3.33.99 $\int \frac{-16e^{x}x-16e^x\log (x)+(e^x(-8-4x+4x^2)+e^x(4+4x)\log (x))\log (9x^2)}{(x^3+3x^2\log (x)+3x{\log }^2(x)+{\log }^3(x)){\log }^3(9x^2)}dx$

Optimal. Leaf size=22 $$-2+\frac{4e^{x}x}{(x+\log (x){)}^2{\log }^2\left(9x^2\right)}$$

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Rubi [F]  time = 2.65, 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 = {} $$\begin{array}{c}\int \frac{-16e^{x}x-16e^x\log (x)+(e^x(-8-4x+4x^2)+e^x(4+4x)\log (x))\log \left(9x^2\right)}{(x^3+3x^2\log (x)+3x{\log }^2(x)+{\log }^3(x)){\log }^3\left(9x^2\right)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(-16*E^x*x - 16*E^x*Log[x] + (E^x*(-8 - 4*x + 4*x^2) + E^x*(4 + 4*x)*Log[x])*Log[9*x^2])/((x^3 + 3*x^2*Log
[x] + 3*x*Log[x]^2 + Log[x]^3)*Log[9*x^2]^3),x]

[Out]

-16*Defer[Int][E^x/((x + Log[x])^2*Log[9*x^2]^3), x] - 8*Defer[Int][E^x/((x + Log[x])^3*Log[9*x^2]^2), x] - 4*
Defer[Int][(E^x*x)/((x + Log[x])^3*Log[9*x^2]^2), x] + 4*Defer[Int][(E^x*x^2)/((x + Log[x])^3*Log[9*x^2]^2), x
] + 4*Defer[Int][(E^x*Log[x])/((x + Log[x])^3*Log[9*x^2]^2), x] + 4*Defer[Int][(E^x*x*Log[x])/((x + Log[x])^3*
Log[9*x^2]^2), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{4e^x(-4x+(-2-x+x^2)\log \left(9x^2\right)+\log (x)(-4+(1+x)\log \left(9x^2\right)))}{(x+\log (x){)}^3{\log }^3\left(9x^2\right)}dx\\ & =4\int \frac{e^x(-4x+(-2-x+x^2)\log \left(9x^2\right)+\log (x)(-4+(1+x)\log \left(9x^2\right)))}{(x+\log (x){)}^3{\log }^3\left(9x^2\right)}dx\\ & =4\int (-\frac{4e^x}{(x+\log (x){)}^2{\log }^3\left(9x^2\right)}+\frac{e^x(1+x)(-2+x+\log (x))}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)})dx\\ & =4\int \frac{e^x(1+x)(-2+x+\log (x))}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx-16\int \frac{e^x}{(x+\log (x){)}^2{\log }^3\left(9x^2\right)}dx\\ & =4\int (\frac{e^x(-2+x+\log (x))}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}+\frac{e^{x}x(-2+x+\log (x))}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)})dx-16\int \frac{e^x}{(x+\log (x){)}^2{\log }^3\left(9x^2\right)}dx\\ & =4\int \frac{e^x(-2+x+\log (x))}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx+4\int \frac{e^{x}x(-2+x+\log (x))}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx-16\int \frac{e^x}{(x+\log (x){)}^2{\log }^3\left(9x^2\right)}dx\\ & =4\int (-\frac{2e^x}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}+\frac{e^{x}x}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}+\frac{e^x\log (x)}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)})dx+4\int (-\frac{2e^{x}x}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}+\frac{e^x{x}^2}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}+\frac{e^{x}x\log (x)}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)})dx-16\int \frac{e^x}{(x+\log (x){)}^2{\log }^3\left(9x^2\right)}dx\\ & =4\int \frac{e^{x}x}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx+4\int \frac{e^x{x}^2}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx+4\int \frac{e^x\log (x)}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx+4\int \frac{e^{x}x\log (x)}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx-8\int \frac{e^x}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx-8\int \frac{e^{x}x}{(x+\log (x){)}^3{\log }^2\left(9x^2\right)}dx-16\int \frac{e^x}{(x+\log (x){)}^2{\log }^3\left(9x^2\right)}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.48, size = 20, normalized size = 0.91 $$\begin{array}{c}\frac{4e^{x}x}{(x+\log (x){)}^2{\log }^2\left(9x^2\right)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(-16*E^x*x - 16*E^x*Log[x] + (E^x*(-8 - 4*x + 4*x^2) + E^x*(4 + 4*x)*Log[x])*Log[9*x^2])/((x^3 + 3*x
^2*Log[x] + 3*x*Log[x]^2 + Log[x]^3)*Log[9*x^2]^3),x]

[Out]

(4*E^x*x)/((x + Log[x])^2*Log[9*x^2]^2)

