Integrand size = 116, antiderivative size = 28 \[ \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{49 x} \, dx=2+\frac {1}{2} \left (25+e^{x \left (e^5+x\right )}-\frac {x}{7}+\log \left (x^2\right )\right )^2 \]
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{49 x} \, dx=\frac {1}{98} \left (-175-7 e^{x \left (e^5+x\right )}+x-7 \log \left (x^2\right )\right )^2 \]
Integrate[(2450 - 189*x + x^2 + E^(2*E^5*x + 2*x^2)*(49*E^5*x + 98*x^2) + E^(E^5*x + x^2)*(98 - 7*x + 2450*x^2 - 14*x^3 + E^5*(1225*x - 7*x^2)) + (9 8 - 7*x + E^(E^5*x + x^2)*(49*E^5*x + 98*x^2))*Log[x^2])/(49*x),x]
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.98 (sec) , antiderivative size = 247, normalized size of antiderivative = 8.82, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {27, 2010, 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.
\(\displaystyle \int \frac {x^2+e^{2 x^2+2 e^5 x} \left (98 x^2+49 e^5 x\right )+\left (e^{x^2+e^5 x} \left (98 x^2+49 e^5 x\right )-7 x+98\right ) \log \left (x^2\right )+e^{x^2+e^5 x} \left (-14 x^3+2450 x^2+e^5 \left (1225 x-7 x^2\right )-7 x+98\right )-189 x+2450}{49 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{49} \int \frac {x^2-189 x+49 e^{2 x^2+2 e^5 x} \left (2 x^2+e^5 x\right )+7 e^{x^2+e^5 x} \left (-2 x^3+350 x^2-x+e^5 \left (175 x-x^2\right )+14\right )+7 \left (-x+7 e^{x^2+e^5 x} \left (2 x^2+e^5 x\right )+14\right ) \log \left (x^2\right )+2450}{x}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{49} \int \left (49 e^{2 x \left (x+e^5\right )} \left (2 x+e^5\right )+\frac {(x-14) \left (x-7 \log \left (x^2\right )-175\right )}{x}+\frac {7 e^{x^2+e^5 x} \left (-2 x^3+14 \log \left (x^2\right ) x^2+350 \left (1-\frac {e^5}{350}\right ) x^2+7 e^5 \log \left (x^2\right ) x-\left (1-175 e^5\right ) x+14\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{49} \left (-\frac {7}{4} e^{5-\frac {e^{10}}{4}} \left (350-e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )-\frac {7}{2} e^{-\frac {e^{10}}{4}} \left (1-175 e^5\right ) \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )-\frac {7}{4} e^{10-\frac {e^{10}}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )+\frac {7}{2} e^{-\frac {e^{10}}{4}} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (2 x+e^5\right )\right )+\frac {7}{2} e^{x^2+e^5 x+5}+\frac {49}{2} e^{2 x^2+2 e^5 x}-7 e^{x^2+e^5 x} x+\frac {7}{2} \left (350-e^5\right ) e^{x^2+e^5 x}+\frac {1}{2} \left (7 \log \left (x^2\right )-x+175\right )^2+49 e^{x^2+e^5 x} \log \left (x^2\right )\right )\) |
Int[(2450 - 189*x + x^2 + E^(2*E^5*x + 2*x^2)*(49*E^5*x + 98*x^2) + E^(E^5 *x + x^2)*(98 - 7*x + 2450*x^2 - 14*x^3 + E^5*(1225*x - 7*x^2)) + (98 - 7* x + E^(E^5*x + x^2)*(49*E^5*x + 98*x^2))*Log[x^2])/(49*x),x]
((7*E^(5 + E^5*x + x^2))/2 + (49*E^(2*E^5*x + 2*x^2))/2 + (7*E^(E^5*x + x^ 2)*(350 - E^5))/2 - 7*E^(E^5*x + x^2)*x + (7*Sqrt[Pi]*Erfi[(E^5 + 2*x)/2]) /(2*E^(E^10/4)) - (7*E^(10 - E^10/4)*Sqrt[Pi]*Erfi[(E^5 + 2*x)/2])/4 - (7* (1 - 175*E^5)*Sqrt[Pi]*Erfi[(E^5 + 2*x)/2])/(2*E^(E^10/4)) - (7*E^(5 - E^1 0/4)*(350 - E^5)*Sqrt[Pi]*Erfi[(E^5 + 2*x)/2])/4 + 49*E^(E^5*x + x^2)*Log[ x^2] + (175 - x + 7*Log[x^2])^2/2)/49
3.20.14.3.1 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_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Leaf count of result is larger than twice the leaf count of optimal. \(72\) vs. \(2(22)=44\).
