Integrand size = 53, antiderivative size = 25 \[ \int e^{\frac {127-250 x+125 x^2+5 \log (2)+\left (25-50 x+25 x^2+\log (2)\right ) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx=2 e^{25 (-1+x)^2+\frac {2}{5+\log (3) \log ^2(4)}} \]
Leaf count is larger than twice the leaf count of optimal. \(76\) vs. \(2(25)=50\).
Time = 0.98 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.04 \[ \int e^{\frac {127-250 x+125 x^2+5 \log (2)+\left (25-50 x+25 x^2+\log (2)\right ) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx=\frac {e^{\frac {127+(25+\log (2)) \log (3) \log ^2(4)-50 x \left (5+\log (3) \log ^2(4)\right )+25 x^2 \left (5+\log (3) \log ^2(4)\right )+\log (32)}{5+\log (3) \log ^2(4)}} \left (5+\log ^2(2) \log (81)\right )}{5+\log (3) \log ^2(4)} \]
Integrate[E^((127 - 250*x + 125*x^2 + 5*Log[2] + (25 - 50*x + 25*x^2 + Log [2])*Log[3]*Log[4]^2)/(5 + Log[3]*Log[4]^2))*(-50 + 50*x),x]
(E^((127 + (25 + Log[2])*Log[3]*Log[4]^2 - 50*x*(5 + Log[3]*Log[4]^2) + 25 *x^2*(5 + Log[3]*Log[4]^2) + Log[32])/(5 + Log[3]*Log[4]^2))*(5 + Log[2]^2 *Log[81]))/(5 + Log[3]*Log[4]^2)
Time = 0.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {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.
\(\displaystyle \int (50 x-50) \exp \left (\frac {125 x^2+\log (3) \log ^2(4) \left (25 x^2-50 x+25+\log (2)\right )-250 x+127+5 \log (2)}{5+\log (3) \log ^2(4)}\right ) \, dx\) |
\(\Big \downarrow \) 2674 |
\(\displaystyle \int (50 x-50) \exp \left (25 x^2-50 x+\frac {127+(25+\log (2)) \log (3) \log ^2(4)+\log (32)}{5+\log (3) \log ^2(4)}\right )dx\) |
\(\Big \downarrow \) 2666 |
\(\displaystyle \exp \left (25 x^2-50 x+\frac {127+(25+\log (2)) \log (3) \log ^2(4)+\log (32)}{5+\log (3) \log ^2(4)}\right )\) |
Int[E^((127 - 250*x + 125*x^2 + 5*Log[2] + (25 - 50*x + 25*x^2 + Log[2])*L og[3]*Log[4]^2)/(5 + Log[3]*Log[4]^2))*(-50 + 50*x),x]
3.15.42.3.1 Defintions of rubi rules used
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])
Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.96
method | result | size |
derivativedivides | \({\mathrm e}^{\frac {4 \left (\ln \left (2\right )+25 x^{2}-50 x +25\right ) \ln \left (3\right ) \ln \left (2\right )^{2}+5 \ln \left (2\right )+125 x^{2}-250 x +127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}\) | \(49\) |
norman | \({\mathrm e}^{\frac {4 \left (\ln \left (2\right )+25 x^{2}-50 x +25\right ) \ln \left (3\right ) \ln \left (2\right )^{2}+5 \ln \left (2\right )+125 x^{2}-250 x +127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}\) | \(49\) |
parallelrisch | \({\mathrm e}^{\frac {4 \left (\ln \left (2\right )+25 x^{2}-50 x +25\right ) \ln \left (3\right ) \ln \left (2\right )^{2}+5 \ln \left (2\right )+125 x^{2}-250 x +127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}\) | \(49\) |
gosper | \({\mathrm e}^{\frac {100 x^{2} \ln \left (3\right ) \ln \left (2\right )^{2}+4 \ln \left (3\right ) \ln \left (2\right )^{3}-200 \ln \left (3\right ) \ln \left (2\right )^{2} x +100 \ln \left (2\right )^{2} \ln \left (3\right )+125 x^{2}+5 \ln \left (2\right )-250 x +127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}\) | \(65\) |
risch | \({\mathrm e}^{\frac {100 x^{2} \ln \left (3\right ) \ln \left (2\right )^{2}+4 \ln \left (3\right ) \ln \left (2\right )^{3}-200 \ln \left (3\right ) \ln \left (2\right )^{2} x +100 \ln \left (2\right )^{2} \ln \left (3\right )+125 x^{2}+5 \ln \left (2\right )-250 x +127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}\) | \(65\) |
