\(\int \frac {x^2}{a+b x^3+c x^6} \, dx\) [140]

Optimal result

Integrand size = 18, antiderivative size = 38 \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 \sqrt {b^2-4 a c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1366, 632, 212} \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=-\frac {2 \text {arctanh}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 \sqrt {b^2-4 a c}} \]

[In]

Int[x^2/(a + b*x^3 + c*x^6),x]

[Out]

(-2*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*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 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=\frac {2 \arctan \left (\frac {b+2 c x^3}{\sqrt {-b^2+4 a c}}\right )}{3 \sqrt {-b^2+4 a c}} \]

[In]

Integrate[x^2/(a + b*x^3 + c*x^6),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {2 \arctan \left (\frac {2 c \,x^{3}+b}{\sqrt {4 a c -b^{2}}}\right )}{3 \sqrt {4 a c -b^{2}}}\)	\(37\)
risch	\(-\frac {\ln \left (\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{3}-2 a \right )}{3 \sqrt {-4 a c +b^{2}}}+\frac {\ln \left (\left (b +\sqrt {-4 a c +b^{2}}\right ) x^{3}+2 a \right )}{3 \sqrt {-4 a c +b^{2}}}\)	\(70\)

[In]

int(x^2/(c*x^6+b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 3.39 \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=\left [\frac {\log \left (\frac {2 \, c^{2} x^{6} + 2 \, b c x^{3} + b^{2} - 2 \, a c - {\left (2 \, c x^{3} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{6} + b x^{3} + a}\right )}{3 \, \sqrt {b^{2} - 4 \, a c}}, -\frac {2 \, \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{3} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{3 \, {\left (b^{2} - 4 \, a c\right )}}\right ] \]

[In]

integrate(x^2/(c*x^6+b*x^3+a),x, algorithm="fricas")

[Out]

[1/3*log((2*c^2*x^6 + 2*b*c*x^3 + b^2 - 2*a*c - (2*c*x^3 + b)*sqrt(b^2 - 4*a*c))/(c*x^6 + b*x^3 + a))/sqrt(b^2
- 4*a*c), -2/3*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^3 + b)*sqrt(-b^2 + 4*a*c)/(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. 131 vs. \(2 (37) = 74\).

Time = 0.38 (sec) , antiderivative size = 131, normalized size of antiderivative = 3.45 \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=- \frac {\sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x^{3} + \frac {- 4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} + b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )}}{3} + \frac {\sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x^{3} + \frac {4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} - b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )}}{3} \]

[In]

integrate(x**2/(c*x**6+b*x**3+a),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(x^2/(c*x^6+b*x^3+a),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 = 0.36 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=\frac {2 \, \arctan \left (\frac {2 \, c x^{3} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{3 \, \sqrt {-b^{2} + 4 \, a c}} \]

[In]

integrate(x^2/(c*x^6+b*x^3+a),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 4.58 \[ \int \frac {x^2}{a+b x^3+c x^6} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {\frac {x^3\,{\left (4\,a\,c-b^2\right )}^4}{2}+a\,b\,{\left (4\,a\,c-b^2\right )}^3+a\,b^3\,{\left (4\,a\,c-b^2\right )}^2+b^2\,x^3\,{\left (4\,a\,c-b^2\right )}^3+\frac {b^4\,x^3\,{\left (4\,a\,c-b^2\right )}^2}{2}}{b^2\,\left (32\,a^3\,c^2\,\sqrt {4\,a\,c-b^2}-4\,a^2\,b^2\,c\,\sqrt {4\,a\,c-b^2}\right )-64\,a^4\,c^3\,\sqrt {4\,a\,c-b^2}}\right )}{3\,\sqrt {4\,a\,c-b^2}} \]

[In]

int(x^2/(a + b*x^3 + c*x^6),x)

[Out]

-(2*atan(((x^3*(4*a*c - b^2)^4)/2 + a*b*(4*a*c - b^2)^3 + a*b^3*(4*a*c - b^2)^2 + b^2*x^3*(4*a*c - b^2)^3 + (b
^4*x^3*(4*a*c - b^2)^2)/2)/(b^2*(32*a^3*c^2*(4*a*c - b^2)^(1/2) - 4*a^2*b^2*c*(4*a*c - b^2)^(1/2)) - 64*a^4*c^
3*(4*a*c - b^2)^(1/2))))/(3*(4*a*c - b^2)^(1/2))