\(\int \frac {x^2}{a x^2+b x^3+c x^4} \, dx\) [14]

Optimal result

Integrand size = 22, antiderivative size = 34 \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1599, 632, 212} \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[In]

Int[x^2/(a*x^2 + b*x^3 + c*x^4),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 632

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 1599

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=\frac {2 \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {2 \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}\)	\(35\)
risch	\(-\frac {\ln \left (2 c x +\sqrt {-4 a c +b^{2}}+b \right )}{\sqrt {-4 a c +b^{2}}}+\frac {\ln \left (-2 c x +\sqrt {-4 a c +b^{2}}-b \right )}{\sqrt {-4 a c +b^{2}}}\)	\(61\)

[In]

int(x^2/(c*x^4+b*x^3+a*x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 3.53 \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=\left [\frac {\log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right )}{\sqrt {b^{2} - 4 \, a c}}, -\frac {2 \, \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right )}{b^{2} - 4 \, a c}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (34) = 68\).

Time = 0.11 (sec) , antiderivative size = 124, normalized size of antiderivative = 3.65 \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=- \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x + \frac {- 4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} + b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )} + \sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x + \frac {4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} - b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )} \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^2/(c*x^4+b*x^3+a*x^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 deta

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=\frac {2 \, \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.35 \[ \int \frac {x^2}{a x^2+b x^3+c x^4} \, dx=\frac {2\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{\sqrt {4\,a\,c-b^2}} \]

[In]

int(x^2/(a*x^2 + b*x^3 + c*x^4),x)

[Out]

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