Integrand size = 66, antiderivative size = 27 \begin {dmath*} \int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} \left (-8+6 x-24 x^2 \log (5)\right )+\left (32-24 x+5^{4 x} (16-32 x \log (5))\right ) \log (x)-32 \log ^2(x)}{x^3} \, dx=\left (-3+\frac {5^{4 x}-x}{x}+\frac {x-4 \log (x)}{x}\right )^2 \end {dmath*}
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \begin {dmath*} \int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} \left (-8+6 x-24 x^2 \log (5)\right )+\left (32-24 x+5^{4 x} (16-32 x \log (5))\right ) \log (x)-32 \log ^2(x)}{x^3} \, dx=\frac {4 \left (625^x \left (625^x-6 x\right ) \log (5)+\left (-4 625^x \log (25)+6 x \log (625)\right ) \log (x)+4 \log (625) \log ^2(x)\right )}{x^2 \log (625)} \end {dmath*}
Integrate[(24*x + 5^(8*x)*(-2 + 8*x*Log[5]) + 5^(4*x)*(-8 + 6*x - 24*x^2*L og[5]) + (32 - 24*x + 5^(4*x)*(16 - 32*x*Log[5]))*Log[x] - 32*Log[x]^2)/x^ 3,x]
(4*(625^x*(625^x - 6*x)*Log[5] + (-4*625^x*Log[25] + 6*x*Log[625])*Log[x] + 4*Log[625]*Log[x]^2))/(x^2*Log[625])
Time = 0.36 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, 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.
\(\displaystyle \int \frac {5^{4 x} \left (-24 x^2 \log (5)+6 x-8\right )+24 x-32 \log ^2(x)+\left (-24 x+5^{4 x} (16-32 x \log (5))+32\right ) \log (x)+5^{8 x} (8 x \log (5)-2)}{x^3} \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (\frac {2\ 5^{8 x} (4 x \log (5)-1)}{x^3}-\frac {8 (\log (x)-1) (3 x+4 \log (x))}{x^3}-\frac {2\ 625^x \left (12 x^2 \log (5)-3 x+8 x \log (25) \log (x)-8 \log (x)+4\right )}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5^{8 x}}{x^2}-\frac {8\ 625^x \left (3 x^2 \log (5)+2 x \log (25) \log (x)\right )}{x^3 \log (625)}+\left (\frac {4 \log (x)}{x}+3\right )^2\) |
Int[(24*x + 5^(8*x)*(-2 + 8*x*Log[5]) + 5^(4*x)*(-8 + 6*x - 24*x^2*Log[5]) + (32 - 24*x + 5^(4*x)*(16 - 32*x*Log[5]))*Log[x] - 32*Log[x]^2)/x^3,x]
5^(8*x)/x^2 + (3 + (4*Log[x])/x)^2 - (8*625^x*(3*x^2*Log[5] + 2*x*Log[25]* Log[x]))/(x^3*Log[625])
3.2.45.3.1 Defintions of rubi rules used
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 = 0.54 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63
method | result | size |
risch | \(\frac {16 \ln \left (x \right )^{2}}{x^{2}}+\frac {8 \left (3 x -625^{x}\right ) \ln \left (x \right )}{x^{2}}-\frac {625^{x} \left (6 x -625^{x}\right )}{x^{2}}\) | \(44\) |
parallelrisch | \(\frac {24 x \ln \left (x \right )-6 \,{\mathrm e}^{4 x \ln \left (5\right )} x +16 \ln \left (x \right )^{2}-8 \ln \left (x \right ) {\mathrm e}^{4 x \ln \left (5\right )}+{\mathrm e}^{8 x \ln \left (5\right )}}{x^{2}}\) | \(44\) |
default | \(\frac {-6 \,{\mathrm e}^{4 x \ln \left (5\right )} x -8 \ln \left (x \right ) {\mathrm e}^{4 x \ln \left (5\right )}}{x^{2}}+\frac {16 \ln \left (x \right )^{2}}{x^{2}}+\frac {24 \ln \left (x \right )}{x}+\frac {{\mathrm e}^{8 x \ln \left (5\right )}}{x^{2}}\) | \(54\) |
parts | \(\frac {-6 \,{\mathrm e}^{4 x \ln \left (5\right )} x -8 \ln \left (x \right ) {\mathrm e}^{4 x \ln \left (5\right )}}{x^{2}}+\frac {16 \ln \left (x \right )^{2}}{x^{2}}+\frac {24 \ln \left (x \right )}{x}+\frac {{\mathrm e}^{8 x \ln \left (5\right )}}{x^{2}}\) | \(54\) |
int((-32*ln(x)^2+((-32*x*ln(5)+16)*exp(4*x*ln(5))-24*x+32)*ln(x)+(8*x*ln(5 )-2)*exp(4*x*ln(5))^2+(-24*x^2*ln(5)+6*x-8)*exp(4*x*ln(5))+24*x)/x^3,x,met hod=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \begin {dmath*} \int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} \left (-8+6 x-24 x^2 \log (5)\right )+\left (32-24 x+5^{4 x} (16-32 x \log (5))\right ) \log (x)-32 \log ^2(x)}{x^3} \, dx=-\frac {6 \cdot 5^{4 \, x} x + 8 \, {\left (5^{4 \, x} - 3 \, x\right )} \log \left (x\right ) - 16 \, \log \left (x\right )^{2} - 5^{8 \, x}}{x^{2}} \end {dmath*}
