\(\int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+(3 e^8 x+4 x^2) \log (\frac {3 e^8+4 x}{x})} \, dx\) [9608]

Optimal result

Integrand size = 59, antiderivative size = 16 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (12 \left (x+\log \left (4+\frac {3 e^8}{x}\right )\right )\right ) \]

[Out]

ln(12*x+12*ln(3/x*exp(4)^2+4))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6873, 6816} \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x+\log \left (\frac {3 e^8}{x}+4\right )\right ) \]

[In]

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

[Out]

Log[x + Log[4 + (3*E^8)/x]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6873

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

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x+\log \left (4+\frac {3 e^8}{x}\right )\right ) \]

[In]

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

[Out]

Log[x + Log[4 + (3*E^8)/x]]

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.06

method	result	size
risch	\(\ln \left (\ln \left (\frac {3 \,{\mathrm e}^{8}+4 x}{x}\right )+x \right )\)	\(17\)
norman	\(\ln \left (\ln \left (\frac {3 \,{\mathrm e}^{8}+4 x}{x}\right )+x \right )\)	\(19\)
parallelrisch	\(\ln \left (\ln \left (\frac {3 \,{\mathrm e}^{8}+4 x}{x}\right )+x \right )\)	\(19\)
derivativedivides	\(-\ln \left (\frac {3 \,{\mathrm e}^{8}}{x}\right )+\ln \left (3 \,{\mathrm e}^{8}+\ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right ) \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )-4 \ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )\right )\)	\(60\)
default	\(-\ln \left (\frac {3 \,{\mathrm e}^{8}}{x}\right )+\ln \left (3 \,{\mathrm e}^{8}+\ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right ) \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )-4 \ln \left (4+\frac {3 \,{\mathrm e}^{8}}{x}\right )\right )\)	\(60\)

[In]

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

[Out]

ln(ln((3*exp(8)+4*x)/x)+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x + \log \left (\frac {4 \, x + 3 \, e^{8}}{x}\right )\right ) \]

[In]

integrate(((-3+3*x)*exp(4)^2+4*x^2)/((3*x*exp(4)^2+4*x^2)*log((3*exp(4)^2+4*x)/x)+3*x^2*exp(4)^2+4*x^3),x, alg
orithm="fricas")

[Out]

log(x + log((4*x + 3*e^8)/x))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log {\left (x + \log {\left (\frac {4 x + 3 e^{8}}{x} \right )} \right )} \]

[In]

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

[Out]

log(x + log((4*x + 3*exp(8))/x))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\log \left (x + \log \left (4 \, x + 3 \, e^{8}\right ) - \log \left (x\right )\right ) \]

[In]

integrate(((-3+3*x)*exp(4)^2+4*x^2)/((3*x*exp(4)^2+4*x^2)*log((3*exp(4)^2+4*x)/x)+3*x^2*exp(4)^2+4*x^3),x, alg
orithm="maxima")

[Out]

log(x + log(4*x + 3*e^8) - log(x))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (17) = 34\).

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 4.50 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx={\left (e^{8} \log \left (\frac {{\left (4 \, x + 3 \, e^{8}\right )} \log \left (\frac {4 \, x + 3 \, e^{8}}{x}\right )}{x} + 3 \, e^{8} - 4 \, \log \left (\frac {4 \, x + 3 \, e^{8}}{x}\right )\right ) - e^{8} \log \left (\frac {4 \, x + 3 \, e^{8}}{x} - 4\right )\right )} e^{\left (-8\right )} \]

[In]

integrate(((-3+3*x)*exp(4)^2+4*x^2)/((3*x*exp(4)^2+4*x^2)*log((3*exp(4)^2+4*x)/x)+3*x^2*exp(4)^2+4*x^3),x, alg
orithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 16.44 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81 \[ \int \frac {4 x^2+e^8 (-3+3 x)}{3 e^8 x^2+4 x^3+\left (3 e^8 x+4 x^2\right ) \log \left (\frac {3 e^8+4 x}{x}\right )} \, dx=\ln \left (x+\ln \left (\frac {3\,{\mathrm {e}}^8}{x}+4\right )\right ) \]

[In]

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

[Out]

log(x + log((3*exp(8))/x + 4))