3.60.48 \(\int \frac {(20+4 x) \log (\frac {1}{25} e^{2 x} (576+288 x+36 x^2))}{4+x} \, dx\)

Optimal. Leaf size=19 \[ -3+\log ^2\left (\frac {36}{25} e^{2 x} (4+x)^2\right ) \]

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Rubi [F]  time = 0.14, 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+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((20 + 4*x)*Log[(E^(2*x)*(576 + 288*x + 36*x^2))/25])/(4 + x),x]

[Out]

-8*x - 4*x^2 + 32*Log[4 + x] + 4*x*Log[(36*E^(2*x)*(4 + x)^2)/25] + 4*Defer[Int][Log[(36*E^(2*x)*(4 + x)^2)/25
]/(4 + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (4 \log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )+\frac {4 \log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )}{4+x}\right ) \, dx\\ &=4 \int \log \left (\frac {36}{25} e^{2 x} (4+x)^2\right ) \, dx+4 \int \frac {\log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )}{4+x} \, dx\\ &=4 x \log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )-4 \int \frac {2 x (5+x)}{4+x} \, dx+4 \int \frac {\log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )}{4+x} \, dx\\ &=4 x \log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )+4 \int \frac {\log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )}{4+x} \, dx-8 \int \frac {x (5+x)}{4+x} \, dx\\ &=4 x \log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )+4 \int \frac {\log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )}{4+x} \, dx-8 \int \left (1+x-\frac {4}{4+x}\right ) \, dx\\ &=-8 x-4 x^2+32 \log (4+x)+4 x \log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )+4 \int \frac {\log \left (\frac {36}{25} e^{2 x} (4+x)^2\right )}{4+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 17, normalized size = 0.89 \begin {gather*} \log ^2\left (\frac {36}{25} e^{2 x} (4+x)^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((20 + 4*x)*Log[(E^(2*x)*(576 + 288*x + 36*x^2))/25])/(4 + x),x]

[Out]

Log[(36*E^(2*x)*(4 + x)^2)/25]^2

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fricas [A]  time = 0.70, size = 17, normalized size = 0.89 \begin {gather*} \log \left (\frac {36}{25} \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (2 \, x\right )}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)*log(1/25*(36*x^2+288*x+576)*exp(x)^2)/(4+x),x, algorithm="fricas")

[Out]

log(36/25*(x^2 + 8*x + 16)*e^(2*x))^2

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giac [B]  time = 0.15, size = 33, normalized size = 1.74 \begin {gather*} 4 \, x^{2} + 4 \, x \log \left (\frac {36}{25} \, x^{2} + \frac {288}{25} \, x + \frac {576}{25}\right ) + \log \left (\frac {36}{25} \, x^{2} + \frac {288}{25} \, x + \frac {576}{25}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)*log(1/25*(36*x^2+288*x+576)*exp(x)^2)/(4+x),x, algorithm="giac")

[Out]

4*x^2 + 4*x*log(36/25*x^2 + 288/25*x + 576/25) + log(36/25*x^2 + 288/25*x + 576/25)^2

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maple [A]  time = 0.70, size = 20, normalized size = 1.05

