3.32.13 \(\int \frac {-10 x+2 x^2+6 x^3+e^x (-3-x+3 x^2)+(2 e^x x+4 x^2) \log (x)+(-4-4 e^x-4 x) \log (e^x+2 x)}{e^x+2 x} \, dx\)

Optimal. Leaf size=27 \[ 2+x (-3+x+x (x+\log (x)))-\left (x+\log \left (e^x+2 x\right )\right )^2 \]

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Rubi [F]  time = 1.44, 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 {-10 x+2 x^2+6 x^3+e^x \left (-3-x+3 x^2\right )+\left (2 e^x x+4 x^2\right ) \log (x)+\left (-4-4 e^x-4 x\right ) \log \left (e^x+2 x\right )}{e^x+2 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x + 2*x^2 + 6*x^3 + E^x*(-3 - x + 3*x^2) + (2*E^x*x + 4*x^2)*Log[x] + (-4 - 4*E^x - 4*x)*Log[E^x + 2*
x])/(E^x + 2*x),x]

[Out]

-3*x + x^2 + x^3 + x^2*Log[x] - 4*x*Log[E^x + 2*x] - 4*Log[E^x + 2*x]*Defer[Int][(E^x + 2*x)^(-1), x] + 4*Defe
r[Int][x/(E^x + 2*x), x] + 4*Log[E^x + 2*x]*Defer[Int][x/(E^x + 2*x), x] - 4*Defer[Int][x^2/(E^x + 2*x), x] +
4*Defer[Int][Defer[Int][(E^x + 2*x)^(-1), x], x] + 8*Defer[Int][Defer[Int][(E^x + 2*x)^(-1), x]/(E^x + 2*x), x
] - 8*Defer[Int][(x*Defer[Int][(E^x + 2*x)^(-1), x])/(E^x + 2*x), x] - 4*Defer[Int][Defer[Int][x/(E^x + 2*x),
x], x] - 8*Defer[Int][Defer[Int][x/(E^x + 2*x), x]/(E^x + 2*x), x] + 8*Defer[Int][(x*Defer[Int][x/(E^x + 2*x),
x])/(E^x + 2*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-3-x+3 x^2+2 x \log (x)-4 \log \left (e^x+2 x\right )+\frac {4 (-1+x) \left (x+\log \left (e^x+2 x\right )\right )}{e^x+2 x}\right ) \, dx\\ &=-3 x-\frac {x^2}{2}+x^3+2 \int x \log (x) \, dx-4 \int \log \left (e^x+2 x\right ) \, dx+4 \int \frac {(-1+x) \left (x+\log \left (e^x+2 x\right )\right )}{e^x+2 x} \, dx\\ &=-3 x-x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )+4 \int \frac {\left (2+e^x\right ) x}{e^x+2 x} \, dx+4 \int \left (-\frac {x+\log \left (e^x+2 x\right )}{e^x+2 x}+\frac {x \left (x+\log \left (e^x+2 x\right )\right )}{e^x+2 x}\right ) \, dx\\ &=-3 x-x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )+4 \int \left (x-\frac {2 (-1+x) x}{e^x+2 x}\right ) \, dx-4 \int \frac {x+\log \left (e^x+2 x\right )}{e^x+2 x} \, dx+4 \int \frac {x \left (x+\log \left (e^x+2 x\right )\right )}{e^x+2 x} \, dx\\ &=-3 x+x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )-4 \int \left (\frac {x}{e^x+2 x}+\frac {\log \left (e^x+2 x\right )}{e^x+2 x}\right ) \, dx+4 \int \left (\frac {x^2}{e^x+2 x}+\frac {x \log \left (e^x+2 x\right )}{e^x+2 x}\right ) \, dx-8 \int \frac {(-1+x) x}{e^x+2 x} \, dx\\ &=-3 x+x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )-4 \int \frac {x}{e^x+2 x} \, dx+4 \int \frac {x^2}{e^x+2 x} \, dx-4 \int \frac {\log \left (e^x+2 x\right )}{e^x+2 x} \, dx+4 \int \frac {x \log \left (e^x+2 x\right )}{e^x+2 x} \, dx-8 \int \left (-\frac {x}{e^x+2 x}+\frac {x^2}{e^x+2 x}\right ) \, dx\\ &=-3 x+x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )-4 \int \frac {x}{e^x+2 x} \, dx+4 \int \frac {x^2}{e^x+2 x} \, dx+4 \int \frac {\left (2+e^x\right ) \int \frac {1}{e^x+2 x} \, dx}{e^x+2 x} \, dx-4 \int \frac {\left (2+e^x\right ) \int \frac {x}{e^x+2 x} \, dx}{e^x+2 x} \, dx+8 \int \frac {x}{e^x+2 x} \, dx-8 \int \frac {x^2}{e^x+2 x} \, dx-\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {1}{e^x+2 x} \, dx+\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {x}{e^x+2 x} \, dx\\ &=-3 x+x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )-4 \int \frac {x}{e^x+2 x} \, dx+4 \int \frac {x^2}{e^x+2 x} \, dx+4 \int \left (\int \frac {1}{e^x+2 x} \, dx-\frac {2 (-1+x) \int \frac {1}{e^x+2 x} \, dx}{e^x+2 x}\right ) \, dx-4 \int \left (\int \frac {x}{e^x+2 x} \, dx-\frac {2 (-1+x) \int \frac {x}{e^x+2 x} \, dx}{e^x+2 x}\right ) \, dx+8 \int \frac {x}{e^x+2 x} \, dx-8 \int \frac {x^2}{e^x+2 x} \, dx-\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {1}{e^x+2 x} \, dx+\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {x}{e^x+2 x} \, dx\\ &=-3 x+x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )-4 \int \frac {x}{e^x+2 x} \, dx+4 \int \frac {x^2}{e^x+2 x} \, dx+4 \int \left (\int \frac {1}{e^x+2 x} \, dx\right ) \, dx-4 \int \left (\int \frac {x}{e^x+2 x} \, dx\right ) \, dx+8 \int \frac {x}{e^x+2 x} \, dx-8 \int \frac {x^2}{e^x+2 x} \, dx-8 \int \frac {(-1+x) \int \frac {1}{e^x+2 x} \, dx}{e^x+2 x} \, dx+8 \int \frac {(-1+x) \int \frac {x}{e^x+2 x} \, dx}{e^x+2 x} \, dx-\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {1}{e^x+2 x} \, dx+\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {x}{e^x+2 x} \, dx\\ &=-3 x+x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )-4 \int \frac {x}{e^x+2 x} \, dx+4 \int \frac {x^2}{e^x+2 x} \, dx+4 \int \left (\int \frac {1}{e^x+2 x} \, dx\right ) \, dx-4 \int \left (\int \frac {x}{e^x+2 x} \, dx\right ) \, dx+8 \int \frac {x}{e^x+2 x} \, dx-8 \int \frac {x^2}{e^x+2 x} \, dx-8 \int \left (-\frac {\int \frac {1}{e^x+2 x} \, dx}{e^x+2 x}+\frac {x \int \frac {1}{e^x+2 x} \, dx}{e^x+2 x}\right ) \, dx+8 \int \left (-\frac {\int \frac {x}{e^x+2 x} \, dx}{e^x+2 x}+\frac {x \int \frac {x}{e^x+2 x} \, dx}{e^x+2 x}\right ) \, dx-\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {1}{e^x+2 x} \, dx+\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {x}{e^x+2 x} \, dx\\ &=-3 x+x^2+x^3+x^2 \log (x)-4 x \log \left (e^x+2 x\right )-4 \int \frac {x}{e^x+2 x} \, dx+4 \int \frac {x^2}{e^x+2 x} \, dx+4 \int \left (\int \frac {1}{e^x+2 x} \, dx\right ) \, dx-4 \int \left (\int \frac {x}{e^x+2 x} \, dx\right ) \, dx+8 \int \frac {x}{e^x+2 x} \, dx-8 \int \frac {x^2}{e^x+2 x} \, dx+8 \int \frac {\int \frac {1}{e^x+2 x} \, dx}{e^x+2 x} \, dx-8 \int \frac {x \int \frac {1}{e^x+2 x} \, dx}{e^x+2 x} \, dx-8 \int \frac {\int \frac {x}{e^x+2 x} \, dx}{e^x+2 x} \, dx+8 \int \frac {x \int \frac {x}{e^x+2 x} \, dx}{e^x+2 x} \, dx-\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {1}{e^x+2 x} \, dx+\left (4 \log \left (e^x+2 x\right )\right ) \int \frac {x}{e^x+2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 36, normalized size = 1.33 \begin {gather*} -3 x+x^3+x^2 \log (x)-2 x \log \left (e^x+2 x\right )-\log ^2\left (e^x+2 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x + 2*x^2 + 6*x^3 + E^x*(-3 - x + 3*x^2) + (2*E^x*x + 4*x^2)*Log[x] + (-4 - 4*E^x - 4*x)*Log[E^
x + 2*x])/(E^x + 2*x),x]

