Integrand size = 63, antiderivative size = 18 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=12 e^{(5+x)^2+\frac {x}{-3+\log (3)}} \] Output:
exp(ln(12*x)+(5+x)^2+x/(ln(3)-3))/x
Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(18)=36\).
Time = 1.42 (sec) , antiderivative size = 60, normalized size of antiderivative = 3.33 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\frac {2\ 3^{\frac {22+10 x+x^2+\log (3)}{-3+\log (3)}} e^{\frac {75+29 x+3 x^2}{3-\log (3)}} (10 \log (3)+x \log (9))}{(5+x) \log (3)} \] Input:
Integrate[(E^((-75 - 29*x - 3*x^2 + (25 + 10*x + x^2)*Log[3] + (-3 + Log[3 ])*Log[12*x])/(-3 + Log[3]))*(-29 - 6*x + (10 + 2*x)*Log[3]))/(-3*x + x*Lo g[3]),x]
Output:
(2*3^((22 + 10*x + x^2 + Log[3])/(-3 + Log[3]))*E^((75 + 29*x + 3*x^2)/(3 - Log[3]))*(10*Log[3] + x*Log[9]))/((5 + x)*Log[3])
Time = 0.46 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6, 27, 27, 2725, 27, 2674, 2666}
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.
\(\displaystyle \int \frac {(-6 x+(2 x+10) \log (3)-29) \exp \left (\frac {-3 x^2+\left (x^2+10 x+25\right ) \log (3)-29 x+(\log (3)-3) \log (12 x)-75}{\log (3)-3}\right )}{x \log (3)-3 x} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {(-6 x+(2 x+10) \log (3)-29) \exp \left (\frac {-3 x^2+\left (x^2+10 x+25\right ) \log (3)-29 x+(\log (3)-3) \log (12 x)-75}{\log (3)-3}\right )}{x (\log (3)-3)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int -4 3^{1-\frac {x^2+10 x+25}{3-\log (3)}} e^{\frac {3 x^2+29 x+75}{3-\log (3)}} (6 x-2 (x+5) \log (3)+29)dx}{3-\log (3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \int 3^{1-\frac {x^2+10 x+25}{3-\log (3)}} e^{\frac {3 x^2+29 x+75}{3-\log (3)}} (6 x-2 (x+5) \log (3)+29)dx}{3-\log (3)}\) |
\(\Big \downarrow \) 2725 |
\(\displaystyle \frac {4 \int 3 e^{x^2+\frac {(29-10 \log (3)) x}{3-\log (3)}+25} (6 x-2 (x+5) \log (3)+29)dx}{3-\log (3)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {12 \int e^{x^2+\frac {(29-10 \log (3)) x}{3-\log (3)}+25} (6 x-2 (x+5) \log (3)+29)dx}{3-\log (3)}\) |
\(\Big \downarrow \) 2674 |
\(\displaystyle \frac {12 \int e^{x^2+\frac {(29-10 \log (3)) x}{3-\log (3)}+25} (2 (3-\log (3)) x-10 \log (3)+29)dx}{3-\log (3)}\) |
\(\Big \downarrow \) 2666 |
\(\displaystyle 12 e^{x^2+\frac {x (29-10 \log (3))}{3-\log (3)}+25}\) |
Input:
Int[(E^((-75 - 29*x - 3*x^2 + (25 + 10*x + x^2)*Log[3] + (-3 + Log[3])*Log [12*x])/(-3 + Log[3]))*(-29 - 6*x + (10 + 2*x)*Log[3]))/(-3*x + x*Log[3]), x]
Output:
12*E^(25 + x^2 + (x*(29 - 10*Log[3]))/(3 - Log[3]))
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol ] :> Simp[e*(F^(a + b*x + c*x^2)/(2*c*Log[F])), x] /; FreeQ[{F, a, b, c, d, e}, x] && EqQ[b*e - 2*c*d, 0]
Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSu m[v, x], x] /; FreeQ[{F, m}, x] && LinearQ[u, x] && QuadraticQ[v, x] && !( LinearMatchQ[u, x] && QuadraticMatchQ[v, x])
Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x], x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]
Leaf count of result is larger than twice the leaf count of optimal. \(50\) vs. \(2(23)=46\).
Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.83
method | result | size |
gosper | \(\frac {{\mathrm e}^{\frac {x^{2} \ln \left (3\right )+\ln \left (12 x \right ) \ln \left (3\right )+10 x \ln \left (3\right )-3 x^{2}+25 \ln \left (3\right )-3 \ln \left (12 x \right )-29 x -75}{\ln \left (3\right )-3}}}{x}\) | \(51\) |
risch | \(\frac {{\mathrm e}^{\frac {x^{2} \ln \left (3\right )+\ln \left (12 x \right ) \ln \left (3\right )+10 x \ln \left (3\right )-3 x^{2}+25 \ln \left (3\right )-3 \ln \left (12 x \right )-29 x -75}{\ln \left (3\right )-3}}}{x}\) | \(51\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {\left (\ln \left (3\right )-3\right ) \ln \left (12 x \right )+\left (x^{2}+10 x +25\right ) \ln \left (3\right )-3 x^{2}-29 x -75}{\ln \left (3\right )-3}} \ln \left (3\right )-3 \,{\mathrm e}^{\frac {\left (\ln \left (3\right )-3\right ) \ln \left (12 x \right )+\left (x^{2}+10 x +25\right ) \ln \left (3\right )-3 x^{2}-29 x -75}{\ln \left (3\right )-3}}}{x \left (\ln \left (3\right )-3\right )}\) | \(93\) |
Input:
int(((2*x+10)*ln(3)-6*x-29)*exp(((ln(3)-3)*ln(12*x)+(x^2+10*x+25)*ln(3)-3* x^2-29*x-75)/(ln(3)-3))/(x*ln(3)-3*x),x,method=_RETURNVERBOSE)
Output:
exp((x^2*ln(3)+ln(12*x)*ln(3)+10*x*ln(3)-3*x^2+25*ln(3)-3*ln(12*x)-29*x-75 )/(ln(3)-3))/x
Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.50 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\frac {e^{\left (-\frac {3 \, x^{2} - {\left (x^{2} + 10 \, x + 25\right )} \log \left (3\right ) - {\left (\log \left (3\right ) - 3\right )} \log \left (12 \, x\right ) + 29 \, x + 75}{\log \left (3\right ) - 3}\right )}}{x} \] Input:
integrate(((2*x+10)*log(3)-6*x-29)*exp(((log(3)-3)*log(12*x)+(x^2+10*x+25) *log(3)-3*x^2-29*x-75)/(log(3)-3))/(x*log(3)-3*x),x, algorithm="fricas")
Output:
e^(-(3*x^2 - (x^2 + 10*x + 25)*log(3) - (log(3) - 3)*log(12*x) + 29*x + 75 )/(log(3) - 3))/x
Timed out. \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\text {Timed out} \] Input:
integrate(((2*x+10)*ln(3)-6*x-29)*exp(((ln(3)-3)*ln(12*x)+(x**2+10*x+25)*l n(3)-3*x**2-29*x-75)/(ln(3)-3))/(x*ln(3)-3*x),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 6.67 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\frac {3^{\frac {22}{\log \left (3\right ) - 3}} 2^{\frac {2 \, \log \left (3\right )}{\log \left (3\right ) - 3} - \frac {6}{\log \left (3\right ) - 3}} e^{\left (\frac {x^{2} \log \left (3\right )}{\log \left (3\right ) - 3} - \frac {3 \, x^{2}}{\log \left (3\right ) - 3} + \frac {10 \, x \log \left (3\right )}{\log \left (3\right ) - 3} + \frac {\log \left (3\right )^{2}}{\log \left (3\right ) - 3} + \frac {\log \left (3\right ) \log \left (x\right )}{\log \left (3\right ) - 3} - \frac {29 \, x}{\log \left (3\right ) - 3} - \frac {3 \, \log \left (x\right )}{\log \left (3\right ) - 3} - \frac {75}{\log \left (3\right ) - 3}\right )}}{x} \] Input:
integrate(((2*x+10)*log(3)-6*x-29)*exp(((log(3)-3)*log(12*x)+(x^2+10*x+25) *log(3)-3*x^2-29*x-75)/(log(3)-3))/(x*log(3)-3*x),x, algorithm="maxima")
Output:
3^(22/(log(3) - 3))*2^(2*log(3)/(log(3) - 3) - 6/(log(3) - 3))*e^(x^2*log( 3)/(log(3) - 3) - 3*x^2/(log(3) - 3) + 10*x*log(3)/(log(3) - 3) + log(3)^2 /(log(3) - 3) + log(3)*log(x)/(log(3) - 3) - 29*x/(log(3) - 3) - 3*log(x)/ (log(3) - 3) - 75/(log(3) - 3))/x
Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.22 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=4 \, e^{\left (\frac {x^{2} \log \left (3\right ) - 3 \, x^{2} + 10 \, x \log \left (3\right ) + \log \left (3\right )^{2} - 29 \, x - 3 \, \log \left (3\right )}{\log \left (3\right ) - 3} + 25\right )} \] Input:
integrate(((2*x+10)*log(3)-6*x-29)*exp(((log(3)-3)*log(12*x)+(x^2+10*x+25) *log(3)-3*x^2-29*x-75)/(log(3)-3))/(x*log(3)-3*x),x, algorithm="giac")
Output:
4*e^((x^2*log(3) - 3*x^2 + 10*x*log(3) + log(3)^2 - 29*x - 3*log(3))/(log( 3) - 3) + 25)
Time = 3.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.28 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx={\left (\frac {1}{64}\right )}^{\frac {1}{\ln \left (3\right )-3}}\,3^{\frac {x^2+10\,x+\ln \left (12\right )+22}{\ln \left (3\right )-3}}\,{\mathrm {e}}^{-\frac {29\,x}{\ln \left (3\right )-3}}\,{\mathrm {e}}^{-\frac {3\,x^2}{\ln \left (3\right )-3}}\,{\mathrm {e}}^{-\frac {75}{\ln \left (3\right )-3}} \] Input:
int((exp(-(29*x - log(12*x)*(log(3) - 3) + 3*x^2 - log(3)*(10*x + x^2 + 25 ) + 75)/(log(3) - 3))*(6*x - log(3)*(2*x + 10) + 29))/(3*x - x*log(3)),x)
Output:
(1/64)^(1/(log(3) - 3))*3^((10*x + log(12) + x^2 + 22)/(log(3) - 3))*exp(- (29*x)/(log(3) - 3))*exp(-(3*x^2)/(log(3) - 3))*exp(-75/(log(3) - 3))
Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.89 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=12 e^{\frac {\mathrm {log}\left (3\right ) x^{2}+10 \,\mathrm {log}\left (3\right ) x -3 x^{2}-29 x}{\mathrm {log}\left (3\right )-3}} e^{25} \] Input:
int(((2*x+10)*log(3)-6*x-29)*exp(((log(3)-3)*log(12*x)+(x^2+10*x+25)*log(3 )-3*x^2-29*x-75)/(log(3)-3))/(x*log(3)-3*x),x)
Output:
12*e**((log(3)*x**2 + 10*log(3)*x - 3*x**2 - 29*x)/(log(3) - 3))*e**25