3.4.48 \(\int \frac {(-x^3-25 x^5) \log (4)+(3 x^3+15 x^5) \log (4) \log (x+10 x^3+25 x^5)+(-1-5 x^2) \log ^2(x+10 x^3+25 x^5)+(1+5 x^2) \log (4^{-\frac {x^3}{\log (x+10 x^3+25 x^5)}} x) \log ^2(x+10 x^3+25 x^5)}{(1+5 x^2) \log ^2(4^{-\frac {x^3}{\log (x+10 x^3+25 x^5)}} x) \log ^2(x+10 x^3+25 x^5)} \, dx\) [348]

3.4.48.1 Optimal result

Integrand size = 165, antiderivative size = 28 \[ \int \frac {\left (-x^3-25 x^5\right ) \log (4)+\left (3 x^3+15 x^5\right ) \log (4) \log \left (x+10 x^3+25 x^5\right )+\left (-1-5 x^2\right ) \log ^2\left (x+10 x^3+25 x^5\right )+\left (1+5 x^2\right ) \log \left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )}{\left (1+5 x^2\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )} \, dx=\frac {x}{\log \left (4^{-\frac {x^3}{\log \left (x \left (1+5 x^2\right )^2\right )}} x\right )} \]

output

x/ln(x/exp(2*x^3/ln(x*(5*x^2+1)^2)*ln(2)))
 

3.4.48.2 Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-x^3-25 x^5\right ) \log (4)+\left (3 x^3+15 x^5\right ) \log (4) \log \left (x+10 x^3+25 x^5\right )+\left (-1-5 x^2\right ) \log ^2\left (x+10 x^3+25 x^5\right )+\left (1+5 x^2\right ) \log \left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )}{\left (1+5 x^2\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )} \, dx=\frac {x}{\log \left (4^{-\frac {x^3}{\log \left (x \left (1+5 x^2\right )^2\right )}} x\right )} \]

input

Integrate[((-x^3 - 25*x^5)*Log[4] + (3*x^3 + 15*x^5)*Log[4]*Log[x + 10*x^3 
 + 25*x^5] + (-1 - 5*x^2)*Log[x + 10*x^3 + 25*x^5]^2 + (1 + 5*x^2)*Log[x/4 
^(x^3/Log[x + 10*x^3 + 25*x^5])]*Log[x + 10*x^3 + 25*x^5]^2)/((1 + 5*x^2)* 
Log[x/4^(x^3/Log[x + 10*x^3 + 25*x^5])]^2*Log[x + 10*x^3 + 25*x^5]^2),x]
 

output

x/Log[x/4^(x^3/Log[x*(1 + 5*x^2)^2])]
 

3.4.48.3 Rubi [F]

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.

\(\displaystyle \int \frac {\left (15 x^5+3 x^3\right ) \log (4) \log \left (25 x^5+10 x^3+x\right )+\left (-25 x^5-x^3\right ) \log (4)+\left (-5 x^2-1\right ) \log ^2\left (25 x^5+10 x^3+x\right )+\left (5 x^2+1\right ) \log \left (x 4^{-\frac {x^3}{\log \left (25 x^5+10 x^3+x\right )}}\right ) \log ^2\left (25 x^5+10 x^3+x\right )}{\left (5 x^2+1\right ) \log ^2\left (x 4^{-\frac {x^3}{\log \left (25 x^5+10 x^3+x\right )}}\right ) \log ^2\left (25 x^5+10 x^3+x\right )} \, dx\)

