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

Optimal. Leaf size=162 \[ -\frac{\left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{a^3 \left (b^2-4 a c\right )^{3/2}}-\frac{b^2-3 a c}{a^2 x^2 \left (b^2-4 a c\right )}+\frac{b \log \left (a+b x^2+c x^4\right )}{2 a^3}-\frac{2 b \log (x)}{a^3}+\frac{-2 a c+b^2+b c x^2}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

-((b^2 - 3*a*c)/(a^2*(b^2 - 4*a*c)*x^2)) + (b^2 - 2*a*c + b*c*x^2)/(2*a*(b^2 - 4*a*c)*x^2*(a + b*x^2 + c*x^4))
- ((b^4 - 6*a*b^2*c + 6*a^2*c^2)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(a^3*(b^2 - 4*a*c)^(3/2)) - (2*b*L
og[x])/a^3 + (b*Log[a + b*x^2 + c*x^4])/(2*a^3)

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Rubi [A]  time = 0.25055, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {1114, 740, 800, 634, 618, 206, 628} \[ -\frac{\left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{a^3 \left (b^2-4 a c\right )^{3/2}}-\frac{b^2-3 a c}{a^2 x^2 \left (b^2-4 a c\right )}+\frac{b \log \left (a+b x^2+c x^4\right )}{2 a^3}-\frac{2 b \log (x)}{a^3}+\frac{-2 a c+b^2+b c x^2}{2 a x^2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2 + c*x^4)^2),x]

[Out]

-((b^2 - 3*a*c)/(a^2*(b^2 - 4*a*c)*x^2)) + (b^2 - 2*a*c + b*c*x^2)/(2*a*(b^2 - 4*a*c)*x^2*(a + b*x^2 + c*x^4))
- ((b^4 - 6*a*b^2*c + 6*a^2*c^2)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(a^3*(b^2 - 4*a*c)^(3/2)) - (2*b*L
og[x])/a^3 + (b*Log[a + b*x^2 + c*x^4])/(2*a^3)

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 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 )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x+c x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \frac{-2 \left (b^2-3 a c\right )-2 b c x}{x^2 \left (a+b x+c x^2\right )} \, dx,x,x^2\right )}{2 a \left (b^2-4 a c\right )}\\ &=\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{2 \left (-b^2+3 a c\right )}{a x^2}-\frac{2 b \left (-b^2+4 a c\right )}{a^2 x}+\frac{2 \left (-b^4+5 a b^2 c-3 a^2 c^2-b c \left (b^2-4 a c\right ) x\right )}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx,x,x^2\right )}{2 a \left (b^2-4 a c\right )}\\ &=-\frac{b^2-3 a c}{a^2 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{2 b \log (x)}{a^3}-\frac{\operatorname{Subst}\left (\int \frac{-b^4+5 a b^2 c-3 a^2 c^2-b c \left (b^2-4 a c\right ) x}{a+b x+c x^2} \, dx,x,x^2\right )}{a^3 \left (b^2-4 a c\right )}\\ &=-\frac{b^2-3 a c}{a^2 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{2 b \log (x)}{a^3}+\frac{b \operatorname{Subst}\left (\int \frac{b+2 c x}{a+b x+c x^2} \, dx,x,x^2\right )}{2 a^3}+\frac{\left (b^4-6 a b^2 c+6 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{a+b x+c x^2} \, dx,x,x^2\right )}{2 a^3 \left (b^2-4 a c\right )}\\ &=-\frac{b^2-3 a c}{a^2 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{2 b \log (x)}{a^3}+\frac{b \log \left (a+b x^2+c x^4\right )}{2 a^3}-\frac{\left (b^4-6 a b^2 c+6 a^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{b^2-4 a c-x^2} \, dx,x,b+2 c x^2\right )}{a^3 \left (b^2-4 a c\right )}\\ &=-\frac{b^2-3 a c}{a^2 \left (b^2-4 a c\right ) x^2}+\frac{b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) x^2 \left (a+b x^2+c x^4\right )}-\frac{\left (b^4-6 a b^2 c+6 a^2 c^2\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{a^3 \left (b^2-4 a c\right )^{3/2}}-\frac{2 b \log (x)}{a^3}+\frac{b \log \left (a+b x^2+c x^4\right )}{2 a^3}\\ \end{align*}

