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

Optimal. Leaf size=122 \[ \frac{b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^{3/2}}-\frac{\log \left (a+b x^2+c x^4\right )}{4 a^2}+\frac{\log (x)}{a^2}+\frac{-2 a c+b^2+b c x^2}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.197928, antiderivative size = 122, 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} \[ \frac{b \left (b^2-6 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x^2}{\sqrt{b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^{3/2}}-\frac{\log \left (a+b x^2+c x^4\right )}{4 a^2}+\frac{\log (x)}{a^2}+\frac{-2 a c+b^2+b c x^2}{2 a \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2 + c*x^4)^2),x]

[Out]

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

Mathematica [A]  time = 0.311731, size = 207, normalized size = 1.7 \[ \frac{\frac{2 a \left (-2 a c+b^2+b c x^2\right )}{\left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac{\left (b^2 \sqrt{b^2-4 a c}-4 a c \sqrt{b^2-4 a c}-6 a b c+b^3\right ) \log \left (-\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+\frac{\left (-b^2 \sqrt{b^2-4 a c}+4 a c \sqrt{b^2-4 a c}-6 a b c+b^3\right ) \log \left (\sqrt{b^2-4 a c}+b+2 c x^2\right )}{\left (b^2-4 a c\right )^{3/2}}+4 \log (x)}{4 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.184, size = 253, normalized size = 2.1 \begin{align*}{\frac{\ln \left ( x \right ) }{{a}^{2}}}-{\frac{bc{x}^{2}}{2\,a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{c}{ \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{{b}^{2}}{2\,a \left ( c{x}^{4}+b{x}^{2}+a \right ) \left ( 4\,ac-{b}^{2} \right ) }}-{\frac{c\ln \left ( c{x}^{4}+b{x}^{2}+a \right ) }{a \left ( 4\,ac-{b}^{2} \right ) }}+{\frac{\ln \left ( c{x}^{4}+b{x}^{2}+a \right ){b}^{2}}{4\,{a}^{2} \left ( 4\,ac-{b}^{2} \right ) }}-3\,{\frac{bc}{a \left ( 4\,ac-{b}^{2} \right ) ^{3/2}}\arctan \left ({\frac{2\,c{x}^{2}+b}{\sqrt{4\,ac-{b}^{2}}}} \right ) }+{\frac{{b}^{3}}{2\,{a}^{2}}\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/(c*x^4+b*x^2+a)^2,x)

[Out]

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

________________________________________________________________________________________

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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.47028, size = 1728, normalized size = 14.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B]  time = 45.2521, size = 772, normalized size = 6.33 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*a**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) -
1/(4*a**2))*log(x**2 + (-32*a**4*c**2*(-b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*a**2*(64*a**3*c**3 - 48*
a**2*b**2*c**2 + 12*a*b**4*c - b**6)) - 1/(4*a**2)) + 16*a**3*b**2*c*(-b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**
2)/(4*a**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) - 1/(4*a**2)) - 2*a**2*b**4*(-b*sqrt(-(4*a
*c - b**2)**3)*(6*a*c - b**2)/(4*a**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) - 1/(4*a**2)) -
8*a**2*c**2 + 7*a*b**2*c - b**4)/(6*a*b*c**2 - b**3*c)) + (b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*a**2*
(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) - 1/(4*a**2))*log(x**2 + (-32*a**4*c**2*(b*sqrt(-(4*a
*c - b**2)**3)*(6*a*c - b**2)/(4*a**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) - 1/(4*a**2)) +
16*a**3*b**2*c*(b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*a**2*(64*a**3*c**3 - 48*a**2*b**2*c**2 + 12*a*b*
*4*c - b**6)) - 1/(4*a**2)) - 2*a**2*b**4*(b*sqrt(-(4*a*c - b**2)**3)*(6*a*c - b**2)/(4*a**2*(64*a**3*c**3 - 4
8*a**2*b**2*c**2 + 12*a*b**4*c - b**6)) - 1/(4*a**2)) - 8*a**2*c**2 + 7*a*b**2*c - b**4)/(6*a*b*c**2 - b**3*c)
) - (-2*a*c + b**2 + b*c*x**2)/(8*a**3*c - 2*a**2*b**2 + x**4*(8*a**2*c**2 - 2*a*b**2*c) + x**2*(8*a**2*b*c -
2*a*b**3)) + log(x)/a**2

________________________________________________________________________________________

Giac [A]  time = 21.4641, size = 224, normalized size = 1.84 \begin{align*} -\frac{{\left (b^{3} - 6 \, a b c\right )} \arctan \left (\frac{2 \, c x^{2} + b}{\sqrt{-b^{2} + 4 \, a c}}\right )}{2 \,{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} \sqrt{-b^{2} + 4 \, a c}} + \frac{b^{2} c x^{4} - 4 \, a c^{2} x^{4} + b^{3} x^{2} - 2 \, a b c x^{2} + 3 \, a b^{2} - 8 \, a^{2} c}{4 \,{\left (c x^{4} + b x^{2} + a\right )}{\left (a^{2} b^{2} - 4 \, a^{3} c\right )}} - \frac{\log \left (c x^{4} + b x^{2} + a\right )}{4 \, a^{2}} + \frac{\log \left (x^{2}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^3 - 6*a*b*c)*arctan((2*c*x^2 + b)/sqrt(-b^2 + 4*a*c))/((a^2*b^2 - 4*a^3*c)*sqrt(-b^2 + 4*a*c)) + 1/4*(
b^2*c*x^4 - 4*a*c^2*x^4 + b^3*x^2 - 2*a*b*c*x^2 + 3*a*b^2 - 8*a^2*c)/((c*x^4 + b*x^2 + a)*(a^2*b^2 - 4*a^3*c))
- 1/4*log(c*x^4 + b*x^2 + a)/a^2 + 1/2*log(x^2)/a^2