\(\int \frac {e^4 (-2+x-2 x^2)}{25 x-10 x^2+11 x^3-2 x^4+x^5+(10 x-2 x^2+2 x^3) \log (2 x^2)+x \log ^2(2 x^2)} \, dx\) [5983]

Optimal result

Integrand size = 67, antiderivative size = 20 \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=\frac {e^4}{5-x+x^2+\log \left (2 x^2\right )} \]

[Out]

exp(4)/(x^2-x+5+ln(2*x^2))

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {12, 6820, 6818} \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=\frac {e^4}{x^2+\log \left (2 x^2\right )-x+5} \]

[In]

Int[(E^4*(-2 + x - 2*x^2))/(25*x - 10*x^2 + 11*x^3 - 2*x^4 + x^5 + (10*x - 2*x^2 + 2*x^3)*Log[2*x^2] + x*Log[2
*x^2]^2),x]

[Out]

E^4/(5 - x + x^2 + Log[2*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = e^4 \int \frac {-2+x-2 x^2}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx \\ & = e^4 \int \frac {-2+x-2 x^2}{x \left (5-x+x^2+\log \left (2 x^2\right )\right )^2} \, dx \\ & = \frac {e^4}{5-x+x^2+\log \left (2 x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=\frac {e^4}{5-x+x^2+\log \left (2 x^2\right )} \]

[In]

Integrate[(E^4*(-2 + x - 2*x^2))/(25*x - 10*x^2 + 11*x^3 - 2*x^4 + x^5 + (10*x - 2*x^2 + 2*x^3)*Log[2*x^2] + x
*Log[2*x^2]^2),x]

[Out]

E^4/(5 - x + x^2 + Log[2*x^2])

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {{\mathrm e}^{4}}{x^{2}-x +5+\ln \left (2 x^{2}\right )}\)	\(20\)
risch	\(\frac {{\mathrm e}^{4}}{x^{2}-x +5+\ln \left (2 x^{2}\right )}\)	\(20\)
parallelrisch	\(\frac {{\mathrm e}^{4}}{x^{2}-x +5+\ln \left (2 x^{2}\right )}\)	\(20\)

[In]

int((-2*x^2+x-2)*exp(4)/(x*ln(2*x^2)^2+(2*x^3-2*x^2+10*x)*ln(2*x^2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x,method=_RE
TURNVERBOSE)

[Out]

exp(4)/(x^2-x+5+ln(2*x^2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=\frac {e^{4}}{x^{2} - x + \log \left (2 \, x^{2}\right ) + 5} \]

[In]

integrate((-2*x^2+x-2)*exp(4)/(x*log(2*x^2)^2+(2*x^3-2*x^2+10*x)*log(2*x^2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x, a
lgorithm="fricas")

[Out]

e^4/(x^2 - x + log(2*x^2) + 5)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=\frac {e^{4}}{x^{2} - x + \log {\left (2 x^{2} \right )} + 5} \]

[In]

integrate((-2*x**2+x-2)*exp(4)/(x*ln(2*x**2)**2+(2*x**3-2*x**2+10*x)*ln(2*x**2)+x**5-2*x**4+11*x**3-10*x**2+25
*x),x)

[Out]

exp(4)/(x**2 - x + log(2*x**2) + 5)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=\frac {e^{4}}{x^{2} - x + \log \left (2\right ) + 2 \, \log \left (x\right ) + 5} \]

[In]

integrate((-2*x^2+x-2)*exp(4)/(x*log(2*x^2)^2+(2*x^3-2*x^2+10*x)*log(2*x^2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x, a
lgorithm="maxima")

[Out]

e^4/(x^2 - x + log(2) + 2*log(x) + 5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=\frac {e^{4}}{x^{2} - x + \log \left (2 \, x^{2}\right ) + 5} \]

[In]

integrate((-2*x^2+x-2)*exp(4)/(x*log(2*x^2)^2+(2*x^3-2*x^2+10*x)*log(2*x^2)+x^5-2*x^4+11*x^3-10*x^2+25*x),x, a
lgorithm="giac")

[Out]

e^4/(x^2 - x + log(2*x^2) + 5)

Mupad [B] (verification not implemented)

Time = 11.72 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.70 \[ \int \frac {e^4 \left (-2+x-2 x^2\right )}{25 x-10 x^2+11 x^3-2 x^4+x^5+\left (10 x-2 x^2+2 x^3\right ) \log \left (2 x^2\right )+x \log ^2\left (2 x^2\right )} \, dx=-\frac {x^2\,\left (\frac {{\mathrm {e}}^4\,\ln \left (2\,x^2\right )}{5}-\frac {{\mathrm {e}}^4\,\left (\ln \left (2\,x^2\right )+5\right )}{5}\right )}{5\,x^2-x^3+x^4+x^2\,\ln \left (2\,x^2\right )} \]

[In]

int(-(exp(4)*(2*x^2 - x + 2))/(25*x + log(2*x^2)*(10*x - 2*x^2 + 2*x^3) - 10*x^2 + 11*x^3 - 2*x^4 + x^5 + x*lo
g(2*x^2)^2),x)

[Out]

-(x^2*((exp(4)*log(2*x^2))/5 - (exp(4)*(log(2*x^2) + 5))/5))/(5*x^2 - x^3 + x^4 + x^2*log(2*x^2))