Integrand size = 83, antiderivative size = 27 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {4 x \log (x) \left (7-x-\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \]
Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=-\frac {4 x \log (x) \left (-7+x+\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \]
Integrate[(56 - 8*x + 28*x^2 - 4*x^3 + (-8 - 4*x^2)*Log[(2*x)/(2 + x^2)] + Log[x]*(48 - 16*x + 32*x^2 - 8*x^3 + (-8 - 4*x^2)*Log[(2*x)/(2 + x^2)]))/ ((2 + x^2)*Log[25]),x]
Time = 0.70 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {27, 27, 7276, 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.
\(\displaystyle \int \frac {-4 x^3+28 x^2+\left (-4 x^2-8\right ) \log \left (\frac {2 x}{x^2+2}\right )+\log (x) \left (-8 x^3+32 x^2+\left (-4 x^2-8\right ) \log \left (\frac {2 x}{x^2+2}\right )-16 x+48\right )-8 x+56}{\left (x^2+2\right ) \log (25)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 \left (-x^3+7 x^2-2 x-\left (x^2+2\right ) \log \left (\frac {2 x}{x^2+2}\right )+\log (x) \left (-2 x^3+8 x^2-4 x-\left (x^2+2\right ) \log \left (\frac {2 x}{x^2+2}\right )+12\right )+14\right )}{x^2+2}dx}{\log (25)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {4 \int \frac {-x^3+7 x^2-2 x-\left (x^2+2\right ) \log \left (\frac {2 x}{x^2+2}\right )+\log (x) \left (-2 x^3+8 x^2-4 x-\left (x^2+2\right ) \log \left (\frac {2 x}{x^2+2}\right )+12\right )+14}{x^2+2}dx}{\log (25)}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {4 \int \left (-\frac {2 \log (x) x^3}{x^2+2}-\frac {x^3}{x^2+2}+\frac {8 \log (x) x^2}{x^2+2}+\frac {7 x^2}{x^2+2}-\frac {4 \log (x) x}{x^2+2}-\frac {2 x}{x^2+2}+\frac {12 \log (x)}{x^2+2}-(\log (x)+1) \log \left (\frac {2 x}{x^2+2}\right )+\frac {14}{x^2+2}\right )dx}{\log (25)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \left (x^2 (-\log (x))-x \log (x) \log \left (\frac {2 x}{x^2+2}\right )+7 x \log (x)\right )}{\log (25)}\) |
Int[(56 - 8*x + 28*x^2 - 4*x^3 + (-8 - 4*x^2)*Log[(2*x)/(2 + x^2)] + Log[x ]*(48 - 16*x + 32*x^2 - 8*x^3 + (-8 - 4*x^2)*Log[(2*x)/(2 + x^2)]))/((2 + x^2)*Log[25]),x]
3.15.60.3.1 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_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.46 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33
method | result | size |
parallelrisch | \(\frac {-4 x^{2} \ln \left (x \right )-4 x \ln \left (x \right ) \ln \left (\frac {2 x}{x^{2}+2}\right )+28 x \ln \left (x \right )}{2 \ln \left (5\right )}\) | \(36\) |
risch | \(\frac {2 \ln \left (x \right ) x \ln \left (x^{2}+2\right )}{\ln \left (5\right )}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x \ln \left (2\right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x^{2} \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x \ln \left (x \right )^{2}}{\ln \left (5\right )}+\frac {14 x \ln \left (x \right )}{\ln \left (5\right )}\) | \(189\) |
default | \(\frac {-4 x \ln \left (2\right ) \ln \left (x \right )-4 \left (\ln \left (x \right )-1\right ) x \ln \left (x \right )+4 \ln \left (x \right ) x \ln \left (x^{2}+2\right )-4 x \ln \left (x^{2}+2\right )-4 x^{2} \ln \left (x \right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x +2 i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )+2 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x -2 i \pi \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} x +2 i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \ln \left (x \right )-2 i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \operatorname {csgn}\left (i x \right ) \ln \left (x \right )-2 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) x -2 i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )+28 x \ln \left (x \right )-4 x \ln \left (\frac {x}{x^{2}+2}\right )}{2 \ln \left (5\right )}\) | \(289\) |
