Integrand size = 78, antiderivative size = 29 \[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=e^{-2 x+\frac {e^x (5+x) \log (5)}{3 (-5+2 x)}}-x \] Output:
exp(1/3*(5+x)*exp(x)/(-5+2*x)*ln(5)-2*x)-x
Time = 1.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=\frac {1}{3} \left (3\ 5^{\frac {e^x (5+x)}{-15+6 x}} e^{-2 x}-3 x\right ) \] Input:
Integrate[(-75 + 60*x - 12*x^2 + E^((30*x - 12*x^2 + E^x*(5 + x)*Log[5])/( -15 + 6*x))*(-150 + 120*x - 24*x^2 + E^x*(-40 + 5*x + 2*x^2)*Log[5]))/(75 - 60*x + 12*x^2),x]
Output:
((3*5^((E^x*(5 + x))/(-15 + 6*x)))/E^(2*x) - 3*x)/3
Leaf count is larger than twice the leaf count of optimal. \(104\) vs. \(2(29)=58\).
Time = 2.87 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.59, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {7277, 27, 7293, 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.
\(\displaystyle \int \frac {\left (-24 x^2+e^x \left (2 x^2+5 x-40\right ) \log (5)+120 x-150\right ) \exp \left (\frac {-12 x^2+30 x+e^x (x+5) \log (5)}{6 x-15}\right )-12 x^2+60 x-75}{12 x^2-60 x+75} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 48 \int -\frac {12 x^2-60 x+5^{-\frac {e^x (x+5)}{15-6 x}} e^{-\frac {2 \left (5 x-2 x^2\right )}{5-2 x}} \left (24 x^2-120 x+e^x \left (-2 x^2-5 x+40\right ) \log (5)+150\right )+75}{144 (5-2 x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{3} \int \frac {12 x^2-60 x+5^{-\frac {e^x (x+5)}{3 (5-2 x)}} e^{-\frac {2 \left (5 x-2 x^2\right )}{5-2 x}} \left (24 x^2-120 x+e^x \left (-2 x^2-5 x+40\right ) \log (5)+150\right )+75}{(5-2 x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{3} \int \left (3-\frac {5^{\frac {e^x (x+5)}{6 x-15}} e^{-2 x} \left (2 e^x \log (5) x^2-24 x^2+5 e^x \log (5) x+120 x-40 e^x \log (5)-150\right )}{(2 x-5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\frac {3\ 5^{-\frac {e^x (x+5)}{3 (5-2 x)}} e^{-2 x} \left (-2 e^x x^2-5 e^x x+40 e^x\right )}{(5-2 x)^2 \left (\frac {e^x (x+5)}{5-2 x}+\frac {2 e^x (x+5)}{(5-2 x)^2}+\frac {e^x}{5-2 x}\right )}-3 x\right )\) |
Input:
Int[(-75 + 60*x - 12*x^2 + E^((30*x - 12*x^2 + E^x*(5 + x)*Log[5])/(-15 + 6*x))*(-150 + 120*x - 24*x^2 + E^x*(-40 + 5*x + 2*x^2)*Log[5]))/(75 - 60*x + 12*x^2),x]
Output:
(-3*x + (3*(40*E^x - 5*E^x*x - 2*E^x*x^2))/(5^((E^x*(5 + x))/(3*(5 - 2*x)) )*E^(2*x)*(5 - 2*x)^2*(E^x/(5 - 2*x) + (2*E^x*(5 + x))/(5 - 2*x)^2 + (E^x* (5 + x))/(5 - 2*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_)*((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]
Time = 0.93 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(-x +{\mathrm e}^{\frac {\left (5+x \right ) \ln \left (5\right ) {\mathrm e}^{x}-12 x^{2}+30 x}{6 x -15}}-10\) | \(33\) |
risch | \(-x +5^{\frac {{\mathrm e}^{x} x}{6 x -15}} 5^{\frac {5 \,{\mathrm e}^{x}}{3 \left (-5+2 x \right )}} {\mathrm e}^{-2 x}\) | \(37\) |
norman | \(\frac {-2 x^{2}+2 x \,{\mathrm e}^{\frac {\left (5+x \right ) \ln \left (5\right ) {\mathrm e}^{x}-12 x^{2}+30 x}{6 x -15}}-5 \,{\mathrm e}^{\frac {\left (5+x \right ) \ln \left (5\right ) {\mathrm e}^{x}-12 x^{2}+30 x}{6 x -15}}+\frac {25}{2}}{-5+2 x}\) | \(73\) |
Input:
int((((2*x^2+5*x-40)*ln(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*ln(5)*exp(x )-12*x^2+30*x)/(6*x-15))-12*x^2+60*x-75)/(12*x^2-60*x+75),x,method=_RETURN VERBOSE)
Output:
-x+exp(1/3/(-5+2*x)*((5+x)*ln(5)*exp(x)-12*x^2+30*x))-10
