Integrand size = 115, antiderivative size = 31 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-\left (\left (-e^{\log ^2(x)}-x\right ) \left (-4^{25 x}+x\right ) \left (e^{\log ^2(x)}+x\right )\right ) \] Output:
-(x-exp(50*x*ln(2)))*(x+exp(ln(x)^2))*(-x-exp(ln(x)^2))
Time = 0.21 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-\left (\left (2^{50 x}-x\right ) \left (e^{\log ^2(x)}+x\right )^2\right ) \] Input:
Integrate[(3*x^3 + 4^(25*x)*(-2*x^2 - 25*x^3*Log[4]) + E^(2*Log[x]^2)*(x - 25*4^(25*x)*x*Log[4] + (-4^(1 + 25*x) + 4*x)*Log[x]) + E^Log[x]^2*(4*x^2 + 4^(25*x)*(-2*x - 50*x^2*Log[4]) + (-(4^(1 + 25*x)*x) + 4*x^2)*Log[x]))/x ,x]
Output:
-((2^(50*x) - x)*(E^Log[x]^2 + x)^2)
Leaf count is larger than twice the leaf count of optimal. \(147\) vs. \(2(31)=62\).
Time = 0.61 (sec) , antiderivative size = 147, normalized size of antiderivative = 4.74, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {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.
\(\displaystyle \int \frac {3 x^3+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-50 x^2 \log (4)-2 x\right )+\left (4 x^2-4^{25 x+1} x\right ) \log (x)\right )+4^{25 x} \left (-25 x^3 \log (4)-2 x^2\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (4 x-4^{25 x+1}\right ) \log (x)\right )}{x} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (-\frac {e^{2 \log ^2(x)} \left (-x-4 x \log (x)+25\ 4^{25 x} x \log (4)+4^{25 x+1} \log (x)\right )}{x}-2 e^{\log ^2(x)} \left (-2 x+2^{50 x}-2 x \log (x)+25\ 2^{50 x} x \log (4)+2^{50 x+1} \log (x)\right )-x \left (-3 x+2^{50 x+1}+25\ 2^{50 x} x \log (4)\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^3-\frac {2^{50 x-1} x^2 \log (4)}{\log (2)}-\frac {2^{50 x-2} \log (4)}{625 \log ^3(2)}-\frac {x e^{\log ^2(x)} \left (2^{50 x+1} \log (x)-2 x \log (x)\right )}{\log (x)}+\frac {2^{50 x-1} x \log (4)}{25 \log ^2(2)}-\frac {e^{2 \log ^2(x)} \left (4^{25 x+1} \log (x)-4 x \log (x)\right )}{4 \log (x)}+\frac {2^{50 x-1}}{625 \log ^2(2)}-\frac {2^{50 x} x}{25 \log (2)}\) |
Input:
Int[(3*x^3 + 4^(25*x)*(-2*x^2 - 25*x^3*Log[4]) + E^(2*Log[x]^2)*(x - 25*4^ (25*x)*x*Log[4] + (-4^(1 + 25*x) + 4*x)*Log[x]) + E^Log[x]^2*(4*x^2 + 4^(2 5*x)*(-2*x - 50*x^2*Log[4]) + (-(4^(1 + 25*x)*x) + 4*x^2)*Log[x]))/x,x]
Output:
x^3 + 2^(-1 + 50*x)/(625*Log[2]^2) - (2^(50*x)*x)/(25*Log[2]) - (2^(-2 + 5 0*x)*Log[4])/(625*Log[2]^3) + (2^(-1 + 50*x)*x*Log[4])/(25*Log[2]^2) - (2^ (-1 + 50*x)*x^2*Log[4])/Log[2] - (E^(2*Log[x]^2)*(4^(1 + 25*x)*Log[x] - 4* x*Log[x]))/(4*Log[x]) - (E^Log[x]^2*x*(2^(1 + 50*x)*Log[x] - 2*x*Log[x]))/ Log[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]]
Time = 1.46 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68
method | result | size |
risch | \(x^{3}-x^{2} {\mathrm e}^{50 x \ln \left (2\right )}+\left (x -{\mathrm e}^{50 x \ln \left (2\right )}\right ) {\mathrm e}^{2 \ln \left (x \right )^{2}}+2 \left (x -{\mathrm e}^{50 x \ln \left (2\right )}\right ) x \,{\mathrm e}^{\ln \left (x \right )^{2}}\) | \(52\) |
default | \({\mathrm e}^{2 \ln \left (x \right )^{2}} x -{\mathrm e}^{2 \ln \left (x \right )^{2}} {\mathrm e}^{50 x \ln \left (2\right )}+2 \,{\mathrm e}^{\ln \left (x \right )^{2}} x^{2}-2 \,{\mathrm e}^{50 x \ln \left (2\right )} {\mathrm e}^{\ln \left (x \right )^{2}} x +x^{3}-x^{2} {\mathrm e}^{50 x \ln \left (2\right )}\) | \(64\) |
