Integrand size = 56, antiderivative size = 31 \[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=-e^x-x+x^2 \left (1-\log \left (e^x (x+x (4+2 x))\right )\right ) \] Output:
(1-ln(exp(x)*(x+x*(4+2*x))))*x^2-exp(x)-x
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=-e^x-x+x^2-x^2 \log \left (e^x x (5+2 x)\right ) \] Input:
Integrate[(-5 + E^x*(-5 - 2*x) + 3*x - 5*x^2 - 2*x^3 + (-10*x - 4*x^2)*Log [E^x*(5*x + 2*x^2)])/(5 + 2*x),x]
Output:
-E^x - x + x^2 - x^2*Log[E^x*x*(5 + 2*x)]
Time = 0.45 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {-2 x^3-5 x^2+\left (-4 x^2-10 x\right ) \log \left (e^x \left (2 x^2+5 x\right )\right )+3 x+e^x (-2 x-5)-5}{2 x+5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-2 x^3-5 x^2-4 x^2 \log \left (e^x x (2 x+5)\right )+3 x-10 x \log \left (e^x x (2 x+5)\right )-5}{2 x+5}-e^x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2+x^2 \left (-\log \left (e^x x (2 x+5)\right )\right )-x-e^x\) |
Input:
Int[(-5 + E^x*(-5 - 2*x) + 3*x - 5*x^2 - 2*x^3 + (-10*x - 4*x^2)*Log[E^x*( 5*x + 2*x^2)])/(5 + 2*x),x]
Output:
-E^x - x + x^2 - x^2*Log[E^x*x*(5 + 2*x)]
Time = 0.38 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(-\ln \left (x \left (5+2 x \right ) {\mathrm e}^{x}\right ) x^{2}-\frac {5}{4}+x^{2}-x -{\mathrm e}^{x}\) | \(28\) |
default | \(-\ln \left (\left (2 x^{2}+5 x \right ) {\mathrm e}^{x}\right ) x^{2}-x +x^{2}-{\mathrm e}^{x}\) | \(30\) |
norman | \(-\ln \left (\left (2 x^{2}+5 x \right ) {\mathrm e}^{x}\right ) x^{2}-x +x^{2}-{\mathrm e}^{x}\) | \(30\) |
parts | \(-\ln \left (\left (2 x^{2}+5 x \right ) {\mathrm e}^{x}\right ) x^{2}-x +x^{2}-{\mathrm e}^{x}\) | \(30\) |
risch | \(-x^{2} \ln \left ({\mathrm e}^{x}\right )-x^{2} \ln \left (\frac {5}{2}+x \right )-x^{2} \ln \left (x \right )+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i \left (\frac {5}{2}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right )^{3}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right ) \operatorname {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i x \right )}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right ) \operatorname {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right )^{2}}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (\frac {5}{2}+x \right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )^{2}}{2}+\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )^{3}}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i x \left (\frac {5}{2}+x \right ) {\mathrm e}^{x}\right )^{2} \operatorname {csgn}\left (i x \right )}{2}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x} \left (\frac {5}{2}+x \right )\right )^{2}}{2}-x^{2} \ln \left (2\right )+x^{2}-x -{\mathrm e}^{x}\) | \(241\) |
Input:
int(((-4*x^2-10*x)*ln((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x^2+3*x- 5)/(5+2*x),x,method=_RETURNVERBOSE)
Output:
-ln(x*(5+2*x)*exp(x))*x^2-5/4+x^2-x-exp(x)
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=-x^{2} \log \left ({\left (2 \, x^{2} + 5 \, x\right )} e^{x}\right ) + x^{2} - x - e^{x} \] Input:
integrate(((-4*x^2-10*x)*log((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x ^2+3*x-5)/(5+2*x),x, algorithm="fricas")
Output:
-x^2*log((2*x^2 + 5*x)*e^x) + x^2 - x - e^x
Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=x^{2} - \frac {49 x}{24} + \left (\frac {25}{24} - x^{2}\right ) \log {\left (\left (2 x^{2} + 5 x\right ) e^{x} \right )} - e^{x} - \frac {25 \log {\left (2 x^{2} + 5 x \right )}}{24} \] Input:
integrate(((-4*x**2-10*x)*ln((2*x**2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x**3-5 *x**2+3*x-5)/(5+2*x),x)
Output:
x**2 - 49*x/24 + (25/24 - x**2)*log((2*x**2 + 5*x)*exp(x)) - exp(x) - 25*l og(2*x**2 + 5*x)/24
\[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=\int { -\frac {2 \, x^{3} + 5 \, x^{2} + {\left (2 \, x + 5\right )} e^{x} + 2 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \left ({\left (2 \, x^{2} + 5 \, x\right )} e^{x}\right ) - 3 \, x + 5}{2 \, x + 5} \,d x } \] Input:
integrate(((-4*x^2-10*x)*log((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x ^2+3*x-5)/(5+2*x),x, algorithm="maxima")
Output:
-x^3 - x^2*log(x) + x^2 + 5/2*e^(-5/2)*exp_integral_e(1, -x - 5/2) - 1/4*( 4*x^2 - 25)*log(2*x + 5) - x - 2*integrate(x*e^x/(2*x + 5), x) - 25/4*log( 2*x + 5)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=-x^{3} - x^{2} \log \left (2 \, x^{2} + 5 \, x\right ) + x^{2} - x - e^{x} \] Input:
integrate(((-4*x^2-10*x)*log((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x ^2+3*x-5)/(5+2*x),x, algorithm="giac")
Output:
-x^3 - x^2*log(2*x^2 + 5*x) + x^2 - x - e^x
Time = 3.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=x^2-{\mathrm {e}}^x-x^2\,\ln \left ({\mathrm {e}}^x\,\left (2\,x^2+5\,x\right )\right )-x \] Input:
int(-(exp(x)*(2*x + 5) - 3*x + 5*x^2 + 2*x^3 + log(exp(x)*(5*x + 2*x^2))*( 10*x + 4*x^2) + 5)/(2*x + 5),x)
Output:
x^2 - exp(x) - x^2*log(exp(x)*(5*x + 2*x^2)) - x
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.03 \[ \int \frac {-5+e^x (-5-2 x)+3 x-5 x^2-2 x^3+\left (-10 x-4 x^2\right ) \log \left (e^x \left (5 x+2 x^2\right )\right )}{5+2 x} \, dx=-e^{x}-\mathrm {log}\left (2 e^{x} x^{2}+5 e^{x} x \right ) x^{2}-\frac {5 \,\mathrm {log}\left (2 e^{x} x^{2}+5 e^{x} x \right )}{2}+\frac {5 \,\mathrm {log}\left (2 x +5\right )}{2}+\frac {5 \,\mathrm {log}\left (x \right )}{2}+x^{2}+\frac {3 x}{2} \] Input:
int(((-4*x^2-10*x)*log((2*x^2+5*x)*exp(x))+(-2*x-5)*exp(x)-2*x^3-5*x^2+3*x -5)/(5+2*x),x)
Output:
( - 2*e**x - 2*log(2*e**x*x**2 + 5*e**x*x)*x**2 - 5*log(2*e**x*x**2 + 5*e* *x*x) + 5*log(2*x + 5) + 5*log(x) + 2*x**2 + 3*x)/2