\(\int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log (2+e^{15+x})+(-18-9 e^{15+x}) \log ^2(2+e^{15+x})}{16 x^2+8 e^{15+x} x^2} \, dx\) [735]

Optimal result

Integrand size = 64, antiderivative size = 24 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {18 \left (2+\frac {1}{16} \left (-3+x+\log ^2\left (2+e^{15+x}\right )\right )\right )}{x} \]

[Out]

18*(29/16+1/16*ln(exp(x+15)+2)^2+1/16*x)/x

Rubi [F]

\[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx \]

[In]

Int[(-522 - 261*E^(15 + x) + 18*E^(15 + x)*x*Log[2 + E^(15 + x)] + (-18 - 9*E^(15 + x))*Log[2 + E^(15 + x)]^2)
/(16*x^2 + 8*E^(15 + x)*x^2),x]

[Out]

261/(8*x) - (9*Log[2 + E^(15 + x)]*Defer[Int][1/((2 + E^(15 + x))*x), x])/2 + (9*Defer[Int][Log[2 + E^(15 + x)
]/x, x])/4 - (9*Defer[Int][Log[2 + E^(15 + x)]^2/x^2, x])/8 + (9*Defer[Int][(E^(15 + x)*Defer[Int][(2*x + E^(1
5 + x)*x)^(-1), x])/(2 + E^(15 + x)), x])/2

Rubi steps \begin{align*} \text {integral}& = \int \frac {9 \left (-29+\frac {2 e^{15+x} x \log \left (2+e^{15+x}\right )}{2+e^{15+x}}-\log ^2\left (2+e^{15+x}\right )\right )}{8 x^2} \, dx \\ & = \frac {9}{8} \int \frac {-29+\frac {2 e^{15+x} x \log \left (2+e^{15+x}\right )}{2+e^{15+x}}-\log ^2\left (2+e^{15+x}\right )}{x^2} \, dx \\ & = \frac {9}{8} \int \left (-\frac {4 \log \left (2+e^{15+x}\right )}{\left (2+e^{15+x}\right ) x}+\frac {-29+2 x \log \left (2+e^{15+x}\right )-\log ^2\left (2+e^{15+x}\right )}{x^2}\right ) \, dx \\ & = \frac {9}{8} \int \frac {-29+2 x \log \left (2+e^{15+x}\right )-\log ^2\left (2+e^{15+x}\right )}{x^2} \, dx-\frac {9}{2} \int \frac {\log \left (2+e^{15+x}\right )}{\left (2+e^{15+x}\right ) x} \, dx \\ & = \frac {9}{8} \int \left (-\frac {29}{x^2}+\frac {2 \log \left (2+e^{15+x}\right )}{x}-\frac {\log ^2\left (2+e^{15+x}\right )}{x^2}\right ) \, dx+\frac {9}{2} \int \frac {e^{15+x} \int \frac {1}{2 x+e^{15+x} x} \, dx}{2+e^{15+x}} \, dx-\frac {1}{2} \left (9 \log \left (2+e^{15+x}\right )\right ) \int \frac {1}{\left (2+e^{15+x}\right ) x} \, dx \\ & = \frac {261}{8 x}-\frac {9}{8} \int \frac {\log ^2\left (2+e^{15+x}\right )}{x^2} \, dx+\frac {9}{4} \int \frac {\log \left (2+e^{15+x}\right )}{x} \, dx+\frac {9}{2} \int \frac {e^{15+x} \int \frac {1}{2 x+e^{15+x} x} \, dx}{2+e^{15+x}} \, dx-\frac {1}{2} \left (9 \log \left (2+e^{15+x}\right )\right ) \int \frac {1}{\left (2+e^{15+x}\right ) x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9}{8} \left (\frac {29}{x}+\frac {\log ^2\left (2+e^{15+x}\right )}{x}\right ) \]

[In]

Integrate[(-522 - 261*E^(15 + x) + 18*E^(15 + x)*x*Log[2 + E^(15 + x)] + (-18 - 9*E^(15 + x))*Log[2 + E^(15 +
x)]^2)/(16*x^2 + 8*E^(15 + x)*x^2),x]

[Out]

(9*(29/x + Log[2 + E^(15 + x)]^2/x))/8

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
norman	\(\frac {\frac {261}{8}+\frac {9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8}}{x}\)	\(18\)
parallelrisch	\(\frac {261+9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8 x}\)	\(19\)
risch	\(\frac {9 \ln \left ({\mathrm e}^{x +15}+2\right )^{2}}{8 x}+\frac {261}{8 x}\)	\(21\)

[In]

int(((-9*exp(x+15)-18)*ln(exp(x+15)+2)^2+18*x*exp(x+15)*ln(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(x+15)+16
*x^2),x,method=_RETURNVERBOSE)

[Out]

(261/8+9/8*ln(exp(x+15)+2)^2)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \]

[In]

integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(
x+15)+16*x^2),x, algorithm="fricas")

[Out]

9/8*(log(e^(x + 15) + 2)^2 + 29)/x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \log {\left (e^{x + 15} + 2 \right )}^{2}}{8 x} + \frac {261}{8 x} \]

[In]

integrate(((-9*exp(x+15)-18)*ln(exp(x+15)+2)**2+18*x*exp(x+15)*ln(exp(x+15)+2)-261*exp(x+15)-522)/(8*x**2*exp(
x+15)+16*x**2),x)

[Out]

9*log(exp(x + 15) + 2)**2/(8*x) + 261/(8*x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \]

[In]

integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(
x+15)+16*x^2),x, algorithm="maxima")

[Out]

9/8*(log(e^(x + 15) + 2)^2 + 29)/x

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9 \, {\left (\log \left (e^{\left (x + 15\right )} + 2\right )^{2} + 29\right )}}{8 \, x} \]

[In]

integrate(((-9*exp(x+15)-18)*log(exp(x+15)+2)^2+18*x*exp(x+15)*log(exp(x+15)+2)-261*exp(x+15)-522)/(8*x^2*exp(
x+15)+16*x^2),x, algorithm="giac")

[Out]

9/8*(log(e^(x + 15) + 2)^2 + 29)/x

Mupad [B] (verification not implemented)

Time = 8.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-522-261 e^{15+x}+18 e^{15+x} x \log \left (2+e^{15+x}\right )+\left (-18-9 e^{15+x}\right ) \log ^2\left (2+e^{15+x}\right )}{16 x^2+8 e^{15+x} x^2} \, dx=\frac {9\,\left ({\ln \left ({\mathrm {e}}^{15}\,{\mathrm {e}}^x+2\right )}^2+29\right )}{8\,x} \]

[In]

int(-(261*exp(x + 15) + log(exp(x + 15) + 2)^2*(9*exp(x + 15) + 18) - 18*x*log(exp(x + 15) + 2)*exp(x + 15) +
522)/(8*x^2*exp(x + 15) + 16*x^2),x)

[Out]

(9*(log(exp(15)*exp(x) + 2)^2 + 29))/(8*x)