parallelrisch | \({\mathrm e}^{2 \ln \left (x \right )^{2}} x -{\mathrm e}^{2 \ln \left (x \right )^{2}} {\mathrm e}^{50 x \ln \left (2\right )}+2 \,{\mathrm e}^{\ln \left (x \right )^{2}} x^{2}-2 \,{\mathrm e}^{50 x \ln \left (2\right )} {\mathrm e}^{\ln \left (x \right )^{2}} x +x^{3}-x^{2} {\mathrm e}^{50 x \ln \left (2\right )}\) | \(64\) |
Input:
int((((-4*exp(50*x*ln(2))+4*x)*ln(x)-50*x*ln(2)*exp(50*x*ln(2))+x)*exp(ln( x)^2)^2+((-4*x*exp(50*x*ln(2))+4*x^2)*ln(x)+(-100*x^2*ln(2)-2*x)*exp(50*x* ln(2))+4*x^2)*exp(ln(x)^2)+(-50*x^3*ln(2)-2*x^2)*exp(50*x*ln(2))+3*x^3)/x, x,method=_RETURNVERBOSE)
Output:
x^3-x^2*exp(50*x*ln(2))+(x-exp(50*x*ln(2)))*exp(ln(x)^2)^2+2*(x-exp(50*x*l n(2)))*x*exp(ln(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (20) = 40\).
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-2^{50 \, x} x^{2} + x^{3} - {\left (2^{50 \, x} - x\right )} e^{\left (2 \, \log \left (x\right )^{2}\right )} - 2 \, {\left (2^{50 \, x} x - x^{2}\right )} e^{\left (\log \left (x\right )^{2}\right )} \] Input:
integrate((((-4*exp(50*x*log(2))+4*x)*log(x)-50*x*log(2)*exp(50*x*log(2))+ x)*exp(log(x)^2)^2+((-4*x*exp(50*x*log(2))+4*x^2)*log(x)+(-100*x^2*log(2)- 2*x)*exp(50*x*log(2))+4*x^2)*exp(log(x)^2)+(-50*x^3*log(2)-2*x^2)*exp(50*x *log(2))+3*x^3)/x,x, algorithm="fricas")
Output:
-2^(50*x)*x^2 + x^3 - (2^(50*x) - x)*e^(2*log(x)^2) - 2*(2^(50*x)*x - x^2) *e^(log(x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (27) = 54\).
Time = 92.83 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=x^{3} + 2 x^{2} e^{\log {\left (x \right )}^{2}} + x e^{2 \log {\left (x \right )}^{2}} + \left (- x^{2} - 2 x e^{\log {\left (x \right )}^{2}} - e^{2 \log {\left (x \right )}^{2}}\right ) e^{50 x \log {\left (2 \right )}} \] Input:
integrate((((-4*exp(50*x*ln(2))+4*x)*ln(x)-50*x*ln(2)*exp(50*x*ln(2))+x)*e xp(ln(x)**2)**2+((-4*x*exp(50*x*ln(2))+4*x**2)*ln(x)+(-100*x**2*ln(2)-2*x) *exp(50*x*ln(2))+4*x**2)*exp(ln(x)**2)+(-50*x**3*ln(2)-2*x**2)*exp(50*x*ln (2))+3*x**3)/x,x)
Output:
x**3 + 2*x**2*exp(log(x)**2) + x*exp(2*log(x)**2) + (-x**2 - 2*x*exp(log(x )**2) - exp(2*log(x)**2))*exp(50*x*log(2))
\[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=\int { \frac {3 \, x^{3} - 2 \, {\left (25 \, x^{3} \log \left (2\right ) + x^{2}\right )} 2^{50 \, x} - {\left (50 \cdot 2^{50 \, x} x \log \left (2\right ) + 4 \, {\left (2^{50 \, x} - x\right )} \log \left (x\right ) - x\right )} e^{\left (2 \, \log \left (x\right )^{2}\right )} - 2 \, {\left ({\left (50 \, x^{2} \log \left (2\right ) + x\right )} 2^{50 \, x} - 2 \, x^{2} + 2 \, {\left (2^{50 \, x} x - x^{2}\right )} \log \left (x\right )\right )} e^{\left (\log \left (x\right )^{2}\right )}}{x} \,d x } \] Input:
integrate((((-4*exp(50*x*log(2))+4*x)*log(x)-50*x*log(2)*exp(50*x*log(2))+ x)*exp(log(x)^2)^2+((-4*x*exp(50*x*log(2))+4*x^2)*log(x)+(-100*x^2*log(2)- 2*x)*exp(50*x*log(2))+4*x^2)*exp(log(x)^2)+(-50*x^3*log(2)-2*x^2)*exp(50*x *log(2))+3*x^3)/x,x, algorithm="maxima")
Output:
x^3 - 2*I*sqrt(pi)*erf(I*log(x) + I)*e^(-1) - 2*x*e^(50*x*log(2) + log(x)^ 2) - (2^(50*x) - x)*e^(2*log(x)^2) - 1/1250*(1250*x^2*log(2)^2 - 50*x*log( 2) + 1)*2^(50*x)/log(2)^2 - 1/1250*(50*x*log(2) - 1)*2^(50*x)/log(2)^2 + 4 *integrate(x*e^(log(x)^2)*log(x), x)
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (20) = 40\).
