\(\int \frac {e^x (-x^2+x^3)+e^{\frac {3 x^2+30 \log (x)}{x}} (-120-12 x^2+4 x^3+120 \log (x))}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx\) [5004]

Optimal result

Integrand size = 79, antiderivative size = 30 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {x^2}{2}-\log \left (e^x+4 e^{3 x+\frac {30 \log (x)}{x}}\right ) \]

[Out]

1/2*x^2-ln(exp(x)+4*exp(30*ln(x)/x+3*x))

Rubi [F]

\[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx \]

[In]

Int[(E^x*(-x^2 + x^3) + E^((3*x^2 + 30*Log[x])/x)*(-120 - 12*x^2 + 4*x^3 + 120*Log[x]))/(E^x*x^2 + 4*E^((3*x^2
+ 30*Log[x])/x)*x^2),x]

[Out]

-3*x + x^2/2 - (30*Log[x])/x + 2*Defer[Int][(1 + 4*E^(2*x)*x^(30/x))^(-1), x] + 30*Defer[Int][1/(x^2*(1 + 4*E^
(2*x)*x^(30/x))), x] - 30*Log[x]*Defer[Int][1/(x^2*(1 + 4*E^(2*x)*x^(30/x))), x] + 30*Defer[Int][Defer[Int][(x
^2 + 4*E^(2*x)*x^(2 + 30/x))^(-1), x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-x^2+x^3+4 e^{2 x} x^{30/x} \left (-30-3 x^2+x^3+30 \log (x)\right )}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx \\ & = \int \left (\frac {2 \left (15+x^2-15 \log (x)\right )}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )}+\frac {-30-3 x^2+x^3+30 \log (x)}{x^2}\right ) \, dx \\ & = 2 \int \frac {15+x^2-15 \log (x)}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+\int \frac {-30-3 x^2+x^3+30 \log (x)}{x^2} \, dx \\ & = 2 \int \left (\frac {1}{1+4 e^{2 x} x^{30/x}}+\frac {15}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )}-\frac {15 \log (x)}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )}\right ) \, dx+\int \left (\frac {-30-3 x^2+x^3}{x^2}+\frac {30 \log (x)}{x^2}\right ) \, dx \\ & = 2 \int \frac {1}{1+4 e^{2 x} x^{30/x}} \, dx+30 \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+30 \int \frac {\log (x)}{x^2} \, dx-30 \int \frac {\log (x)}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+\int \frac {-30-3 x^2+x^3}{x^2} \, dx \\ & = -\frac {30}{x}-\frac {30 \log (x)}{x}+2 \int \frac {1}{1+4 e^{2 x} x^{30/x}} \, dx+30 \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+30 \int \frac {\int \frac {1}{x^2+4 e^{2 x} x^{2+\frac {30}{x}}} \, dx}{x} \, dx-(30 \log (x)) \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+\int \left (-3-\frac {30}{x^2}+x\right ) \, dx \\ & = -3 x+\frac {x^2}{2}-\frac {30 \log (x)}{x}+2 \int \frac {1}{1+4 e^{2 x} x^{30/x}} \, dx+30 \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx+30 \int \frac {\int \frac {1}{x^2+4 e^{2 x} x^{2+\frac {30}{x}}} \, dx}{x} \, dx-(30 \log (x)) \int \frac {1}{x^2 \left (1+4 e^{2 x} x^{30/x}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=-x+\frac {x^2}{2}-\log \left (1+4 e^{2 x} x^{30/x}\right ) \]

[In]

Integrate[(E^x*(-x^2 + x^3) + E^((3*x^2 + 30*Log[x])/x)*(-120 - 12*x^2 + 4*x^3 + 120*Log[x]))/(E^x*x^2 + 4*E^(
(3*x^2 + 30*Log[x])/x)*x^2),x]

[Out]

-x + x^2/2 - Log[1 + 4*E^(2*x)*x^(30/x)]

