Integrand size = 116, antiderivative size = 28 \[ \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx=\frac {1}{5} e^{x \left (x-\frac {3}{x-5 x^2}\right )} x \left (e^{16}+\log (x)\right ) \]
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx=\frac {1}{5} e^{x^2+\frac {3}{-1+5 x}} x \left (e^{16}+\log (x)\right ) \]
Integrate[(E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*(1 - 10*x + 25*x^2 + E^16*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)) + E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)*Log[x])/(5 - 50*x + 125*x^2),x]
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.
\(\displaystyle \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}} \left (25 x^2+e^{16} \left (50 x^4-20 x^3+27 x^2-25 x+1\right )-10 x+1\right )+e^{\frac {5 x^3-x^2+3}{5 x-1}} \left (50 x^4-20 x^3+27 x^2-25 x+1\right ) \log (x)}{125 x^2-50 x+5} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 500 \int \frac {e^{-\frac {5 x^3-x^2+3}{1-5 x}} \left (25 x^2-10 x+e^{16} \left (50 x^4-20 x^3+27 x^2-25 x+1\right )+1\right )+e^{-\frac {5 x^3-x^2+3}{1-5 x}} \left (50 x^4-20 x^3+27 x^2-25 x+1\right ) \log (x)}{2500 (1-5 x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {e^{-\frac {5 x^3-x^2+3}{1-5 x}} \left (25 x^2-10 x+e^{16} \left (50 x^4-20 x^3+27 x^2-25 x+1\right )+1\right )+e^{-\frac {5 x^3-x^2+3}{1-5 x}} \left (50 x^4-20 x^3+27 x^2-25 x+1\right ) \log (x)}{(1-5 x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {50 e^{\frac {5 x^3-x^2+3}{5 x-1}} \log (x) x^4}{(1-5 x)^2}+\frac {50 e^{\frac {5 x^3-x^2+80 x-13}{5 x-1}} x^4}{(1-5 x)^2}-\frac {20 e^{\frac {5 x^3-x^2+3}{5 x-1}} \log (x) x^3}{(1-5 x)^2}-\frac {20 e^{\frac {5 x^3-x^2+80 x-13}{5 x-1}} x^3}{(1-5 x)^2}+\frac {27 e^{\frac {5 x^3-x^2+3}{5 x-1}} \log (x) x^2}{(1-5 x)^2}+\frac {25 e^{\frac {5 x^3-x^2+3}{5 x-1}} \left (1+\frac {27 e^{16}}{25}\right ) x^2}{(1-5 x)^2}-\frac {25 e^{\frac {5 x^3-x^2+3}{5 x-1}} \log (x) x}{(1-5 x)^2}-\frac {10 e^{\frac {5 x^3-x^2+3}{5 x-1}} \left (1+\frac {5 e^{16}}{2}\right ) x}{(1-5 x)^2}+\frac {e^{\frac {5 x^3-x^2+3}{5 x-1}} \log (x)}{(1-5 x)^2}+\frac {e^{\frac {5 x^3-x^2+3}{5 x-1}} \left (1+e^{16}\right )}{(1-5 x)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {1}{25} \left (25+27 e^{16}\right ) \int e^{\frac {5 x^3-x^2+3}{5 x-1}}dx-\frac {2}{25} \int e^{\frac {5 x^3-x^2+80 x-13}{5 x-1}}dx+\left (1+e^{16}\right ) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{(1-5 x)^2}dx+2 \int e^{\frac {5 x^3-x^2+80 x-13}{5 x-1}} x^2dx+\frac {1}{25} \left (25+27 e^{16}\right ) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{(5 x-1)^2}dx-\left (2+5 e^{16}\right ) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{(5 x-1)^2}dx-\frac {2}{25} \int \frac {e^{\frac {5 x^3-x^2+80 x-13}{5 x-1}}}{(5 x-1)^2}dx+\frac {2}{25} \left (25+27 e^{16}\right ) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{5 x-1}dx-\left (2+5 e^{16}\right ) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{5 x-1}dx-\frac {4}{25} \int \frac {e^{\frac {5 x^3-x^2+80 x-13}{5 x-1}}}{5 x-1}dx-\int \frac {\int e^{\frac {5 x^3-x^2+3}{5 x-1}}dx}{x}dx+3 \int \frac {\int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{(1-5 x)^2}dx}{x}dx-2 \int \frac {\int e^{\frac {5 x^3-x^2+3}{5 x-1}} x^2dx}{x}dx+3 \int \frac {\int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{5 x-1}dx}{x}dx+\log (x) \int e^{\frac {5 x^3-x^2+3}{5 x-1}}dx+\log (x) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{(1-5 x)^2}dx+2 \log (x) \int e^{\frac {5 x^3-x^2+3}{5 x-1}} x^2dx-4 \log (x) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{(5 x-1)^2}dx-3 \log (x) \int \frac {e^{\frac {5 x^3-x^2+3}{5 x-1}}}{5 x-1}dx\right )\) |
Int[(E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*(1 - 10*x + 25*x^2 + E^16*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)) + E^((3 - x^2 + 5*x^3)/(-1 + 5*x))*(1 - 25*x + 27*x^2 - 20*x^3 + 50*x^4)*Log[x])/(5 - 50*x + 125*x^2),x]
3.13.7.3.1 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_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(53\) vs. \(2(24)=48\).
