\(\int \frac {56-8 x+28 x^2-4 x^3+(-8-4 x^2) \log (\frac {2 x}{2+x^2})+\log (x) (48-16 x+32 x^2-8 x^3+(-8-4 x^2) \log (\frac {2 x}{2+x^2}))}{(2+x^2) \log (25)} \, dx\) [1460]

Optimal result

Integrand size = 83, antiderivative size = 27 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {4 x \log (x) \left (7-x-\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \] Output:

2*x*ln(x)*(7-ln(2*x/(x^2+2))-x)/ln(5)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=-\frac {4 x \log (x) \left (-7+x+\log \left (\frac {2 x}{2+x^2}\right )\right )}{\log (25)} \] Input:

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]
 

Output:

(-4*x*Log[x]*(-7 + x + Log[(2*x)/(2 + x^2)]))/Log[25]
 

Rubi [A] (verified)

Time = 0.76 (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

\(\displaystyle \frac {4 \left (x^2 (-\log (x))-x \log (x) \log \left (\frac {2 x}{x^2+2}\right )+7 x \log (x)\right )}{\log (25)}\)

Input:

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]
 

Output:

(4*(7*x*Log[x] - x^2*Log[x] - x*Log[x]*Log[(2*x)/(2 + x^2)]))/Log[25]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7276

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]
 

Maple [A] (verified)

Time = 163.86 (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	\(\frac {2 \ln \left (x \right ) x \ln \left (x^{2}+2\right )}{\ln \left (5\right )}+\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )}{\ln \left (5\right )}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )}{\ln \left (5\right )}+\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x \ln \left (2\right ) \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x^{2} \ln \left (x \right )}{\ln \left (5\right )}-\frac {2 x \ln \left (x \right )^{2}}{\ln \left (5\right )}+\frac {14 x \ln \left (x \right )}{\ln \left (5\right )}\)	\(189\)
default	\(\frac {-4 x \ln \left (2\right ) \ln \left (x \right )-4 \left (\ln \left (x \right )-1\right ) x \ln \left (x \right )+4 \ln \left (x \right ) x \ln \left (x^{2}+2\right )-4 x \ln \left (x^{2}+2\right )-4 x^{2} \ln \left (x \right )-2 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )-2 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) x +2 i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x +2 i \pi x \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} \ln \left (x \right )-2 i \pi \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{3} x -2 i \pi x \,\operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} \ln \left (x \right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x +2 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right ) \ln \left (x \right )+28 x \ln \left (x \right )-4 x \ln \left (\frac {x}{x^{2}+2}\right )}{2 \ln \left (5\right )}\)	\(289\)
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 (i x \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 (\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 {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 (i x \right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\right )^{2} x}{\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 {4 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, x}{2}\right )}{\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 x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{x^{2}+2}\right ) \operatorname {csgn}\left (\frac {i x}{x^{2}+2}\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 {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 )}\)	\(411\)
orering	\(\text {Expression too large to display}\)	\(3274\)

Input:

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)
 

Output:

1/2/ln(5)*(-4*x^2*ln(x)-4*x*ln(x)*ln(2*x/(x^2+2))+28*x*ln(x))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=-\frac {2 \, {\left (x^{2} + x \log \left (\frac {2 \, x}{x^{2} + 2}\right ) - 7 \, x\right )} \log \left (x\right )}{\log \left (5\right )} \] Input:

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="fricas")
 

Output:

-2*(x^2 + x*log(2*x/(x^2 + 2)) - 7*x)*log(x)/log(5)
 

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=- \frac {2 x \log {\left (x \right )} \log {\left (\frac {2 x}{x^{2} + 2} \right )}}{\log {\left (5 \right )}} + \frac {\left (- 2 x^{2} + 14 x\right ) \log {\left (x \right )}}{\log {\left (5 \right )}} \] Input:

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)
 

Output:

-2*x*log(x)*log(2*x/(x**2 + 2))/log(5) + (-2*x**2 + 14*x)*log(x)/log(5)
 

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \left (x\right ) - x \log \left (x\right )^{2} - {\left (x^{2} + x {\left (\log \left (2\right ) - 7\right )}\right )} \log \left (x\right )\right )}}{\log \left (5\right )} \] Input:

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="maxima")
 

Output:

2*(x*log(x^2 + 2)*log(x) - x*log(x)^2 - (x^2 + x*(log(2) - 7))*log(x))/log 
(5)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {2 \, {\left (x \log \left (x^{2} + 2\right ) \log \left (x\right ) - x \log \left (x\right )^{2} - {\left (x^{2} + x {\left (\log \left (2\right ) - 7\right )}\right )} \log \left (x\right )\right )}}{\log \left (5\right )} \] Input:

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="giac")
 

Output:

2*(x*log(x^2 + 2)*log(x) - x*log(x)^2 - (x^2 + x*(log(2) - 7))*log(x))/log 
(5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\int -\frac {4\,x+\frac {\ln \left (x\right )\,\left (16\,x+\ln \left (\frac {2\,x}{x^2+2}\right )\,\left (4\,x^2+8\right )-32\,x^2+8\,x^3-48\right )}{2}+\frac {\ln \left (\frac {2\,x}{x^2+2}\right )\,\left (4\,x^2+8\right )}{2}-14\,x^2+2\,x^3-28}{\ln \left (5\right )\,\left (x^2+2\right )} \,d x \] Input:

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)
 

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {56-8 x+28 x^2-4 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )+\log (x) \left (48-16 x+32 x^2-8 x^3+\left (-8-4 x^2\right ) \log \left (\frac {2 x}{2+x^2}\right )\right )}{\left (2+x^2\right ) \log (25)} \, dx=\frac {2 \,\mathrm {log}\left (x \right ) x \left (-\mathrm {log}\left (\frac {2 x}{x^{2}+2}\right )-x +7\right )}{\mathrm {log}\left (5\right )} \] Input:

int(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)
 

Output:

(2*log(x)*x*( - log((2*x)/(x**2 + 2)) - x + 7))/log(5)