Integrand size = 80, antiderivative size = 22 \[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=2 e^{x-(3+x)^x} (1-2 x)-3 x \] Output:
2*exp(-exp(x*ln(3+x))+ln(1-2*x)+x)-3*x
Timed out. \[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=\text {\$Aborted} \] Input:
Integrate[(9 - 15*x - 6*x^2 + E^(x - (3 + x)^x)*(1 - 2*x)*(6 + 14*x + 4*x^ 2 + (3 + x)^x*(2*x - 4*x^2 + (6 - 10*x - 4*x^2)*Log[3 + x])))/(-3 + 5*x + 2*x^2),x]
Output:
$Aborted
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {-6 x^2+e^{x-(x+3)^x} (1-2 x) \left (4 x^2+(x+3)^x \left (-4 x^2+\left (-4 x^2-10 x+6\right ) \log (x+3)+2 x\right )+14 x+6\right )-15 x+9}{2 x^2+5 x-3} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (-\frac {6 x^2}{2 x^2+5 x-3}-\frac {15 x}{2 x^2+5 x-3}-\frac {4 e^{x-(x+3)^x} x^2}{x+3}-\frac {6 e^{x-(x+3)^x}}{x+3}-\frac {14 e^{x-(x+3)^x} x}{x+3}+\frac {9}{(2 x-1) (x+3)}+2 e^{x-(x+3)^x} (2 x-1) (x+3)^{x-1} (x+x \log (x+3)+3 \log (x+3))\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int e^{x-(x+3)^x} x^2 (x+3)^{x-1}dx-4 \int \frac {\int e^{x-(x+3)^x} x^2 (x+3)^{x-1}dx}{x+3}dx+4 \log (x+3) \int e^{x-(x+3)^x} x^2 (x+3)^{x-1}dx-2 \int e^{x-(x+3)^x}dx-4 \int e^{x-(x+3)^x} xdx-2 \int e^{x-(x+3)^x} x (x+3)^{x-1}dx+6 \int \frac {\int e^{x-(x+3)^x} (x+3)^{x-1}dx}{x+3}dx-10 \int \frac {\int e^{x-(x+3)^x} x (x+3)^{x-1}dx}{x+3}dx-6 \log (x+3) \int e^{x-(x+3)^x} (x+3)^{x-1}dx+10 \log (x+3) \int e^{x-(x+3)^x} x (x+3)^{x-1}dx-3 x\) |
Input:
Int[(9 - 15*x - 6*x^2 + E^(x - (3 + x)^x)*(1 - 2*x)*(6 + 14*x + 4*x^2 + (3 + x)^x*(2*x - 4*x^2 + (6 - 10*x - 4*x^2)*Log[3 + x])))/(-3 + 5*x + 2*x^2) ,x]
Output:
$Aborted
Time = 5.98 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \(2 \left (1-2 x \right ) {\mathrm e}^{-\left (3+x \right )^{x}+x}-3 x\) | \(22\) |
parallelrisch | \(\frac {39}{20}-3 x +2 \,{\mathrm e}^{-{\mathrm e}^{x \ln \left (3+x \right )}+\ln \left (1-2 x \right )+x}\) | \(26\) |
Input:
int(((((-4*x^2-10*x+6)*ln(3+x)-4*x^2+2*x)*exp(x*ln(3+x))+4*x^2+14*x+6)*exp (-exp(x*ln(3+x))+ln(1-2*x)+x)-6*x^2-15*x+9)/(2*x^2+5*x-3),x,method=_RETURN VERBOSE)
Output:
2*(1-2*x)*exp(-(3+x)^x+x)-3*x
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=-3 \, x + 2 \, e^{\left (-{\left (x + 3\right )}^{x} + x + \log \left (-2 \, x + 1\right )\right )} \] Input:
integrate(((((-4*x^2-10*x+6)*log(3+x)-4*x^2+2*x)*exp(x*log(3+x))+4*x^2+14* x+6)*exp(-exp(x*log(3+x))+log(1-2*x)+x)-6*x^2-15*x+9)/(2*x^2+5*x-3),x, alg orithm="fricas")
Output:
-3*x + 2*e^(-(x + 3)^x + x + log(-2*x + 1))
Time = 3.90 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=- 3 x + \left (2 - 4 x\right ) e^{x - e^{x \log {\left (x + 3 \right )}}} \] Input:
integrate(((((-4*x**2-10*x+6)*ln(3+x)-4*x**2+2*x)*exp(x*ln(3+x))+4*x**2+14 *x+6)*exp(-exp(x*ln(3+x))+ln(1-2*x)+x)-6*x**2-15*x+9)/(2*x**2+5*x-3),x)
Output:
-3*x + (2 - 4*x)*exp(x - exp(x*log(x + 3)))
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=-2 \, {\left (2 \, x - 1\right )} e^{\left (-{\left (x + 3\right )}^{x} + x\right )} - 3 \, x \] Input:
integrate(((((-4*x^2-10*x+6)*log(3+x)-4*x^2+2*x)*exp(x*log(3+x))+4*x^2+14* x+6)*exp(-exp(x*log(3+x))+log(1-2*x)+x)-6*x^2-15*x+9)/(2*x^2+5*x-3),x, alg orithm="maxima")
Output:
-2*(2*x - 1)*e^(-(x + 3)^x + x) - 3*x
\[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=\int { -\frac {6 \, x^{2} + 2 \, {\left ({\left (2 \, x^{2} + {\left (2 \, x^{2} + 5 \, x - 3\right )} \log \left (x + 3\right ) - x\right )} {\left (x + 3\right )}^{x} - 2 \, x^{2} - 7 \, x - 3\right )} e^{\left (-{\left (x + 3\right )}^{x} + x + \log \left (-2 \, x + 1\right )\right )} + 15 \, x - 9}{2 \, x^{2} + 5 \, x - 3} \,d x } \] Input:
integrate(((((-4*x^2-10*x+6)*log(3+x)-4*x^2+2*x)*exp(x*log(3+x))+4*x^2+14* x+6)*exp(-exp(x*log(3+x))+log(1-2*x)+x)-6*x^2-15*x+9)/(2*x^2+5*x-3),x, alg orithm="giac")
Output:
integrate(-(6*x^2 + 2*((2*x^2 + (2*x^2 + 5*x - 3)*log(x + 3) - x)*(x + 3)^ x - 2*x^2 - 7*x - 3)*e^(-(x + 3)^x + x + log(-2*x + 1)) + 15*x - 9)/(2*x^2 + 5*x - 3), x)
Time = 0.60 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=2\,{\mathrm {e}}^{-{\left (x+3\right )}^x}\,{\mathrm {e}}^x-3\,x-4\,x\,{\mathrm {e}}^{-{\left (x+3\right )}^x}\,{\mathrm {e}}^x \] Input:
int(-(15*x + 6*x^2 - exp(x - exp(x*log(x + 3)) + log(1 - 2*x))*(14*x - exp (x*log(x + 3))*(log(x + 3)*(10*x + 4*x^2 - 6) - 2*x + 4*x^2) + 4*x^2 + 6) - 9)/(5*x + 2*x^2 - 3),x)
Output:
2*exp(-(x + 3)^x)*exp(x) - 3*x - 4*x*exp(-(x + 3)^x)*exp(x)
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {9-15 x-6 x^2+e^{x-(3+x)^x} (1-2 x) \left (6+14 x+4 x^2+(3+x)^x \left (2 x-4 x^2+\left (6-10 x-4 x^2\right ) \log (3+x)\right )\right )}{-3+5 x+2 x^2} \, dx=\frac {-3 e^{\left (x +3\right )^{x}} x -4 e^{x} x +2 e^{x}}{e^{\left (x +3\right )^{x}}} \] Input:
int(((((-4*x^2-10*x+6)*log(3+x)-4*x^2+2*x)*exp(x*log(3+x))+4*x^2+14*x+6)*e xp(-exp(x*log(3+x))+log(1-2*x)+x)-6*x^2-15*x+9)/(2*x^2+5*x-3),x)
Output:
( - 3*e**((x + 3)**x)*x - 4*e**x*x + 2*e**x)/e**((x + 3)**x)