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

Optimal. Leaf size=24 \[ \frac {1}{-e^x+\frac {\log \left (x^2\right )}{2}+\log \left (-e^x+x\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.57, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, integrand size = 154, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6688, 6686} \begin {gather*} -\frac {2}{-\log \left (x^2\right )+2 e^x-2 \log \left (x-e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(2*E^x - Log[x^2] - 2*Log[-E^x + 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} &=\int \frac {8 x+4 e^{2 x} x-4 e^x \left (1+x+x^2\right )}{\left (e^x-x\right ) x \left (2 e^x-\log \left (x^2\right )-2 \log \left (-e^x+x\right )\right )^2} \, dx\\ &=-\frac {2}{2 e^x-\log \left (x^2\right )-2 \log \left (-e^x+x\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 24, normalized size = 1.00 \begin {gather*} \frac {2}{-2 e^x+\log \left (x^2\right )+2 \log \left (-e^x+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/(-2*E^x + Log[x^2] + 2*Log[-E^x + x])

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 24, normalized size = 1.00 \begin {gather*} -\frac {2}{2 \, e^{x} - \log \left (x^{2}\right ) - 2 \, \log \left (x - e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*log(x-exp(x))^2+((4*exp(x)*x-4*x^2)*log
(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*log(x-exp(x))+(exp(x)*x-x^2)*log(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*log(x^2)
+4*x*exp(x)^3-4*exp(x)^2*x^2),x, algorithm="fricas")

[Out]

-2/(2*e^x - log(x^2) - 2*log(x - e^x))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, {\left (x e^{\left (2 \, x\right )} - {\left (x^{2} + x + 1\right )} e^{x} + 2 \, x\right )}}{4 \, x^{2} e^{\left (2 \, x\right )} + {\left (x^{2} - x e^{x}\right )} \log \left (x^{2}\right )^{2} + 4 \, {\left (x^{2} - x e^{x}\right )} \log \left (x - e^{x}\right )^{2} - 4 \, x e^{\left (3 \, x\right )} - 4 \, {\left (x^{2} e^{x} - x e^{\left (2 \, x\right )}\right )} \log \left (x^{2}\right ) - 4 \, {\left (2 \, x^{2} e^{x} - 2 \, x e^{\left (2 \, x\right )} - {\left (x^{2} - x e^{x}\right )} \log \left (x^{2}\right )\right )} \log \left (x - e^{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*log(x-exp(x))^2+((4*exp(x)*x-4*x^2)*log
(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*log(x-exp(x))+(exp(x)*x-x^2)*log(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*log(x^2)
+4*x*exp(x)^3-4*exp(x)^2*x^2),x, algorithm="giac")

[Out]

integrate(-4*(x*e^(2*x) - (x^2 + x + 1)*e^x + 2*x)/(4*x^2*e^(2*x) + (x^2 - x*e^x)*log(x^2)^2 + 4*(x^2 - x*e^x)
*log(x - e^x)^2 - 4*x*e^(3*x) - 4*(x^2*e^x - x*e^(2*x))*log(x^2) - 4*(2*x^2*e^x - 2*x*e^(2*x) - (x^2 - x*e^x)*
log(x^2))*log(x - e^x)), x)

________________________________________________________________________________________

maple [C]  time = 0.12, size = 71, normalized size = 2.96




method result size



risch \(\frac {4 i}{\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-4 i {\mathrm e}^{x}+4 i \ln \relax (x )+4 i \ln \left (x -{\mathrm e}^{x}\right )}\) \(71\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*ln(x-exp(x))^2+((4*exp(x)*x-4*x^2)*ln(x^2)-8*
x*exp(x)^2+8*exp(x)*x^2)*ln(x-exp(x))+(exp(x)*x-x^2)*ln(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*ln(x^2)+4*x*exp(x)
^3-4*exp(x)^2*x^2),x,method=_RETURNVERBOSE)

[Out]

4*I/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x^2)^3-4*I*exp(x)+4*I*ln(x)+4*I*ln(x-ex
p(x)))

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 20, normalized size = 0.83 \begin {gather*} -\frac {1}{e^{x} - \log \left (x - e^{x}\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)^2+(-4*x^2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x^2)*log(x-exp(x))^2+((4*exp(x)*x-4*x^2)*log
(x^2)-8*x*exp(x)^2+8*exp(x)*x^2)*log(x-exp(x))+(exp(x)*x-x^2)*log(x^2)^2+(-4*x*exp(x)^2+4*exp(x)*x^2)*log(x^2)
+4*x*exp(x)^3-4*exp(x)^2*x^2),x, algorithm="maxima")

[Out]

-1/(e^x - log(x - e^x) - log(x))

________________________________________________________________________________________

mupad [B]  time = 3.25, size = 22, normalized size = 0.92 \begin {gather*} \frac {2}{\ln \left (x^2\right )+2\,\ln \left (x-{\mathrm {e}}^x\right )-2\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(log(x^2) + 2*log(x - exp(x)) - 2*exp(x))

________________________________________________________________________________________

sympy [A]  time = 0.43, size = 19, normalized size = 0.79 \begin {gather*} \frac {2}{- 2 e^{x} + \log {\left (x^{2} \right )} + 2 \log {\left (x - e^{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*exp(x)**2+(-4*x**2-4*x-4)*exp(x)+8*x)/((4*exp(x)*x-4*x**2)*ln(x-exp(x))**2+((4*exp(x)*x-4*x**2)
*ln(x**2)-8*x*exp(x)**2+8*exp(x)*x**2)*ln(x-exp(x))+(exp(x)*x-x**2)*ln(x**2)**2+(-4*x*exp(x)**2+4*exp(x)*x**2)
*ln(x**2)+4*x*exp(x)**3-4*exp(x)**2*x**2),x)

[Out]

2/(-2*exp(x) + log(x**2) + 2*log(x - exp(x)))

________________________________________________________________________________________