3.68.42 \(\int \frac {20+25 e^x+5 x}{e^x+x} \, dx\)

Optimal. Leaf size=27 \[ x+x \left (4+\frac {5 \left (-3+2 \log \left (\frac {1}{4} \left (e^x+x\right )^2\right )\right )}{x}\right ) \]

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Rubi [F]  time = 0.17, 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 {20+25 e^x+5 x}{e^x+x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20 + 25*E^x + 5*x)/(E^x + x),x]

[Out]

25*x + 20*Defer[Int][(E^x + x)^(-1), x] - 20*Defer[Int][x/(E^x + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (4+5 e^x+x\right )}{e^x+x} \, dx\\ &=5 \int \frac {4+5 e^x+x}{e^x+x} \, dx\\ &=5 \int \left (5-\frac {4 (-1+x)}{e^x+x}\right ) \, dx\\ &=25 x-20 \int \frac {-1+x}{e^x+x} \, dx\\ &=25 x-20 \int \left (-\frac {1}{e^x+x}+\frac {x}{e^x+x}\right ) \, dx\\ &=25 x+20 \int \frac {1}{e^x+x} \, dx-20 \int \frac {x}{e^x+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 12, normalized size = 0.44 \begin {gather*} 5 \left (x+4 \log \left (e^x+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + 25*E^x + 5*x)/(E^x + x),x]

[Out]

5*(x + 4*Log[E^x + x])

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fricas [A]  time = 0.50, size = 11, normalized size = 0.41 \begin {gather*} 5 \, x + 20 \, \log \left (x + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(x)+20+5*x)/(exp(x)+x),x, algorithm="fricas")

[Out]

5*x + 20*log(x + e^x)

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giac [A]  time = 4.00, size = 11, normalized size = 0.41 \begin {gather*} 5 \, x + 20 \, \log \left (x + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(x)+20+5*x)/(exp(x)+x),x, algorithm="giac")

[Out]

5*x + 20*log(x + e^x)

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maple [A]  time = 0.02, size = 12, normalized size = 0.44

method	result	size
norman	\(5 x +20 \ln \left ({\mathrm e}^{x}+x \right )\)	\(12\)
risch	\(5 x +20 \ln \left ({\mathrm e}^{x}+x \right )\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(x)+20+5*x)/(exp(x)+x),x,method=_RETURNVERBOSE)

[Out]

5*x+20*ln(exp(x)+x)

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maxima [A]  time = 0.40, size = 11, normalized size = 0.41 \begin {gather*} 5 \, x + 20 \, \log \left (x + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(x)+20+5*x)/(exp(x)+x),x, algorithm="maxima")

[Out]

5*x + 20*log(x + e^x)

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mupad [B]  time = 3.95, size = 11, normalized size = 0.41 \begin {gather*} 5\,x+20\,\ln \left (x+{\mathrm {e}}^x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 25*exp(x) + 20)/(x + exp(x)),x)

[Out]

5*x + 20*log(x + exp(x))

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sympy [A]  time = 0.09, size = 10, normalized size = 0.37 \begin {gather*} 5 x + 20 \log {\left (x + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(x)+20+5*x)/(exp(x)+x),x)

[Out]

5*x + 20*log(x + exp(x))

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