\(\int \frac {3 x^2+24 x^4+e^2 (2-x^2-8 x^4)+(6 x^2-2 e^2 x^2) \log (x)}{2 e^2 x} \, dx\) [6195]

Optimal result

Integrand size = 54, antiderivative size = 29 \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=\frac {3 x \left (x-\frac {e^2 x}{3}\right ) \left (x^2+\frac {\log (x)}{2}\right )}{e^2}+\log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 14, 2341} \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=\frac {\left (3-e^2\right ) x^4}{e^2}+\frac {\left (3-e^2\right ) x^2 \log (x)}{2 e^2}+\log (x) \]

[In]

Int[(3*x^2 + 24*x^4 + E^2*(2 - x^2 - 8*x^4) + (6*x^2 - 2*E^2*x^2)*Log[x])/(2*E^2*x),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2341

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{x} \, dx}{2 e^2} \\ & = \frac {\int \left (\frac {2 e^2+\left (3-e^2\right ) x^2+8 \left (3-e^2\right ) x^4}{x}-2 \left (-3+e^2\right ) x \log (x)\right ) \, dx}{2 e^2} \\ & = \frac {\int \frac {2 e^2+\left (3-e^2\right ) x^2+8 \left (3-e^2\right ) x^4}{x} \, dx}{2 e^2}+\frac {\left (3-e^2\right ) \int x \log (x) \, dx}{e^2} \\ & = -\frac {\left (3-e^2\right ) x^2}{4 e^2}+\frac {\left (3-e^2\right ) x^2 \log (x)}{2 e^2}+\frac {\int \left (\frac {2 e^2}{x}+\left (3-e^2\right ) x+8 \left (3-e^2\right ) x^3\right ) \, dx}{2 e^2} \\ & = \frac {\left (3-e^2\right ) x^4}{e^2}+\log (x)+\frac {\left (3-e^2\right ) x^2 \log (x)}{2 e^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=-\frac {\left (-3+e^2\right ) x^4}{e^2}+\log (x)+\frac {\left (3-e^2\right ) x^2 \log (x)}{2 e^2} \]

[In]

Integrate[(3*x^2 + 24*x^4 + E^2*(2 - x^2 - 8*x^4) + (6*x^2 - 2*E^2*x^2)*Log[x])/(2*E^2*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00

method	result	size
risch	\(-\frac {\left ({\mathrm e}^{2}-3\right ) {\mathrm e}^{-2} x^{2} \ln \left (x \right )}{2}-x^{4}+3 \,{\mathrm e}^{-2} x^{4}+\ln \left (x \right )\)	\(29\)
norman	\(\ln \left (x \right )-\left ({\mathrm e}^{2}-3\right ) {\mathrm e}^{-2} x^{4}-\frac {\left ({\mathrm e}^{2}-3\right ) {\mathrm e}^{-2} x^{2} \ln \left (x \right )}{2}\)	\(32\)
parallelrisch	\(\frac {{\mathrm e}^{-2} \left (-2 x^{4} {\mathrm e}^{2}-x^{2} {\mathrm e}^{2} \ln \left (x \right )+6 x^{4}+3 x^{2} \ln \left (x \right )+2 \,{\mathrm e}^{2} \ln \left (x \right )\right )}{2}\)	\(42\)
parts	\(-x^{4}+3 \,{\mathrm e}^{-2} x^{4}-\frac {x^{2}}{4}+\frac {3 \,{\mathrm e}^{-2} x^{2}}{4}+\ln \left (x \right )-\left ({\mathrm e}^{2}-3\right ) {\mathrm e}^{-2} \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )\)	\(55\)
default	\(\frac {{\mathrm e}^{-2} \left (-2 x^{4} {\mathrm e}^{2}-2 \,{\mathrm e}^{2} \left (\frac {x^{2} \ln \left (x \right )}{2}-\frac {x^{2}}{4}\right )+6 x^{4}-\frac {x^{2} {\mathrm e}^{2}}{2}+3 x^{2} \ln \left (x \right )+2 \,{\mathrm e}^{2} \ln \left (x \right )\right )}{2}\)	\(57\)

[In]

int(1/2*((-2*x^2*exp(2)+6*x^2)*ln(x)+(-8*x^4-x^2+2)*exp(2)+24*x^4+3*x^2)/exp(2)/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=-\frac {1}{2} \, {\left (2 \, x^{4} e^{2} - 6 \, x^{4} - {\left (3 \, x^{2} - {\left (x^{2} - 2\right )} e^{2}\right )} \log \left (x\right )\right )} e^{\left (-2\right )} \]

[In]

integrate(1/2*((-2*x^2*exp(2)+6*x^2)*log(x)+(-8*x^4-x^2+2)*exp(2)+24*x^4+3*x^2)/exp(2)/x,x, algorithm="fricas"
)

[Out]

-1/2*(2*x^4*e^2 - 6*x^4 - (3*x^2 - (x^2 - 2)*e^2)*log(x))*e^(-2)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=\frac {\left (- x^{2} e^{2} + 3 x^{2}\right ) \log {\left (x \right )}}{2 e^{2}} + \frac {x^{4} \cdot \left (3 - e^{2}\right ) + e^{2} \log {\left (x \right )}}{e^{2}} \]

[In]

integrate(1/2*((-2*x**2*exp(2)+6*x**2)*ln(x)+(-8*x**4-x**2+2)*exp(2)+24*x**4+3*x**2)/exp(2)/x,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).

Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79 \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=-\frac {1}{4} \, {\left (4 \, x^{4} e^{2} - 12 \, x^{4} + x^{2} e^{2} - 6 \, x^{2} \log \left (x\right ) + {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} e^{2} - 4 \, e^{2} \log \left (x\right )\right )} e^{\left (-2\right )} \]

[In]

integrate(1/2*((-2*x^2*exp(2)+6*x^2)*log(x)+(-8*x^4-x^2+2)*exp(2)+24*x^4+3*x^2)/exp(2)/x,x, algorithm="maxima"
)

[Out]

-1/4*(4*x^4*e^2 - 12*x^4 + x^2*e^2 - 6*x^2*log(x) + (2*x^2*log(x) - x^2)*e^2 - 4*e^2*log(x))*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=-\frac {1}{2} \, {\left (2 \, x^{4} e^{2} - 6 \, x^{4} + x^{2} e^{2} \log \left (x\right ) - 3 \, x^{2} \log \left (x\right ) - 2 \, e^{2} \log \left (x\right )\right )} e^{\left (-2\right )} \]

[In]

integrate(1/2*((-2*x^2*exp(2)+6*x^2)*log(x)+(-8*x^4-x^2+2)*exp(2)+24*x^4+3*x^2)/exp(2)/x,x, algorithm="giac")

[Out]

-1/2*(2*x^4*e^2 - 6*x^4 + x^2*e^2*log(x) - 3*x^2*log(x) - 2*e^2*log(x))*e^(-2)

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {3 x^2+24 x^4+e^2 \left (2-x^2-8 x^4\right )+\left (6 x^2-2 e^2 x^2\right ) \log (x)}{2 e^2 x} \, dx=\ln \left (x\right )+x^4\,\left (3\,{\mathrm {e}}^{-2}-1\right )+\frac {x^2\,\ln \left (x\right )\,\left (3\,{\mathrm {e}}^{-2}-1\right )}{2} \]

[In]

int(-(exp(-2)*((exp(2)*(x^2 + 8*x^4 - 2))/2 + (log(x)*(2*x^2*exp(2) - 6*x^2))/2 - (3*x^2)/2 - 12*x^4))/x,x)

[Out]

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