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

Optimal. Leaf size=20 \[ \frac {e^4}{5-x+x^2+\log \left (2 x^2\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {12, 6688, 6686} \begin {gather*} \frac {e^4}{x^2+\log \left (2 x^2\right )-x+5} \end {gather*}

Antiderivative was successfully verified.

[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 6686

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {e^4}{5-x+x^2+\log \left (2 x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[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])

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 19, normalized size = 0.95 \begin {gather*} \frac {e^{4}}{x^{2} - x + \log \left (2 \, x^{2}\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 19, normalized size = 0.95 \begin {gather*} \frac {e^{4}}{x^{2} - x + \log \left (2 \, x^{2}\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [A]  time = 0.09, size = 20, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

maxima [A]  time = 0.47, size = 19, normalized size = 0.95 \begin {gather*} \frac {e^{4}}{x^{2} - x + \log \relax (2) + 2 \, \log \relax (x) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 4.34, size = 54, normalized size = 2.70 \begin {gather*} -\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 0.13, size = 15, normalized size = 0.75 \begin {gather*} \frac {e^{4}}{x^{2} - x + \log {\left (2 x^{2} \right )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________