Integrand size = 56, antiderivative size = 27 \[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\left (5+\frac {2}{3} \left (1-e^x \left (e^5+x\right ) \log (2)\right )\right ) (-2-\log (x)) \] Output:
(-ln(x)-2)*(17/3-2/3*exp(x)*ln(2)*(exp(5)+x))
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\frac {1}{3} \left (e^x \left (e^5+x\right ) \log (16)+\left (-17+e^{5+x} \log (4)+e^x x \log (4)\right ) \log (x)\right ) \] Input:
Integrate[(-17 + E^x*(6*x + 4*x^2 + E^5*(2 + 4*x))*Log[2] + E^x*(2*x + 2*E ^5*x + 2*x^2)*Log[2]*Log[x])/(3*x),x]
Output:
(E^x*(E^5 + x)*Log[16] + (-17 + E^(5 + x)*Log[4] + E^x*x*Log[4])*Log[x])/3
Leaf count is larger than twice the leaf count of optimal. \(62\) vs. \(2(27)=54\).
Time = 0.53 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.30, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {e^x \left (4 x^2+6 x+e^5 (4 x+2)\right ) \log (2)+e^x \left (2 x^2+2 e^5 x+2 x\right ) \log (2) \log (x)-17}{3 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {-2 e^x \log (2) \left (2 x^2+3 x+e^5 (2 x+1)\right )-2 e^x \left (x^2+e^5 x+x\right ) \log (2) \log (x)+17}{x}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{3} \int \frac {-2 e^x \log (2) \left (2 x^2+3 x+e^5 (2 x+1)\right )-2 e^x \left (x^2+e^5 x+x\right ) \log (2) \log (x)+17}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{3} \int \left (\frac {2 e^x \log (2) \left (-\log (x) x^2-2 x^2-\left (1+e^5\right ) \log (x) x-3 \left (1+\frac {2 e^5}{3}\right ) x-e^5\right )}{x}+\frac {17}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (4 e^x x \log (2)-2 e^x \log (2) \log (x)+2 e^x \left (x+e^5+1\right ) \log (2) \log (x)-17 \log (x)-6 e^x \log (2)+2 \left (3+2 e^5\right ) e^x \log (2)\right )\) |
Input:
Int[(-17 + E^x*(6*x + 4*x^2 + E^5*(2 + 4*x))*Log[2] + E^x*(2*x + 2*E^5*x + 2*x^2)*Log[2]*Log[x])/(3*x),x]
Output:
(-6*E^x*Log[2] + 2*E^x*(3 + 2*E^5)*Log[2] + 4*E^x*x*Log[2] - 17*Log[x] - 2 *E^x*Log[2]*Log[x] + 2*E^x*(1 + E^5 + x)*Log[2]*Log[x])/3
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 = 0.65 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22
method | result | size |
risch | \(\frac {2 \ln \left (2\right ) \left ({\mathrm e}^{5}+x \right ) {\mathrm e}^{x} \ln \left (x \right )}{3}+\frac {4 \ln \left (2\right ) {\mathrm e}^{5+x}}{3}+\frac {4 x \ln \left (2\right ) {\mathrm e}^{x}}{3}-\frac {17 \ln \left (x \right )}{3}\) | \(33\) |
default | \(\frac {4 x \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {2 x \ln \left (2\right ) {\mathrm e}^{x} \ln \left (x \right )}{3}+\frac {2 \,{\mathrm e}^{x} {\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )}{3}-\frac {17 \ln \left (x \right )}{3}\) | \(40\) |
norman | \(\frac {4 x \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {2 x \ln \left (2\right ) {\mathrm e}^{x} \ln \left (x \right )}{3}+\frac {2 \,{\mathrm e}^{x} {\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )}{3}-\frac {17 \ln \left (x \right )}{3}\) | \(40\) |
parallelrisch | \(\frac {4 x \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {2 x \ln \left (2\right ) {\mathrm e}^{x} \ln \left (x \right )}{3}+\frac {2 \,{\mathrm e}^{x} {\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )}{3}-\frac {17 \ln \left (x \right )}{3}\) | \(40\) |
parts | \(\frac {4 x \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {4 \,{\mathrm e}^{5} \ln \left (2\right ) {\mathrm e}^{x}}{3}+\frac {2 x \ln \left (2\right ) {\mathrm e}^{x} \ln \left (x \right )}{3}+\frac {2 \,{\mathrm e}^{x} {\mathrm e}^{5} \ln \left (2\right ) \ln \left (x \right )}{3}-\frac {17 \ln \left (x \right )}{3}\) | \(40\) |
Input:
int(1/3*((2*x*exp(5)+2*x^2+2*x)*ln(2)*exp(x)*ln(x)+((4*x+2)*exp(5)+4*x^2+6 *x)*ln(2)*exp(x)-17)/x,x,method=_RETURNVERBOSE)
Output:
2/3*ln(2)*(exp(5)+x)*exp(x)*ln(x)+4/3*ln(2)*exp(5+x)+4/3*x*ln(2)*exp(x)-17 /3*ln(x)
