3.8.50 \(\int \frac {25}{18} x^{2 x} (5+10 x-\frac {9 x^{-2 x}}{5}+10 x \log (x)) \, dx\)

Optimal. Leaf size=24 \[ \frac {5 \left (-x^2+\frac {25}{9} x^{2+2 x}\right )}{2 x} \]

________________________________________________________________________________________

Rubi [F]  time = 0.13, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {25}{18} x^{2 x} \left (5+10 x-\frac {9 x^{-2 x}}{5}+10 x \log (x)\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*x^(2*x)*(5 + 10*x - 9/(5*x^(2*x)) + 10*x*Log[x]))/18,x]

[Out]

(-5*x)/2 + (125*Defer[Int][x^(2*x), x])/18 + (125*Defer[Int][x^(1 + 2*x), x])/9 + (125*Log[x]*Defer[Int][x^(1
+ 2*x), x])/9 - (125*Defer[Int][Defer[Int][x^(1 + 2*x), x]/x, x])/9

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {25}{18} \int x^{2 x} \left (5+10 x-\frac {9 x^{-2 x}}{5}+10 x \log (x)\right ) \, dx\\ &=\frac {25}{18} \int \left (-\frac {9}{5}+5 x^{2 x}+10 x^{1+2 x}+10 x^{1+2 x} \log (x)\right ) \, dx\\ &=-\frac {5 x}{2}+\frac {125}{18} \int x^{2 x} \, dx+\frac {125}{9} \int x^{1+2 x} \, dx+\frac {125}{9} \int x^{1+2 x} \log (x) \, dx\\ &=-\frac {5 x}{2}+\frac {125}{18} \int x^{2 x} \, dx+\frac {125}{9} \int x^{1+2 x} \, dx-\frac {125}{9} \int \frac {\int x^{1+2 x} \, dx}{x} \, dx+\frac {1}{9} (125 \log (x)) \int x^{1+2 x} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 19, normalized size = 0.79 \begin {gather*} \frac {25}{18} \left (-\frac {9 x}{5}+5 x^{1+2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^(2*x)*(5 + 10*x - 9/(5*x^(2*x)) + 10*x*Log[x]))/18,x]

[Out]

(25*((-9*x)/5 + 5*x^(1 + 2*x)))/18

________________________________________________________________________________________

fricas [A]  time = 0.76, size = 18, normalized size = 0.75 \begin {gather*} \frac {5}{2} \, x e^{\left (2 \, x \log \relax (x) - 2 \, \log \left (\frac {3}{5}\right )\right )} - \frac {5}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-5*exp(-x*log(x)+log(3/5))^2+10*x*log(x)+10*x+5)/exp(-x*log(x)+log(3/5))^2,x, algorithm="fricas
")

[Out]

5/2*x*e^(2*x*log(x) - 2*log(3/5)) - 5/2*x

________________________________________________________________________________________

giac [A]  time = 0.22, size = 12, normalized size = 0.50 \begin {gather*} \frac {125}{18} \, x x^{2 \, x} - \frac {5}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-5*exp(-x*log(x)+log(3/5))^2+10*x*log(x)+10*x+5)/exp(-x*log(x)+log(3/5))^2,x, algorithm="giac")

[Out]

125/18*x*x^(2*x) - 5/2*x

________________________________________________________________________________________

maple [A]  time = 0.06, size = 15, normalized size = 0.62

method	result	size
risch	\(-\frac {5 x}{2}+\frac {125 x \,x^{2 x}}{18}\)	\(15\)
default	\(-\frac {5 x}{2}+\frac {125 x \,{\mathrm e}^{2 x \ln \relax (x )}}{18}\)	\(19\)
norman	\(\frac {25 \left (\frac {5 x}{2}-\frac {9 x \,{\mathrm e}^{-2 x \ln \relax (x )}}{10}\right ) {\mathrm e}^{2 x \ln \relax (x )}}{9}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2*x+125/18*x/(x^(-x))^2

________________________________________________________________________________________

maxima [A]  time = 0.57, size = 12, normalized size = 0.50 \begin {gather*} \frac {125}{18} \, x x^{2 \, x} - \frac {5}{2} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-5*exp(-x*log(x)+log(3/5))^2+10*x*log(x)+10*x+5)/exp(-x*log(x)+log(3/5))^2,x, algorithm="maxima
")

[Out]

125/18*x*x^(2*x) - 5/2*x

________________________________________________________________________________________

mupad [B]  time = 0.57, size = 12, normalized size = 0.50 \begin {gather*} \frac {5\,x\,\left (25\,x^{2\,x}-9\right )}{18} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*x*(25*x^(2*x) - 9))/18

________________________________________________________________________________________

sympy [A]  time = 0.35, size = 17, normalized size = 0.71 \begin {gather*} \frac {125 x e^{2 x \log {\relax (x )}}}{18} - \frac {5 x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125*x*exp(2*x*log(x))/18 - 5*x/2

________________________________________________________________________________________