Time = 0.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.61
method | result | size |
parallelrisch | \(\frac {x^{2}}{98}-\frac {x \ln \left (x^{2}\right )}{7}-\frac {{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} x}{7}+\frac {\ln \left (x^{2}\right )^{2}}{2}+\ln \left (x^{2}\right ) {\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x}+\frac {{\mathrm e}^{2 \left ({\mathrm e}^{5}+x \right ) x}}{2}-\frac {25 x}{7}+25 \ln \left (x^{2}\right )+25 \,{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x}\) | \(73\) |
default | \(\frac {{\mathrm e}^{2 x \,{\mathrm e}^{5}+2 x^{2}}}{2}+\frac {\left (1225+49 \ln \left (x^{2}\right )-98 \ln \left (x \right )\right ) {\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}}}{49}-\frac {{\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}} x}{7}+2 \ln \left (x \right ) {\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}}+\frac {x^{2}}{98}-\frac {25 x}{7}+50 \ln \left (x \right )+2 \ln \left (x \right ) \ln \left (x^{2}\right )-2 \ln \left (x \right )^{2}-\frac {x \ln \left (x^{2}\right )}{7}\) | \(96\) |
parts | \(\frac {{\mathrm e}^{2 x \,{\mathrm e}^{5}+2 x^{2}}}{2}+\frac {\left (175+7 \ln \left (x^{2}\right )-14 \ln \left (x \right )\right ) {\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}}}{7}-\frac {{\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}} x}{7}+2 \ln \left (x \right ) {\mathrm e}^{x \,{\mathrm e}^{5}+x^{2}}+\frac {x^{2}}{98}-\frac {25 x}{7}+50 \ln \left (x \right )+2 \ln \left (x \right ) \ln \left (x^{2}\right )-2 \ln \left (x \right )^{2}-\frac {x \ln \left (x^{2}\right )}{7}\) | \(96\) |
risch | \(2 \ln \left (x \right )^{2}+\frac {\left (-14 x +98 \,{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x}\right ) \ln \left (x \right )}{49}+2 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}}{7}+\frac {i \pi x \operatorname {csgn}\left (i x^{2}\right )^{3}}{14}+\frac {x^{2}}{98}-\frac {25 x}{7}-\frac {i {\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{2}+i {\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\frac {i \pi x \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )}{14}+50 \ln \left (x \right )+\frac {{\mathrm e}^{2 \left ({\mathrm e}^{5}+x \right ) x}}{2}-\frac {i {\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}-i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x^{2}\right )^{3}-i \pi \ln \left (x \right ) \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-\frac {{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x} x}{7}+25 \,{\mathrm e}^{\left ({\mathrm e}^{5}+x \right ) x}\) | \(243\) |
int(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*ln(x^2)+(49*x*ex p(5)+98*x^2)*exp(x*exp(5)+x^2)^2+((-7*x^2+1225*x)*exp(5)-14*x^3+2450*x^2-7 *x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x,method=_RETURNVERBOSE)
1/98*x^2-1/7*x*ln(x^2)-1/7*exp((exp(5)+x)*x)*x+1/2*ln(x^2)^2+ln(x^2)*exp(( exp(5)+x)*x)+1/2*exp((exp(5)+x)*x)^2-25/7*x+25*ln(x^2)+25*exp((exp(5)+x)*x )
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.32 \[ \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{49 x} \, dx=\frac {1}{98} \, x^{2} - \frac {1}{7} \, {\left (x - 175\right )} e^{\left (x^{2} + x e^{5}\right )} - \frac {1}{7} \, {\left (x - 7 \, e^{\left (x^{2} + x e^{5}\right )} - 175\right )} \log \left (x^{2}\right ) + \frac {1}{2} \, \log \left (x^{2}\right )^{2} - \frac {25}{7} \, x + \frac {1}{2} \, e^{\left (2 \, x^{2} + 2 \, x e^{5}\right )} \]
integrate(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*log(x^2)+( 49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)^2+((-7*x^2+1225*x)*exp(5)-14*x^3+245 0*x^2-7*x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x, algorithm=\
1/98*x^2 - 1/7*(x - 175)*e^(x^2 + x*e^5) - 1/7*(x - 7*e^(x^2 + x*e^5) - 17 5)*log(x^2) + 1/2*log(x^2)^2 - 25/7*x + 1/2*e^(2*x^2 + 2*x*e^5)
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (22) = 44\).
