Integrand size = 40, antiderivative size = 30 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=\log \left (\frac {e^{4+e^{-3+x+\left (4-x+x^2\right ) \log (4)}-4 (4+x)}}{x}\right ) \] Output:
ln(exp(-12-4*x+exp(2*(x^2-x+4)*ln(2)+x-3))/x)
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=-4 x+\frac {2^{7+2 x^2} e^{-3+x (1-\log (4))} \log (16)}{\log (4)}-\log (x) \] Input:
Integrate[(-1 - 4*x + E^(-3 + x + (4 - x + x^2)*Log[4])*(x + (-x + 2*x^2)* Log[4]))/x,x]
Output:
-4*x + (2^(7 + 2*x^2)*E^(-3 + x*(1 - Log[4]))*Log[16])/Log[4] - Log[x]
Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2010, 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.
\(\displaystyle \int \frac {e^{\left (x^2-x+4\right ) \log (4)+x-3} \left (\left (2 x^2-x\right ) \log (4)+x\right )-4 x-1}{x} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {-4 x-1}{x}-4^{x^2-x+4} e^{x-3} (x (-\log (16))-1+\log (4))\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2^{2 x^2+7} \log (16) e^{x (1-\log (4))-3}}{\log (4)}-4 x-\log (x)\) |
Input:
Int[(-1 - 4*x + E^(-3 + x + (4 - x + x^2)*Log[4])*(x + (-x + 2*x^2)*Log[4] ))/x,x]
Output:
-4*x + (2^(7 + 2*x^2)*E^(-3 + x*(1 - Log[4]))*Log[16])/Log[4] - Log[x]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83
method | result | size |
norman | \(-4 x +{\mathrm e}^{2 \left (x^{2}-x +4\right ) \ln \left (2\right )+x -3}-\ln \left (x \right )\) | \(25\) |
parallelrisch | \(-4 x +{\mathrm e}^{2 \left (x^{2}-x +4\right ) \ln \left (2\right )+x -3}-\ln \left (x \right )\) | \(25\) |
risch | \(-4 x +2^{2 x^{2}-2 x +8} {\mathrm e}^{-3+x}-\ln \left (x \right )\) | \(26\) |
parts | \({\mathrm e}^{2 x^{2} \ln \left (2\right )+\left (-2 \ln \left (2\right )+1\right ) x +8 \ln \left (2\right )-3}-4 x -\ln \left (x \right )\) | \(31\) |
default | \({\mathrm e}^{2 x^{2} \ln \left (2\right )+\left (-2 \ln \left (2\right )+1\right ) x +8 \ln \left (2\right )-3}+\frac {i \left (-2 \ln \left (2\right )+1\right ) \sqrt {\pi }\, {\mathrm e}^{8 \ln \left (2\right )-3-\frac {\left (-2 \ln \left (2\right )+1\right )^{2}}{8 \ln \left (2\right )}} \sqrt {2}\, \operatorname {erf}\left (i \sqrt {2}\, \sqrt {\ln \left (2\right )}\, x +\frac {i \left (-2 \ln \left (2\right )+1\right ) \sqrt {2}}{4 \sqrt {\ln \left (2\right )}}\right )}{4 \sqrt {\ln \left (2\right )}}+\frac {i \sqrt {\ln \left (2\right )}\, \sqrt {\pi }\, {\mathrm e}^{8 \ln \left (2\right )-3-\frac {\left (-2 \ln \left (2\right )+1\right )^{2}}{8 \ln \left (2\right )}} \sqrt {2}\, \operatorname {erf}\left (i \sqrt {2}\, \sqrt {\ln \left (2\right )}\, x +\frac {i \left (-2 \ln \left (2\right )+1\right ) \sqrt {2}}{4 \sqrt {\ln \left (2\right )}}\right )}{2}-\frac {i \sqrt {\pi }\, {\mathrm e}^{8 \ln \left (2\right )-3-\frac {\left (-2 \ln \left (2\right )+1\right )^{2}}{8 \ln \left (2\right )}} \sqrt {2}\, \operatorname {erf}\left (i \sqrt {2}\, \sqrt {\ln \left (2\right )}\, x +\frac {i \left (-2 \ln \left (2\right )+1\right ) \sqrt {2}}{4 \sqrt {\ln \left (2\right )}}\right )}{4 \sqrt {\ln \left (2\right )}}-4 x -\ln \left (x \right )\) | \(226\) |
Input:
int(((2*(2*x^2-x)*ln(2)+x)*exp(2*(x^2-x+4)*ln(2)+x-3)-4*x-1)/x,x,method=_R ETURNVERBOSE)
Output:
-4*x+exp(2*(x^2-x+4)*ln(2)+x-3)-ln(x)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=-4 \, x + e^{\left (2 \, {\left (x^{2} - x + 4\right )} \log \left (2\right ) + x - 3\right )} - \log \left (x\right ) \] Input:
integrate(((2*(2*x^2-x)*log(2)+x)*exp(2*(x^2-x+4)*log(2)+x-3)-4*x-1)/x,x, algorithm="fricas")
Output:
-4*x + e^(2*(x^2 - x + 4)*log(2) + x - 3) - log(x)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=- 4 x + e^{x + \left (2 x^{2} - 2 x + 8\right ) \log {\left (2 \right )} - 3} - \log {\left (x \right )} \] Input:
integrate(((2*(2*x**2-x)*ln(2)+x)*exp(2*(x**2-x+4)*ln(2)+x-3)-4*x-1)/x,x)
Output:
-4*x + exp(x + (2*x**2 - 2*x + 8)*log(2) - 3) - log(x)
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.15 (sec) , antiderivative size = 269, normalized size of antiderivative = 8.97 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=-\frac {128 \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} x \sqrt {-\log \left (2\right )} + \frac {\sqrt {2} {\left (2 \, \log \left (2\right ) - 1\right )}}{4 \, \sqrt {-\log \left (2\right )}}\right ) e^{\left (-\frac {{\left (2 \, \log \left (2\right ) - 1\right )}^{2}}{8 \, \log \left (2\right )} - 3\right )} \log \left (2\right )}{\sqrt {-\log \left (2\right )}} + 64 \, \sqrt {2} {\left (\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\pi } {\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {\frac {1}{2}} \sqrt {-\frac {{\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )}^{2}}{\log \left (2\right )}}\right ) - 1\right )} {\left (2 \, \log \left (2\right ) - 1\right )}}{\sqrt {-\frac {{\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )}^{2}}{\log \left (2\right )}} \log \left (2\right )^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} e^{\left (\frac {{\left (4 \, x \log \left (2\right ) - 2 \, \log \left (2\right ) + 1\right )}^{2}}{8 \, \log \left (2\right )}\right )}}{\sqrt {\log \left (2\right )}}\right )} e^{\left (-\frac {{\left (2 \, \log \left (2\right ) - 1\right )}^{2}}{8 \, \log \left (2\right )} - 3\right )} \sqrt {\log \left (2\right )} + \frac {64 \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} x \sqrt {-\log \left (2\right )} + \frac {\sqrt {2} {\left (2 \, \log \left (2\right ) - 1\right )}}{4 \, \sqrt {-\log \left (2\right )}}\right ) e^{\left (-\frac {{\left (2 \, \log \left (2\right ) - 1\right )}^{2}}{8 \, \log \left (2\right )} - 3\right )}}{\sqrt {-\log \left (2\right )}} - 4 \, x - \log \left (x\right ) \] Input:
integrate(((2*(2*x^2-x)*log(2)+x)*exp(2*(x^2-x+4)*log(2)+x-3)-4*x-1)/x,x, algorithm="maxima")
Output:
-128*sqrt(2)*sqrt(pi)*erf(sqrt(2)*x*sqrt(-log(2)) + 1/4*sqrt(2)*(2*log(2) - 1)/sqrt(-log(2)))*e^(-1/8*(2*log(2) - 1)^2/log(2) - 3)*log(2)/sqrt(-log( 2)) + 64*sqrt(2)*(sqrt(2)*sqrt(1/2)*sqrt(pi)*(4*x*log(2) - 2*log(2) + 1)*( erf(1/2*sqrt(1/2)*sqrt(-(4*x*log(2) - 2*log(2) + 1)^2/log(2))) - 1)*(2*log (2) - 1)/(sqrt(-(4*x*log(2) - 2*log(2) + 1)^2/log(2))*log(2)^(3/2)) + 2*sq rt(2)*e^(1/8*(4*x*log(2) - 2*log(2) + 1)^2/log(2))/sqrt(log(2)))*e^(-1/8*( 2*log(2) - 1)^2/log(2) - 3)*sqrt(log(2)) + 64*sqrt(2)*sqrt(pi)*erf(sqrt(2) *x*sqrt(-log(2)) + 1/4*sqrt(2)*(2*log(2) - 1)/sqrt(-log(2)))*e^(-1/8*(2*lo g(2) - 1)^2/log(2) - 3)/sqrt(-log(2)) - 4*x - log(x)
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=-{\left (4 \, x e^{3} + e^{3} \log \left (x\right ) - 256 \, e^{\left (2 \, x^{2} \log \left (2\right ) - 2 \, x \log \left (2\right ) + x\right )}\right )} e^{\left (-3\right )} \] Input:
integrate(((2*(2*x^2-x)*log(2)+x)*exp(2*(x^2-x+4)*log(2)+x-3)-4*x-1)/x,x, algorithm="giac")
Output:
-(4*x*e^3 + e^3*log(x) - 256*e^(2*x^2*log(2) - 2*x*log(2) + x))*e^(-3)
Time = 2.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=\frac {256\,2^{2\,x^2}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}{2^{2\,x}}-\ln \left (x\right )-4\,x \] Input:
int(-(4*x - exp(x + 2*log(2)*(x^2 - x + 4) - 3)*(x - 2*log(2)*(x - 2*x^2)) + 1)/x,x)
Output:
(256*2^(2*x^2)*exp(-3)*exp(x))/2^(2*x) - log(x) - 4*x
Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.57 \[ \int \frac {-1-4 x+e^{-3+x+\left (4-x+x^2\right ) \log (4)} \left (x+\left (-x+2 x^2\right ) \log (4)\right )}{x} \, dx=\frac {256 e^{x} 2^{2 x^{2}}-2^{2 x} \mathrm {log}\left (x \right ) e^{3}-4 \,2^{2 x} e^{3} x}{2^{2 x} e^{3}} \] Input:
int(((2*(2*x^2-x)*log(2)+x)*exp(2*(x^2-x+4)*log(2)+x-3)-4*x-1)/x,x)
Output:
(256*e**x*2**(2*x**2) - 2**(2*x)*log(x)*e**3 - 4*2**(2*x)*e**3*x)/(2**(2*x )*e**3)