\(\int \frac {e^{33-3 e^5+3 e^x+96 x^2} (1+(3 e^x x+192 x^2) \log (x))}{x} \, dx\) [9190]

   Optimal result
   Rubi [B] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 40, antiderivative size = 22 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{3 \left (11-e^5+e^x+32 x^2\right )} \log (x) \]

[Out]

ln(x)*exp(3*exp(x)-3*exp(5)+96*x^2+33)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(22)=44\).

Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.18, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {2326} \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=\frac {e^{96 x^2+3 e^x+3 \left (11-e^5\right )} \left (64 x^2+e^x x\right ) \log (x)}{x \left (64 x+e^x\right )} \]

[In]

Int[(E^(33 - 3*E^5 + 3*E^x + 96*x^2)*(1 + (3*E^x*x + 192*x^2)*Log[x]))/x,x]

[Out]

(E^(3*E^x + 3*(11 - E^5) + 96*x^2)*(E^x*x + 64*x^2)*Log[x])/(x*(E^x + 64*x))

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}& = \frac {e^{3 e^x+3 \left (11-e^5\right )+96 x^2} \left (e^x x+64 x^2\right ) \log (x)}{x \left (e^x+64 x\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{33-3 e^5+3 e^x+96 x^2} \log (x) \]

[In]

Integrate[(E^(33 - 3*E^5 + 3*E^x + 96*x^2)*(1 + (3*E^x*x + 192*x^2)*Log[x]))/x,x]

[Out]

E^(33 - 3*E^5 + 3*E^x + 96*x^2)*Log[x]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method result size
risch \(\ln \left (x \right ) {\mathrm e}^{3 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{5}+96 x^{2}+33}\) \(20\)
parallelrisch \(\ln \left (x \right ) {\mathrm e}^{3 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{5}+96 x^{2}+33}\) \(20\)

[In]

int(((3*exp(x)*x+192*x^2)*ln(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)*exp(3*exp(x)-3*exp(5)+96*x^2+33)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \left (x\right ) \]

[In]

integrate(((3*exp(x)*x+192*x^2)*log(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x, algorithm="fricas")

[Out]

e^(96*x^2 - 3*e^5 + 3*e^x + 33)*log(x)

Sympy [A] (verification not implemented)

Time = 2.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{96 x^{2} + 3 e^{x} - 3 e^{5} + 33} \log {\left (x \right )} \]

[In]

integrate(((3*exp(x)*x+192*x**2)*ln(x)+1)*exp(3*exp(x)-3*exp(5)+96*x**2+33)/x,x)

[Out]

exp(96*x**2 + 3*exp(x) - 3*exp(5) + 33)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \left (x\right ) \]

[In]

integrate(((3*exp(x)*x+192*x^2)*log(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x, algorithm="maxima")

[Out]

e^(96*x^2 - 3*e^5 + 3*e^x + 33)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \left (x\right ) \]

[In]

integrate(((3*exp(x)*x+192*x^2)*log(x)+1)*exp(3*exp(x)-3*exp(5)+96*x^2+33)/x,x, algorithm="giac")

[Out]

e^(96*x^2 - 3*e^5 + 3*e^x + 33)*log(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=\int \frac {{\mathrm {e}}^{3\,{\mathrm {e}}^x-3\,{\mathrm {e}}^5+96\,x^2+33}\,\left (\ln \left (x\right )\,\left (3\,x\,{\mathrm {e}}^x+192\,x^2\right )+1\right )}{x} \,d x \]

[In]

int((exp(3*exp(x) - 3*exp(5) + 96*x^2 + 33)*(log(x)*(3*x*exp(x) + 192*x^2) + 1))/x,x)

[Out]

int((exp(3*exp(x) - 3*exp(5) + 96*x^2 + 33)*(log(x)*(3*x*exp(x) + 192*x^2) + 1))/x, x)