parts | \(-\frac {2 x \ln \left (x \right )^{2}}{\ln \left (5\right )}+\frac {16 x \ln \left (x \right )}{\ln \left (5\right )}+\frac {2 \ln \left (x \right ) x \ln \left (x^{2}+2\right )}{\ln \left (5\right )}-\frac {2 x \ln \left (x^{2}+2\right )}{\ln \left (5\right )}+\frac {x^{2}}{\ln \left (5\right )}-\frac {2 x^{2} \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x}{\ln \left (5\right )}+\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x}{\ln \left (5\right )}-\frac {i \pi \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} x}{\ln \left (5\right )}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) x}{\ln \left (5\right )}-\frac {4 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )}{\ln \left (5\right )}+\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \operatorname {csgn}\left (i x \right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {12 x}{\ln \left (5\right )}-\frac {2 \ln \left (2\right ) \left (x \ln \left (x \right )-x \right )}{\ln \left (5\right )}-\frac {2 \left (-7 x +\frac {1}{2} x^{2}\right )}{\ln \left (5\right )}-\frac {2 \left (x \ln \left (2\right )+x \ln \left (\frac {x}{x^{2}+2}\right )+x -2 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )\right )}{\ln \left (5\right )}\) | \(411\) |
int(1/2*(((-4*x^2-8)*ln(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*ln(x)+(-4*x^2-8 )*ln(2*x/(x^2+2))-4*x^3+28*x^2-8*x+56)/(x^2+2)/ln(5),x,method=_RETURNVERBO SE)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=-\frac {2 \, {\left (x^{2} + x \log \left (\frac {2 \, x}{x^{2} + 2}\right ) - 7 \, x\right )} \log \left (x\right )}{\log \left (5\right )} \]
integrate(1/2*(((-4*x^2-8)*log(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*log(x)+( -4*x^2-8)*log(2*x/(x^2+2))-4*x^3+28*x^2-8*x+56)/(x^2+2)/log(5),x, algorith m=\
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=- \frac {2 x \log {\left (x \right )} \log {\left (\frac {2 x}{x^{2} + 2} \right )}}{\log {\left (5 \right )}} + \frac {\left (- 2 x^{2} + 14 x\right ) \log {\left (x \right )}}{\log {\left (5 \right )}} \]
integrate(1/2*(((-4*x**2-8)*ln(2*x/(x**2+2))-8*x**3+32*x**2-16*x+48)*ln(x) +(-4*x**2-8)*ln(2*x/(x**2+2))-4*x**3+28*x**2-8*x+56)/(x**2+2)/ln(5),x)
Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \left (x\right ) - x \log \left (x\right )^{2} - {\left (x^{2} + x {\left (\log \left (2\right ) - 7\right )}\right )} \log \left (x\right )\right )}}{\log \left (5\right )} \]
integrate(1/2*(((-4*x^2-8)*log(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*log(x)+( -4*x^2-8)*log(2*x/(x^2+2))-4*x^3+28*x^2-8*x+56)/(x^2+2)/log(5),x, algorith m=\
Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \left (x\right ) - x \log \left (x\right )^{2} - {\left (x^{2} + x {\left (\log \left (2\right ) - 7\right )}\right )} \log \left (x\right )\right )}}{\log \left (5\right )} \]
integrate(1/2*(((-4*x^2-8)*log(2*x/(x^2+2))-8*x^3+32*x^2-16*x+48)*log(x)+( -4*x^2-8)*log(2*x/(x^2+2))-4*x^3+28*x^2-8*x+56)/(x^2+2)/log(5),x, algorith m=\
Timed out. \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\int -\frac {4\,x+\frac {\ln \left (x\right )\,\left (16\,x+\ln \left (\frac {2\,x}{x^2+2}\right )\,\left (4\,x^2+8\right )-32\,x^2+8\,x^3-48\right )}{2}+\frac {\ln \left (\frac {2\,x}{x^2+2}\right )\,\left (4\,x^2+8\right )}{2}-14\,x^2+2\,x^3-28}{\ln \left (5\right )\,\left (x^2+2\right )} \,d x \]
int(-(4*x + (log(x)*(16*x + log((2*x)/(x^2 + 2))*(4*x^2 + 8) - 32*x^2 + 8* x^3 - 48))/2 + (log((2*x)/(x^2 + 2))*(4*x^2 + 8))/2 - 14*x^2 + 2*x^3 - 28) /(log(5)*(x^2 + 2)),x)