3.48.39 \(\int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x (16 x-16 x^3+e^2 (-32+16 x^2))+e^x (e^2 (16-16 x)-16 x+16 x^2) \log (e^2-x)}{e^{2 x} (e^2-x)+e^2 x^2-x^3+e^x (2 e^2 x-2 x^2)} \, dx\)

Optimal. Leaf size=25 \[ 3+\frac {16 x \left (-2-x+\log \left (e^2-x\right )\right )}{e^x+x} \]

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Rubi [F]  time = 4.49, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-16 x^2-16 e^2 x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-16*x^2 - 16*E^2*x^2 + 16*x^3 + E^x*(16*x - 16*x^3 + E^2*(-32 + 16*x^2)) + E^x*(E^2*(16 - 16*x) - 16*x +
16*x^2)*Log[E^2 - x])/(E^(2*x)*(E^2 - x) + E^2*x^2 - x^3 + E^x*(2*E^2*x - 2*x^2)),x]

[Out]

32/(E^x + x) - (16*Log[E^2 - x])/(E^x + x) + 32*Defer[Int][(E^x + x)^(-2), x] - 16*Log[E^2 - x]*Defer[Int][(E^
x + x)^(-2), x] - 16*Defer[Int][x^2/(E^x + x)^2, x] + 16*Log[E^2 - x]*Defer[Int][x^2/(E^x + x)^2, x] - 16*Defe
r[Int][x^3/(E^x + x)^2, x] + 16*Defer[Int][(E^x + x)^(-1), x] - 16*Log[E^2 - x]*Defer[Int][(E^x + x)^(-1), x]
- 16*E^2*Log[E^2 - x]*Defer[Int][(E^x + x)^(-1), x] + 16*(1 + E^2)*Log[E^2 - x]*Defer[Int][(E^x + x)^(-1), x]
- 16*Defer[Int][1/((E^2 - x)*(E^x + x)), x] - 16*E^2*Defer[Int][1/((E^2 - x)*(E^x + x)), x] + 16*E^2*Log[E^2 -
x]*Defer[Int][1/((E^2 - x)*(E^x + x)), x] + 16*E^4*Log[E^2 - x]*Defer[Int][1/((E^2 - x)*(E^x + x)), x] - 16*E
^2*(1 + E^2)*Log[E^2 - x]*Defer[Int][1/((E^2 - x)*(E^x + x)), x] - 16*Log[E^2 - x]*Defer[Int][x/(E^x + x), x]
+ 16*Defer[Int][x^2/(E^x + x), x] - 16*Defer[Int][Defer[Int][(E^x + x)^(-2), x]/(E^2 - x), x] + 16*Defer[Int][
Defer[Int][x^2/(E^x + x)^2, x]/(E^2 - x), x] - 16*Defer[Int][Defer[Int][(E^x + x)^(-1), x]/(E^2 - x), x] - 16*
E^2*Defer[Int][Defer[Int][(E^x + x)^(-1), x]/(E^2 - x), x] + 16*(1 + E^2)*Defer[Int][Defer[Int][(E^x + x)^(-1)
, x]/(E^2 - x), x] + 16*E^2*Defer[Int][Defer[Int][1/((E^2 - x)*(E^x + x)), x]/(E^2 - x), x] + 16*E^4*Defer[Int
][Defer[Int][1/((E^2 - x)*(E^x + x)), x]/(E^2 - x), x] - 16*E^2*(1 + E^2)*Defer[Int][Defer[Int][1/((E^2 - x)*(
E^x + x)), x]/(E^2 - x), x] - 16*Defer[Int][Defer[Int][x/(E^x + x), x]/(E^2 - x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-16-16 e^2\right ) x^2+16 x^3+e^x \left (16 x-16 x^3+e^2 \left (-32+16 x^2\right )\right )+e^x \left (e^2 (16-16 x)-16 x+16 x^2\right ) \log \left (e^2-x\right )}{e^{2 x} \left (e^2-x\right )+e^2 x^2-x^3+e^x \left (2 e^2 x-2 x^2\right )} \, dx\\ &=\int \frac {16 \left (-e^2 x^2+(-1+x) x^2+e^{2+x} \left (-2+x^2\right )+e^x \left (x-x^3\right )-e^x \left (e^2-x\right ) (-1+x) \log \left (e^2-x\right )\right )}{\left (e^2-x\right ) \left (e^x+x\right )^2} \, dx\\ &=16 \int \frac {-e^2 x^2+(-1+x) x^2+e^{2+x} \left (-2+x^2\right )+e^x \left (x-x^3\right )-e^x \left (e^2-x\right ) (-1+x) \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )^2} \, dx\\ &=16 \int \left (-\frac {(-1+x) x \left (2+x-\log \left (e^2-x\right )\right )}{\left (e^x+x\right )^2}+\frac {-2 e^2+x+e^2 x^2-x^3+e^2 \log \left (e^2-x\right )-\left (1+e^2\right ) x \log \left (e^2-x\right )+x^2 \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )}\right ) \, dx\\ &=-\left (16 \int \frac {(-1+x) x \left (2+x-\log \left (e^2-x\right )\right )}{\left (e^x+x\right )^2} \, dx\right )+16 \int \frac {-2 e^2+x+e^2 x^2-x^3+e^2 \log \left (e^2-x\right )-\left (1+e^2\right ) x \log \left (e^2-x\right )+x^2 \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx\\ &=-\left (16 \int \left (-\frac {x \left (2+x-\log \left (e^2-x\right )\right )}{\left (e^x+x\right )^2}+\frac {x^2 \left (2+x-\log \left (e^2-x\right )\right )}{\left (e^x+x\right )^2}\right ) \, dx\right )+16 \int \frac {x-x^3+e^2 \left (-2+x^2\right )-\left (e^2-x\right ) (-1+x) \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx\\ &=16 \int \frac {x \left (2+x-\log \left (e^2-x\right )\right )}{\left (e^x+x\right )^2} \, dx-16 \int \frac {x^2 \left (2+x-\log \left (e^2-x\right )\right )}{\left (e^x+x\right )^2} \, dx+16 \int \left (-\frac {2 e^2}{\left (e^2-x\right ) \left (e^x+x\right )}+\frac {x}{\left (e^2-x\right ) \left (e^x+x\right )}+\frac {e^2 x^2}{\left (e^2-x\right ) \left (e^x+x\right )}-\frac {x^3}{\left (e^2-x\right ) \left (e^x+x\right )}+\frac {e^2 \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )}-\frac {\left (1+e^2\right ) x \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )}+\frac {x^2 \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )}\right ) \, dx\\ &=16 \int \frac {x}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx-16 \int \frac {x^3}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+16 \int \frac {x^2 \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+16 \int \left (\frac {2 x}{\left (e^x+x\right )^2}+\frac {x^2}{\left (e^x+x\right )^2}-\frac {x \log \left (e^2-x\right )}{\left (e^x+x\right )^2}\right ) \, dx-16 \int \left (\frac {2 x^2}{\left (e^x+x\right )^2}+\frac {x^3}{\left (e^x+x\right )^2}-\frac {x^2 \log \left (e^2-x\right )}{\left (e^x+x\right )^2}\right ) \, dx+\left (16 e^2\right ) \int \frac {x^2}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+\left (16 e^2\right ) \int \frac {\log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx-\left (32 e^2\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx-\left (16 \left (1+e^2\right )\right ) \int \frac {x \log \left (e^2-x\right )}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx\\ &=16 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx-16 \int \frac {x^3}{\left (e^x+x\right )^2} \, dx+16 \int \left (-\frac {1}{e^x+x}+\frac {e^2}{\left (e^2-x\right ) \left (e^x+x\right )}\right ) \, dx-16 \int \left (-\frac {e^4}{e^x+x}+\frac {e^6}{\left (e^2-x\right ) \left (e^x+x\right )}-\frac {e^2 x}{e^x+x}-\frac {x^2}{e^x+x}\right ) \, dx-16 \int \frac {x \log \left (e^2-x\right )}{\left (e^x+x\right )^2} \, dx+16 \int \frac {x^2 \log \left (e^2-x\right )}{\left (e^x+x\right )^2} \, dx+16 \int \frac {-e^2 \int \frac {1}{e^x+x} \, dx+e^4 \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx-\int \frac {x}{e^x+x} \, dx}{e^2-x} \, dx+32 \int \frac {x}{\left (e^x+x\right )^2} \, dx-32 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx+\left (16 e^2\right ) \int \left (-\frac {e^2}{e^x+x}+\frac {e^4}{\left (e^2-x\right ) \left (e^x+x\right )}-\frac {x}{e^x+x}\right ) \, dx+\left (16 e^2\right ) \int \frac {\int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx}{e^2-x} \, dx-\left (32 e^2\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx-\left (16 \left (1+e^2\right )\right ) \int \frac {-\int \frac {1}{e^x+x} \, dx+e^2 \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx}{e^2-x} \, dx-\left (16 \log \left (e^2-x\right )\right ) \int \frac {x}{e^x+x} \, dx-\left (16 e^2 \log \left (e^2-x\right )\right ) \int \frac {1}{e^x+x} \, dx+\left (16 e^2 \log \left (e^2-x\right )\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+\left (16 e^4 \log \left (e^2-x\right )\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+\left (16 \left (1+e^2\right ) \log \left (e^2-x\right )\right ) \int \frac {1}{e^x+x} \, dx-\left (16 e^2 \left (1+e^2\right ) \log \left (e^2-x\right )\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx\\ &=\frac {32}{e^x+x}-\frac {16 \log \left (e^2-x\right )}{e^x+x}+16 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx-16 \int \frac {x^3}{\left (e^x+x\right )^2} \, dx-16 \int \frac {1}{e^x+x} \, dx+16 \int \frac {x^2}{e^x+x} \, dx+16 \int \frac {\int \frac {x^2}{\left (e^x+x\right )^2} \, dx}{e^2-x} \, dx-16 \int \frac {\frac {1}{e^x+x}+\int \frac {1}{\left (e^x+x\right )^2} \, dx+\int \frac {1}{e^x+x} \, dx}{e^2-x} \, dx+16 \int \left (\frac {e^2 \left (-\int \frac {1}{e^x+x} \, dx+e^2 \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx\right )}{e^2-x}-\frac {\int \frac {x}{e^x+x} \, dx}{e^2-x}\right ) \, dx+32 \int \frac {1}{\left (e^x+x\right )^2} \, dx-32 \int \frac {x^2}{\left (e^x+x\right )^2} \, dx+32 \int \frac {1}{e^x+x} \, dx+\left (16 e^2\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+\left (16 e^2\right ) \int \frac {\int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx}{e^2-x} \, dx-\left (32 e^2\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx-\left (16 \left (1+e^2\right )\right ) \int \left (-\frac {\int \frac {1}{e^x+x} \, dx}{e^2-x}+\frac {e^2 \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx}{e^2-x}\right ) \, dx-\left (16 \log \left (e^2-x\right )\right ) \int \frac {1}{\left (e^x+x\right )^2} \, dx+\left (16 \log \left (e^2-x\right )\right ) \int \frac {x^2}{\left (e^x+x\right )^2} \, dx-\left (16 \log \left (e^2-x\right )\right ) \int \frac {1}{e^x+x} \, dx-\left (16 \log \left (e^2-x\right )\right ) \int \frac {x}{e^x+x} \, dx-\left (16 e^2 \log \left (e^2-x\right )\right ) \int \frac {1}{e^x+x} \, dx+\left (16 e^2 \log \left (e^2-x\right )\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+\left (16 e^4 \log \left (e^2-x\right )\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx+\left (16 \left (1+e^2\right ) \log \left (e^2-x\right )\right ) \int \frac {1}{e^x+x} \, dx-\left (16 e^2 \left (1+e^2\right ) \log \left (e^2-x\right )\right ) \int \frac {1}{\left (e^2-x\right ) \left (e^x+x\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.73, size = 23, normalized size = 0.92 \begin {gather*} -\frac {16 x \left (2+x-\log \left (e^2-x\right )\right )}{e^x+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 - 16*E^2*x^2 + 16*x^3 + E^x*(16*x - 16*x^3 + E^2*(-32 + 16*x^2)) + E^x*(E^2*(16 - 16*x) - 1
6*x + 16*x^2)*Log[E^2 - x])/(E^(2*x)*(E^2 - x) + E^2*x^2 - x^3 + E^x*(2*E^2*x - 2*x^2)),x]