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\frac {4}{3} \, {\left (x + e^{5}\right )} e^{x} \log \left (2\right ) + \frac {1}{3} \, {\left (2 \, {\left (x + e^{5}\right )} e^{x} \log \left (2\right ) - 17\right )} \log \left (x\right ) \] Input:
integrate(1/3*((2*x*exp(5)+2*x^2+2*x)*log(2)*exp(x)*log(x)+((2+4*x)*exp(5) +4*x^2+6*x)*log(2)*exp(x)-17)/x,x, algorithm="fricas")
Output:
4/3*(x + e^5)*e^x*log(2) + 1/3*(2*(x + e^5)*e^x*log(2) - 17)*log(x)
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\frac {\left (2 x \log {\left (2 \right )} \log {\left (x \right )} + 4 x \log {\left (2 \right )} + 2 e^{5} \log {\left (2 \right )} \log {\left (x \right )} + 4 e^{5} \log {\left (2 \right )}\right ) e^{x}}{3} - \frac {17 \log {\left (x \right )}}{3} \] Input:
integrate(1/3*((2*x*exp(5)+2*x**2+2*x)*ln(2)*exp(x)*ln(x)+((2+4*x)*exp(5)+ 4*x**2+6*x)*ln(2)*exp(x)-17)/x,x)
Output:
(2*x*log(2)*log(x) + 4*x*log(2) + 2*exp(5)*log(2)*log(x) + 4*exp(5)*log(2) )*exp(x)/3 - 17*log(x)/3
\[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\int { \frac {2 \, {\left (x^{2} + x e^{5} + x\right )} e^{x} \log \left (2\right ) \log \left (x\right ) + 2 \, {\left (2 \, x^{2} + {\left (2 \, x + 1\right )} e^{5} + 3 \, x\right )} e^{x} \log \left (2\right ) - 17}{3 \, x} \,d x } \] Input:
integrate(1/3*((2*x*exp(5)+2*x^2+2*x)*log(2)*exp(x)*log(x)+((2+4*x)*exp(5) +4*x^2+6*x)*log(2)*exp(x)-17)/x,x, algorithm="maxima")
Output:
2/3*Ei(x)*e^5*log(2) + 4/3*(x - 1)*e^x*log(2) + 2/3*(x*log(2) + e^5*log(2) - log(2))*e^x*log(x) + 2/3*e^x*log(2)*log(x) - 2/3*Ei(x)*log(2) + 4/3*e^( x + 5)*log(2) + 2*e^x*log(2) - 1/3*integrate(2*(x*log(2) + e^5*log(2) - lo g(2))*e^x/x, x) - 17/3*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (18) = 36\).
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\frac {2}{3} \, x e^{x} \log \left (2\right ) \log \left (x\right ) + \frac {4}{3} \, x e^{x} \log \left (2\right ) + \frac {2}{3} \, e^{\left (x + 5\right )} \log \left (2\right ) \log \left (x\right ) + \frac {4}{3} \, e^{\left (x + 5\right )} \log \left (2\right ) - \frac {17}{3} \, \log \left (x\right ) \] Input:
integrate(1/3*((2*x*exp(5)+2*x^2+2*x)*log(2)*exp(x)*log(x)+((2+4*x)*exp(5) +4*x^2+6*x)*log(2)*exp(x)-17)/x,x, algorithm="giac")
Output:
2/3*x*e^x*log(2)*log(x) + 4/3*x*e^x*log(2) + 2/3*e^(x + 5)*log(2)*log(x) + 4/3*e^(x + 5)*log(2) - 17/3*log(x)
Time = 7.51 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\frac {4\,{\mathrm {e}}^{x+5}\,\ln \left (2\right )}{3}-\frac {17\,\ln \left (x\right )}{3}+\frac {4\,x\,{\mathrm {e}}^x\,\ln \left (2\right )}{3}+\frac {2\,{\mathrm {e}}^{x+5}\,\ln \left (2\right )\,\ln \left (x\right )}{3}+\frac {2\,x\,{\mathrm {e}}^x\,\ln \left (2\right )\,\ln \left (x\right )}{3} \] Input:
int(((exp(x)*log(2)*(6*x + 4*x^2 + exp(5)*(4*x + 2)))/3 + (exp(x)*log(2)*l og(x)*(2*x + 2*x*exp(5) + 2*x^2))/3 - 17/3)/x,x)
Output:
(4*exp(x + 5)*log(2))/3 - (17*log(x))/3 + (4*x*exp(x)*log(2))/3 + (2*exp(x + 5)*log(2)*log(x))/3 + (2*x*exp(x)*log(2)*log(x))/3
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.67 \[ \int \frac {-17+e^x \left (6 x+4 x^2+e^5 (2+4 x)\right ) \log (2)+e^x \left (2 x+2 e^5 x+2 x^2\right ) \log (2) \log (x)}{3 x} \, dx=\frac {2 e^{x} \mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) e^{5}}{3}+\frac {2 e^{x} \mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) x}{3}+\frac {4 e^{x} \mathrm {log}\left (2\right ) e^{5}}{3}+\frac {4 e^{x} \mathrm {log}\left (2\right ) x}{3}-\frac {17 \,\mathrm {log}\left (x \right )}{3} \] Input:
int(1/3*((2*x*exp(5)+2*x^2+2*x)*log(2)*exp(x)*log(x)+((2+4*x)*exp(5)+4*x^2 +6*x)*log(2)*exp(x)-17)/x,x)
Output:
(2*e**x*log(x)*log(2)*e**5 + 2*e**x*log(x)*log(2)*x + 4*e**x*log(2)*e**5 + 4*e**x*log(2)*x - 17*log(x))/3