\(\int \frac {(20+4 x) \log (\frac {1}{25} e^{2 x} (576+288 x+36 x^2))}{4+x} \, dx\) [5948]

Optimal result

Integrand size = 31, antiderivative size = 19 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=-3+\log ^2\left (\frac {36}{25} e^{2 x} (4+x)^2\right ) \]

[Out]

ln(36/25*(4+x)^2*exp(x)^2)^2-3

Rubi [F]

\[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=\int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=\log ^2\left (\frac {36}{25} e^{2 x} (4+x)^2\right ) \]

[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

Maple [A] (verified)

Time = 1.87 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\({\ln \left (\frac {36 \left (x^{2}+8 x +16\right ) {\mathrm e}^{2 x}}{25}\right )}^{2}\)	\(18\)
default	\({\ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right )}^{2}\)	\(20\)
norman	\({\ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right )}^{2}\)	\(20\)
parts	\(4 \ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right ) \ln \left (4+x \right )+4 \ln \left (\frac {\left (36 x^{2}+288 x +576\right ) {\mathrm e}^{2 x}}{25}\right ) x -4 x^{2}+32 \ln \left (4+x \right )-8 \left (4+x \right ) \ln \left (4+x \right )+32-4 \ln \left (4+x \right )^{2}\)	\(74\)
risch	\(4 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+4 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-2 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}+2 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+2 i \pi x \,\operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{2}+4 \ln \left (4+x \right )^{2}+8 \ln \left (4+x \right ) \ln \left (2\right )+8 \ln \left (4+x \right ) \ln \left (3\right )-8 \ln \left (4+x \right ) \ln \left (5\right )+\left (8 x +8 \ln \left (4+x \right )\right ) \ln \left ({\mathrm e}^{x}\right )-8 x \ln \left (5\right )+8 x \ln \left (2\right )+8 x \ln \left (3\right )-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{3}-2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 i \pi x \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )^{3}-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-2 i \pi x \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{3}-4 x^{2}-2 i \pi x \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )-2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{2 x} \left (4+x \right )^{2}\right )-2 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}-2 i \ln \left (4+x \right ) \pi \operatorname {csgn}\left (i \left (4+x \right )\right )^{2} \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )+4 i \ln \left (4+x \right ) \pi \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{2}-2 i \pi x \operatorname {csgn}\left (i \left (4+x \right )\right )^{2} \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )+4 i \pi x \,\operatorname {csgn}\left (i \left (4+x \right )\right ) \operatorname {csgn}\left (i \left (4+x \right )^{2}\right )^{2}\)	\(560\)

[In]

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

[Out]

ln(36/25*(x^2+8*x+16)*exp(x)^2)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=\log \left (\frac {36}{25} \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (2 \, x\right )}\right )^{2} \]

[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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=\log {\left (\left (\frac {36 x^{2}}{25} + \frac {288 x}{25} + \frac {576}{25}\right ) e^{2 x} \right )}^{2} \]

[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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (16) = 32\).

Time = 0.21 (sec) , antiderivative size = 184, normalized size of antiderivative = 9.68 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=4 \, {\left (x + \log \left (x + 4\right )\right )} \log \left (\frac {36}{25} \, {\left (x^{2} + 8 \, x + 16\right )} e^{\left (2 \, x\right )}\right ) - 8 \, {\left (\frac {x^{2} + 4 \, x - 16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - 72 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (x + 4\right ) - \frac {4 \, {\left (x^{3} - 12 \, x^{2} - 64 \, x + 128\right )}}{x + 4} - \frac {8 \, {\left (4 \, {\left (x + 4\right )} \log \left (x + 4\right )^{2} - x^{2} + 4 \, {\left (x + 4\right )} \log \left (x + 4\right ) - 4 \, x - 16\right )}}{x + 4} + \frac {36 \, {\left ({\left (x + 4\right )} \log \left (x + 4\right )^{2} - 8\right )}}{x + 4} - \frac {72 \, {\left (x^{2} + 4 \, x - 16\right )}}{x + 4} + \frac {160 \, \log \left (x + 4\right )}{x + 4} - \frac {480}{x + 4} + 32 \, \log \left (x + 4\right ) \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (16) = 32\).

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx=4 \, x^{2} + 4 \, x \log \left (\frac {36}{25} \, x^{2} + \frac {288}{25} \, x + \frac {576}{25}\right ) + \log \left (\frac {36}{25} \, x^{2} + \frac {288}{25} \, x + \frac {576}{25}\right )^{2} \]

[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

Mupad [B] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {(20+4 x) \log \left (\frac {1}{25} e^{2 x} \left (576+288 x+36 x^2\right )\right )}{4+x} \, dx={\left (2\,x+\ln \left (\frac {36\,x^2}{25}+\frac {288\,x}{25}+\frac {576}{25}\right )\right )}^2 \]

[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