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

Optimal. Leaf size=74 \[ \frac {2 c \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1107, 614, 618, 206} \[ \frac {2 c \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2}}-\frac {b+2 c x^2}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x^2 + c*x^4)^2,x]

[Out]

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

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 614

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^(p + 1))/((p +
1)*(b^2 - 4*a*c)), x] - Dist[(2*c*(2*p + 3))/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && IntegerQ[4*p]

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 1107

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 79, normalized size = 1.07 \[ -\frac {\frac {4 c \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\frac {b+2 c x^2}{a+b x^2+c x^4}}{2 \left (b^2-4 a c\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x^2 + c*x^4)^2,x]

[Out]

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

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fricas [B]  time = 0.87, size = 361, normalized size = 4.88 \[ \left [-\frac {b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} + 2 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}, -\frac {b^{3} - 4 \, a b c + 2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} - 4 \, {\left (c^{2} x^{4} + b c x^{2} + a c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{2 \, {\left (a b^{4} - 8 \, a^{2} b^{2} c + 16 \, a^{3} c^{2} + {\left (b^{4} c - 8 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{4} + {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.58, size = 82, normalized size = 1.11 \[ -\frac {2 \, c \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (b^{2} - 4 \, a c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, c x^{2} + b}{2 \, {\left (c x^{4} + b x^{2} + a\right )} {\left (b^{2} - 4 \, a c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 75, normalized size = 1.01 \[ \frac {2 c \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}}}+\frac {2 c \,x^{2}+b}{2 \left (4 a c -b^{2}\right ) \left (c \,x^{4}+b \,x^{2}+a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 4.31, size = 172, normalized size = 2.32 \[ \frac {\frac {b}{2\,\left (4\,a\,c-b^2\right )}+\frac {c\,x^2}{4\,a\,c-b^2}}{c\,x^4+b\,x^2+a}-\frac {2\,c\,\mathrm {atan}\left (\frac {b^3-4\,a\,b\,c}{{\left (4\,a\,c-b^2\right )}^{3/2}}-\frac {x^2\,{\left (4\,a\,c-b^2\right )}^4\,\left (\frac {4\,c^4}{a\,{\left (4\,a\,c-b^2\right )}^{7/2}}+\frac {4\,c^2\,\left (b^3\,c^2-4\,a\,b\,c^3\right )\,\left (b^3-4\,a\,b\,c\right )}{a\,{\left (4\,a\,c-b^2\right )}^{13/2}}\right )}{8\,c^4}\right )}{{\left (4\,a\,c-b^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^2 + c*x^4)^2,x)

[Out]

(b/(2*(4*a*c - b^2)) + (c*x^2)/(4*a*c - b^2))/(a + b*x^2 + c*x^4) - (2*c*atan((b^3 - 4*a*b*c)/(4*a*c - b^2)^(3
/2) - (x^2*(4*a*c - b^2)^4*((4*c^4)/(a*(4*a*c - b^2)^(7/2)) + (4*c^2*(b^3*c^2 - 4*a*b*c^3)*(b^3 - 4*a*b*c))/(a
*(4*a*c - b^2)^(13/2))))/(8*c^4)))/(4*a*c - b^2)^(3/2)

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sympy [B]  time = 2.78, size = 267, normalized size = 3.61 \[ - c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {- 16 a^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + 8 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b c}{2 c^{2}} \right )} + c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} \log {\left (x^{2} + \frac {16 a^{2} c^{3} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} - 8 a b^{2} c^{2} \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b^{4} c \sqrt {- \frac {1}{\left (4 a c - b^{2}\right )^{3}}} + b c}{2 c^{2}} \right )} + \frac {b + 2 c x^{2}}{8 a^{2} c - 2 a b^{2} + x^{4} \left (8 a c^{2} - 2 b^{2} c\right ) + x^{2} \left (8 a b c - 2 b^{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*sqrt(-1/(4*a*c - b**2)**3)*log(x**2 + (-16*a**2*c**3*sqrt(-1/(4*a*c - b**2)**3) + 8*a*b**2*c**2*sqrt(-1/(4*
a*c - b**2)**3) - b**4*c*sqrt(-1/(4*a*c - b**2)**3) + b*c)/(2*c**2)) + c*sqrt(-1/(4*a*c - b**2)**3)*log(x**2 +
(16*a**2*c**3*sqrt(-1/(4*a*c - b**2)**3) - 8*a*b**2*c**2*sqrt(-1/(4*a*c - b**2)**3) + b**4*c*sqrt(-1/(4*a*c -
b**2)**3) + b*c)/(2*c**2)) + (b + 2*c*x**2)/(8*a**2*c - 2*a*b**2 + x**4*(8*a*c**2 - 2*b**2*c) + x**2*(8*a*b*c
- 2*b**3))

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