\(\int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} (4 x^2+(4 e^8+8 x^2) \log (x)+4 e^8 \log (3 x))}{x} \, dx\) [6596]

Optimal result

Integrand size = 57, antiderivative size = 21 \[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=-4+x+x^{4 x \left (x+\frac {e^8 \log (3 x)}{x}\right )} \]

[Out]

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

Rubi [F]

\[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=\int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx \]

[In]

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

[Out]

x + 4*E^8*Log[3]*Defer[Int][x^(-1 + 4*x^2 + 4*E^8*Log[3*x]), x] + 4*Defer[Int][x^(1 + 4*x^2 + 4*E^8*Log[3*x]),
x] + 8*E^8*Defer[Int][x^(-1 + 4*x^2 + 4*E^8*Log[3*x])*Log[x], x] + 8*Defer[Int][x^(1 + 4*x^2 + 4*E^8*Log[3*x]
)*Log[x], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (1+4 x^{-1+4 x^2+4 e^8 \log (3 x)} \left (x^2+e^8 \log (3)+2 e^8 \log (x)+2 x^2 \log (x)\right )\right ) \, dx \\ & = x+4 \int x^{-1+4 x^2+4 e^8 \log (3 x)} \left (x^2+e^8 \log (3)+2 e^8 \log (x)+2 x^2 \log (x)\right ) \, dx \\ & = x+4 \int \left (x^{1+4 x^2+4 e^8 \log (3 x)}+e^8 x^{-1+4 x^2+4 e^8 \log (3 x)} \log (3)+2 e^8 x^{-1+4 x^2+4 e^8 \log (3 x)} \log (x)+2 x^{1+4 x^2+4 e^8 \log (3 x)} \log (x)\right ) \, dx \\ & = x+4 \int x^{1+4 x^2+4 e^8 \log (3 x)} \, dx+8 \int x^{1+4 x^2+4 e^8 \log (3 x)} \log (x) \, dx+\left (8 e^8\right ) \int x^{-1+4 x^2+4 e^8 \log (3 x)} \log (x) \, dx+\left (4 e^8 \log (3)\right ) \int x^{-1+4 x^2+4 e^8 \log (3 x)} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=\int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(x +{\mathrm e}^{4 \ln \left (x \right ) \left ({\mathrm e}^{8} \ln \left (3 x \right )+x^{2}\right )}\)	\(21\)
risch	\(x +x^{4 \,{\mathrm e}^{8} \left (\ln \left (x \right )+\ln \left (3\right )\right )} x^{4 x^{2}}\)	\(22\)

[In]

int(((4*exp(4)^2*ln(3*x)+(4*exp(4)^2+8*x^2)*ln(x)+4*x^2)*exp(4*exp(4)^2*ln(x)*ln(3*x)+4*x^2*ln(x))+x)/x,x,meth
od=_RETURNVERBOSE)

[Out]

x+exp(4*ln(x)*(exp(4)^2*ln(3*x)+x^2))

Fricas [A] (verification not implemented)

none

Time = 0.66 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=x + e^{\left (4 \, e^{8} \log \left (x\right )^{2} + 4 \, {\left (x^{2} + e^{8} \log \left (3\right )\right )} \log \left (x\right )\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=x + e^{4 x^{2} \log {\left (x \right )} + 4 \left (\log {\left (x \right )} + \log {\left (3 \right )}\right ) e^{8} \log {\left (x \right )}} \]

[In]

integrate(((4*exp(4)**2*ln(3*x)+(4*exp(4)**2+8*x**2)*ln(x)+4*x**2)*exp(4*exp(4)**2*ln(x)*ln(3*x)+4*x**2*ln(x))
+x)/x,x)

[Out]

x + exp(4*x**2*log(x) + 4*(log(x) + log(3))*exp(8)*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=x + e^{\left (4 \, x^{2} \log \left (x\right ) + 4 \, e^{8} \log \left (3\right ) \log \left (x\right ) + 4 \, e^{8} \log \left (x\right )^{2}\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=x + e^{\left (4 \, x^{2} \log \left (x\right ) + 4 \, e^{8} \log \left (3 \, x\right ) \log \left (x\right )\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.46 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {x+e^{4 x^2 \log (x)+4 e^8 \log (x) \log (3 x)} \left (4 x^2+\left (4 e^8+8 x^2\right ) \log (x)+4 e^8 \log (3 x)\right )}{x} \, dx=x+x^{4\,{\mathrm {e}}^8\,\ln \left (3\right )}\,x^{4\,x^2}\,{\mathrm {e}}^{4\,{\mathrm {e}}^8\,{\ln \left (x\right )}^2} \]

[In]

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

[Out]

x + x^(4*exp(8)*log(3))*x^(4*x^2)*exp(4*exp(8)*log(x)^2)