3.13.88 \(\int \frac {4 x^2+e^x (-1+5 x)+(e^x+x) \log (e^x+x)}{e^x+x} \, dx\)

Optimal. Leaf size=18 \[ 1-x+x^2+x \left (x+\log \left (e^x+x\right )\right ) \]

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Rubi [A]  time = 0.48, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 9, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6742, 2548} \begin {gather*} 2 x^2-x+x \log \left (x+e^x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x^2 + E^x*(-1 + 5*x) + (E^x + x)*Log[E^x + x])/(E^x + x),x]

[Out]

-x + 2*x^2 + x*Log[E^x + x]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1+5 x-\frac {(-1+x) x}{e^x+x}+\log \left (e^x+x\right )\right ) \, dx\\ &=-x+\frac {5 x^2}{2}-\int \frac {(-1+x) x}{e^x+x} \, dx+\int \log \left (e^x+x\right ) \, dx\\ &=-x+\frac {5 x^2}{2}+x \log \left (e^x+x\right )-\int \frac {\left (1+e^x\right ) x}{e^x+x} \, dx-\int \left (-\frac {x}{e^x+x}+\frac {x^2}{e^x+x}\right ) \, dx\\ &=-x+\frac {5 x^2}{2}+x \log \left (e^x+x\right )+\int \frac {x}{e^x+x} \, dx-\int \frac {x^2}{e^x+x} \, dx-\int \left (x-\frac {(-1+x) x}{e^x+x}\right ) \, dx\\ &=-x+2 x^2+x \log \left (e^x+x\right )+\int \frac {x}{e^x+x} \, dx+\int \frac {(-1+x) x}{e^x+x} \, dx-\int \frac {x^2}{e^x+x} \, dx\\ &=-x+2 x^2+x \log \left (e^x+x\right )+\int \frac {x}{e^x+x} \, dx-\int \frac {x^2}{e^x+x} \, dx+\int \left (-\frac {x}{e^x+x}+\frac {x^2}{e^x+x}\right ) \, dx\\ &=-x+2 x^2+x \log \left (e^x+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 17, normalized size = 0.94 \begin {gather*} -x+2 x^2+x \log \left (e^x+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + E^x*(-1 + 5*x) + (E^x + x)*Log[E^x + x])/(E^x + x),x]

[Out]

-x + 2*x^2 + x*Log[E^x + x]

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fricas [A]  time = 1.06, size = 16, normalized size = 0.89 \begin {gather*} 2 \, x^{2} + x \log \left (x + e^{x}\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+x)*log(exp(x)+x)+(5*x-1)*exp(x)+4*x^2)/(exp(x)+x),x, algorithm="fricas")

[Out]

2*x^2 + x*log(x + e^x) - x

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giac [A]  time = 0.25, size = 16, normalized size = 0.89 \begin {gather*} 2 \, x^{2} + x \log \left (x + e^{x}\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+x)*log(exp(x)+x)+(5*x-1)*exp(x)+4*x^2)/(exp(x)+x),x, algorithm="giac")

[Out]

2*x^2 + x*log(x + e^x) - x

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maple [A]  time = 0.02, size = 17, normalized size = 0.94

method	result	size
norman	\(\ln \left ({\mathrm e}^{x}+x \right ) x -x +2 x^{2}\)	\(17\)
risch	\(\ln \left ({\mathrm e}^{x}+x \right ) x -x +2 x^{2}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(x)+x)*ln(exp(x)+x)+(5*x-1)*exp(x)+4*x^2)/(exp(x)+x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.56, size = 16, normalized size = 0.89 \begin {gather*} 2 \, x^{2} + x \log \left (x + e^{x}\right ) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+x)*log(exp(x)+x)+(5*x-1)*exp(x)+4*x^2)/(exp(x)+x),x, algorithm="maxima")

[Out]

2*x^2 + x*log(x + e^x) - x

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mupad [B]  time = 1.02, size = 12, normalized size = 0.67 \begin {gather*} x\,\left (2\,x+\ln \left (x+{\mathrm {e}}^x\right )-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + exp(x))*(x + exp(x)) + exp(x)*(5*x - 1) + 4*x^2)/(x + exp(x)),x)

[Out]

x*(2*x + log(x + exp(x)) - 1)

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sympy [A]  time = 0.27, size = 14, normalized size = 0.78 \begin {gather*} 2 x^{2} + x \log {\left (x + e^{x} \right )} - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(x)+x)*ln(exp(x)+x)+(5*x-1)*exp(x)+4*x**2)/(exp(x)+x),x)

[Out]

2*x**2 + x*log(x + exp(x)) - x

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