default | \(\frac {25 i \sqrt {\pi }\, {\mathrm e}^{\frac {4 \left (\ln \left (2\right )+25\right ) \ln \left (3\right ) \ln \left (2\right )^{2}+5 \ln \left (2\right )+127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}-\frac {\left (-200 \ln \left (2\right )^{2} \ln \left (3\right )-250\right )^{2}}{4 \left (4 \ln \left (2\right )^{2} \ln \left (3\right )+5\right ) \left (100 \ln \left (2\right )^{2} \ln \left (3\right )+125\right )}} \operatorname {erf}\left (i \sqrt {\frac {100 \ln \left (2\right )^{2} \ln \left (3\right )+125}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}\, x +\frac {i \left (-200 \ln \left (2\right )^{2} \ln \left (3\right )-250\right )}{2 \left (4 \ln \left (2\right )^{2} \ln \left (3\right )+5\right ) \sqrt {\frac {100 \ln \left (2\right )^{2} \ln \left (3\right )+125}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}}\right )}{\sqrt {\frac {100 \ln \left (2\right )^{2} \ln \left (3\right )+125}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}}+\frac {25 \left (4 \ln \left (2\right )^{2} \ln \left (3\right )+5\right ) {\mathrm e}^{\frac {\left (100 \ln \left (2\right )^{2} \ln \left (3\right )+125\right ) x^{2}}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}+\frac {\left (-200 \ln \left (2\right )^{2} \ln \left (3\right )-250\right ) x}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}+\frac {4 \left (\ln \left (2\right )+25\right ) \ln \left (3\right ) \ln \left (2\right )^{2}+5 \ln \left (2\right )+127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}}{100 \ln \left (2\right )^{2} \ln \left (3\right )+125}+\frac {25 i \left (-200 \ln \left (2\right )^{2} \ln \left (3\right )-250\right ) \sqrt {\pi }\, {\mathrm e}^{\frac {4 \left (\ln \left (2\right )+25\right ) \ln \left (3\right ) \ln \left (2\right )^{2}+5 \ln \left (2\right )+127}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}-\frac {\left (-200 \ln \left (2\right )^{2} \ln \left (3\right )-250\right )^{2}}{4 \left (4 \ln \left (2\right )^{2} \ln \left (3\right )+5\right ) \left (100 \ln \left (2\right )^{2} \ln \left (3\right )+125\right )}} \operatorname {erf}\left (i \sqrt {\frac {100 \ln \left (2\right )^{2} \ln \left (3\right )+125}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}\, x +\frac {i \left (-200 \ln \left (2\right )^{2} \ln \left (3\right )-250\right )}{2 \left (4 \ln \left (2\right )^{2} \ln \left (3\right )+5\right ) \sqrt {\frac {100 \ln \left (2\right )^{2} \ln \left (3\right )+125}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}}\right )}{2 \left (100 \ln \left (2\right )^{2} \ln \left (3\right )+125\right ) \sqrt {\frac {100 \ln \left (2\right )^{2} \ln \left (3\right )+125}{4 \ln \left (2\right )^{2} \ln \left (3\right )+5}}}\) | \(497\) |
parts | \(\text {Expression too large to display}\) | \(719\) |
int((50*x-50)*exp((4*(ln(2)+25*x^2-50*x+25)*ln(3)*ln(2)^2+5*ln(2)+125*x^2- 250*x+127)/(4*ln(2)^2*ln(3)+5)),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (25) = 50\).
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int e^{\frac {127-250 x+125 x^2+5 \log (2)+\left (25-50 x+25 x^2+\log (2)\right ) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx=e^{\left (\frac {125 \, x^{2} + 4 \, {\left (25 \, {\left (x^{2} - 2 \, x + 1\right )} \log \left (2\right )^{2} + \log \left (2\right )^{3}\right )} \log \left (3\right ) - 250 \, x + 5 \, \log \left (2\right ) + 127}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5}\right )} \]
integrate((50*x-50)*exp((4*(log(2)+25*x^2-50*x+25)*log(3)*log(2)^2+5*log(2 )+125*x^2-250*x+127)/(4*log(2)^2*log(3)+5)),x, algorithm=\
e^((125*x^2 + 4*(25*(x^2 - 2*x + 1)*log(2)^2 + log(2)^3)*log(3) - 250*x + 5*log(2) + 127)/(4*log(3)*log(2)^2 + 5))
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.04 \[ \int e^{\frac {127-250 x+125 x^2+5 \log (2)+\left (25-50 x+25 x^2+\log (2)\right ) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx=e^{\frac {125 x^{2} - 250 x + \left (100 x^{2} - 200 x + 4 \log {\left (2 \right )} + 100\right ) \log {\left (2 \right )}^{2} \log {\left (3 \right )} + 5 \log {\left (2 \right )} + 127}{4 \log {\left (2 \right )}^{2} \log {\left (3 \right )} + 5}} \]
integrate((50*x-50)*exp((4*(ln(2)+25*x**2-50*x+25)*ln(3)*ln(2)**2+5*ln(2)+ 125*x**2-250*x+127)/(4*ln(2)**2*ln(3)+5)),x)
exp((125*x**2 - 250*x + (100*x**2 - 200*x + 4*log(2) + 100)*log(2)**2*log( 3) + 5*log(2) + 127)/(4*log(2)**2*log(3) + 5))
