3.76.46 \(\int \frac {-24+16 x-56 x^2+12 x^3+e^x (-3-13 x+3 x^2)+(e^x+4 x) \log (e^x+4 x)}{e^x+4 x} \, dx\)

Optimal. Leaf size=26 \[ x \left (-3+(-6+x) \left (x+\frac {-x+\log \left (e^x+4 x\right )}{x}\right )\right ) \]

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Rubi [F]  time = 0.53, 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 {-24+16 x-56 x^2+12 x^3+e^x \left (-3-13 x+3 x^2\right )+\left (e^x+4 x\right ) \log \left (e^x+4 x\right )}{e^x+4 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-24 + 16*x - 56*x^2 + 12*x^3 + E^x*(-3 - 13*x + 3*x^2) + (E^x + 4*x)*Log[E^x + 4*x])/(E^x + 4*x),x]

[Out]

-3*x - 7*x^2 + x^3 + x*Log[E^x + 4*x] - 24*Defer[Int][(E^x + 4*x)^(-1), x] + 24*Defer[Int][x/(E^x + 4*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-3-13 x+3 x^2-\frac {4 \left (6-7 x+x^2\right )}{e^x+4 x}+\log \left (e^x+4 x\right )\right ) \, dx\\ &=-3 x-\frac {13 x^2}{2}+x^3-4 \int \frac {6-7 x+x^2}{e^x+4 x} \, dx+\int \log \left (e^x+4 x\right ) \, dx\\ &=-3 x-\frac {13 x^2}{2}+x^3+x \log \left (e^x+4 x\right )-4 \int \left (\frac {6}{e^x+4 x}-\frac {7 x}{e^x+4 x}+\frac {x^2}{e^x+4 x}\right ) \, dx-\int \frac {\left (4+e^x\right ) x}{e^x+4 x} \, dx\\ &=-3 x-\frac {13 x^2}{2}+x^3+x \log \left (e^x+4 x\right )-4 \int \frac {x^2}{e^x+4 x} \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx-\int \left (x-\frac {4 (-1+x) x}{e^x+4 x}\right ) \, dx\\ &=-3 x-7 x^2+x^3+x \log \left (e^x+4 x\right )+4 \int \frac {(-1+x) x}{e^x+4 x} \, dx-4 \int \frac {x^2}{e^x+4 x} \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx\\ &=-3 x-7 x^2+x^3+x \log \left (e^x+4 x\right )-4 \int \frac {x^2}{e^x+4 x} \, dx+4 \int \left (-\frac {x}{e^x+4 x}+\frac {x^2}{e^x+4 x}\right ) \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx\\ &=-3 x-7 x^2+x^3+x \log \left (e^x+4 x\right )-4 \int \frac {x}{e^x+4 x} \, dx-24 \int \frac {1}{e^x+4 x} \, dx+28 \int \frac {x}{e^x+4 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 32, normalized size = 1.23 \begin {gather*} 3 x-7 x^2+x^3-6 \log \left (e^x+4 x\right )+x \log \left (e^x+4 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24 + 16*x - 56*x^2 + 12*x^3 + E^x*(-3 - 13*x + 3*x^2) + (E^x + 4*x)*Log[E^x + 4*x])/(E^x + 4*x),x]

[Out]

3*x - 7*x^2 + x^3 - 6*Log[E^x + 4*x] + x*Log[E^x + 4*x]

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fricas [A]  time = 1.02, size = 23, normalized size = 0.88 \begin {gather*} x^{3} - 7 \, x^{2} + {\left (x - 6\right )} \log \left (4 \, x + e^{x}\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+exp(x))*log(4*x+exp(x))+(3*x^2-13*x-3)*exp(x)+12*x^3-56*x^2+16*x-24)/(4*x+exp(x)),x, algorithm
="fricas")

[Out]

x^3 - 7*x^2 + (x - 6)*log(4*x + e^x) + 3*x

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giac [A]  time = 0.19, size = 32, normalized size = 1.23 \begin {gather*} x^{3} - 7 \, x^{2} + x \log \left (4 \, x + e^{x}\right ) + 3 \, x - 6 \, \log \left (-4 \, x - e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+exp(x))*log(4*x+exp(x))+(3*x^2-13*x-3)*exp(x)+12*x^3-56*x^2+16*x-24)/(4*x+exp(x)),x, algorithm
="giac")

[Out]

x^3 - 7*x^2 + x*log(4*x + e^x) + 3*x - 6*log(-4*x - e^x)

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maple [A]  time = 0.03, size = 31, normalized size = 1.19

method	result	size
norman	\(x^{3}-6 \ln \left (4 x +{\mathrm e}^{x}\right )+\ln \left (4 x +{\mathrm e}^{x}\right ) x +3 x -7 x^{2}\)	\(31\)
risch	\(x^{3}-6 \ln \left (4 x +{\mathrm e}^{x}\right )+\ln \left (4 x +{\mathrm e}^{x}\right ) x +3 x -7 x^{2}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x+exp(x))*ln(4*x+exp(x))+(3*x^2-13*x-3)*exp(x)+12*x^3-56*x^2+16*x-24)/(4*x+exp(x)),x,method=_RETURNVER
BOSE)

[Out]

x^3-6*ln(4*x+exp(x))+ln(4*x+exp(x))*x+3*x-7*x^2

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maxima [A]  time = 0.37, size = 23, normalized size = 0.88 \begin {gather*} x^{3} - 7 \, x^{2} + {\left (x - 6\right )} \log \left (4 \, x + e^{x}\right ) + 3 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+exp(x))*log(4*x+exp(x))+(3*x^2-13*x-3)*exp(x)+12*x^3-56*x^2+16*x-24)/(4*x+exp(x)),x, algorithm
="maxima")

[Out]

x^3 - 7*x^2 + (x - 6)*log(4*x + e^x) + 3*x

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mupad [B]  time = 4.76, size = 30, normalized size = 1.15 \begin {gather*} 3\,x-6\,\ln \left (4\,x+{\mathrm {e}}^x\right )+x\,\ln \left (4\,x+{\mathrm {e}}^x\right )-7\,x^2+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x - exp(x)*(13*x - 3*x^2 + 3) - 56*x^2 + 12*x^3 + log(4*x + exp(x))*(4*x + exp(x)) - 24)/(4*x + exp(x)
),x)

[Out]

3*x - 6*log(4*x + exp(x)) + x*log(4*x + exp(x)) - 7*x^2 + x^3

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sympy [A]  time = 0.34, size = 31, normalized size = 1.19 \begin {gather*} x^{3} - 7 x^{2} + x \log {\left (4 x + e^{x} \right )} + 3 x - 6 \log {\left (4 x + e^{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x+exp(x))*ln(4*x+exp(x))+(3*x**2-13*x-3)*exp(x)+12*x**3-56*x**2+16*x-24)/(4*x+exp(x)),x)

[Out]

x**3 - 7*x**2 + x*log(4*x + exp(x)) + 3*x - 6*log(4*x + exp(x))

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