Mathematica [A]  time = 0.26558, size = 248, normalized size = 1.53 \[ \frac{\frac{\left (6 a^2 c^2+b^3 \sqrt{b^2-4 a c}-6 a b^2 c-4 a b c \sqrt{b^2-4 a c}+b^4\right ) \log \left (-\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{\left (-6 a^2 c^2+b^3 \sqrt{b^2-4 a c}+6 a b^2 c-4 a b c \sqrt{b^2-4 a c}-b^4\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac{a \left (-3 a b c-2 a c^2 x^2+b^2 c x^2+b^3\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{a}{x^2}-4 b \log (x)}{2 a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a/x^2) - (a*(b^3 - 3*a*b*c + b^2*c*x^2 - 2*a*c^2*x^2))/((b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - 4*b*Log[x] + (
(b^4 - 6*a*b^2*c + 6*a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 4*a*b*c*Sqrt[b^2 - 4*a*c])*Log[b - Sqrt[b^2 - 4*a*c] +
2*c*x^2])/(b^2 - 4*a*c)^(3/2) + ((-b^4 + 6*a*b^2*c - 6*a^2*c^2 + b^3*Sqrt[b^2 - 4*a*c] - 4*a*b*c*Sqrt[b^2 - 4*
a*c])*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(b^2 - 4*a*c)^(3/2))/(2*a^3)

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Maple [B]  time = 0.184, size = 352, normalized size = 2.2 \begin{align*} -{\frac{1}{2\,{a}^{2}{x}^{2}}}-2\,{\frac{b\ln \left ( x \right ) }{{a}^{3}}}-{\frac{{c}^{2}{x}^{2}}{a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{{b}^{2}c{x}^{2}}{2\,{a}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{3\,bc}{2\,a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{{b}^{3}}{2\,{a}^{2} \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+2\,{\frac{c\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) b}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{3}}{2\,{a}^{3} \left ( 4\,ac-{b}^{2} \right ) }}-6\,{\frac{{c}^{2}}{a \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+6\,{\frac{{b}^{2}c}{{a}^{2} \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }-{\frac{{b}^{4}}{{a}^{3}}\arctan \left ({(2\,c{x}^{2}+b){\frac{1}{\sqrt{4\,ac-{b}^{2}}}}} \right ) \left ( 4\,ac-{b}^{2} \right ) ^{-{\frac{3}{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)^2,x)

[Out]

-1/2/a^2/x^2-2*b*ln(x)/a^3-1/a/(c*x^4+b*x^2+a)*c^2/(4*a*c-b^2)*x^2+1/2/a^2/(c*x^4+b*x^2+a)*c/(4*a*c-b^2)*x^2*b
^2-3/2/a/(c*x^4+b*x^2+a)*b/(4*a*c-b^2)*c+1/2/a^2/(c*x^4+b*x^2+a)*b^3/(4*a*c-b^2)+2/a^2/(4*a*c-b^2)*c*ln(c*x^4+
b*x^2+a)*b-1/2/a^3/(4*a*c-b^2)*ln(c*x^4+b*x^2+a)*b^3-6/a/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2
))*c^2+6/a^2/(4*a*c-b^2)^(3/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*c*b^2-1/a^3/(4*a*c-b^2)^(3/2)*arctan((2*c
*x^2+b)/(4*a*c-b^2)^(1/2))*b^4

