\(\int \frac {4+x^2+3 x^4+5 x^6}{x^5 (2+3 x^2+x^4)^2} \, dx\) [81]

Optimal result

Integrand size = 31, antiderivative size = 64 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=-\frac {1}{4 x^4}+\frac {11}{8 x^2}-\frac {5+9 x^2}{8 \left (2+3 x^2+x^4\right )}+\frac {23 \log (x)}{4}-\frac {11}{2} \log \left (1+x^2\right )+\frac {21}{8} \log \left (2+x^2\right ) \]

[Out]

-1/4/x^4+11/8/x^2+1/8*(-9*x^2-5)/(x^4+3*x^2+2)+23/4*ln(x)-11/2*ln(x^2+1)+21/8*ln(x^2+2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 64, 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} \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=-\frac {1}{4 x^4}+\frac {11}{8 x^2}-\frac {11}{2} \log \left (x^2+1\right )+\frac {21}{8} \log \left (x^2+2\right )-\frac {9 x^2+5}{8 \left (x^4+3 x^2+2\right )}+\frac {23 \log (x)}{4} \]

[In]

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

[Out]

-1/4*1/x^4 + 11/(8*x^2) - (5 + 9*x^2)/(8*(2 + 3*x^2 + x^4)) + (23*Log[x])/4 - (11*Log[1 + x^2])/2 + (21*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^3 \left (2+3 x+x^2\right )^2} \, dx,x,x^2\right ) \\ & = -\frac {5+9 x^2}{8 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \text {Subst}\left (\int \frac {-2+\frac {5 x}{2}-\frac {17 x^2}{4}+\frac {9 x^3}{4}}{x^3 \left (2+3 x+x^2\right )} \, dx,x,x^2\right ) \\ & = -\frac {5+9 x^2}{8 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \text {Subst}\left (\int \left (-\frac {1}{x^3}+\frac {11}{4 x^2}-\frac {23}{4 x}+\frac {11}{1+x}-\frac {21}{4 (2+x)}\right ) \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^4}+\frac {11}{8 x^2}-\frac {5+9 x^2}{8 \left (2+3 x^2+x^4\right )}+\frac {23 \log (x)}{4}-\frac {11}{2} \log \left (1+x^2\right )+\frac {21}{8} \log \left (2+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {1}{8} \left (-\frac {2}{x^4}+\frac {11}{x^2}-\frac {5+9 x^2}{2+3 x^2+x^4}+46 \log (x)-44 \log \left (1+x^2\right )+21 \log \left (2+x^2\right )\right ) \]

[In]

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

[Out]

(-2/x^4 + 11/x^2 - (5 + 9*x^2)/(2 + 3*x^2 + x^4) + 46*Log[x] - 44*Log[1 + x^2] + 21*Log[2 + x^2])/8

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {21 \ln \left (x^{2}+2\right )}{8}-\frac {13}{8 \left (x^{2}+2\right )}-\frac {1}{4 x^{4}}+\frac {11}{8 x^{2}}+\frac {23 \ln \left (x \right )}{4}-\frac {11 \ln \left (x^{2}+1\right )}{2}+\frac {1}{2 x^{2}+2}\)	\(50\)
norman	\(\frac {-\frac {1}{2}+\frac {1}{4} x^{6}+\frac {13}{4} x^{4}+2 x^{2}}{x^{4} \left (x^{4}+3 x^{2}+2\right )}+\frac {23 \ln \left (x \right )}{4}-\frac {11 \ln \left (x^{2}+1\right )}{2}+\frac {21 \ln \left (x^{2}+2\right )}{8}\)	\(55\)
risch	\(\frac {-\frac {1}{2}+\frac {1}{4} x^{6}+\frac {13}{4} x^{4}+2 x^{2}}{x^{4} \left (x^{4}+3 x^{2}+2\right )}+\frac {23 \ln \left (x \right )}{4}-\frac {11 \ln \left (x^{2}+1\right )}{2}+\frac {21 \ln \left (x^{2}+2\right )}{8}\)	\(55\)
parallelrisch	\(\frac {46 \ln \left (x \right ) x^{8}-44 \ln \left (x^{2}+1\right ) x^{8}+21 \ln \left (x^{2}+2\right ) x^{8}-4+138 \ln \left (x \right ) x^{6}-132 \ln \left (x^{2}+1\right ) x^{6}+63 \ln \left (x^{2}+2\right ) x^{6}+2 x^{6}+92 \ln \left (x \right ) x^{4}-88 \ln \left (x^{2}+1\right ) x^{4}+42 \ln \left (x^{2}+2\right ) x^{4}+26 x^{4}+16 x^{2}}{8 x^{4} \left (x^{4}+3 x^{2}+2\right )}\)	\(122\)