[Out]

-3*x + x^3 + x^2*Log[x] - 2*x*Log[E^x + 2*x] - Log[E^x + 2*x]^2

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fricas [A]  time = 1.02, size = 34, normalized size = 1.26 \begin {gather*} x^{3} + x^{2} \log \relax (x) - 2 \, x \log \left (2 \, x + e^{x}\right ) - \log \left (2 \, x + e^{x}\right )^{2} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-4*x-4)*log(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*log(x)+(3*x^2-x-3)*exp(x)+6*x^3+2*x^2-10*x)/(e
xp(x)+2*x),x, algorithm="fricas")

[Out]

x^3 + x^2*log(x) - 2*x*log(2*x + e^x) - log(2*x + e^x)^2 - 3*x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {6 \, x^{3} + 2 \, x^{2} + {\left (3 \, x^{2} - x - 3\right )} e^{x} - 4 \, {\left (x + e^{x} + 1\right )} \log \left (2 \, x + e^{x}\right ) + 2 \, {\left (2 \, x^{2} + x e^{x}\right )} \log \relax (x) - 10 \, x}{2 \, x + e^{x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-4*x-4)*log(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*log(x)+(3*x^2-x-3)*exp(x)+6*x^3+2*x^2-10*x)/(e
xp(x)+2*x),x, algorithm="giac")

[Out]

integrate((6*x^3 + 2*x^2 + (3*x^2 - x - 3)*e^x - 4*(x + e^x + 1)*log(2*x + e^x) + 2*(2*x^2 + x*e^x)*log(x) - 1
0*x)/(2*x + e^x), x)

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maple [A]  time = 0.03, size = 35, normalized size = 1.30

method	result	size
risch	\(x^{3}+x^{2} \ln \relax (x )-2 \ln \left ({\mathrm e}^{x}+2 x \right ) x -\ln \left ({\mathrm e}^{x}+2 x \right )^{2}-3 x\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(x)-4*x-4)*ln(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*ln(x)+(3*x^2-x-3)*exp(x)+6*x^3+2*x^2-10*x)/(exp(x)+2*
x),x,method=_RETURNVERBOSE)

[Out]

x^3+x^2*ln(x)-2*ln(exp(x)+2*x)*x-ln(exp(x)+2*x)^2-3*x

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maxima [A]  time = 0.50, size = 34, normalized size = 1.26 \begin {gather*} x^{3} + x^{2} \log \relax (x) - 2 \, x \log \left (2 \, x + e^{x}\right ) - \log \left (2 \, x + e^{x}\right )^{2} - 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-4*x-4)*log(exp(x)+2*x)+(2*exp(x)*x+4*x^2)*log(x)+(3*x^2-x-3)*exp(x)+6*x^3+2*x^2-10*x)/(e
xp(x)+2*x),x, algorithm="maxima")

[Out]

x^3 + x^2*log(x) - 2*x*log(2*x + e^x) - log(2*x + e^x)^2 - 3*x

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mupad [B]  time = 1.89, size = 34, normalized size = 1.26 \begin {gather*} x^2\,\ln \relax (x)-3\,x-2\,x\,\ln \left (2\,x+{\mathrm {e}}^x\right )+x^3-{\ln \left (2\,x+{\mathrm {e}}^x\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + exp(x)*(x - 3*x^2 + 3) + log(2*x + exp(x))*(4*x + 4*exp(x) + 4) - log(x)*(2*x*exp(x) + 4*x^2) - 2
*x^2 - 6*x^3)/(2*x + exp(x)),x)

[Out]

x^2*log(x) - 3*x - 2*x*log(2*x + exp(x)) + x^3 - log(2*x + exp(x))^2

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sympy [A]  time = 0.44, size = 34, normalized size = 1.26 \begin {gather*} x^{3} + x^{2} \log {\relax (x )} - 2 x \log {\left (2 x + e^{x} \right )} - 3 x - \log {\left (2 x + e^{x} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-4*x-4)*ln(exp(x)+2*x)+(2*exp(x)*x+4*x**2)*ln(x)+(3*x**2-x-3)*exp(x)+6*x**3+2*x**2-10*x)/
(exp(x)+2*x),x)

[Out]

x**3 + x**2*log(x) - 2*x*log(2*x + exp(x)) - 3*x - log(2*x + exp(x))**2

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