\(\int (-1+5^{-x} (3 x-4 x \log (5)+x \log ^2(5))^x (1+x+x \log (\frac {1}{5} (3 x-4 x \log (5)+x \log ^2(5))))) \, dx\) [4488]

Optimal result

Integrand size = 50, antiderivative size = 22 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x \left (-1+5^{-x} \left (x \left (-1+(-2+\log (5))^2\right )\right )^x\right ) \]

[Out]

x*(exp(x*ln(1/5*x*((ln(5)-2)^2-1)))-1)

Rubi [F]

\[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=\int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx \]

[In]

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

[Out]

-x + Defer[Int][(x*(1 - Log[5]))^x*((3 - Log[5])/5)^x, x] + Defer[Int][(x*(1 - Log[5]))^(1 + x)*((3 - Log[5])/
5)^x, x]/(1 - Log[5]) + (Log[(x*(1 - Log[5])*(3 - Log[5]))/5]*Defer[Int][(x*(1 - Log[5]))^(1 + x)*((3 - Log[5]
)/5)^x, x])/(1 - Log[5]) - Defer[Int][Defer[Int][(x*(1 - Log[5]))^(1 + x)*((3 - Log[5])/5)^x, x]/x, x]/(1 - Lo
g[5])

Rubi steps \begin{align*} \text {integral}& = -x+\int 5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int 5^{-x} \left (x (3-4 \log (5))+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int 5^{-x} \left (x \left (3-4 \log (5)+\log ^2(5)\right )\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int (x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right ) \, dx \\ & = -x+\int \left ((x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x+\frac {(x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x}{1-\log (5)}+\frac {(x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \log \left (\frac {1}{5} x \left (3-4 \log (5)+\log ^2(5)\right )\right )}{1-\log (5)}\right ) \, dx \\ & = -x+\frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{1-\log (5)}+\frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \log \left (\frac {1}{5} x \left (3-4 \log (5)+\log ^2(5)\right )\right ) \, dx}{1-\log (5)}+\int (x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x \, dx \\ & = -x+\frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{1-\log (5)}-\frac {\int \frac {\int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{x} \, dx}{1-\log (5)}+\frac {\log \left (\frac {1}{5} x \left (3-4 \log (5)+\log ^2(5)\right )\right ) \int (x (1-\log (5)))^{1+x} \left (\frac {1}{5} (3-\log (5))\right )^x \, dx}{1-\log (5)}+\int (x (1-\log (5)))^x \left (\frac {1}{5} (3-\log (5))\right )^x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=-x+\frac {5^{-x} \left (x \left (3-4 \log (5)+\log ^2(5)\right )\right )^{1+x}}{3-4 \log (5)+\log ^2(5)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09

method	result	size
parallelrisch	\({\mathrm e}^{x \ln \left (\frac {x \left (\ln \left (5\right )^{2}-4 \ln \left (5\right )+3\right )}{5}\right )} x -x\)	\(24\)
risch	\(-x +\left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )^{x} x\)	\(25\)
default	\(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\)	\(27\)
norman	\(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\)	\(27\)
parts	\(-x +{\mathrm e}^{x \ln \left (\frac {x \ln \left (5\right )^{2}}{5}-\frac {4 x \ln \left (5\right )}{5}+\frac {3 x}{5}\right )} x\)	\(27\)

[In]

int((x*ln(1/5*x*ln(5)^2-4/5*x*ln(5)+3/5*x)+x+1)*exp(x*ln(1/5*x*ln(5)^2-4/5*x*ln(5)+3/5*x))-1,x,method=_RETURNV
ERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx={\left (\frac {1}{5} \, x \log \left (5\right )^{2} - \frac {4}{5} \, x \log \left (5\right ) + \frac {3}{5} \, x\right )}^{x} x - x \]

[In]

integrate((x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x)+x+1)*exp(x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x))-1,x, al
gorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x e^{x \log {\left (- \frac {4 x \log {\left (5 \right )}}{5} + \frac {x \log {\left (5 \right )}^{2}}{5} + \frac {3 x}{5} \right )}} - x \]

[In]

integrate((x*ln(1/5*x*ln(5)**2-4/5*x*ln(5)+3/5*x)+x+1)*exp(x*ln(1/5*x*ln(5)**2-4/5*x*ln(5)+3/5*x))-1,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x e^{\left (-x \log \left (5\right ) + x \log \left (x\right ) + x \log \left (\log \left (5\right ) - 1\right ) + x \log \left (\log \left (5\right ) - 3\right )\right )} - x \]

[In]

integrate((x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x)+x+1)*exp(x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x))-1,x, al
gorithm="maxima")

[Out]

x*e^(-x*log(5) + x*log(x) + x*log(log(5) - 1) + x*log(log(5) - 3)) - x

Giac [F(-1)]

Timed out. \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate((x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x)+x+1)*exp(x*log(1/5*x*log(5)^2-4/5*x*log(5)+3/5*x))-1,x, al
gorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 14.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \left (-1+5^{-x} \left (3 x-4 x \log (5)+x \log ^2(5)\right )^x \left (1+x+x \log \left (\frac {1}{5} \left (3 x-4 x \log (5)+x \log ^2(5)\right )\right )\right )\right ) \, dx=x\,\left ({\left (\frac {3\,x}{5}-\frac {4\,x\,\ln \left (5\right )}{5}+\frac {x\,{\ln \left (5\right )}^2}{5}\right )}^x-1\right ) \]

[In]

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

[Out]

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