\(\int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} (4 e x \log (x)+e (-2-2 x) \log ^2(x)+(-4 e \log (x)+4 e \log ^2(x)) \log (2 x))}{x^3} \, dx\) [5362]

Optimal result

Integrand size = 71, antiderivative size = 25 \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=e^{1+\frac {2 \log ^2(x) (x-\log (2 x))}{x^2}}+x^2 \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(25)=50\).

Time = 0.91 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.88, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {14, 2326} \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=x^2+\frac {2^{-\frac {2 \log ^2(x)}{x^2}} e^{\frac {2 \log ^2(x)}{x}+1} \log (x) (2 x-x \log (x)) x^{-\frac {2 \log ^2(x)}{x^2}-3}}{\frac {2 \log (x)}{x^2}-\frac {\log ^2(x)}{x^2}} \]

[In]

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

[Out]

x^2 + (E^(1 + (2*Log[x]^2)/x)*x^(-3 - (2*Log[x]^2)/x^2)*Log[x]*(2*x - x*Log[x]))/(2^((2*Log[x]^2)/x^2)*((2*Log
[x])/x^2 - Log[x]^2/x^2))

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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (2 x-2^{1-\frac {2 \log ^2(x)}{x^2}} e^{1+\frac {2 \log ^2(x)}{x}} x^{-3-\frac {2 \log ^2(x)}{x^2}} \log (x) (-2 x+\log (x)+x \log (x)+2 \log (2 x)-2 \log (x) \log (2 x))\right ) \, dx \\ & = x^2-\int 2^{1-\frac {2 \log ^2(x)}{x^2}} e^{1+\frac {2 \log ^2(x)}{x}} x^{-3-\frac {2 \log ^2(x)}{x^2}} \log (x) (-2 x+\log (x)+x \log (x)+2 \log (2 x)-2 \log (x) \log (2 x)) \, dx \\ & = x^2+\frac {2^{-\frac {2 \log ^2(x)}{x^2}} e^{1+\frac {2 \log ^2(x)}{x}} x^{-3-\frac {2 \log ^2(x)}{x^2}} \log (x) (2 x-x \log (x))}{\frac {2 \log (x)}{x^2}-\frac {\log ^2(x)}{x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.14 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=e^{1+\frac {2 \log ^2(x) (x-\log (2 x))}{x^2}}+x^2 \]

[In]

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

[Out]

E^(1 + (2*Log[x]^2*(x - Log[2*x]))/x^2) + x^2

Maple [A] (verified)

Time = 4.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\({\mathrm e} \,{\mathrm e}^{-\frac {2 \ln \left (x \right )^{2} \left (\ln \left (2 x \right )-x \right )}{x^{2}}}+x^{2}\)	\(26\)
risch	\(\left (\frac {1}{4}\right )^{\frac {\ln \left (x \right )^{2}}{x^{2}}} {\mathrm e}^{-\frac {2 \ln \left (x \right )^{3}-2 x \ln \left (x \right )^{2}-x^{2}}{x^{2}}}+x^{2}\)	\(41\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=x^{2} + e^{\left (\frac {2 \, {\left ({\left (x - \log \left (2\right )\right )} \log \left (x\right )^{2} - \log \left (x\right )^{3}\right )}}{x^{2}} + 1\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=x^{2} + e e^{\frac {2 x \log {\left (x \right )}^{2} - 2 \left (\log {\left (x \right )} + \log {\left (2 \right )}\right ) \log {\left (x \right )}^{2}}{x^{2}}} \]

[In]

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

[Out]

x**2 + E*exp((2*x*log(x)**2 - 2*(log(x) + log(2))*log(x)**2)/x**2)

Maxima [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=x^{2} + e^{\left (\frac {2 \, \log \left (x\right )^{2}}{x} - \frac {2 \, \log \left (2\right ) \log \left (x\right )^{2}}{x^{2}} - \frac {2 \, \log \left (x\right )^{3}}{x^{2}} + 1\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=x^{2} + e^{\left (\frac {2 \, x \log \left (x\right )^{2} - 2 \, \log \left (2\right ) \log \left (x\right )^{2} - 2 \, \log \left (x\right )^{3} + x^{2}}{x^{2}}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.85 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60 \[ \int \frac {2 x^4+e^{\frac {2 x \log ^2(x)-2 \log ^2(x) \log (2 x)}{x^2}} \left (4 e x \log (x)+e (-2-2 x) \log ^2(x)+\left (-4 e \log (x)+4 e \log ^2(x)\right ) \log (2 x)\right )}{x^3} \, dx=x^2+\frac {\mathrm {e}\,{\mathrm {e}}^{\frac {2\,{\ln \left (x\right )}^2}{x}}\,{\mathrm {e}}^{-\frac {2\,{\ln \left (x\right )}^3}{x^2}}}{2^{\frac {2\,{\ln \left (x\right )}^2}{x^2}}} \]

[In]

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

[Out]

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