\(\int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx\) [3219]

Optimal result

Integrand size = 57, antiderivative size = 23 \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\log \left (-2+x+6 \left (4+\frac {\left (x+e^{-2+x} x\right )^2}{x}\right )\right ) \]

[Out]

ln(22+6*(x+x/exp(2)*exp(x))^2/x+x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6816} \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\log \left (6 e^{2 x} x+12 e^{x+2} x+e^4 (7 x+22)\right ) \]

[In]

Int[(7*E^4 + E^(2*x)*(6 + 12*x) + E^(2 + x)*(12 + 12*x))/(6*E^(2*x)*x + 12*E^(2 + x)*x + E^4*(22 + 7*x)),x]

[Out]

Log[6*E^(2*x)*x + 12*E^(2 + x)*x + E^4*(22 + 7*x)]

Rule 6816

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

Rubi steps \begin{align*} \text {integral}& = \log \left (6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\log \left (6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)\right ) \]

[In]

Integrate[(7*E^4 + E^(2*x)*(6 + 12*x) + E^(2 + x)*(12 + 12*x))/(6*E^(2*x)*x + 12*E^(2 + x)*x + E^4*(22 + 7*x))
,x]

[Out]

Log[6*E^(2*x)*x + 12*E^(2 + x)*x + E^4*(22 + 7*x)]

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22

method	result	size
risch	\(\ln \left (x \right )+\ln \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{2+x}+\frac {\left (7 x +22\right ) {\mathrm e}^{4}}{6 x}\right )\)	\(28\)
parallelrisch	\(\ln \left (\frac {7 x \,{\mathrm e}^{4}}{6}+2 x \,{\mathrm e}^{2} {\mathrm e}^{x}+x \,{\mathrm e}^{2 x}+\frac {11 \,{\mathrm e}^{4}}{3}\right )\)	\(29\)
norman	\(\ln \left (7 x \,{\mathrm e}^{4}+12 x \,{\mathrm e}^{2} {\mathrm e}^{x}+6 x \,{\mathrm e}^{2 x}+22 \,{\mathrm e}^{4}\right )\)	\(30\)

[In]

int(((12*x+6)*exp(x)^2+(12*x+12)*exp(2)*exp(x)+7*exp(2)^2)/(6*x*exp(x)^2+12*x*exp(2)*exp(x)+(7*x+22)*exp(2)^2)
,x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(exp(2*x)+2*exp(2+x)+1/6*(7*x+22)*exp(4)/x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\log \left (x\right ) + \log \left (\frac {{\left (7 \, x + 22\right )} e^{8} + 6 \, x e^{\left (2 \, x + 4\right )} + 12 \, x e^{\left (x + 6\right )}}{x}\right ) \]

[In]

integrate(((12*x+6)*exp(x)^2+(12*x+12)*exp(2)*exp(x)+7*exp(2)^2)/(6*x*exp(x)^2+12*x*exp(2)*exp(x)+(7*x+22)*exp
(2)^2),x, algorithm="fricas")

[Out]

log(x) + log(((7*x + 22)*e^8 + 6*x*e^(2*x + 4) + 12*x*e^(x + 6))/x)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\log {\left (x \right )} + \log {\left (e^{2 x} + 2 e^{2} e^{x} + \frac {7 x e^{4} + 22 e^{4}}{6 x} \right )} \]

[In]

integrate(((12*x+6)*exp(x)**2+(12*x+12)*exp(2)*exp(x)+7*exp(2)**2)/(6*x*exp(x)**2+12*x*exp(2)*exp(x)+(7*x+22)*
exp(2)**2),x)

[Out]

log(x) + log(exp(2*x) + 2*exp(2)*exp(x) + (7*x*exp(4) + 22*exp(4))/(6*x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\log \left (x\right ) + \log \left (\frac {7 \, x e^{4} + 6 \, x e^{\left (2 \, x\right )} + 12 \, x e^{\left (x + 2\right )} + 22 \, e^{4}}{6 \, x}\right ) \]

[In]

integrate(((12*x+6)*exp(x)^2+(12*x+12)*exp(2)*exp(x)+7*exp(2)^2)/(6*x*exp(x)^2+12*x*exp(2)*exp(x)+(7*x+22)*exp
(2)^2),x, algorithm="maxima")

[Out]

log(x) + log(1/6*(7*x*e^4 + 6*x*e^(2*x) + 12*x*e^(x + 2) + 22*e^4)/x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\log \left (7 \, {\left (x + 2\right )} e^{8} + 6 \, {\left (x + 2\right )} e^{\left (2 \, x + 4\right )} + 12 \, {\left (x + 2\right )} e^{\left (x + 6\right )} + 8 \, e^{8} - 12 \, e^{\left (2 \, x + 4\right )} - 24 \, e^{\left (x + 6\right )}\right ) \]

[In]

integrate(((12*x+6)*exp(x)^2+(12*x+12)*exp(2)*exp(x)+7*exp(2)^2)/(6*x*exp(x)^2+12*x*exp(2)*exp(x)+(7*x+22)*exp
(2)^2),x, algorithm="giac")

[Out]

log(7*(x + 2)*e^8 + 6*(x + 2)*e^(2*x + 4) + 12*(x + 2)*e^(x + 6) + 8*e^8 - 12*e^(2*x + 4) - 24*e^(x + 6))

Mupad [B] (verification not implemented)

Time = 9.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {7 e^4+e^{2 x} (6+12 x)+e^{2+x} (12+12 x)}{6 e^{2 x} x+12 e^{2+x} x+e^4 (22+7 x)} \, dx=\ln \left (22\,{\mathrm {e}}^8+12\,x\,{\mathrm {e}}^{x+6}+7\,x\,{\mathrm {e}}^8+6\,x\,{\mathrm {e}}^{2\,x+4}\right ) \]

[In]

int((7*exp(4) + exp(2*x)*(12*x + 6) + exp(2)*exp(x)*(12*x + 12))/(6*x*exp(2*x) + exp(4)*(7*x + 22) + 12*x*exp(
2)*exp(x)),x)

[Out]

log(22*exp(8) + 12*x*exp(x + 6) + 7*x*exp(8) + 6*x*exp(2*x + 4))