method	result	size
norman	\(\ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right )^{2}\)	\(20\)
default	\(4 \ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right ) \ln \left (4+x \right )+4 \ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right ) x -4 x^{2}+32 \ln \left (4+x \right )-8 \left (4+x \right ) \ln \left (4+x \right )+32-4 \ln \left (4+x \right )^{2}\)	\(74\)
risch	\(-4 x^{2}-2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (4+x \right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-2 i \pi x \,\mathrm {csgn}\left (i \left (4+x \right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-8 x \ln \relax (5)+8 x \ln \relax (3)+8 x \ln \relax (2)-2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{3}-2 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 i \pi x \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{3}-8 \ln \relax (5) \ln \left (4+x \right )+8 \ln \relax (2) \ln \left (4+x \right )+8 \ln \relax (3) \ln \left (4+x \right )+\left (8 x +8 \ln \left (4+x \right )\right ) \ln \left ({\mathrm e}^{x}\right )+4 \ln \left (4+x \right )^{2}-2 i \pi x \mathrm {csgn}\left (i \left (4+x \right )\right )^{2} \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )+4 i \pi x \,\mathrm {csgn}\left (i \left (4+x \right )\right ) \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{2}+2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (4+x \right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+4 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}+2 i \pi x \,\mathrm {csgn}\left (i \left (4+x \right )^{2}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}-2 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (4+x \right )\right )^{2} \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )+4 i \pi \ln \left (4+x \right ) \mathrm {csgn}\left (i \left (4+x \right )\right ) \mathrm {csgn}\left (i \left (4+x \right )^{2}\right )^{2}+2 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )-2 i \pi x \mathrm {csgn}\left (i {\mathrm e}^{x}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )+4 i \pi x \,\mathrm {csgn}\left (i {\mathrm e}^{x}\right ) \mathrm {csgn}\left (i {\mathrm e}^{2 x}\right )^{2}\)	\(560\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20+4*x)*ln(1/25*(36*x^2+288*x+576)*exp(x)^2)/(4+x),x,method=_RETURNVERBOSE)

[Out]

ln(1/25*(36*x^2+288*x+576)*exp(x)^2)^2

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maxima [B]  time = 0.38, size = 184, normalized size = 9.68 \begin {gather*} 4 \, {\left (x + \log \left (x + 4\right )\right )} \log \left (\frac {36}{25} \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (2 \, x\right )}\right ) - 8 \, {\left (\frac {x^{2} + 4 \, x - 16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - 72 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - \frac {4 \, {\left (x^{3} - 12 \, x^{2} - 64 \, x + 128\right )}}{x + 4} - \frac {8 \, {\left (4 \, {\left (x + 4\right )} \log \left (x + 4\right )^{2} - x^{2} + 4 \, {\left (x + 4\right )} \log \left (x + 4\right ) - 4 \, x - 16\right )}}{x + 4} + \frac {36 \, {\left ({\left (x + 4\right )} \log \left (x + 4\right )^{2} - 8\right )}}{x + 4} - \frac {72 \, {\left (x^{2} + 4 \, x - 16\right )}}{x + 4} + \frac {160 \, \log \left (x + 4\right )}{x + 4} - \frac {480}{x + 4} + 32 \, \log \left (x + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)*log(1/25*(36*x^2+288*x+576)*exp(x)^2)/(4+x),x, algorithm="maxima")

[Out]

4*(x + log(x + 4))*log(36/25*(x^2 + 8*x + 16)*e^(2*x)) - 8*((x^2 + 4*x - 16)/(x + 4) - 8*log(x + 4))*log(x + 4
) - 72*(4/(x + 4) + log(x + 4))*log(x + 4) - 4*(x^3 - 12*x^2 - 64*x + 128)/(x + 4) - 8*(4*(x + 4)*log(x + 4)^2
- x^2 + 4*(x + 4)*log(x + 4) - 4*x - 16)/(x + 4) + 36*((x + 4)*log(x + 4)^2 - 8)/(x + 4) - 72*(x^2 + 4*x - 16
)/(x + 4) + 160*log(x + 4)/(x + 4) - 480/(x + 4) + 32*log(x + 4)

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mupad [B]  time = 0.65, size = 17, normalized size = 0.89 \begin {gather*} {\left (2\,x+\ln \left (\frac {36\,x^2}{25}+\frac {288\,x}{25}+\frac {576}{25}\right )\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((exp(2*x)*(288*x + 36*x^2 + 576))/25)*(4*x + 20))/(x + 4),x)

[Out]

(2*x + log((288*x)/25 + (36*x^2)/25 + 576/25))^2

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sympy [A]  time = 0.16, size = 22, normalized size = 1.16 \begin {gather*} \log {\left (\left (\frac {36 x^{2}}{25} + \frac {288 x}{25} + \frac {576}{25}\right ) e^{2 x} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20+4*x)*ln(1/25*(36*x**2+288*x+576)*exp(x)**2)/(4+x),x)

[Out]

log((36*x**2/25 + 288*x/25 + 576/25)*exp(2*x))**2

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