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

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

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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} $$\begin{array}{c}-\frac{2}{-\log \left(x^2\right)+2e^x-2\log (x-e^x)}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{8x+4e^{2x}x-4e^x(1+x+x^2)}{(e^x-x)x{(2e^x-\log \left(x^2\right)-2\log (-e^x+x))}^2}dx\\ & =-\frac{2}{2e^x-\log \left(x^2\right)-2\log (-e^x+x)}\end{array}\end{array}$$

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

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])

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

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))

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

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)

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

method	result	size
risch	$\frac{4i}{\pi \mathrm{csgn}{\left(ix\right)}^2\mathrm{csgn}\left(i{x}^2\right)-2\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(i{x}^2\right)}^2+\pi \mathrm{csgn}{\left(i{x}^2\right)}^3-4i{\mathrm{e}}^x+4i\ln (x)+4i\ln (x-{\mathrm{e}}^x)}$	$71$

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)))

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maxima [A]  time = 0.51, size = 20, normalized size = 0.83 $$\begin{array}{c}-\frac{1}{e^x-\log (x-e^x)-\log (x)}\end{array}$$

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))

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

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))

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

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)))

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