Integrand size = 120, antiderivative size = 28 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=\frac {4}{2+e^{16+2 x}}-\left (-x^2+\log (5 x)\right )^2 \] Output:
4/(exp(4)^4*exp(x)^2+2)-(ln(5*x)-x^2)^2
Time = 1.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=\frac {4}{2+e^{2 (8+x)}}-x^4+2 x^2 \log (5 x)-\log ^2(5 x) \] Input:
Integrate[(8*x^2 - 16*x^4 + E^(16 + 2*x)*(-8*x + 8*x^2 - 16*x^4) + E^(32 + 4*x)*(2*x^2 - 4*x^4) + (-8 + 16*x^2 + E^(32 + 4*x)*(-2 + 4*x^2) + E^(16 + 2*x)*(-8 + 16*x^2))*Log[5*x])/(4*x + 4*E^(16 + 2*x)*x + E^(32 + 4*x)*x),x ]
Output:
4/(2 + E^(2*(8 + x))) - x^4 + 2*x^2*Log[5*x] - Log[5*x]^2
Time = 1.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {7292, 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.
\(\displaystyle \int \frac {-16 x^4+8 x^2+\left (16 x^2+e^{4 x+32} \left (4 x^2-2\right )+e^{2 x+16} \left (16 x^2-8\right )-8\right ) \log (5 x)+e^{2 x+16} \left (-16 x^4+8 x^2-8 x\right )+e^{4 x+32} \left (2 x^2-4 x^4\right )}{4 e^{2 x+16} x+e^{4 x+32} x+4 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-16 x^4+8 x^2+\left (16 x^2+e^{4 x+32} \left (4 x^2-2\right )+e^{2 x+16} \left (16 x^2-8\right )-8\right ) \log (5 x)+e^{2 x+16} \left (-16 x^4+8 x^2-8 x\right )+e^{4 x+32} \left (2 x^2-4 x^4\right )}{\left (e^{2 x+16}+2\right )^2 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 \left (2 x^2-1\right ) \left (x^2-\log (5 x)\right )}{x}-\frac {8}{e^{2 x+16}+2}+\frac {16}{\left (e^{2 x+16}+2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4}{e^{2 x+16}+2}-\left (x^2-\log (5 x)\right )^2\) |
Input:
Int[(8*x^2 - 16*x^4 + E^(16 + 2*x)*(-8*x + 8*x^2 - 16*x^4) + E^(32 + 4*x)* (2*x^2 - 4*x^4) + (-8 + 16*x^2 + E^(32 + 4*x)*(-2 + 4*x^2) + E^(16 + 2*x)* (-8 + 16*x^2))*Log[5*x])/(4*x + 4*E^(16 + 2*x)*x + E^(32 + 4*x)*x),x]
Output:
4/(2 + E^(16 + 2*x)) - (x^2 - Log[5*x])^2
Time = 75.60 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
method | result | size |
risch | \(-\ln \left (5 x \right )^{2}+2 \ln \left (5 x \right ) x^{2}-\frac {{\mathrm e}^{16+2 x} x^{4}+2 x^{4}-4}{{\mathrm e}^{16+2 x}+2}\) | \(48\) |
parallelrisch | \(\frac {-2 \,{\mathrm e}^{16} {\mathrm e}^{2 x} x^{4}+4 \,{\mathrm e}^{16} {\mathrm e}^{2 x} \ln \left (5 x \right ) x^{2}-2 \,{\mathrm e}^{16} \ln \left (5 x \right )^{2} {\mathrm e}^{2 x}-4 \,{\mathrm e}^{16} {\mathrm e}^{2 x}-4 x^{4}+8 \ln \left (5 x \right ) x^{2}-4 \ln \left (5 x \right )^{2}}{2 \,{\mathrm e}^{16} {\mathrm e}^{2 x}+4}\) | \(95\) |
Input:
int((((4*x^2-2)*exp(4)^8*exp(x)^4+(16*x^2-8)*exp(4)^4*exp(x)^2+16*x^2-8)*l n(5*x)+(-4*x^4+2*x^2)*exp(4)^8*exp(x)^4+(-16*x^4+8*x^2-8*x)*exp(4)^4*exp(x )^2-16*x^4+8*x^2)/(x*exp(4)^8*exp(x)^4+4*x*exp(4)^4*exp(x)^2+4*x),x,method =_RETURNVERBOSE)
