Integrand size = 56, antiderivative size = 24 \begin {dmath*} \int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3 \log (25)} \, dx=(-3+x) \left (-e^{\frac {1-\frac {1}{\log (25)}}{x^2}}+2 x\right ) \end {dmath*}
Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \begin {dmath*} \int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3 \log (25)} \, dx=e^{-\frac {1}{x^2 \log (25)}} (-3+x) \left (-e^{\frac {1}{x^2}}+2 e^{\frac {1}{x^2 \log (25)}} x\right ) \end {dmath*}
Integrate[((-6*x^3 + 4*x^4)*Log[25] + E^((-1 + Log[25])/(x^2*Log[25]))*(6 - 2*x + (-6 + 2*x - x^3)*Log[25]))/(x^3*Log[25]),x]
Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(24)=48\).
Time = 0.31 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.67, 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.
\(\displaystyle \int \frac {\left (4 x^4-6 x^3\right ) \log (25)+e^{\frac {\log (25)-1}{x^2 \log (25)}} \left (\left (-x^3+2 x-6\right ) \log (25)-2 x+6\right )}{x^3 \log (25)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {2 \left (3 x^3-2 x^4\right ) \log (25)-\left (\frac {25}{e}\right )^{\frac {1}{x^2 \log (25)}} \left (-2 x-\left (x^3-2 x+6\right ) \log (25)+6\right )}{x^3}dx}{\log (25)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {2 \left (3 x^3-2 x^4\right ) \log (25)-\left (\frac {25}{e}\right )^{\frac {1}{x^2 \log (25)}} \left (-2 x-\left (x^3-2 x+6\right ) \log (25)+6\right )}{x^3}dx}{\log (25)}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (\frac {\left (\frac {25}{e}\right )^{\frac {1}{x^2 \log (25)}} \left (\log (25) x^3+2 (1-\log (25)) x-6 (1-\log (25))\right )}{x^3}-2 (2 x-3) \log (25)\right )dx}{\log (25)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {\log (25) \left (\frac {25}{e}\right )^{\frac {1}{x^2 \log (25)}} (3 (1-\log (25))-x (1-\log (25)))}{1-\log (25)}-\frac {1}{2} (3-2 x)^2 \log (25)}{\log (25)}\) |
Int[((-6*x^3 + 4*x^4)*Log[25] + E^((-1 + Log[25])/(x^2*Log[25]))*(6 - 2*x + (-6 + 2*x - x^3)*Log[25]))/(x^3*Log[25]),x]
-((-1/2*((3 - 2*x)^2*Log[25]) - ((25/E)^(1/(x^2*Log[25]))*(3*(1 - Log[25]) - x*(1 - Log[25]))*Log[25])/(1 - Log[25]))/Log[25])
3.3.93.3.1 Defintions of rubi rules used
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 = 0.73 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75
method | result | size |
risch | \(2 x^{2}-6 x +\frac {\left (-2 x \ln \left (5\right )+6 \ln \left (5\right )\right ) {\mathrm e}^{\frac {2 \ln \left (5\right )-1}{2 x^{2} \ln \left (5\right )}}}{2 \ln \left (5\right )}\) | \(42\) |
norman | \(\frac {-6 x^{3}+2 x^{4}+3 x^{2} {\mathrm e}^{\frac {2 \ln \left (5\right )-1}{2 x^{2} \ln \left (5\right )}}-x^{3} {\mathrm e}^{\frac {2 \ln \left (5\right )-1}{2 x^{2} \ln \left (5\right )}}}{x^{2}}\) | \(58\) |
parallelrisch | \(\frac {4 x^{2} \ln \left (5\right )-2 \ln \left (5\right ) {\mathrm e}^{\frac {2 \ln \left (5\right )-1}{2 x^{2} \ln \left (5\right )}} x -12 x \ln \left (5\right )+6 \ln \left (5\right ) {\mathrm e}^{\frac {2 \ln \left (5\right )-1}{2 x^{2} \ln \left (5\right )}}}{2 \ln \left (5\right )}\) | \(61\) |
parts | \(2 x^{2}-6 x -x \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}-\frac {3 \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}}{2 \ln \left (5\right ) \left (1-\frac {1}{2 \ln \left (5\right )}\right )}+\frac {3 \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}}{1-\frac {1}{2 \ln \left (5\right )}}\) | \(80\) |
derivativedivides | \(-\frac {\frac {2 i \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {4-\frac {2}{\ln \left (5\right )}}}{2 x}\right )}{\sqrt {4-\frac {2}{\ln \left (5\right )}}}-4 x^{2} \ln \left (5\right )+12 x \ln \left (5\right )+\frac {3 \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}}{1-\frac {1}{2 \ln \left (5\right )}}-\frac {4 i \ln \left (5\right ) \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {4-\frac {2}{\ln \left (5\right )}}}{2 x}\right )}{\sqrt {4-\frac {2}{\ln \left (5\right )}}}-\frac {6 \ln \left (5\right ) {\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}}{1-\frac {1}{2 \ln \left (5\right )}}-2 \ln \left (5\right ) \left (-x \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}-\frac {2 i \left (1-\frac {1}{2 \ln \left (5\right )}\right ) \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {4-\frac {2}{\ln \left (5\right )}}}{2 x}\right )}{\sqrt {4-\frac {2}{\ln \left (5\right )}}}\right )}{2 \ln \left (5\right )}\) | \(202\) |