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=-x + e^{\left (\frac {{\left (x + 5\right )} e^{x} \log \left (5\right ) - 12 \, x^{2} + 30 \, x}{3 \, {\left (2 \, x - 5\right )}}\right )} \] Input:
integrate((((2*x^2+5*x-40)*log(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*log( 5)*exp(x)-12*x^2+30*x)/(6*x-15))-12*x^2+60*x-75)/(12*x^2-60*x+75),x, algor ithm="fricas")
Output:
-x + e^(1/3*((x + 5)*e^x*log(5) - 12*x^2 + 30*x)/(2*x - 5))
Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=- x + e^{\frac {- 12 x^{2} + 30 x + \left (x + 5\right ) e^{x} \log {\left (5 \right )}}{6 x - 15}} \] Input:
integrate((((2*x**2+5*x-40)*ln(5)*exp(x)-24*x**2+120*x-150)*exp(((5+x)*ln( 5)*exp(x)-12*x**2+30*x)/(6*x-15))-12*x**2+60*x-75)/(12*x**2-60*x+75),x)
Output:
-x + exp((-12*x**2 + 30*x + (x + 5)*exp(x)*log(5))/(6*x - 15))
\[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=\int { -\frac {12 \, x^{2} - {\left ({\left (2 \, x^{2} + 5 \, x - 40\right )} e^{x} \log \left (5\right ) - 24 \, x^{2} + 120 \, x - 150\right )} e^{\left (\frac {{\left (x + 5\right )} e^{x} \log \left (5\right ) - 12 \, x^{2} + 30 \, x}{3 \, {\left (2 \, x - 5\right )}}\right )} - 60 \, x + 75}{3 \, {\left (4 \, x^{2} - 20 \, x + 25\right )}} \,d x } \] Input:
integrate((((2*x^2+5*x-40)*log(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*log( 5)*exp(x)-12*x^2+30*x)/(6*x-15))-12*x^2+60*x-75)/(12*x^2-60*x+75),x, algor ithm="maxima")
Output:
-x + 1/3*integrate(-(24*x^2 - (2*x^2*log(5) + 5*x*log(5) - 40*log(5))*e^x - 120*x + 150)*e^(1/6*e^x*log(5) - 2*x + 5/2*e^x*log(5)/(2*x - 5))/(4*x^2 - 20*x + 25), x)
\[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=\int { -\frac {12 \, x^{2} - {\left ({\left (2 \, x^{2} + 5 \, x - 40\right )} e^{x} \log \left (5\right ) - 24 \, x^{2} + 120 \, x - 150\right )} e^{\left (\frac {{\left (x + 5\right )} e^{x} \log \left (5\right ) - 12 \, x^{2} + 30 \, x}{3 \, {\left (2 \, x - 5\right )}}\right )} - 60 \, x + 75}{3 \, {\left (4 \, x^{2} - 20 \, x + 25\right )}} \,d x } \] Input:
integrate((((2*x^2+5*x-40)*log(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*log( 5)*exp(x)-12*x^2+30*x)/(6*x-15))-12*x^2+60*x-75)/(12*x^2-60*x+75),x, algor ithm="giac")
Output:
integrate(-1/3*(12*x^2 - ((2*x^2 + 5*x - 40)*e^x*log(5) - 24*x^2 + 120*x - 150)*e^(1/3*((x + 5)*e^x*log(5) - 12*x^2 + 30*x)/(2*x - 5)) - 60*x + 75)/ (4*x^2 - 20*x + 25), x)
Time = 7.16 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.90 \[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=5^{\frac {5\,{\mathrm {e}}^x}{6\,x-15}}\,5^{\frac {x\,{\mathrm {e}}^x}{6\,x-15}}\,{\mathrm {e}}^{-\frac {12\,x^2}{6\,x-15}}\,{\mathrm {e}}^{\frac {30\,x}{6\,x-15}}-x \] Input:
int((60*x + exp((30*x - 12*x^2 + exp(x)*log(5)*(x + 5))/(6*x - 15))*(120*x - 24*x^2 + exp(x)*log(5)*(5*x + 2*x^2 - 40) - 150) - 12*x^2 - 75)/(12*x^2 - 60*x + 75),x)
Output:
5^((5*exp(x))/(6*x - 15))*5^((x*exp(x))/(6*x - 15))*exp(-(12*x^2)/(6*x - 1 5))*exp((30*x)/(6*x - 15)) - x
Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {-75+60 x-12 x^2+e^{\frac {30 x-12 x^2+e^x (5+x) \log (5)}{-15+6 x}} \left (-150+120 x-24 x^2+e^x \left (-40+5 x+2 x^2\right ) \log (5)\right )}{75-60 x+12 x^2} \, dx=\frac {e^{\frac {e^{x} \mathrm {log}\left (5\right ) x +5 e^{x} \mathrm {log}\left (5\right )}{6 x -15}}-e^{2 x} x}{e^{2 x}} \] Input:
int((((2*x^2+5*x-40)*log(5)*exp(x)-24*x^2+120*x-150)*exp(((5+x)*log(5)*exp (x)-12*x^2+30*x)/(6*x-15))-12*x^2+60*x-75)/(12*x^2-60*x+75),x)
Output:
(e**((e**x*log(5)*x + 5*e**x*log(5))/(6*x - 15)) - e**(2*x)*x)/e**(2*x)