[Out]

(-16*x*(2 + x - Log[E^2 - x]))/(E^x + x)

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fricas [A]  time = 0.80, size = 25, normalized size = 1.00 \begin {gather*} -\frac {16 \, {\left (x^{2} - x \log \left (-x + e^{2}\right ) + 2 \, x\right )}}{x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*log(exp(2)-x)+((16*x^2-32)*exp(2)-16*x^3+16*x)*exp(x)-16*x^2
*exp(2)+16*x^3-16*x^2)/((exp(2)-x)*exp(x)^2+(2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x, algorithm="fricas")

[Out]

-16*(x^2 - x*log(-x + e^2) + 2*x)/(x + e^x)

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giac [A]  time = 1.65, size = 25, normalized size = 1.00 \begin {gather*} -\frac {16 \, {\left (x^{2} - x \log \left (-x + e^{2}\right ) + 2 \, x\right )}}{x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*log(exp(2)-x)+((16*x^2-32)*exp(2)-16*x^3+16*x)*exp(x)-16*x^2
*exp(2)+16*x^3-16*x^2)/((exp(2)-x)*exp(x)^2+(2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x, algorithm="giac")

[Out]

-16*(x^2 - x*log(-x + e^2) + 2*x)/(x + e^x)

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maple [A]  time = 0.05, size = 30, normalized size = 1.20

method	result	size
risch	\(\frac {16 x \ln \left ({\mathrm e}^{2}-x \right )}{{\mathrm e}^{x}+x}-\frac {16 x \left (2+x \right )}{{\mathrm e}^{x}+x}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*ln(exp(2)-x)+((16*x^2-32)*exp(2)-16*x^3+16*x)*exp(x)-16*x^2*exp(2)
+16*x^3-16*x^2)/((exp(2)-x)*exp(x)^2+(2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x,method=_RETURNVERBOSE)

[Out]

16*x/(exp(x)+x)*ln(exp(2)-x)-16*x*(2+x)/(exp(x)+x)

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maxima [A]  time = 0.41, size = 25, normalized size = 1.00 \begin {gather*} -\frac {16 \, {\left (x^{2} - x \log \left (-x + e^{2}\right ) + 2 \, x\right )}}{x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(2)+16*x^2-16*x)*exp(x)*log(exp(2)-x)+((16*x^2-32)*exp(2)-16*x^3+16*x)*exp(x)-16*x^2
*exp(2)+16*x^3-16*x^2)/((exp(2)-x)*exp(x)^2+(2*exp(2)*x-2*x^2)*exp(x)+x^2*exp(2)-x^3),x, algorithm="maxima")

[Out]

-16*(x^2 - x*log(-x + e^2) + 2*x)/(x + e^x)

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mupad [B]  time = 3.62, size = 21, normalized size = 0.84 \begin {gather*} -\frac {16\,x\,\left (x-\ln \left ({\mathrm {e}}^2-x\right )+2\right )}{x+{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*x^2*exp(2) - exp(x)*(16*x + exp(2)*(16*x^2 - 32) - 16*x^3) + 16*x^2 - 16*x^3 + log(exp(2) - x)*exp(x)
*(16*x - 16*x^2 + exp(2)*(16*x - 16)))/(exp(x)*(2*x*exp(2) - 2*x^2) - exp(2*x)*(x - exp(2)) + x^2*exp(2) - x^3
),x)

[Out]

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

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sympy [A]  time = 0.33, size = 22, normalized size = 0.88 \begin {gather*} \frac {- 16 x^{2} + 16 x \log {\left (- x + e^{2} \right )} - 32 x}{x + e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-16*x+16)*exp(2)+16*x**2-16*x)*exp(x)*ln(exp(2)-x)+((16*x**2-32)*exp(2)-16*x**3+16*x)*exp(x)-16*x
**2*exp(2)+16*x**3-16*x**2)/((exp(2)-x)*exp(x)**2+(2*exp(2)*x-2*x**2)*exp(x)+x**2*exp(2)-x**3),x)

[Out]

(-16*x**2 + 16*x*log(-x + exp(2)) - 32*x)/(x + exp(x))

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