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97

method	result	size
parallelrisch	\(\frac {x^{2}}{2}-\ln \left ({\mathrm e}^{x}+4 \,{\mathrm e}^{\frac {30 \ln \left (x \right )+3 x^{2}}{x}}\right )\)	\(29\)
risch	\(-\frac {30 \ln \left (x \right )}{x}-3 x +\frac {x^{2}}{2}+\frac {30 \ln \left (x \right )+3 x^{2}}{x}-\ln \left (\frac {{\mathrm e}^{x}}{4}+x^{\frac {30}{x}} {\mathrm e}^{3 x}\right )\)	\(51\)

[In]

int(((120*ln(x)+4*x^3-12*x^2-120)*exp((30*ln(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*ln(x)+3*x^2)/x)+exp
(x)*x^2),x,method=_RETURNVERBOSE)

[Out]

1/2*x^2-ln(4*exp(3*(x^2+10*ln(x))/x)+exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {1}{2} \, x^{2} - \log \left (e^{x} + 4 \, e^{\left (\frac {3 \, {\left (x^{2} + 10 \, \log \left (x\right )\right )}}{x}\right )}\right ) \]

[In]

integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*log(x)+3*x^
2)/x)+exp(x)*x^2),x, algorithm="fricas")

[Out]

1/2*x^2 - log(e^x + 4*e^(3*(x^2 + 10*log(x))/x))

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {x^{2}}{2} - \log {\left (\frac {e^{x}}{4} + e^{\frac {3 x^{2} + 30 \log {\left (x \right )}}{x}} \right )} \]

[In]

integrate(((120*ln(x)+4*x**3-12*x**2-120)*exp((30*ln(x)+3*x**2)/x)+(x**3-x**2)*exp(x))/(4*x**2*exp((30*ln(x)+3
*x**2)/x)+exp(x)*x**2),x)

[Out]

x**2/2 - log(exp(x)/4 + exp((3*x**2 + 30*log(x))/x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {1}{2} \, x^{2} - 3 \, x - \log \left (\frac {1}{4} \, {\left (4 \, e^{\left (2 \, x + \frac {30 \, \log \left (x\right )}{x}\right )} + 1\right )} e^{\left (-2 \, x\right )}\right ) \]

[In]

integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*log(x)+3*x^
2)/x)+exp(x)*x^2),x, algorithm="maxima")

[Out]

1/2*x^2 - 3*x - log(1/4*(4*e^(2*x + 30*log(x)/x) + 1)*e^(-2*x))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {1}{2} \, x^{2} - \log \left (e^{x} + 4 \, e^{\left (\frac {3 \, {\left (x^{2} + 10 \, \log \left (x\right )\right )}}{x}\right )}\right ) \]

[In]

integrate(((120*log(x)+4*x^3-12*x^2-120)*exp((30*log(x)+3*x^2)/x)+(x^3-x^2)*exp(x))/(4*x^2*exp((30*log(x)+3*x^
2)/x)+exp(x)*x^2),x, algorithm="giac")

[Out]

1/2*x^2 - log(e^x + 4*e^(3*(x^2 + 10*log(x))/x))

Mupad [B] (verification not implemented)

Time = 11.49 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^x \left (-x^2+x^3\right )+e^{\frac {3 x^2+30 \log (x)}{x}} \left (-120-12 x^2+4 x^3+120 \log (x)\right )}{e^x x^2+4 e^{\frac {3 x^2+30 \log (x)}{x}} x^2} \, dx=\frac {x^2}{2}-\ln \left (\frac {{\mathrm {e}}^x}{4}+x^{30/x}\,{\mathrm {e}}^{3\,x}\right ) \]

[In]

int(-(exp(x)*(x^2 - x^3) - exp((30*log(x) + 3*x^2)/x)*(120*log(x) - 12*x^2 + 4*x^3 - 120))/(x^2*exp(x) + 4*x^2
*exp((30*log(x) + 3*x^2)/x)),x)

[Out]

x^2/2 - log(exp(x)/4 + x^(30/x)*exp(3*x))