Integrand size = 60, antiderivative size = 25 \[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=1-\frac {\left (2^{5 (1+x)}-2 x\right ) (4+x) \log (x)}{2 x} \] Output:
1-1/2*(4+x)/x*ln(x)*(exp(5*ln(2)*(1+x))-2*x)
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=\frac {(4+x) \left (-2^{4+5 x}+x\right ) \log (x)}{x} \] Input:
Integrate[(2^(5 + 5*x)*(-4 - x) + 8*x + 2*x^2 + (2*x^2 + 2^(5 + 5*x)*(4 + (-20*x - 5*x^2)*Log[2]))*Log[x])/(2*x^2),x]
Output:
((4 + x)*(-2^(4 + 5*x) + x)*Log[x])/x
Time = 0.31 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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.
\(\displaystyle \int \frac {2 x^2+\left (2 x^2+2^{5 x+5} \left (\left (-5 x^2-20 x\right ) \log (2)+4\right )\right ) \log (x)+8 x+2^{5 x+5} (-x-4)}{2 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {2 x^2+8 x-2^{5 x+5} (x+4)+\left (2 x^2+2^{5 x+5} \left (4-5 \left (x^2+4 x\right ) \log (2)\right )\right ) \log (x)}{x^2}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{2} \int \left (\frac {2 (\log (x) x+x+4)}{x}-\frac {32^{x+1} \left (\log (32) \log (x) x^2+20 \log (2) \log (x) x+x-4 \log (x)+4\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {32^{x+1} \left (x^2 \log (32) \log (x)+20 x \log (2) \log (x)\right )}{x^2 \log (32)}+2 x \log (x)+8 \log (x)\right )\) |
Input:
Int[(2^(5 + 5*x)*(-4 - x) + 8*x + 2*x^2 + (2*x^2 + 2^(5 + 5*x)*(4 + (-20*x - 5*x^2)*Log[2]))*Log[x])/(2*x^2),x]
Output:
(8*Log[x] + 2*x*Log[x] - (32^(1 + x)*(20*x*Log[2]*Log[x] + x^2*Log[32]*Log [x]))/(x^2*Log[32]))/2
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]]
Time = 6.76 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52
method | result | size |
risch | \(\frac {\left (2 x^{2}-x 2^{5 x +5}-4 \,2^{5 x +5}\right ) \ln \left (x \right )}{2 x}+4 \ln \left (x \right )\) | \(38\) |
parts | \(4 \ln \left (x \right )+\frac {-2 \ln \left (x \right ) {\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )}-\frac {x \,{\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )} \ln \left (x \right )}{2}}{x}+x \ln \left (x \right )\) | \(42\) |
default | \(\frac {-4 \ln \left (x \right ) {\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )}-x \,{\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )} \ln \left (x \right )}{2 x}+4 \ln \left (x \right )+x \ln \left (x \right )\) | \(43\) |
norman | \(\frac {x^{2} \ln \left (x \right )+4 x \ln \left (x \right )-2 \ln \left (x \right ) {\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )}-\frac {x \,{\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )} \ln \left (x \right )}{2}}{x}\) | \(44\) |
parallelrisch | \(\frac {2 x^{2} \ln \left (x \right )-x \,{\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )} \ln \left (x \right )+8 x \ln \left (x \right )-4 \ln \left (x \right ) {\mathrm e}^{\left (5 x +5\right ) \ln \left (2\right )}}{2 x}\) | \(46\) |
Input:
int(1/2*((((-5*x^2-20*x)*ln(2)+4)*exp((5*x+5)*ln(2))+2*x^2)*ln(x)+(-4-x)*e xp((5*x+5)*ln(2))+2*x^2+8*x)/x^2,x,method=_RETURNVERBOSE)
Output:
1/2*(2*x^2-x*2^(5*x+5)-4*2^(5*x+5))/x*ln(x)+4*ln(x)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=-\frac {{\left (2^{5 \, x + 5} {\left (x + 4\right )} - 2 \, x^{2} - 8 \, x\right )} \log \left (x\right )}{2 \, x} \] Input:
integrate(1/2*((((-5*x^2-20*x)*log(2)+4)*exp((5*x+5)*log(2))+2*x^2)*log(x) +(-4-x)*exp((5*x+5)*log(2))+2*x^2+8*x)/x^2,x, algorithm="fricas")
Output:
-1/2*(2^(5*x + 5)*(x + 4) - 2*x^2 - 8*x)*log(x)/x
Time = 0.46 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=x \log {\left (x \right )} + 4 \log {\left (x \right )} + \frac {\left (- x \log {\left (x \right )} - 4 \log {\left (x \right )}\right ) e^{\left (5 x + 5\right ) \log {\left (2 \right )}}}{2 x} \] Input:
integrate(1/2*((((-5*x**2-20*x)*ln(2)+4)*exp((5*x+5)*ln(2))+2*x**2)*ln(x)+ (-4-x)*exp((5*x+5)*ln(2))+2*x**2+8*x)/x**2,x)
Output:
x*log(x) + 4*log(x) + (-x*log(x) - 4*log(x))*exp((5*x + 5)*log(2))/(2*x)
\[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=\int { -\frac {2^{5 \, x + 5} {\left (x + 4\right )} - 2 \, x^{2} + {\left ({\left (5 \, {\left (x^{2} + 4 \, x\right )} \log \left (2\right ) - 4\right )} 2^{5 \, x + 5} - 2 \, x^{2}\right )} \log \left (x\right ) - 8 \, x}{2 \, x^{2}} \,d x } \] Input:
integrate(1/2*((((-5*x^2-20*x)*log(2)+4)*exp((5*x+5)*log(2))+2*x^2)*log(x) +(-4-x)*exp((5*x+5)*log(2))+2*x^2+8*x)/x^2,x, algorithm="maxima")
Output:
-320*gamma(-1, -5*x*log(2))*log(2) - 1/2*2^(5*x + 5)*log(x) + x*log(x) - 6 4*2^(5*x)*log(x)/x + 64*integrate(2^(5*x)/x^2, x) + 4*log(x)
Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=-16 \cdot 2^{5 \, x} \log \left (x\right ) + x \log \left (x\right ) - \frac {64 \cdot 2^{5 \, x} \log \left (x\right )}{x} + 4 \, \log \left (x\right ) \] Input:
integrate(1/2*((((-5*x^2-20*x)*log(2)+4)*exp((5*x+5)*log(2))+2*x^2)*log(x) +(-4-x)*exp((5*x+5)*log(2))+2*x^2+8*x)/x^2,x, algorithm="giac")
Output:
-16*2^(5*x)*log(x) + x*log(x) - 64*2^(5*x)*log(x)/x + 4*log(x)
Time = 2.47 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=\frac {\ln \left (x\right )\,\left (x-16\,2^{5\,x}\right )\,\left (x+4\right )}{x} \] Input:
int((4*x - (log(x)*(exp(log(2)*(5*x + 5))*(log(2)*(20*x + 5*x^2) - 4) - 2* x^2))/2 - (exp(log(2)*(5*x + 5))*(x + 4))/2 + x^2)/x^2,x)
Output:
(log(x)*(x - 16*2^(5*x))*(x + 4))/x
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {2^{5+5 x} (-4-x)+8 x+2 x^2+\left (2 x^2+2^{5+5 x} \left (4+\left (-20 x-5 x^2\right ) \log (2)\right )\right ) \log (x)}{2 x^2} \, dx=\frac {\mathrm {log}\left (x \right ) \left (-16 \,2^{5 x} x -64 \,2^{5 x}+x^{2}+4 x \right )}{x} \] Input:
int(1/2*((((-5*x^2-20*x)*log(2)+4)*exp((5*x+5)*log(2))+2*x^2)*log(x)+(-4-x )*exp((5*x+5)*log(2))+2*x^2+8*x)/x^2,x)
Output:
(log(x)*( - 16*2**(5*x)*x - 64*2**(5*x) + x**2 + 4*x))/x