\(\Big \downarrow \) 7276

\(\displaystyle \int \left (\frac {x^3 \log (64)}{\log ^2\left (x 4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}}\right ) \log \left (x \left (5 x^2+1\right )^2\right )}-\frac {\left (25 x^2+1\right ) x^3 \log (4)}{\left (5 x^2+1\right ) \log ^2\left (x 4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}}\right ) \log ^2\left (x \left (5 x^2+1\right )^2\right )}+\frac {\log \left (x 4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}}\right )-1}{\log ^2\left (x 4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\int \frac {1}{\log ^2\left (4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}} x\right )}dx+\frac {4}{5} \log (4) \int \frac {x}{\log ^2\left (4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}} x\right ) \log ^2\left (x \left (5 x^2+1\right )^2\right )}dx-5 \log (4) \int \frac {x^3}{\log ^2\left (4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}} x\right ) \log ^2\left (x \left (5 x^2+1\right )^2\right )}dx+\frac {2 \log (4) \int \frac {1}{\left (i-\sqrt {5} x\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}} x\right ) \log ^2\left (x \left (5 x^2+1\right )^2\right )}dx}{5 \sqrt {5}}-\frac {2 \log (4) \int \frac {1}{\left (\sqrt {5} x+i\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}} x\right ) \log ^2\left (x \left (5 x^2+1\right )^2\right )}dx}{5 \sqrt {5}}+\log (64) \int \frac {x^3}{\log ^2\left (4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}} x\right ) \log \left (x \left (5 x^2+1\right )^2\right )}dx+\int \frac {1}{\log \left (4^{-\frac {x^3}{\log \left (x \left (5 x^2+1\right )^2\right )}} x\right )}dx\)

input

Int[((-x^3 - 25*x^5)*Log[4] + (3*x^3 + 15*x^5)*Log[4]*Log[x + 10*x^3 + 25* 
x^5] + (-1 - 5*x^2)*Log[x + 10*x^3 + 25*x^5]^2 + (1 + 5*x^2)*Log[x/4^(x^3/ 
Log[x + 10*x^3 + 25*x^5])]*Log[x + 10*x^3 + 25*x^5]^2)/((1 + 5*x^2)*Log[x/ 
4^(x^3/Log[x + 10*x^3 + 25*x^5])]^2*Log[x + 10*x^3 + 25*x^5]^2),x]
 

output

$Aborted
 

3.4.48.3.1 Defintions of rubi rules used

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]
 

3.4.48.4 Maple [A] (verified)

Time = 138.70 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(\frac {x}{\ln \left (x \,{\mathrm e}^{-\frac {2 x^{3} \ln \left (2\right )}{\ln \left (25 x^{5}+10 x^{3}+x \right )}}\right )}\)	\(33\)

input

int(((5*x^2+1)*ln(25*x^5+10*x^3+x)^2*ln(x/exp(2*x^3*ln(2)/ln(25*x^5+10*x^3 
+x)))+(-5*x^2-1)*ln(25*x^5+10*x^3+x)^2+2*(15*x^5+3*x^3)*ln(2)*ln(25*x^5+10 
*x^3+x)+2*(-25*x^5-x^3)*ln(2))/(5*x^2+1)/ln(25*x^5+10*x^3+x)^2/ln(x/exp(2* 
x^3*ln(2)/ln(25*x^5+10*x^3+x)))^2,x,method=_RETURNVERBOSE)
 

output

x/ln(x/exp(2*x^3*ln(2)/ln(25*x^5+10*x^3+x)))
 

3.4.48.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {\left (-x^3-25 x^5\right ) \log (4)+\left (3 x^3+15 x^5\right ) \log (4) \log \left (x+10 x^3+25 x^5\right )+\left (-1-5 x^2\right ) \log ^2\left (x+10 x^3+25 x^5\right )+\left (1+5 x^2\right ) \log \left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )}{\left (1+5 x^2\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )} \, dx=\frac {x}{\log \left (\frac {x}{2^{\frac {2 \, x^{3}}{\log \left (25 \, x^{5} + 10 \, x^{3} + x\right )}}}\right )} \]

input

integrate(((5*x^2+1)*log(25*x^5+10*x^3+x)^2*log(x/exp(2*x^3*log(2)/log(25* 
x^5+10*x^3+x)))+(-5*x^2-1)*log(25*x^5+10*x^3+x)^2+2*(15*x^5+3*x^3)*log(2)* 
log(25*x^5+10*x^3+x)+2*(-25*x^5-x^3)*log(2))/(5*x^2+1)/log(25*x^5+10*x^3+x 
)^2/log(x/exp(2*x^3*log(2)/log(25*x^5+10*x^3+x)))^2,x, algorithm=\
 

output

x/log(x/2^(2*x^3/log(25*x^5 + 10*x^3 + x)))
 