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.50 \[ \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{49 x} \, dx=\frac {x^{2}}{98} - \frac {x \log {\left (x^{2} \right )}}{7} - \frac {25 x}{7} + \frac {\left (- 2 x + 14 \log {\left (x^{2} \right )} + 350\right ) e^{x^{2} + x e^{5}}}{14} + \frac {e^{2 x^{2} + 2 x e^{5}}}{2} + 50 \log {\left (x \right )} + \frac {\log {\left (x^{2} \right )}^{2}}{2} \]
integrate(1/49*(((49*x*exp(5)+98*x**2)*exp(x*exp(5)+x**2)-7*x+98)*ln(x**2) +(49*x*exp(5)+98*x**2)*exp(x*exp(5)+x**2)**2+((-7*x**2+1225*x)*exp(5)-14*x **3+2450*x**2-7*x+98)*exp(x*exp(5)+x**2)+x**2-189*x+2450)/x,x)
x**2/98 - x*log(x**2)/7 - 25*x/7 + (-2*x + 14*log(x**2) + 350)*exp(x**2 + x*exp(5))/14 + exp(2*x**2 + 2*x*exp(5))/2 + 50*log(x) + log(x**2)**2/2
\[ \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{49 x} \, dx=\int { \frac {x^{2} + 49 \, {\left (2 \, x^{2} + x e^{5}\right )} e^{\left (2 \, x^{2} + 2 \, x e^{5}\right )} - 7 \, {\left (2 \, x^{3} - 350 \, x^{2} + {\left (x^{2} - 175 \, x\right )} e^{5} + x - 14\right )} e^{\left (x^{2} + x e^{5}\right )} + 7 \, {\left (7 \, {\left (2 \, x^{2} + x e^{5}\right )} e^{\left (x^{2} + x e^{5}\right )} - x + 14\right )} \log \left (x^{2}\right ) - 189 \, x + 2450}{49 \, x} \,d x } \]
integrate(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*log(x^2)+( 49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)^2+((-7*x^2+1225*x)*exp(5)-14*x^3+245 0*x^2-7*x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x, algorithm=\
1/14*I*sqrt(pi)*erf(I*x + 1/2*I*e^5)*e^(-1/4*e^10) + 1/98*x^2 - 2/7*x*log( x) + 2*log(x)^2 - 25/7*x + 1/2*e^(2*x^2 + 2*x*e^5) + 1/49*integrate(-7*(2* x^3 + x^2*(e^5 - 350) - 175*x*e^5 - 14*(2*x^2 + x*e^5)*log(x) - 14)*e^(x^2 + x*e^5)/x, x) + 50*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (24) = 48\).
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.82 \[ \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{49 x} \, dx=\frac {1}{98} \, x^{2} - \frac {1}{7} \, x e^{\left (x^{2} + x e^{5}\right )} - \frac {1}{7} \, x \log \left (x^{2}\right ) + e^{\left (x^{2} + x e^{5}\right )} \log \left (x^{2}\right ) + \frac {1}{2} \, \log \left (x^{2}\right )^{2} - \frac {25}{7} \, x + \frac {1}{2} \, e^{\left (2 \, x^{2} + 2 \, x e^{5}\right )} + 25 \, e^{\left (x^{2} + x e^{5}\right )} + 50 \, \log \left (x\right ) \]
integrate(1/49*(((49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)-7*x+98)*log(x^2)+( 49*x*exp(5)+98*x^2)*exp(x*exp(5)+x^2)^2+((-7*x^2+1225*x)*exp(5)-14*x^3+245 0*x^2-7*x+98)*exp(x*exp(5)+x^2)+x^2-189*x+2450)/x,x, algorithm=\
1/98*x^2 - 1/7*x*e^(x^2 + x*e^5) - 1/7*x*log(x^2) + e^(x^2 + x*e^5)*log(x^ 2) + 1/2*log(x^2)^2 - 25/7*x + 1/2*e^(2*x^2 + 2*x*e^5) + 25*e^(x^2 + x*e^5 ) + 50*log(x)
Time = 10.72 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.89 \[ \int \frac {2450-189 x+x^2+e^{2 e^5 x+2 x^2} \left (49 e^5 x+98 x^2\right )+e^{e^5 x+x^2} \left (98-7 x+2450 x^2-14 x^3+e^5 \left (1225 x-7 x^2\right )\right )+\left (98-7 x+e^{e^5 x+x^2} \left (49 e^5 x+98 x^2\right )\right ) \log \left (x^2\right )}{49 x} \, dx=\frac {{\mathrm {e}}^{2\,x^2+2\,{\mathrm {e}}^5\,x}}{2}-\frac {25\,x}{7}+25\,\ln \left (x^2\right )+25\,{\mathrm {e}}^{x^2+{\mathrm {e}}^5\,x}-\frac {x\,\ln \left (x^2\right )}{7}+\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+{\mathrm {e}}^5\,x}+\frac {{\ln \left (x^2\right )}^2}{2}-\frac {x\,{\mathrm {e}}^{x^2+{\mathrm {e}}^5\,x}}{7}+\frac {x^2}{98} \]
int(((exp(x*exp(5) + x^2)*(exp(5)*(1225*x - 7*x^2) - 7*x + 2450*x^2 - 14*x ^3 + 98))/49 - (27*x)/7 + (exp(2*x*exp(5) + 2*x^2)*(49*x*exp(5) + 98*x^2)) /49 + x^2/49 + (log(x^2)*(exp(x*exp(5) + x^2)*(49*x*exp(5) + 98*x^2) - 7*x + 98))/49 + 50)/x,x)