[In]

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

[Out]

21/8*ln(x^2+2)-13/8/(x^2+2)-1/4/x^4+11/8/x^2+23/4*ln(x)-11/2*ln(x^2+1)+1/2/(x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.52 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {2 \, x^{6} + 26 \, x^{4} + 16 \, x^{2} + 21 \, {\left (x^{8} + 3 \, x^{6} + 2 \, x^{4}\right )} \log \left (x^{2} + 2\right ) - 44 \, {\left (x^{8} + 3 \, x^{6} + 2 \, x^{4}\right )} \log \left (x^{2} + 1\right ) + 46 \, {\left (x^{8} + 3 \, x^{6} + 2 \, x^{4}\right )} \log \left (x\right ) - 4}{8 \, {\left (x^{8} + 3 \, x^{6} + 2 \, x^{4}\right )}} \]

[In]

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

[Out]

1/8*(2*x^6 + 26*x^4 + 16*x^2 + 21*(x^8 + 3*x^6 + 2*x^4)*log(x^2 + 2) - 44*(x^8 + 3*x^6 + 2*x^4)*log(x^2 + 1) +
46*(x^8 + 3*x^6 + 2*x^4)*log(x) - 4)/(x^8 + 3*x^6 + 2*x^4)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {23 \log {\left (x \right )}}{4} - \frac {11 \log {\left (x^{2} + 1 \right )}}{2} + \frac {21 \log {\left (x^{2} + 2 \right )}}{8} + \frac {x^{6} + 13 x^{4} + 8 x^{2} - 2}{4 x^{8} + 12 x^{6} + 8 x^{4}} \]

[In]

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

[Out]

23*log(x)/4 - 11*log(x**2 + 1)/2 + 21*log(x**2 + 2)/8 + (x**6 + 13*x**4 + 8*x**2 - 2)/(4*x**8 + 12*x**6 + 8*x*
*4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {x^{6} + 13 \, x^{4} + 8 \, x^{2} - 2}{4 \, {\left (x^{8} + 3 \, x^{6} + 2 \, x^{4}\right )}} + \frac {21}{8} \, \log \left (x^{2} + 2\right ) - \frac {11}{2} \, \log \left (x^{2} + 1\right ) + \frac {23}{8} \, \log \left (x^{2}\right ) \]

[In]

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

[Out]

1/4*(x^6 + 13*x^4 + 8*x^2 - 2)/(x^8 + 3*x^6 + 2*x^4) + 21/8*log(x^2 + 2) - 11/2*log(x^2 + 1) + 23/8*log(x^2)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.03 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {23 \, x^{4} + 51 \, x^{2} + 36}{16 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac {69 \, x^{4} - 22 \, x^{2} + 4}{16 \, x^{4}} + \frac {21}{8} \, \log \left (x^{2} + 2\right ) - \frac {11}{2} \, \log \left (x^{2} + 1\right ) + \frac {23}{8} \, \log \left (x^{2}\right ) \]

[In]

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

[Out]

1/16*(23*x^4 + 51*x^2 + 36)/(x^4 + 3*x^2 + 2) - 1/16*(69*x^4 - 22*x^2 + 4)/x^4 + 21/8*log(x^2 + 2) - 11/2*log(
x^2 + 1) + 23/8*log(x^2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.86 \[ \int \frac {4+x^2+3 x^4+5 x^6}{x^5 \left (2+3 x^2+x^4\right )^2} \, dx=\frac {21\,\ln \left (x^2+2\right )}{8}-\frac {11\,\ln \left (x^2+1\right )}{2}+\frac {23\,\ln \left (x\right )}{4}+\frac {\frac {x^6}{4}+\frac {13\,x^4}{4}+2\,x^2-\frac {1}{2}}{x^8+3\,x^6+2\,x^4} \]

[In]

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

[Out]

(21*log(x^2 + 2))/8 - (11*log(x^2 + 1))/2 + (23*log(x))/4 + (2*x^2 + (13*x^4)/4 + x^6/4 - 1/2)/(2*x^4 + 3*x^6
+ x^8)