\(\int \frac {x (4+x^2+3 x^4+5 x^6)}{(2+3 x^2+x^4)^2} \, dx\) [78]

Optimal result

Integrand size = 29, antiderivative size = 42 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=\frac {24+25 x^2}{2 \left (2+3 x^2+x^4\right )}+4 \log \left (1+x^2\right )-\frac {3}{2} \log \left (2+x^2\right ) \]

[Out]

1/2*(25*x^2+24)/(x^4+3*x^2+2)+4*ln(x^2+1)-3/2*ln(x^2+2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1677, 1674, 646, 31} \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=4 \log \left (x^2+1\right )-\frac {3}{2} \log \left (x^2+2\right )+\frac {25 x^2+24}{2 \left (x^4+3 x^2+2\right )} \]

[In]

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

[Out]

(24 + 25*x^2)/(2*(2 + 3*x^2 + x^4)) + 4*Log[1 + x^2] - (3*Log[2 + x^2])/2

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 1674

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

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}{\left (2+3 x+x^2\right )^2} \, dx,x,x^2\right ) \\ & = \frac {24+25 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {1}{2} \text {Subst}\left (\int \frac {-13-5 x}{2+3 x+x^2} \, dx,x,x^2\right ) \\ & = \frac {24+25 x^2}{2 \left (2+3 x^2+x^4\right )}-\frac {3}{2} \text {Subst}\left (\int \frac {1}{2+x} \, dx,x,x^2\right )+4 \text {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right ) \\ & = \frac {24+25 x^2}{2 \left (2+3 x^2+x^4\right )}+4 \log \left (1+x^2\right )-\frac {3}{2} \log \left (2+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=\frac {24+25 x^2}{2 \left (2+3 x^2+x^4\right )}+4 \log \left (1+x^2\right )-\frac {3}{2} \log \left (2+x^2\right ) \]

[In]

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

[Out]

(24 + 25*x^2)/(2*(2 + 3*x^2 + x^4)) + 4*Log[1 + x^2] - (3*Log[2 + x^2])/2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86

method	result	size
default	\(-\frac {3 \ln \left (x^{2}+2\right )}{2}+\frac {13}{x^{2}+2}+4 \ln \left (x^{2}+1\right )-\frac {1}{2 \left (x^{2}+1\right )}\)	\(36\)
norman	\(\frac {\frac {25 x^{2}}{2}+12}{x^{4}+3 x^{2}+2}+4 \ln \left (x^{2}+1\right )-\frac {3 \ln \left (x^{2}+2\right )}{2}\)	\(38\)
risch	\(\frac {\frac {25 x^{2}}{2}+12}{x^{4}+3 x^{2}+2}+4 \ln \left (x^{2}+1\right )-\frac {3 \ln \left (x^{2}+2\right )}{2}\)	\(38\)
parallelrisch	\(\frac {8 \ln \left (x^{2}+1\right ) x^{4}-3 \ln \left (x^{2}+2\right ) x^{4}+24+24 \ln \left (x^{2}+1\right ) x^{2}-9 \ln \left (x^{2}+2\right ) x^{2}+25 x^{2}+16 \ln \left (x^{2}+1\right )-6 \ln \left (x^{2}+2\right )}{2 x^{4}+6 x^{2}+4}\)	\(82\)

[In]

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

[Out]

-3/2*ln(x^2+2)+13/(x^2+2)+4*ln(x^2+1)-1/2/(x^2+1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.36 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=\frac {25 \, x^{2} - 3 \, {\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 2\right ) + 8 \, {\left (x^{4} + 3 \, x^{2} + 2\right )} \log \left (x^{2} + 1\right ) + 24}{2 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} \]

[In]

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

[Out]

1/2*(25*x^2 - 3*(x^4 + 3*x^2 + 2)*log(x^2 + 2) + 8*(x^4 + 3*x^2 + 2)*log(x^2 + 1) + 24)/(x^4 + 3*x^2 + 2)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.86 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=\frac {25 x^{2} + 24}{2 x^{4} + 6 x^{2} + 4} + 4 \log {\left (x^{2} + 1 \right )} - \frac {3 \log {\left (x^{2} + 2 \right )}}{2} \]

[In]

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

[Out]

(25*x**2 + 24)/(2*x**4 + 6*x**2 + 4) + 4*log(x**2 + 1) - 3*log(x**2 + 2)/2

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.90 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=\frac {25 \, x^{2} + 24}{2 \, {\left (x^{4} + 3 \, x^{2} + 2\right )}} - \frac {3}{2} \, \log \left (x^{2} + 2\right ) + 4 \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

1/2*(25*x^2 + 24)/(x^4 + 3*x^2 + 2) - 3/2*log(x^2 + 2) + 4*log(x^2 + 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.95 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=\frac {25 \, x^{2} + 24}{2 \, {\left (x^{2} + 2\right )} {\left (x^{2} + 1\right )}} - \frac {3}{2} \, \log \left (x^{2} + 2\right ) + 4 \, \log \left (x^{2} + 1\right ) \]

[In]

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

[Out]

1/2*(25*x^2 + 24)/((x^2 + 2)*(x^2 + 1)) - 3/2*log(x^2 + 2) + 4*log(x^2 + 1)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \frac {x \left (4+x^2+3 x^4+5 x^6\right )}{\left (2+3 x^2+x^4\right )^2} \, dx=4\,\ln \left (x^2+1\right )-\frac {3\,\ln \left (x^2+2\right )}{2}+\frac {\frac {25\,x^2}{2}+12}{x^4+3\,x^2+2} \]

[In]

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

[Out]

4*log(x^2 + 1) - (3*log(x^2 + 2))/2 + ((25*x^2)/2 + 12)/(3*x^2 + x^4 + 2)