\(\int \frac {4+x^2+3 x^4+5 x^6}{x^3 (2+3 x^2+x^4)^2} \, dx\) [80]

Optimal result

Integrand size = 31, antiderivative size = 55 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=-\frac {1}{2 x^2}+\frac {9+11 x^2}{4 \left (2+3 x^2+x^4\right )}-\frac {11 \log (x)}{4}+5 \log \left (1+x^2\right )-\frac {29}{8} \log \left (2+x^2\right ) \]

[Out]

-1/2/x^2+1/4*(11*x^2+9)/(x^4+3*x^2+2)-11/4*ln(x)+5*ln(x^2+1)-29/8*ln(x^2+2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {1677, 1660, 1642} \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=-\frac {1}{2 x^2}+5 \log \left (x^2+1\right )-\frac {29}{8} \log \left (x^2+2\right )+\frac {11 x^2+9}{4 \left (x^4+3 x^2+2\right )}-\frac {11 \log (x)}{4} \]

[In]

Int[(4 + x^2 + 3*x^4 + 5*x^6)/(x^3*(2 + 3*x^2 + x^4)^2),x]

[Out]

-1/2*1/x^2 + (9 + 11*x^2)/(4*(2 + 3*x^2 + x^4)) - (11*Log[x])/4 + 5*Log[1 + x^2] - (29*Log[2 + x^2])/8

Rule 1642

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1660

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*
c*f - b*g)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1677

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)
*SubstFor[x^2, Pq, x]*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x^2] && Inte
gerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {4+x+3 x^2+5 x^3}{x^2 \left (2+3 x+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {9+11 x^2}{4 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \text {Subst}\left (\int \frac {-2+\frac {5 x}{2}-\frac {11 x^2}{2}}{x^2 \left (2+3 x+x^2\right )} \, dx,x,x^2\right ) \\ & = \frac {9+11 x^2}{4 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \text {Subst}\left (\int \left (-\frac {1}{x^2}+\frac {11}{4 x}-\frac {10}{1+x}+\frac {29}{4 (2+x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {1}{2 x^2}+\frac {9+11 x^2}{4 \left (2+3 x^2+x^4\right )}-\frac {11 \log (x)}{4}+5 \log \left (1+x^2\right )-\frac {29}{8} \log \left (2+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {1}{8} \left (-\frac {4}{x^2}+\frac {18+22 x^2}{2+3 x^2+x^4}-22 \log (x)+40 \log \left (1+x^2\right )-29 \log \left (2+x^2\right )\right ) \]

[In]

Integrate[(4 + x^2 + 3*x^4 + 5*x^6)/(x^3*(2 + 3*x^2 + x^4)^2),x]

[Out]

(-4/x^2 + (18 + 22*x^2)/(2 + 3*x^2 + x^4) - 22*Log[x] + 40*Log[1 + x^2] - 29*Log[2 + x^2])/8

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {29 \ln \left (x^{2}+2\right )}{8}+\frac {13}{4 \left (x^{2}+2\right )}-\frac {1}{2 x^{2}}-\frac {11 \ln \left (x \right )}{4}+5 \ln \left (x^{2}+1\right )-\frac {1}{2 \left (x^{2}+1\right )}\)	\(45\)
norman	\(\frac {-1+\frac {3}{4} x^{2}+\frac {9}{4} x^{4}}{x^{2} \left (x^{4}+3 x^{2}+2\right )}-\frac {11 \ln \left (x \right )}{4}+5 \ln \left (x^{2}+1\right )-\frac {29 \ln \left (x^{2}+2\right )}{8}\)	\(50\)
risch	\(\frac {-1+\frac {3}{4} x^{2}+\frac {9}{4} x^{4}}{x^{2} \left (x^{4}+3 x^{2}+2\right )}-\frac {11 \ln \left (x \right )}{4}+5 \ln \left (x^{2}+1\right )-\frac {29 \ln \left (x^{2}+2\right )}{8}\)	\(50\)
parallelrisch	\(-\frac {22 \ln \left (x \right ) x^{6}-40 \ln \left (x^{2}+1\right ) x^{6}+29 \ln \left (x^{2}+2\right ) x^{6}+8+66 \ln \left (x \right ) x^{4}-120 \ln \left (x^{2}+1\right ) x^{4}+87 \ln \left (x^{2}+2\right ) x^{4}-18 x^{4}+44 \ln \left (x \right ) x^{2}-80 \ln \left (x^{2}+1\right ) x^{2}+58 \ln \left (x^{2}+2\right ) x^{2}-6 x^{2}}{8 x^{2} \left (x^{4}+3 x^{2}+2\right )}\)	\(117\)