Output:
-ln(5*x)^2+2*ln(5*x)*x^2-(exp(16+2*x)*x^4+2*x^4-4)/(exp(16+2*x)+2)
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.36 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=-\frac {x^{4} e^{\left (2 \, x + 16\right )} + 2 \, x^{4} + {\left (e^{\left (2 \, x + 16\right )} + 2\right )} \log \left (5 \, x\right )^{2} - 2 \, {\left (x^{2} e^{\left (2 \, x + 16\right )} + 2 \, x^{2}\right )} \log \left (5 \, x\right ) - 4}{e^{\left (2 \, x + 16\right )} + 2} \] Input:
integrate((((4*x^2-2)*exp(4)^8*exp(x)^4+(16*x^2-8)*exp(4)^4*exp(x)^2+16*x^ 2-8)*log(5*x)+(-4*x^4+2*x^2)*exp(4)^8*exp(x)^4+(-16*x^4+8*x^2-8*x)*exp(4)^ 4*exp(x)^2-16*x^4+8*x^2)/(x*exp(4)^8*exp(x)^4+4*x*exp(4)^4*exp(x)^2+4*x),x , algorithm="fricas")
Output:
-(x^4*e^(2*x + 16) + 2*x^4 + (e^(2*x + 16) + 2)*log(5*x)^2 - 2*(x^2*e^(2*x + 16) + 2*x^2)*log(5*x) - 4)/(e^(2*x + 16) + 2)
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=- x^{4} + 2 x^{2} \log {\left (5 x \right )} - \log {\left (5 x \right )}^{2} + \frac {4}{e^{16} e^{2 x} + 2} \] Input:
integrate((((4*x**2-2)*exp(4)**8*exp(x)**4+(16*x**2-8)*exp(4)**4*exp(x)**2 +16*x**2-8)*ln(5*x)+(-4*x**4+2*x**2)*exp(4)**8*exp(x)**4+(-16*x**4+8*x**2- 8*x)*exp(4)**4*exp(x)**2-16*x**4+8*x**2)/(x*exp(4)**8*exp(x)**4+4*x*exp(4) **4*exp(x)**2+4*x),x)
Output:
-x**4 + 2*x**2*log(5*x) - log(5*x)**2 + 4/(exp(16)*exp(2*x) + 2)
Leaf count of result is larger than twice the leaf count of optimal. 89 vs. \(2 (27) = 54\).
Time = 0.15 (sec) , antiderivative size = 89, normalized size of antiderivative = 3.18 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=-\frac {2 \, x^{4} - 4 \, x^{2} \log \left (5\right ) + {\left (x^{4} e^{16} - 2 \, x^{2} e^{16} \log \left (5\right ) + e^{16} \log \left (x\right )^{2} - 2 \, {\left (x^{2} e^{16} - e^{16} \log \left (5\right )\right )} \log \left (x\right )\right )} e^{\left (2 \, x\right )} - 4 \, {\left (x^{2} - \log \left (5\right )\right )} \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 4}{e^{\left (2 \, x + 16\right )} + 2} \] Input:
integrate((((4*x^2-2)*exp(4)^8*exp(x)^4+(16*x^2-8)*exp(4)^4*exp(x)^2+16*x^ 2-8)*log(5*x)+(-4*x^4+2*x^2)*exp(4)^8*exp(x)^4+(-16*x^4+8*x^2-8*x)*exp(4)^ 4*exp(x)^2-16*x^4+8*x^2)/(x*exp(4)^8*exp(x)^4+4*x*exp(4)^4*exp(x)^2+4*x),x , algorithm="maxima")
Output:
-(2*x^4 - 4*x^2*log(5) + (x^4*e^16 - 2*x^2*e^16*log(5) + e^16*log(x)^2 - 2 *(x^2*e^16 - e^16*log(5))*log(x))*e^(2*x) - 4*(x^2 - log(5))*log(x) + 2*lo g(x)^2 - 4)/(e^(2*x + 16) + 2)
Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (27) = 54\).
