\(\int \frac {x^2}{a x+b x^3+c x^5} \, dx\) [84]

Optimal result

Integrand size = 20, antiderivative size = 36 \[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=-\frac {\text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1599, 1121, 632, 212} \[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=-\frac {\text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}} \]

[In]

Int[x^2/(a*x + b*x^3 + c*x^5),x]

[Out]

-(ArcTanh[(b + 2*c*x^2)/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 1121

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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.08 \[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=\frac {\arctan \left (\frac {b+2 c x^2}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}} \]

[In]

Integrate[x^2/(a*x + b*x^3 + c*x^5),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00

method	result	size
default	\(\frac {\arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}\)	\(36\)
risch	\(-\frac {\ln \left (\left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}-2 a \right )}{2 \sqrt {-4 a c +b^{2}}}+\frac {\ln \left (\left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}+2 a \right )}{2 \sqrt {-4 a c +b^{2}}}\)	\(70\)

[In]

int(x^2/(c*x^5+b*x^3+a*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 3.58 \[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=\left [\frac {\log \left (\frac {2 \, c^{2} x^{4} + 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} + b x^{2} + a}\right )}{2 \, \sqrt {b^{2} - 4 \, a c}}, -\frac {\sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right )}{b^{2} - 4 \, a c}\right ] \]

[In]

integrate(x^2/(c*x^5+b*x^3+a*x),x, algorithm="fricas")

[Out]

[1/2*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))/sqrt(b^2
- 4*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))/(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. \(2 (34) = 68\).

Time = 0.26 (sec) , antiderivative size = 131, normalized size of antiderivative = 3.64 \[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=- \frac {\sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x^{2} + \frac {- 4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} + b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )}}{2} + \frac {\sqrt {- \frac {1}{4 a c - b^{2}}} \log {\left (x^{2} + \frac {4 a c \sqrt {- \frac {1}{4 a c - b^{2}}} - b^{2} \sqrt {- \frac {1}{4 a c - b^{2}}} + b}{2 c} \right )}}{2} \]

[In]

integrate(x**2/(c*x**5+b*x**3+a*x),x)

[Out]

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

Maxima [F]

\[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=\int { \frac {x^{2}}{c x^{5} + b x^{3} + a x} \,d x } \]

[In]

integrate(x^2/(c*x^5+b*x^3+a*x),x, algorithm="maxima")

[Out]

integrate(x^2/(c*x^5 + b*x^3 + a*x), x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=\frac {\arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c}} \]

[In]

integrate(x^2/(c*x^5+b*x^3+a*x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.47 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.14 \[ \int \frac {x^2}{a x+b x^3+c x^5} \, dx=\frac {\mathrm {atan}\left (\frac {2\,a\,c\,x^2+a\,b}{a\,\sqrt {4\,a\,c-b^2}}\right )}{\sqrt {4\,a\,c-b^2}} \]

[In]

int(x^2/(a*x + b*x^3 + c*x^5),x)

[Out]

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