default | \(\frac {-\frac {2 i \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {4-\frac {2}{\ln \left (5\right )}}}{2 x}\right )}{\sqrt {4-\frac {2}{\ln \left (5\right )}}}+4 x^{2} \ln \left (5\right )-12 x \ln \left (5\right )-\frac {3 \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}}{1-\frac {1}{2 \ln \left (5\right )}}+\frac {4 i \ln \left (5\right ) \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {4-\frac {2}{\ln \left (5\right )}}}{2 x}\right )}{\sqrt {4-\frac {2}{\ln \left (5\right )}}}+\frac {6 \ln \left (5\right ) {\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}}{1-\frac {1}{2 \ln \left (5\right )}}+2 \ln \left (5\right ) \left (-x \,{\mathrm e}^{\frac {1-\frac {1}{2 \ln \left (5\right )}}{x^{2}}}-\frac {2 i \left (1-\frac {1}{2 \ln \left (5\right )}\right ) \sqrt {\pi }\, \operatorname {erf}\left (\frac {i \sqrt {4-\frac {2}{\ln \left (5\right )}}}{2 x}\right )}{\sqrt {4-\frac {2}{\ln \left (5\right )}}}\right )}{2 \ln \left (5\right )}\) | \(202\) |
int(1/2*((2*(-x^3+2*x-6)*ln(5)+6-2*x)*exp(1/2*(2*ln(5)-1)/x^2/ln(5))+2*(4* x^4-6*x^3)*ln(5))/x^3/ln(5),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \begin {dmath*} \int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3 \log (25)} \, dx=2 \, x^{2} - {\left (x - 3\right )} e^{\left (\frac {2 \, \log \left (5\right ) - 1}{2 \, x^{2} \log \left (5\right )}\right )} - 6 \, x \end {dmath*}
integrate(1/2*((2*(-x^3+2*x-6)*log(5)+6-2*x)*exp(1/2*(2*log(5)-1)/x^2/log( 5))+2*(4*x^4-6*x^3)*log(5))/x^3/log(5),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \begin {dmath*} \int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3 \log (25)} \, dx=2 x^{2} - 6 x + \left (3 - x\right ) e^{\frac {- \frac {1}{2} + \log {\left (5 \right )}}{x^{2} \log {\left (5 \right )}}} \end {dmath*}
integrate(1/2*((2*(-x**3+2*x-6)*ln(5)+6-2*x)*exp(1/2*(2*ln(5)-1)/x**2/ln(5 ))+2*(4*x**4-6*x**3)*ln(5))/x**3/ln(5),x)
\begin {dmath*} \int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3 \log (25)} \, dx=\int { -\frac {{\left ({\left (x^{3} - 2 \, x + 6\right )} \log \left (5\right ) + x - 3\right )} e^{\left (\frac {2 \, \log \left (5\right ) - 1}{2 \, x^{2} \log \left (5\right )}\right )} - 2 \, {\left (2 \, x^{4} - 3 \, x^{3}\right )} \log \left (5\right )}{x^{3} \log \left (5\right )} \,d x } \end {dmath*}
integrate(1/2*((2*(-x^3+2*x-6)*log(5)+6-2*x)*exp(1/2*(2*log(5)-1)/x^2/log( 5))+2*(4*x^4-6*x^3)*log(5))/x^3/log(5),x, algorithm=\
(2*x^2*log(5) - 6*x*log(5) - integrate((x^3*log(5) - x*(2*log(5) - 1) + 6* log(5) - 3)*e^(1/x^2 - 1/2/(x^2*log(5)))/x^3, x))/log(5)
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46 \begin {dmath*} \int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3 \log (25)} \, dx=\frac {2 \, x^{2} \log \left (5\right ) - x e^{\left (\frac {2 \, \log \left (5\right ) - 1}{2 \, x^{2} \log \left (5\right )}\right )} \log \left (5\right ) - 6 \, x \log \left (5\right ) + 3 \, e^{\left (\frac {2 \, \log \left (5\right ) - 1}{2 \, x^{2} \log \left (5\right )}\right )} \log \left (5\right )}{\log \left (5\right )} \end {dmath*}
integrate(1/2*((2*(-x^3+2*x-6)*log(5)+6-2*x)*exp(1/2*(2*log(5)-1)/x^2/log( 5))+2*(4*x^4-6*x^3)*log(5))/x^3/log(5),x, algorithm=\
(2*x^2*log(5) - x*e^(1/2*(2*log(5) - 1)/(x^2*log(5)))*log(5) - 6*x*log(5) + 3*e^(1/2*(2*log(5) - 1)/(x^2*log(5)))*log(5))/log(5)
Time = 12.69 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.17 \begin {dmath*} \int \frac {\left (-6 x^3+4 x^4\right ) \log (25)+e^{\frac {-1+\log (25)}{x^2 \log (25)}} \left (6-2 x+\left (-6+2 x-x^3\right ) \log (25)\right )}{x^3 \log (25)} \, dx=\frac {{\mathrm {e}}^{\frac {1}{x^2}-\frac {1}{2\,x^2\,\ln \left (5\right )}}\,\ln \left (125\right )}{\ln \left (5\right )}-x\,{\mathrm {e}}^{\frac {1}{x^2}-\frac {1}{2\,x^2\,\ln \left (5\right )}}-6\,x+\frac {x^2\,\ln \left (25\right )}{\ln \left (5\right )} \end {dmath*}
int(-(log(5)*(6*x^3 - 4*x^4) + (exp((log(5) - 1/2)/(x^2*log(5)))*(2*x + 2* log(5)*(x^3 - 2*x + 6) - 6))/2)/(x^3*log(5)),x)