3.38.75 \(\int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} (-18+9 x^2)+(32+32 x^2+8 x^4) \log (x)}{4+4 x^2+x^4} \, dx\)

Optimal. Leaf size=31 \[ 2-e^{\frac {9}{\frac {2}{x}+x}}+4 x \left (-1-x^2+2 \log (x)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 28, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {28, 6688, 6706, 2295} \begin {gather*} -4 x^3-e^{\frac {9 x}{x^2+2}}-4 x+8 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16 - 32*x^2 - 44*x^4 - 12*x^6 + E^((9*x)/(2 + x^2))*(-18 + 9*x^2) + (32 + 32*x^2 + 8*x^4)*Log[x])/(4 + 4*
x^2 + x^4),x]

[Out]

-E^((9*x)/(2 + x^2)) - 4*x - 4*x^3 + 8*x*Log[x]

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16-32 x^2-44 x^4-12 x^6+e^{\frac {9 x}{2+x^2}} \left (-18+9 x^2\right )+\left (32+32 x^2+8 x^4\right ) \log (x)}{\left (2+x^2\right )^2} \, dx\\ &=\int \left (4-12 x^2+\frac {9 e^{\frac {9 x}{2+x^2}} \left (-2+x^2\right )}{\left (2+x^2\right )^2}+8 \log (x)\right ) \, dx\\ &=4 x-4 x^3+8 \int \log (x) \, dx+9 \int \frac {e^{\frac {9 x}{2+x^2}} \left (-2+x^2\right )}{\left (2+x^2\right )^2} \, dx\\ &=-e^{\frac {9 x}{2+x^2}}-4 x-4 x^3+8 x \log (x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 28, normalized size = 0.90 \begin {gather*} -e^{\frac {9 x}{2+x^2}}-4 x-4 x^3+8 x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 32*x^2 - 44*x^4 - 12*x^6 + E^((9*x)/(2 + x^2))*(-18 + 9*x^2) + (32 + 32*x^2 + 8*x^4)*Log[x])/(
4 + 4*x^2 + x^4),x]

[Out]

-E^((9*x)/(2 + x^2)) - 4*x - 4*x^3 + 8*x*Log[x]

________________________________________________________________________________________

fricas [A]  time = 0.85, size = 27, normalized size = 0.87 \begin {gather*} -4 \, x^{3} + 8 \, x \log \relax (x) - 4 \, x - e^{\left (\frac {9 \, x}{x^{2} + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4+32*x^2+32)*log(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44*x^4-32*x^2+16)/(x^4+4*x^2+4),x, algo
rithm="fricas")

[Out]

-4*x^3 + 8*x*log(x) - 4*x - e^(9*x/(x^2 + 2))

________________________________________________________________________________________

giac [A]  time = 0.21, size = 27, normalized size = 0.87 \begin {gather*} -4 \, x^{3} + 8 \, x \log \relax (x) - 4 \, x - e^{\left (\frac {9 \, x}{x^{2} + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4+32*x^2+32)*log(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44*x^4-32*x^2+16)/(x^4+4*x^2+4),x, algo
rithm="giac")

[Out]

-4*x^3 + 8*x*log(x) - 4*x - e^(9*x/(x^2 + 2))

________________________________________________________________________________________

maple [A]  time = 0.09, size = 28, normalized size = 0.90

method	result	size
risch	\(-4 x^{3}+8 x \ln \relax (x )-4 x -{\mathrm e}^{\frac {9 x}{x^{2}+2}}\)	\(28\)
default	\(-4 x^{3}-4 x +\frac {-x^{2} {\mathrm e}^{\frac {9 x}{x^{2}+2}}-2 \,{\mathrm e}^{\frac {9 x}{x^{2}+2}}}{x^{2}+2}+8 x \ln \relax (x )\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^4+32*x^2+32)*ln(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44*x^4-32*x^2+16)/(x^4+4*x^2+4),x,method=_RETU
RNVERBOSE)

[Out]

-4*x^3+8*x*ln(x)-4*x-exp(9*x/(x^2+2))

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 27, normalized size = 0.87 \begin {gather*} -4 \, x^{3} + 8 \, x \log \relax (x) - 4 \, x - e^{\left (\frac {9 \, x}{x^{2} + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^4+32*x^2+32)*log(x)+(9*x^2-18)*exp(9*x/(x^2+2))-12*x^6-44*x^4-32*x^2+16)/(x^4+4*x^2+4),x, algo
rithm="maxima")

[Out]

-4*x^3 + 8*x*log(x) - 4*x - e^(9*x/(x^2 + 2))

________________________________________________________________________________________

mupad [B]  time = 2.54, size = 29, normalized size = 0.94 \begin {gather*} 4\,x-{\mathrm {e}}^{\frac {9\,x}{x^2+2}}+8\,x\,\left (\ln \relax (x)-1\right )-4\,x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(32*x^2 + 8*x^4 + 32) + exp((9*x)/(x^2 + 2))*(9*x^2 - 18) - 32*x^2 - 44*x^4 - 12*x^6 + 16)/(4*x^2
+ x^4 + 4),x)

[Out]

4*x - exp((9*x)/(x^2 + 2)) + 8*x*(log(x) - 1) - 4*x^3

________________________________________________________________________________________

sympy [A]  time = 0.38, size = 24, normalized size = 0.77 \begin {gather*} - 4 x^{3} + 8 x \log {\relax (x )} - 4 x - e^{\frac {9 x}{x^{2} + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**4+32*x**2+32)*ln(x)+(9*x**2-18)*exp(9*x/(x**2+2))-12*x**6-44*x**4-32*x**2+16)/(x**4+4*x**2+4)
,x)

[Out]

-4*x**3 + 8*x*log(x) - 4*x - exp(9*x/(x**2 + 2))

________________________________________________________________________________________