\(\int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx\) [5160]

Optimal result

Integrand size = 18, antiderivative size = 21 \[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=i \pi +x^{2 (-5+x)}+\log \left (-\frac {5}{2}+e^6\right ) \]

[Out]

ln(5/2-exp(6))+exp(2*ln(x)*(-5+x))

Rubi [F]

\[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=\int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx \]

[In]

Int[x^(-11 + 2*x)*(-10 + 2*x + 2*x*Log[x]),x]

[Out]

-10*Defer[Int][x^(-11 + 2*x), x] + 2*Defer[Int][x^(-10 + 2*x), x] + 2*Log[x]*Defer[Int][x^(-10 + 2*x), x] - 2*
Defer[Int][Defer[Int][x^(-10 + 2*x), x]/x, x]

Rubi steps \begin{align*} \text {integral}& = \int 2 x^{-11+2 x} (-5+x+x \log (x)) \, dx \\ & = 2 \int x^{-11+2 x} (-5+x+x \log (x)) \, dx \\ & = 2 \int \left (-5 x^{-11+2 x}+x^{-10+2 x}+x^{-10+2 x} \log (x)\right ) \, dx \\ & = 2 \int x^{-10+2 x} \, dx+2 \int x^{-10+2 x} \log (x) \, dx-10 \int x^{-11+2 x} \, dx \\ & = 2 \int x^{-10+2 x} \, dx-2 \int \frac {\int x^{-10+2 x} \, dx}{x} \, dx-10 \int x^{-11+2 x} \, dx+(2 \log (x)) \int x^{-10+2 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.33 \[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=x^{-10+2 x} \]

[In]

Integrate[x^(-11 + 2*x)*(-10 + 2*x + 2*x*Log[x]),x]

[Out]

x^(-10 + 2*x)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38

method	result	size
risch	\(x^{2 x -10}\)	\(8\)
norman	\({\mathrm e}^{\left (2 x -10\right ) \ln \left (x \right )}\)	\(10\)
parallelrisch	\({\mathrm e}^{\left (2 x -10\right ) \ln \left (x \right )}\)	\(10\)

[In]

int((2*x*ln(x)+2*x-10)*exp((2*x-10)*ln(x))/x,x,method=_RETURNVERBOSE)

[Out]

x^(2*x-10)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.33 \[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=x^{2 \, x - 10} \]

[In]

integrate((2*x*log(x)+2*x-10)*exp((2*x-10)*log(x))/x,x, algorithm="fricas")

[Out]

x^(2*x - 10)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.38 \[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=e^{\left (2 x - 10\right ) \log {\left (x \right )}} \]

[In]

integrate((2*x*ln(x)+2*x-10)*exp((2*x-10)*ln(x))/x,x)

[Out]

exp((2*x - 10)*log(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.43 \[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=\frac {x^{2 \, x}}{x^{10}} \]

[In]

integrate((2*x*log(x)+2*x-10)*exp((2*x-10)*log(x))/x,x, algorithm="maxima")

[Out]

x^(2*x)/x^10

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.33 \[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=x^{2 \, x - 10} \]

[In]

integrate((2*x*log(x)+2*x-10)*exp((2*x-10)*log(x))/x,x, algorithm="giac")

[Out]

x^(2*x - 10)

Mupad [B] (verification not implemented)

Time = 11.92 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.33 \[ \int x^{-11+2 x} (-10+2 x+2 x \log (x)) \, dx=x^{2\,x-10} \]

[In]

int((exp(log(x)*(2*x - 10))*(2*x + 2*x*log(x) - 10))/x,x)

[Out]

x^(2*x - 10)