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.00 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-2^{50 \, x} x^{2} + x^{3} + 2 \, x^{2} e^{\left (\log \left (x\right )^{2}\right )} - 2 \, x e^{\left (50 \, x \log \left (2\right ) + \log \left (x\right )^{2}\right )} + x e^{\left (2 \, \log \left (x\right )^{2}\right )} - e^{\left (50 \, x \log \left (2\right ) + 2 \, \log \left (x\right )^{2}\right )} \] Input:
integrate((((-4*exp(50*x*log(2))+4*x)*log(x)-50*x*log(2)*exp(50*x*log(2))+ x)*exp(log(x)^2)^2+((-4*x*exp(50*x*log(2))+4*x^2)*log(x)+(-100*x^2*log(2)- 2*x)*exp(50*x*log(2))+4*x^2)*exp(log(x)^2)+(-50*x^3*log(2)-2*x^2)*exp(50*x *log(2))+3*x^3)/x,x, algorithm="giac")
Output:
-2^(50*x)*x^2 + x^3 + 2*x^2*e^(log(x)^2) - 2*x*e^(50*x*log(2) + log(x)^2) + x*e^(2*log(x)^2) - e^(50*x*log(2) + 2*log(x)^2)
Time = 2.79 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx={\mathrm {e}}^{2\,{\ln \left (x\right )}^2}\,\left (x-2^{50\,x}\right )-2^{50\,x}\,x^2+x^3+2\,x\,{\mathrm {e}}^{{\ln \left (x\right )}^2}\,\left (x-2^{50\,x}\right ) \] Input:
int(-(exp(50*x*log(2))*(50*x^3*log(2) + 2*x^2) - exp(2*log(x)^2)*(x + log( x)*(4*x - 4*exp(50*x*log(2))) - 50*x*exp(50*x*log(2))*log(2)) - 3*x^3 + ex p(log(x)^2)*(log(x)*(4*x*exp(50*x*log(2)) - 4*x^2) - 4*x^2 + exp(50*x*log( 2))*(2*x + 100*x^2*log(2))))/x,x)
Output:
exp(2*log(x)^2)*(x - 2^(50*x)) - 2^(50*x)*x^2 + x^3 + 2*x*exp(log(x)^2)*(x - 2^(50*x))
Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.06 \[ \int \frac {3 x^3+4^{25 x} \left (-2 x^2-25 x^3 \log (4)\right )+e^{2 \log ^2(x)} \left (x-25\ 4^{25 x} x \log (4)+\left (-4^{1+25 x}+4 x\right ) \log (x)\right )+e^{\log ^2(x)} \left (4 x^2+4^{25 x} \left (-2 x-50 x^2 \log (4)\right )+\left (-4^{1+25 x} x+4 x^2\right ) \log (x)\right )}{x} \, dx=-e^{2 \mathrm {log}\left (x \right )^{2}} 2^{50 x}+e^{2 \mathrm {log}\left (x \right )^{2}} x -2 e^{\mathrm {log}\left (x \right )^{2}} 2^{50 x} x +2 e^{\mathrm {log}\left (x \right )^{2}} x^{2}-2^{50 x} x^{2}+x^{3} \] Input:
int((((-4*exp(50*x*log(2))+4*x)*log(x)-50*x*log(2)*exp(50*x*log(2))+x)*exp (log(x)^2)^2+((-4*x*exp(50*x*log(2))+4*x^2)*log(x)+(-100*x^2*log(2)-2*x)*e xp(50*x*log(2))+4*x^2)*exp(log(x)^2)+(-50*x^3*log(2)-2*x^2)*exp(50*x*log(2 ))+3*x^3)/x,x)
Output:
- e**(2*log(x)**2)*2**(50*x) + e**(2*log(x)**2)*x - 2*e**(log(x)**2)*2**( 50*x)*x + 2*e**(log(x)**2)*x**2 - 2**(50*x)*x**2 + x**3