integrate((-32*log(x)^2+((-32*x*log(5)+16)*exp(4*x*log(5))-24*x+32)*log(x) +(8*x*log(5)-2)*exp(4*x*log(5))^2+(-24*x^2*log(5)+6*x-8)*exp(4*x*log(5))+2 4*x)/x^3,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.00 \begin {dmath*} \int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} \left (-8+6 x-24 x^2 \log (5)\right )+\left (32-24 x+5^{4 x} (16-32 x \log (5))\right ) \log (x)-32 \log ^2(x)}{x^3} \, dx=\frac {24 \log {\left (x \right )}}{x} + \frac {16 \log {\left (x \right )}^{2}}{x^{2}} + \frac {x^{2} e^{8 x \log {\left (5 \right )}} + \left (- 6 x^{3} - 8 x^{2} \log {\left (x \right )}\right ) e^{4 x \log {\left (5 \right )}}}{x^{4}} \end {dmath*}
integrate((-32*ln(x)**2+((-32*x*ln(5)+16)*exp(4*x*ln(5))-24*x+32)*ln(x)+(8 *x*ln(5)-2)*exp(4*x*ln(5))**2+(-24*x**2*ln(5)+6*x-8)*exp(4*x*ln(5))+24*x)/ x**3,x)
24*log(x)/x + 16*log(x)**2/x**2 + (x**2*exp(8*x*log(5)) + (-6*x**3 - 8*x** 2*log(x))*exp(4*x*log(5)))/x**4
\begin {dmath*} \int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} \left (-8+6 x-24 x^2 \log (5)\right )+\left (32-24 x+5^{4 x} (16-32 x \log (5))\right ) \log (x)-32 \log ^2(x)}{x^3} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x \log \left (5\right ) - 1\right )} 5^{8 \, x} - {\left (12 \, x^{2} \log \left (5\right ) - 3 \, x + 4\right )} 5^{4 \, x} - 4 \, {\left (2 \, {\left (2 \, x \log \left (5\right ) - 1\right )} 5^{4 \, x} + 3 \, x - 4\right )} \log \left (x\right ) - 16 \, \log \left (x\right )^{2} + 12 \, x\right )}}{x^{3}} \,d x } \end {dmath*}
integrate((-32*log(x)^2+((-32*x*log(5)+16)*exp(4*x*log(5))-24*x+32)*log(x) +(8*x*log(5)-2)*exp(4*x*log(5))^2+(-24*x^2*log(5)+6*x-8)*exp(4*x*log(5))+2 4*x)/x^3,x, algorithm=\
64*gamma(-1, -8*x*log(5))*log(5)^2 + 128*gamma(-2, -4*x*log(5))*log(5)^2 + 128*gamma(-2, -8*x*log(5))*log(5)^2 - 24*Ei(4*x*log(5))*log(5) + 24*gamma (-1, -4*x*log(5))*log(5) + 24*log(x)/x - 8*(5^(4*x)*log(x) - 2*log(x)^2 - 2*log(x) - 1)/x^2 - 16*log(x)/x^2 - 8/x^2 + 8*integrate(5^(4*x)/x^3, x)
\begin {dmath*} \int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} \left (-8+6 x-24 x^2 \log (5)\right )+\left (32-24 x+5^{4 x} (16-32 x \log (5))\right ) \log (x)-32 \log ^2(x)}{x^3} \, dx=\int { \frac {2 \, {\left ({\left (4 \, x \log \left (5\right ) - 1\right )} 5^{8 \, x} - {\left (12 \, x^{2} \log \left (5\right ) - 3 \, x + 4\right )} 5^{4 \, x} - 4 \, {\left (2 \, {\left (2 \, x \log \left (5\right ) - 1\right )} 5^{4 \, x} + 3 \, x - 4\right )} \log \left (x\right ) - 16 \, \log \left (x\right )^{2} + 12 \, x\right )}}{x^{3}} \,d x } \end {dmath*}
integrate((-32*log(x)^2+((-32*x*log(5)+16)*exp(4*x*log(5))-24*x+32)*log(x) +(8*x*log(5)-2)*exp(4*x*log(5))^2+(-24*x^2*log(5)+6*x-8)*exp(4*x*log(5))+2 4*x)/x^3,x, algorithm=\
integrate(2*((4*x*log(5) - 1)*5^(8*x) - (12*x^2*log(5) - 3*x + 4)*5^(4*x) - 4*(2*(2*x*log(5) - 1)*5^(4*x) + 3*x - 4)*log(x) - 16*log(x)^2 + 12*x)/x^ 3, x)
Time = 13.55 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \begin {dmath*} \int \frac {24 x+5^{8 x} (-2+8 x \log (5))+5^{4 x} \left (-8+6 x-24 x^2 \log (5)\right )+\left (32-24 x+5^{4 x} (16-32 x \log (5))\right ) \log (x)-32 \log ^2(x)}{x^3} \, dx=\frac {\left (4\,\ln \left (x\right )-5^{4\,x}\right )\,\left (6\,x+4\,\ln \left (x\right )-5^{4\,x}\right )}{x^2} \end {dmath*}