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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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.58553, size = 2103, normalized size = 12.98 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + 2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^3*c^3)*x^4 + (2*a*b^5 - 15*a^2*b
^3*c + 28*a^3*b*c^2)*x^2 + ((b^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)*x^6 + (b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*x^4 + (a*b
^4 - 6*a^2*b^2*c + 6*a^3*c^2)*x^2)*sqrt(b^2 - 4*a*c)*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^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^6 + (b^6 - 8*a*b^4*c + 16*a^
2*b^2*c^2)*x^4 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x^2)*log(c*x^4 + b*x^2 + a) + 4*((b^5*c - 8*a*b^3*c^2 +
16*a^2*b*c^3)*x^6 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^4 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x^2)*log(x))
/((a^3*b^4*c - 8*a^4*b^2*c^2 + 16*a^5*c^3)*x^6 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^4 + (a^4*b^4 - 8*a^5
*b^2*c + 16*a^6*c^2)*x^2), -1/2*(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + 2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^3*c^3)
*x^4 + (2*a*b^5 - 15*a^2*b^3*c + 28*a^3*b*c^2)*x^2 + 2*((b^4*c - 6*a*b^2*c^2 + 6*a^2*c^3)*x^6 + (b^5 - 6*a*b^3
*c + 6*a^2*b*c^2)*x^4 + (a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*x^2)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-
b^2 + 4*a*c)/(b^2 - 4*a*c)) - ((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^6 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x
^4 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x^2)*log(c*x^4 + b*x^2 + a) + 4*((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3
)*x^6 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^4 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x^2)*log(x))/((a^3*b^4*c
- 8*a^4*b^2*c^2 + 16*a^5*c^3)*x^6 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^4 + (a^4*b^4 - 8*a^5*b^2*c + 16*
a^6*c^2)*x^2)]

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Sympy [B]  time = 78.9888, size = 906, normalized size = 5.59 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b/(2*a**3) - sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*a**3*(64*a**3*c**3 - 48*a**2*b**2*
c**2 + 12*a*b**4*c - b**6)))*log(x**2 + (-16*a**5*c**2*(b/(2*a**3) - sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6
*a*b**2*c + b**4)/(2*a**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 8*a**4*b**2*c*(b/(2*a**3
) - sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*a**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*
a*b**4*c - b**6))) - a**3*b**4*(b/(2*a**3) - sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*a**
3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 11*a**2*b*c**2 - 7*a*b**3*c + b**5)/(6*a**2*c**3
- 6*a*b**2*c**2 + b**4*c)) + (b/(2*a**3) + sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*a**3
*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)))*log(x**2 + (-16*a**5*c**2*(b/(2*a**3) + sqrt(-(4*a*
c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*a**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6
))) + 8*a**4*b**2*c*(b/(2*a**3) + sqrt(-(4*a*c - b**2)**3)*(6*a**2*c**2 - 6*a*b**2*c + b**4)/(2*a**3*(64*a**3*
c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) - a**3*b**4*(b/(2*a**3) + sqrt(-(4*a*c - b**2)**3)*(6*a**2*c*
*2 - 6*a*b**2*c + b**4)/(2*a**3*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6))) + 11*a**2*b*c**2 - 7
*a*b**3*c + b**5)/(6*a**2*c**3 - 6*a*b**2*c**2 + b**4*c)) - (4*a**2*c - a*b**2 + x**4*(6*a*c**2 - 2*b**2*c) +
x**2*(7*a*b*c - 2*b**3))/(x**6*(8*a**3*c**2 - 2*a**2*b**2*c) + x**4*(8*a**3*b*c - 2*a**2*b**3) + x**2*(8*a**4*
c - 2*a**3*b**2)) - 2*b*log(x)/a**3

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Giac [A]  time = 20.868, size = 246, normalized size = 1.52 \begin{align*} \frac{{\left (b^{4} - 6 \, a b^{2} c + 6 \, a^{2} c^{2}\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} \sqrt{-b^{2} + 4 \, a c}} - \frac{2 \, b^{2} c x^{4} - 6 \, a c^{2} x^{4} + 2 \, b^{3} x^{2} - 7 \, a b c x^{2} + a b^{2} - 4 \, a^{2} c}{2 \,{\left (c x^{6} + b x^{4} + a x^{2}\right )}{\left (a^{2} b^{2} - 4 \, a^{3} c\right )}} + \frac{b \log \left (c x^{4} + b x^{2} + a\right )}{2 \, a^{3}} - \frac{b \log \left (x^{2}\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^4 - 6*a*b^2*c + 6*a^2*c^2)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((a^3*b^2 - 4*a^4*c)*sqrt(-b^2 + 4*a*c)
) - 1/2*(2*b^2*c*x^4 - 6*a*c^2*x^4 + 2*b^3*x^2 - 7*a*b*c*x^2 + a*b^2 - 4*a^2*c)/((c*x^6 + b*x^4 + a*x^2)*(a^2*
b^2 - 4*a^3*c)) + 1/2*b*log(c*x^4 + b*x^2 + a)/a^3 - b*log(x^2)/a^3