[In]

int((5*x^6+3*x^4+x^2+4)/x^3/(x^4+3*x^2+2)^2,x,method=_RETURNVERBOSE)

[Out]

-29/8*ln(x^2+2)+13/4/(x^2+2)-1/2/x^2-11/4*ln(x)+5*ln(x^2+1)-1/2/(x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.67 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {18 \, x^{4} + 6 \, x^{2} - 29 \, {\left (x^{6} + 3 \, x^{4} + 2 \, x^{2}\right )} \log \left (x^{2} + 2\right ) + 40 \, {\left (x^{6} + 3 \, x^{4} + 2 \, x^{2}\right )} \log \left (x^{2} + 1\right ) - 22 \, {\left (x^{6} + 3 \, x^{4} + 2 \, x^{2}\right )} \log \left (x\right ) - 8}{8 \, {\left (x^{6} + 3 \, x^{4} + 2 \, x^{2}\right )}} \]

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^3/(x^4+3*x^2+2)^2,x, algorithm="fricas")

[Out]

1/8*(18*x^4 + 6*x^2 - 29*(x^6 + 3*x^4 + 2*x^2)*log(x^2 + 2) + 40*(x^6 + 3*x^4 + 2*x^2)*log(x^2 + 1) - 22*(x^6
+ 3*x^4 + 2*x^2)*log(x) - 8)/(x^6 + 3*x^4 + 2*x^2)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {9 x^{4} + 3 x^{2} - 4}{4 x^{6} + 12 x^{4} + 8 x^{2}} - \frac {11 \log {\left (x \right )}}{4} + 5 \log {\left (x^{2} + 1 \right )} - \frac {29 \log {\left (x^{2} + 2 \right )}}{8} \]

[In]

integrate((5*x**6+3*x**4+x**2+4)/x**3/(x**4+3*x**2+2)**2,x)

[Out]

(9*x**4 + 3*x**2 - 4)/(4*x**6 + 12*x**4 + 8*x**2) - 11*log(x)/4 + 5*log(x**2 + 1) - 29*log(x**2 + 2)/8

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {9 \, x^{4} + 3 \, x^{2} - 4}{4 \, {\left (x^{6} + 3 \, x^{4} + 2 \, x^{2}\right )}} - \frac {29}{8} \, \log \left (x^{2} + 2\right ) + 5 \, \log \left (x^{2} + 1\right ) - \frac {11}{8} \, \log \left (x^{2}\right ) \]

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^3/(x^4+3*x^2+2)^2,x, algorithm="maxima")

[Out]

1/4*(9*x^4 + 3*x^2 - 4)/(x^6 + 3*x^4 + 2*x^2) - 29/8*log(x^2 + 2) + 5*log(x^2 + 1) - 11/8*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {9 \, x^{4} + 3 \, x^{2} - 4}{4 \, {\left (x^{6} + 3 \, x^{4} + 2 \, x^{2}\right )}} - \frac {29}{8} \, \log \left (x^{2} + 2\right ) + 5 \, \log \left (x^{2} + 1\right ) - \frac {11}{8} \, \log \left (x^{2}\right ) \]

[In]

integrate((5*x^6+3*x^4+x^2+4)/x^3/(x^4+3*x^2+2)^2,x, algorithm="giac")

[Out]

1/4*(9*x^4 + 3*x^2 - 4)/(x^6 + 3*x^4 + 2*x^2) - 29/8*log(x^2 + 2) + 5*log(x^2 + 1) - 11/8*log(x^2)

Mupad [B] (verification not implemented)

Time = 8.52 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^3 \left (2+3 x^2+x^4\right )^2} \, dx=5\,\ln \left (x^2+1\right )-\frac {29\,\ln \left (x^2+2\right )}{8}-\frac {11\,\ln \left (x\right )}{4}+\frac {\frac {9\,x^4}{4}+\frac {3\,x^2}{4}-1}{x^6+3\,x^4+2\,x^2} \]

[In]

int((x^2 + 3*x^4 + 5*x^6 + 4)/(x^3*(3*x^2 + x^4 + 2)^2),x)

[Out]

5*log(x^2 + 1) - (29*log(x^2 + 2))/8 - (11*log(x))/4 + ((3*x^2)/4 + (9*x^4)/4 - 1)/(2*x^2 + 3*x^4 + x^6)