3.4.48.6 Sympy [A] (verification not implemented)

Time = 0.92 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {\left (-x^3-25 x^5\right ) \log (4)+\left (3 x^3+15 x^5\right ) \log (4) \log \left (x+10 x^3+25 x^5\right )+\left (-1-5 x^2\right ) \log ^2\left (x+10 x^3+25 x^5\right )+\left (1+5 x^2\right ) \log \left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )}{\left (1+5 x^2\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )} \, dx=\frac {x}{\log {\left (x e^{- \frac {2 x^{3} \log {\left (2 \right )}}{\log {\left (25 x^{5} + 10 x^{3} + x \right )}}} \right )}} \]

input

integrate(((5*x**2+1)*ln(25*x**5+10*x**3+x)**2*ln(x/exp(2*x**3*ln(2)/ln(25 
*x**5+10*x**3+x)))+(-5*x**2-1)*ln(25*x**5+10*x**3+x)**2+2*(15*x**5+3*x**3) 
*ln(2)*ln(25*x**5+10*x**3+x)+2*(-25*x**5-x**3)*ln(2))/(5*x**2+1)/ln(25*x** 
5+10*x**3+x)**2/ln(x/exp(2*x**3*ln(2)/ln(25*x**5+10*x**3+x)))**2,x)
 

output

x/log(x*exp(-2*x**3*log(2)/log(25*x**5 + 10*x**3 + x)))
 

3.4.48.7 Maxima [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-x^3-25 x^5\right ) \log (4)+\left (3 x^3+15 x^5\right ) \log (4) \log \left (x+10 x^3+25 x^5\right )+\left (-1-5 x^2\right ) \log ^2\left (x+10 x^3+25 x^5\right )+\left (1+5 x^2\right ) \log \left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )}{\left (1+5 x^2\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )} \, dx=-\frac {x}{2 \, \log \left (2^{\frac {x^{3}}{2 \, \log \left (5 \, x^{2} + 1\right ) + \log \left (x\right )}}\right ) - \log \left (x\right )} \]

input

integrate(((5*x^2+1)*log(25*x^5+10*x^3+x)^2*log(x/exp(2*x^3*log(2)/log(25* 
x^5+10*x^3+x)))+(-5*x^2-1)*log(25*x^5+10*x^3+x)^2+2*(15*x^5+3*x^3)*log(2)* 
log(25*x^5+10*x^3+x)+2*(-25*x^5-x^3)*log(2))/(5*x^2+1)/log(25*x^5+10*x^3+x 
)^2/log(x/exp(2*x^3*log(2)/log(25*x^5+10*x^3+x)))^2,x, algorithm=\
 

output

-x/(2*log(2^(x^3/(2*log(5*x^2 + 1) + log(x)))) - log(x))
 

3.4.48.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {\left (-x^3-25 x^5\right ) \log (4)+\left (3 x^3+15 x^5\right ) \log (4) \log \left (x+10 x^3+25 x^5\right )+\left (-1-5 x^2\right ) \log ^2\left (x+10 x^3+25 x^5\right )+\left (1+5 x^2\right ) \log \left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )}{\left (1+5 x^2\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )} \, dx=-\frac {2 \, x^{4} \log \left (2\right )}{2 \, x^{3} \log \left (2\right ) \log \left (x\right ) - 2 \, \log \left (5 \, x^{2} + 1\right ) \log \left (x\right )^{2} - \log \left (x\right )^{3}} + \frac {x}{\log \left (x\right )} \]

