3.2.56 \(\int \frac {x^5}{-4+x^2+3 x^4} \, dx\) [156]

Optimal. Leaf size=32 \[ \frac {x^2}{6}+\frac {1}{14} \log \left (1-x^2\right )-\frac {8}{63} \log \left (4+3 x^2\right ) \]

[Out]

1/6*x^2+1/14*ln(-x^2+1)-8/63*ln(3*x^2+4)

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Rubi [A]
time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1128, 717, 646, 31} \begin {gather*} \frac {x^2}{6}+\frac {1}{14} \log \left (1-x^2\right )-\frac {8}{63} \log \left (3 x^2+4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(-4 + x^2 + 3*x^4),x]

[Out]

x^2/6 + Log[1 - x^2]/14 - (8*Log[4 + 3*x^2])/63

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 646

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), 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 717

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m - 1)/(c*(
m - 1))), x] + Dist[1/c, Int[(d + e*x)^(m - 2)*(Simp[c*d^2 - a*e^2 + e*(2*c*d - b*e)*x, x]/(a + b*x + c*x^2)),
x], x] /; FreeQ[{a, b, c, d, e}, 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] && GtQ[m, 1]

Rule 1128

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]

Rubi steps

\begin {align*} \int \frac {x^5}{-4+x^2+3 x^4} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{-4+x+3 x^2} \, dx,x,x^2\right )\\ &=\frac {x^2}{6}+\frac {1}{6} \text {Subst}\left (\int \frac {4-x}{-4+x+3 x^2} \, dx,x,x^2\right )\\ &=\frac {x^2}{6}+\frac {3}{14} \text {Subst}\left (\int \frac {1}{-3+3 x} \, dx,x,x^2\right )-\frac {8}{21} \text {Subst}\left (\int \frac {1}{4+3 x} \, dx,x,x^2\right )\\ &=\frac {x^2}{6}+\frac {1}{14} \log \left (1-x^2\right )-\frac {8}{63} \log \left (4+3 x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 32, normalized size = 1.00 \begin {gather*} \frac {x^2}{6}+\frac {1}{14} \log \left (1-x^2\right )-\frac {8}{63} \log \left (4+3 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(-4 + x^2 + 3*x^4),x]

[Out]

x^2/6 + Log[1 - x^2]/14 - (8*Log[4 + 3*x^2])/63

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Maple [A]
time = 0.02, size = 25, normalized size = 0.78

method	result	size
default	\(\frac {x^{2}}{6}+\frac {\ln \left (x^{2}-1\right )}{14}-\frac {8 \ln \left (3 x^{2}+4\right )}{63}\)	\(25\)
risch	\(\frac {x^{2}}{6}+\frac {\ln \left (x^{2}-1\right )}{14}-\frac {8 \ln \left (3 x^{2}+4\right )}{63}\)	\(25\)
norman	\(\frac {x^{2}}{6}+\frac {\ln \left (-1+x \right )}{14}+\frac {\ln \left (1+x \right )}{14}-\frac {8 \ln \left (3 x^{2}+4\right )}{63}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(3*x^4+x^2-4),x,method=_RETURNVERBOSE)

[Out]

1/6*x^2+1/14*ln(x^2-1)-8/63*ln(3*x^2+4)

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Maxima [A]
time = 2.30, size = 24, normalized size = 0.75 \begin {gather*} \frac {1}{6} \, x^{2} - \frac {8}{63} \, \log \left (3 \, x^{2} + 4\right ) + \frac {1}{14} \, \log \left (x^{2} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(3*x^4+x^2-4),x, algorithm="maxima")

[Out]

1/6*x^2 - 8/63*log(3*x^2 + 4) + 1/14*log(x^2 - 1)

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Fricas [A]
time = 0.43, size = 24, normalized size = 0.75 \begin {gather*} \frac {1}{6} \, x^{2} - \frac {8}{63} \, \log \left (3 \, x^{2} + 4\right ) + \frac {1}{14} \, \log \left (x^{2} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(3*x^4+x^2-4),x, algorithm="fricas")

[Out]

1/6*x^2 - 8/63*log(3*x^2 + 4) + 1/14*log(x^2 - 1)

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Sympy [A]
time = 0.04, size = 24, normalized size = 0.75 \begin {gather*} \frac {x^{2}}{6} + \frac {\log {\left (x^{2} - 1 \right )}}{14} - \frac {8 \log {\left (x^{2} + \frac {4}{3} \right )}}{63} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(3*x**4+x**2-4),x)

[Out]

x**2/6 + log(x**2 - 1)/14 - 8*log(x**2 + 4/3)/63

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Giac [A]
time = 0.51, size = 25, normalized size = 0.78 \begin {gather*} \frac {1}{6} \, x^{2} - \frac {8}{63} \, \log \left (3 \, x^{2} + 4\right ) + \frac {1}{14} \, \log \left ({\left | x^{2} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(3*x^4+x^2-4),x, algorithm="giac")

[Out]

1/6*x^2 - 8/63*log(3*x^2 + 4) + 1/14*log(abs(x^2 - 1))

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Mupad [B]
time = 0.09, size = 22, normalized size = 0.69 \begin {gather*} \frac {\ln \left (x^2-1\right )}{14}-\frac {8\,\ln \left (x^2+\frac {4}{3}\right )}{63}+\frac {x^2}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(x^2 + 3*x^4 - 4),x)

[Out]

log(x^2 - 1)/14 - (8*log(x^2 + 4/3))/63 + x^2/6

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