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.38 (sec) , antiderivative size = 200, normalized size of antiderivative = 8.00 \[ \int e^{\frac {127-250 x+125 x^2+5 \log (2)+\left (25-50 x+25 x^2+\log (2)\right ) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx=10 i \, \sqrt {\pi } 3^{\frac {4 \, \log \left (2\right )^{3}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {100 \, \log \left (2\right )^{2}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5}} 2^{\frac {5}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} - 1} \operatorname {erf}\left (5 i \, x - 5 i\right ) e^{\left (\frac {127}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} - 25\right )} + {\left (\frac {5 \, \sqrt {\pi } {\left (x - 1\right )} {\left (\operatorname {erf}\left (5 \, \sqrt {-{\left (x - 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (x - 1\right )}^{2}}} + e^{\left (25 \, {\left (x - 1\right )}^{2}\right )}\right )} e^{\left (\frac {4 \, \log \left (3\right ) \log \left (2\right )^{3}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {100 \, \log \left (3\right ) \log \left (2\right )^{2}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {5 \, \log \left (2\right )}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {127}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} - 25\right )} \]
integrate((50*x-50)*exp((4*(log(2)+25*x^2-50*x+25)*log(3)*log(2)^2+5*log(2 )+125*x^2-250*x+127)/(4*log(2)^2*log(3)+5)),x, algorithm=\
10*I*sqrt(pi)*3^(4*log(2)^3/(4*log(3)*log(2)^2 + 5) + 100*log(2)^2/(4*log( 3)*log(2)^2 + 5))*2^(5/(4*log(3)*log(2)^2 + 5) - 1)*erf(5*I*x - 5*I)*e^(12 7/(4*log(3)*log(2)^2 + 5) - 25) + (5*sqrt(pi)*(x - 1)*(erf(5*sqrt(-(x - 1) ^2)) - 1)/sqrt(-(x - 1)^2) + e^(25*(x - 1)^2))*e^(4*log(3)*log(2)^3/(4*log (3)*log(2)^2 + 5) + 100*log(3)*log(2)^2/(4*log(3)*log(2)^2 + 5) + 5*log(2) /(4*log(3)*log(2)^2 + 5) + 127/(4*log(3)*log(2)^2 + 5) - 25)
Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 148, normalized size of antiderivative = 5.92 \[ \int e^{\frac {127-250 x+125 x^2+5 \log (2)+\left (25-50 x+25 x^2+\log (2)\right ) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx=e^{\left (\frac {100 \, x^{2} \log \left (3\right ) \log \left (2\right )^{2}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} - \frac {200 \, x \log \left (3\right ) \log \left (2\right )^{2}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {4 \, \log \left (3\right ) \log \left (2\right )^{3}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {100 \, \log \left (3\right ) \log \left (2\right )^{2}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {125 \, x^{2}}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} - \frac {250 \, x}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {5 \, \log \left (2\right )}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5} + \frac {127}{4 \, \log \left (3\right ) \log \left (2\right )^{2} + 5}\right )} \]
integrate((50*x-50)*exp((4*(log(2)+25*x^2-50*x+25)*log(3)*log(2)^2+5*log(2 )+125*x^2-250*x+127)/(4*log(2)^2*log(3)+5)),x, algorithm=\
e^(100*x^2*log(3)*log(2)^2/(4*log(3)*log(2)^2 + 5) - 200*x*log(3)*log(2)^2 /(4*log(3)*log(2)^2 + 5) + 4*log(3)*log(2)^3/(4*log(3)*log(2)^2 + 5) + 100 *log(3)*log(2)^2/(4*log(3)*log(2)^2 + 5) + 125*x^2/(4*log(3)*log(2)^2 + 5) - 250*x/(4*log(3)*log(2)^2 + 5) + 5*log(2)/(4*log(3)*log(2)^2 + 5) + 127/ (4*log(3)*log(2)^2 + 5))
Time = 0.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.24 \[ \int e^{\frac {127-250 x+125 x^2+5 \log (2)+\left (25-50 x+25 x^2+\log (2)\right ) \log (3) \log ^2(4)}{5+\log (3) \log ^2(4)}} (-50+50 x) \, dx={32}^{\frac {1}{4\,{\ln \left (2\right )}^2\,\ln \left (3\right )+5}}\,{81}^{\frac {25\,{\ln \left (2\right )}^2\,x^2-50\,{\ln \left (2\right )}^2\,x+25\,{\ln \left (2\right )}^2+{\ln \left (2\right )}^3}{4\,{\ln \left (2\right )}^2\,\ln \left (3\right )+5}}\,{\mathrm {e}}^{\frac {127}{4\,{\ln \left (2\right )}^2\,\ln \left (3\right )+5}}\,{\mathrm {e}}^{-\frac {250\,x}{4\,{\ln \left (2\right )}^2\,\ln \left (3\right )+5}}\,{\mathrm {e}}^{\frac {125\,x^2}{4\,{\ln \left (2\right )}^2\,\ln \left (3\right )+5}} \]
int(exp((5*log(2) - 250*x + 125*x^2 + 4*log(2)^2*log(3)*(log(2) - 50*x + 2 5*x^2 + 25) + 127)/(4*log(2)^2*log(3) + 5))*(50*x - 50),x)