\(\int \frac {e^{\frac {-75-29 x-3 x^2+(25+10 x+x^2) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx\) [7426]

Optimal result

Integrand size = 63, antiderivative size = 18 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=12 e^{(5+x)^2+\frac {x}{-3+\log (3)}} \]

[Out]

exp(ln(12*x)+(5+x)^2+x/(ln(3)-3))/x

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {6, 12, 2306, 2276, 2268} \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=12 e^{x^2+\frac {x (29-10 \log (3))}{3-\log (3)}+25} \]

[In]

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]

[Out]

12*E^(25 + x^2 + (x*(29 - 10*Log[3]))/(3 - Log[3]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2268

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]

Rule 2276

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rule 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (\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)}\right ) (-29-6 x+(10+2 x) \log (3))}{x (-3+\log (3))} \, dx \\ & = \frac {\int \frac {\exp \left (\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)}\right ) (-29-6 x+(10+2 x) \log (3))}{x} \, dx}{-3+\log (3)} \\ & = \frac {\int 12 \exp \left (\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)}{-3+\log (3)}\right ) (-29-6 x+(10+2 x) \log (3)) \, dx}{-3+\log (3)} \\ & = -\frac {12 \int \exp \left (\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)}{-3+\log (3)}\right ) (-29-6 x+(10+2 x) \log (3)) \, dx}{3-\log (3)} \\ & = -\frac {12 \int e^{25+x^2+\frac {x (29-10 \log (3))}{3-\log (3)}} (-29-2 x (3-\log (3))+10 \log (3)) \, dx}{3-\log (3)} \\ & = 12 e^{25+x^2+\frac {x (29-10 \log (3))}{3-\log (3)}} \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx \]

[In]

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*Log[3]),x]

[Out]

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*Log[3]), x]

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(50\) vs. \(2(23)=46\).

Time = 0.12 (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	\(\frac {{\mathrm e}^{\frac {\left (\ln \left (3\right )-3\right ) \ln \left (12 x \right )+\left (x^{2}+10 x +25\right ) \ln \left (3\right )-3 x^{2}-29 x -75}{\ln \left (3\right )-3}} \ln \left (3\right )-3 \,{\mathrm e}^{\frac {\left (\ln \left (3\right )-3\right ) \ln \left (12 x \right )+\left (x^{2}+10 x +25\right ) \ln \left (3\right )-3 x^{2}-29 x -75}{\ln \left (3\right )-3}}}{x \left (\ln \left (3\right )-3\right )}\)	\(93\)

[In]

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)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 2.50 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\frac {e^{\left (-\frac {3 \, x^{2} - {\left (x^{2} + 10 \, x + 25\right )} \log \left (3\right ) - {\left (\log \left (3\right ) - 3\right )} \log \left (12 \, x\right ) + 29 \, x + 75}{\log \left (3\right ) - 3}\right )}}{x} \]

[In]

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")

[Out]

e^(-(3*x^2 - (x^2 + 10*x + 25)*log(3) - (log(3) - 3)*log(12*x) + 29*x + 75)/(log(3) - 3))/x

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\text {Timed out} \]

[In]

integrate(((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*l
n(3)-3*x),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (23) = 46\).

Time = 0.48 (sec) , antiderivative size = 120, normalized size of antiderivative = 6.67 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=\frac {3^{\frac {22}{\log \left (3\right ) - 3}} 2^{\frac {2 \, \log \left (3\right )}{\log \left (3\right ) - 3} - \frac {6}{\log \left (3\right ) - 3}} e^{\left (\frac {x^{2} \log \left (3\right )}{\log \left (3\right ) - 3} - \frac {3 \, x^{2}}{\log \left (3\right ) - 3} + \frac {10 \, x \log \left (3\right )}{\log \left (3\right ) - 3} + \frac {\log \left (3\right )^{2}}{\log \left (3\right ) - 3} + \frac {\log \left (3\right ) \log \left (x\right )}{\log \left (3\right ) - 3} - \frac {29 \, x}{\log \left (3\right ) - 3} - \frac {3 \, \log \left (x\right )}{\log \left (3\right ) - 3} - \frac {75}{\log \left (3\right ) - 3}\right )}}{x} \]

[In]

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")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.22 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx=4 \, e^{\left (\frac {x^{2} \log \left (3\right ) - 3 \, x^{2} + 10 \, x \log \left (3\right ) + \log \left (3\right )^{2} - 29 \, x - 3 \, \log \left (3\right )}{\log \left (3\right ) - 3} + 25\right )} \]

[In]

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")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 12.57 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.28 \[ \int \frac {e^{\frac {-75-29 x-3 x^2+\left (25+10 x+x^2\right ) \log (3)+(-3+\log (3)) \log (12 x)}{-3+\log (3)}} (-29-6 x+(10+2 x) \log (3))}{-3 x+x \log (3)} \, dx={\left (\frac {1}{64}\right )}^{\frac {1}{\ln \left (3\right )-3}}\,3^{\frac {x^2+10\,x+\ln \left (12\right )+22}{\ln \left (3\right )-3}}\,{\mathrm {e}}^{-\frac {29\,x}{\ln \left (3\right )-3}}\,{\mathrm {e}}^{-\frac {3\,x^2}{\ln \left (3\right )-3}}\,{\mathrm {e}}^{-\frac {75}{\ln \left (3\right )-3}} \]

[In]

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)

[Out]

(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)/(l
og(3) - 3))*exp(-75/(log(3) - 3))