Time = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.71 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=-\frac {x^{4} e^{\left (2 \, x + 16\right )} + 2 \, x^{4} - 2 \, x^{2} e^{\left (2 \, x + 16\right )} \log \left (5\right ) - 2 \, x^{2} e^{\left (2 \, x + 16\right )} \log \left (x\right ) - 4 \, x^{2} \log \left (5\right ) - 4 \, x^{2} \log \left (x\right ) + 2 \, e^{\left (2 \, x + 16\right )} \log \left (5\right ) \log \left (x\right ) + e^{\left (2 \, x + 16\right )} \log \left (x\right )^{2} + 4 \, \log \left (5\right ) \log \left (x\right ) + 2 \, \log \left (x\right )^{2} - 4}{e^{\left (2 \, x + 16\right )} + 2} \] Input:
integrate((((4*x^2-2)*exp(4)^8*exp(x)^4+(16*x^2-8)*exp(4)^4*exp(x)^2+16*x^ 2-8)*log(5*x)+(-4*x^4+2*x^2)*exp(4)^8*exp(x)^4+(-16*x^4+8*x^2-8*x)*exp(4)^ 4*exp(x)^2-16*x^4+8*x^2)/(x*exp(4)^8*exp(x)^4+4*x*exp(4)^4*exp(x)^2+4*x),x , algorithm="giac")
Output:
-(x^4*e^(2*x + 16) + 2*x^4 - 2*x^2*e^(2*x + 16)*log(5) - 2*x^2*e^(2*x + 16 )*log(x) - 4*x^2*log(5) - 4*x^2*log(x) + 2*e^(2*x + 16)*log(5)*log(x) + e^ (2*x + 16)*log(x)^2 + 4*log(5)*log(x) + 2*log(x)^2 - 4)/(e^(2*x + 16) + 2)
Time = 3.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=2\,x^2\,\ln \left (5\,x\right )+\frac {4\,{\mathrm {e}}^{-16}}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^{-16}}-{\ln \left (5\,x\right )}^2-x^4 \] Input:
int((log(5*x)*(16*x^2 + exp(2*x)*exp(16)*(16*x^2 - 8) + exp(4*x)*exp(32)*( 4*x^2 - 2) - 8) + 8*x^2 - 16*x^4 - exp(2*x)*exp(16)*(8*x - 8*x^2 + 16*x^4) + exp(4*x)*exp(32)*(2*x^2 - 4*x^4))/(4*x + 4*x*exp(2*x)*exp(16) + x*exp(4 *x)*exp(32)),x)
Output:
2*x^2*log(5*x) + (4*exp(-16))/(exp(2*x) + 2*exp(-16)) - log(5*x)^2 - x^4
Time = 0.17 (sec) , antiderivative size = 93, normalized size of antiderivative = 3.32 \[ \int \frac {8 x^2-16 x^4+e^{16+2 x} \left (-8 x+8 x^2-16 x^4\right )+e^{32+4 x} \left (2 x^2-4 x^4\right )+\left (-8+16 x^2+e^{32+4 x} \left (-2+4 x^2\right )+e^{16+2 x} \left (-8+16 x^2\right )\right ) \log (5 x)}{4 x+4 e^{16+2 x} x+e^{32+4 x} x} \, dx=\frac {-e^{2 x} \mathrm {log}\left (5 x \right )^{2} e^{16}+2 e^{2 x} \mathrm {log}\left (5 x \right ) e^{16} x^{2}-e^{2 x} e^{16} x^{4}-2 e^{2 x} e^{16}-2 \mathrm {log}\left (5 x \right )^{2}+4 \,\mathrm {log}\left (5 x \right ) x^{2}-2 x^{4}}{e^{2 x} e^{16}+2} \] Input:
int((((4*x^2-2)*exp(4)^8*exp(x)^4+(16*x^2-8)*exp(4)^4*exp(x)^2+16*x^2-8)*l og(5*x)+(-4*x^4+2*x^2)*exp(4)^8*exp(x)^4+(-16*x^4+8*x^2-8*x)*exp(4)^4*exp( x)^2-16*x^4+8*x^2)/(x*exp(4)^8*exp(x)^4+4*x*exp(4)^4*exp(x)^2+4*x),x)
Output:
( - e**(2*x)*log(5*x)**2*e**16 + 2*e**(2*x)*log(5*x)*e**16*x**2 - e**(2*x) *e**16*x**4 - 2*e**(2*x)*e**16 - 2*log(5*x)**2 + 4*log(5*x)*x**2 - 2*x**4) /(e**(2*x)*e**16 + 2)