input

integrate(((5*x^2+1)*log(25*x^5+10*x^3+x)^2*log(x/exp(2*x^3*log(2)/log(25* 
x^5+10*x^3+x)))+(-5*x^2-1)*log(25*x^5+10*x^3+x)^2+2*(15*x^5+3*x^3)*log(2)* 
log(25*x^5+10*x^3+x)+2*(-25*x^5-x^3)*log(2))/(5*x^2+1)/log(25*x^5+10*x^3+x 
)^2/log(x/exp(2*x^3*log(2)/log(25*x^5+10*x^3+x)))^2,x, algorithm=\
 

output

-2*x^4*log(2)/(2*x^3*log(2)*log(x) - 2*log(5*x^2 + 1)*log(x)^2 - log(x)^3) 
 + x/log(x)
 

3.4.48.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-x^3-25 x^5\right ) \log (4)+\left (3 x^3+15 x^5\right ) \log (4) \log \left (x+10 x^3+25 x^5\right )+\left (-1-5 x^2\right ) \log ^2\left (x+10 x^3+25 x^5\right )+\left (1+5 x^2\right ) \log \left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )}{\left (1+5 x^2\right ) \log ^2\left (4^{-\frac {x^3}{\log \left (x+10 x^3+25 x^5\right )}} x\right ) \log ^2\left (x+10 x^3+25 x^5\right )} \, dx=\int -\frac {{\ln \left (25\,x^5+10\,x^3+x\right )}^2\,\left (5\,x^2+1\right )+2\,\ln \left (2\right )\,\left (25\,x^5+x^3\right )-2\,\ln \left (2\right )\,\ln \left (25\,x^5+10\,x^3+x\right )\,\left (15\,x^5+3\,x^3\right )-\ln \left (x\,{\mathrm {e}}^{-\frac {2\,x^3\,\ln \left (2\right )}{\ln \left (25\,x^5+10\,x^3+x\right )}}\right )\,{\ln \left (25\,x^5+10\,x^3+x\right )}^2\,\left (5\,x^2+1\right )}{{\ln \left (x\,{\mathrm {e}}^{-\frac {2\,x^3\,\ln \left (2\right )}{\ln \left (25\,x^5+10\,x^3+x\right )}}\right )}^2\,{\ln \left (25\,x^5+10\,x^3+x\right )}^2\,\left (5\,x^2+1\right )} \,d x \]

input

int(-(log(x + 10*x^3 + 25*x^5)^2*(5*x^2 + 1) + 2*log(2)*(x^3 + 25*x^5) - 2 
*log(2)*log(x + 10*x^3 + 25*x^5)*(3*x^3 + 15*x^5) - log(x*exp(-(2*x^3*log( 
2))/log(x + 10*x^3 + 25*x^5)))*log(x + 10*x^3 + 25*x^5)^2*(5*x^2 + 1))/(lo 
g(x*exp(-(2*x^3*log(2))/log(x + 10*x^3 + 25*x^5)))^2*log(x + 10*x^3 + 25*x 
^5)^2*(5*x^2 + 1)),x)
 

output

int(-(log(x + 10*x^3 + 25*x^5)^2*(5*x^2 + 1) + 2*log(2)*(x^3 + 25*x^5) - 2 
*log(2)*log(x + 10*x^3 + 25*x^5)*(3*x^3 + 15*x^5) - log(x*exp(-(2*x^3*log( 
2))/log(x + 10*x^3 + 25*x^5)))*log(x + 10*x^3 + 25*x^5)^2*(5*x^2 + 1))/(lo 
g(x*exp(-(2*x^3*log(2))/log(x + 10*x^3 + 25*x^5)))^2*log(x + 10*x^3 + 25*x 
^5)^2*(5*x^2 + 1)), x)