*c*(-b/(4*a**2) + sqrt(-4*a*c + b**2)*(2*a*c
- b**2)/(4*a**2*(4*a*c - b**2))) + 2*a**2*b**2*(-b/(4*a**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(4*a**2*(4*a
*c - b**2))) - 3*a*b*c + b**3)/(2*a*c**2 - b**2*c)) - 1/(2*a*x**2) + b*log(x)/a**2

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Giac [A]  time = 1.44124, size = 128, normalized size = 1.44 \begin{align*} -\frac{b \log \left (c x^{4} - b x^{2} + a\right )}{4 \, a^{2}} + \frac{b \log \left (x^{2}\right )}{2 \, a^{2}} + \frac{{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac{2 \, c x^{2} - b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt{-b^{2} + 4 \, a c} a^{2}} - \frac{b x^{2} + a}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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