Time = 1.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.93
method | result | size |
parallelrisch | \(\frac {\ln \left (x \right ) {\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} x}{5}+\frac {{\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} {\mathrm e}^{16} x}{5}\) | \(54\) |
risch | \(\frac {\ln \left (x \right ) {\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} x}{5}+\frac {x \,{\mathrm e}^{\frac {5 x^{3}-x^{2}+80 x -13}{5 x -1}}}{5}\) | \(55\) |
norman | \(\frac {{\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} {\mathrm e}^{16} x^{2}+\ln \left (x \right ) {\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} x^{2}-\frac {{\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} {\mathrm e}^{16} x}{5}-\frac {\ln \left (x \right ) {\mathrm e}^{\frac {5 x^{3}-x^{2}+3}{5 x -1}} x}{5}}{5 x -1}\) | \(116\) |
int(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*ln(x)+((50*x ^4-20*x^3+27*x^2-25*x+1)*exp(16)+25*x^2-10*x+1)*exp((5*x^3-x^2+3)/(5*x-1)) )/(125*x^2-50*x+5),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.89 \[ \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx=\frac {1}{5} \, x e^{\left (\frac {5 \, x^{3} - x^{2} + 3}{5 \, x - 1}\right )} \log \left (x\right ) + \frac {1}{5} \, x e^{\left (\frac {5 \, x^{3} - x^{2} + 3}{5 \, x - 1} + 16\right )} \]
integrate(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*log(x) +((50*x^4-20*x^3+27*x^2-25*x+1)*exp(16)+25*x^2-10*x+1)*exp((5*x^3-x^2+3)/( 5*x-1)))/(125*x^2-50*x+5),x, algorithm=\
Time = 73.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx=\frac {\left (x \log {\left (x \right )} + x e^{16}\right ) e^{\frac {5 x^{3} - x^{2} + 3}{5 x - 1}}}{5} \]
integrate(((50*x**4-20*x**3+27*x**2-25*x+1)*exp((5*x**3-x**2+3)/(5*x-1))*l n(x)+((50*x**4-20*x**3+27*x**2-25*x+1)*exp(16)+25*x**2-10*x+1)*exp((5*x**3 -x**2+3)/(5*x-1)))/(125*x**2-50*x+5),x)
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx=\frac {1}{5} \, {\left (x e^{16} + x \log \left (x\right )\right )} e^{\left (x^{2} + \frac {3}{5 \, x - 1}\right )} \]
integrate(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*log(x) +((50*x^4-20*x^3+27*x^2-25*x+1)*exp(16)+25*x^2-10*x+1)*exp((5*x^3-x^2+3)/( 5*x-1)))/(125*x^2-50*x+5),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (26) = 52\).
Time = 0.39 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx=\frac {1}{5} \, {\left (x e^{\left (\frac {5 \, x^{3} - x^{2} + 15 \, x}{5 \, x - 1}\right )} \log \left (x\right ) + x e^{\left (\frac {5 \, x^{3} - x^{2} + 15 \, x}{5 \, x - 1} + 16\right )}\right )} e^{\left (-3\right )} \]
integrate(((50*x^4-20*x^3+27*x^2-25*x+1)*exp((5*x^3-x^2+3)/(5*x-1))*log(x) +((50*x^4-20*x^3+27*x^2-25*x+1)*exp(16)+25*x^2-10*x+1)*exp((5*x^3-x^2+3)/( 5*x-1)))/(125*x^2-50*x+5),x, algorithm=\
1/5*(x*e^((5*x^3 - x^2 + 15*x)/(5*x - 1))*log(x) + x*e^((5*x^3 - x^2 + 15* x)/(5*x - 1) + 16))*e^(-3)
Timed out. \[ \int \frac {e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-10 x+25 x^2+e^{16} \left (1-25 x+27 x^2-20 x^3+50 x^4\right )\right )+e^{\frac {3-x^2+5 x^3}{-1+5 x}} \left (1-25 x+27 x^2-20 x^3+50 x^4\right ) \log (x)}{5-50 x+125 x^2} \, dx=\int \frac {{\mathrm {e}}^{\frac {5\,x^3-x^2+3}{5\,x-1}}\,\left ({\mathrm {e}}^{16}\,\left (50\,x^4-20\,x^3+27\,x^2-25\,x+1\right )-10\,x+25\,x^2+1\right )+{\mathrm {e}}^{\frac {5\,x^3-x^2+3}{5\,x-1}}\,\ln \left (x\right )\,\left (50\,x^4-20\,x^3+27\,x^2-25\,x+1\right )}{125\,x^2-50\,x+5} \,d x \]
int((exp((5*x^3 - x^2 + 3)/(5*x - 1))*(exp(16)*(27*x^2 - 25*x - 20*x^3 + 5 0*x^4 + 1) - 10*x + 25*x^2 + 1) + exp((5*x^3 - x^2 + 3)/(5*x - 1))*log(x)* (27*x^2 - 25*x - 20*x^3 + 50*x^4 + 1))/(125*x^2 - 50*x + 5),x)