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fricas [B]  time = 0.54, size = 64, normalized size = 2.91 $$\begin{array}{c}\frac{x{e}^x}{x^2\log (3{)}^2+2(x+\log (3))\log (x{)}^3+\log (x{)}^4+(x^2+4x\log (3)+\log (3{)}^2)\log (x{)}^2+2(x^2\log (3)+x\log (3{)}^2)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp(x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x
*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2)^3,x, algorithm="fricas")

[Out]

x*e^x/(x^2*log(3)^2 + 2*(x + log(3))*log(x)^3 + log(x)^4 + (x^2 + 4*x*log(3) + log(3)^2)*log(x)^2 + 2*(x^2*log
(3) + x*log(3)^2)*log(x))

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giac [B]  time = 0.40, size = 78, normalized size = 3.55 $$\begin{array}{c}\frac{x{e}^x}{x^2\log (3{)}^2+2x^2\log (3)\log (x)+2x\log (3{)}^2\log (x)+x^2\log (x{)}^2+4x\log (3)\log (x{)}^2+\log (3{)}^2\log (x{)}^2+2x\log (x{)}^3+2\log (3)\log (x{)}^3+\log (x{)}^4}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp(x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x
*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2)^3,x, algorithm="giac")

[Out]

x*e^x/(x^2*log(3)^2 + 2*x^2*log(3)*log(x) + 2*x*log(3)^2*log(x) + x^2*log(x)^2 + 4*x*log(3)*log(x)^2 + log(3)^
2*log(x)^2 + 2*x*log(x)^3 + 2*log(3)*log(x)^3 + log(x)^4)

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 $$\int \frac{((4x+4){\mathrm{e}}^x\ln (x)+(4x^2-4x-8){\mathrm{e}}^x)\ln \left(9x^2\right)-16{\mathrm{e}}^x\ln (x)-16{\mathrm{e}}^{x}x}{(\ln (x{)}^3+3x\ln (x{)}^2+3x^2\ln (x)+x^3)\ln {\left(9x^2\right)}^3}dx$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4*x+4)*exp(x)*ln(x)+(4*x^2-4*x-8)*exp(x))*ln(9*x^2)-16*exp(x)*ln(x)-16*exp(x)*x)/(ln(x)^3+3*x*ln(x)^2+3
*x^2*ln(x)+x^3)/ln(9*x^2)^3,x)

[Out]

int((((4*x+4)*exp(x)*ln(x)+(4*x^2-4*x-8)*exp(x))*ln(9*x^2)-16*exp(x)*ln(x)-16*exp(x)*x)/(ln(x)^3+3*x*ln(x)^2+3
*x^2*ln(x)+x^3)/ln(9*x^2)^3,x)

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maxima [B]  time = 0.64, size = 64, normalized size = 2.91 $$\begin{array}{c}\frac{x{e}^x}{x^2\log (3{)}^2+2(x+\log (3))\log (x{)}^3+\log (x{)}^4+(x^2+4x\log (3)+\log (3{)}^2)\log (x{)}^2+2(x^2\log (3)+x\log (3{)}^2)\log (x)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*log(x)+(4*x^2-4*x-8)*exp(x))*log(9*x^2)-16*exp(x)*log(x)-16*exp(x)*x)/(log(x)^3+3*x
*log(x)^2+3*x^2*log(x)+x^3)/log(9*x^2)^3,x, algorithm="maxima")

[Out]

x*e^x/(x^2*log(3)^2 + 2*(x + log(3))*log(x)^3 + log(x)^4 + (x^2 + 4*x*log(3) + log(3)^2)*log(x)^2 + 2*(x^2*log
(3) + x*log(3)^2)*log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 $$\begin{array}{c}\int -\frac{16{\mathrm{e}}^x\ln (x)+\ln \left(9x^2\right)({\mathrm{e}}^x(-4x^2+4x+8)-{\mathrm{e}}^x\ln (x)(4x+4))+16x{\mathrm{e}}^x}{{\ln \left(9x^2\right)}^3(x^3+3x^2\ln (x)+3x{\ln (x)}^2+{\ln (x)}^3)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(x)*log(x) + log(9*x^2)*(exp(x)*(4*x - 4*x^2 + 8) - exp(x)*log(x)*(4*x + 4)) + 16*x*exp(x))/(log(9
*x^2)^3*(3*x*log(x)^2 + 3*x^2*log(x) + log(x)^3 + x^3)),x)

[Out]

int(-(16*exp(x)*log(x) + log(9*x^2)*(exp(x)*(4*x - 4*x^2 + 8) - exp(x)*log(x)*(4*x + 4)) + 16*x*exp(x))/(log(9
*x^2)^3*(3*x*log(x)^2 + 3*x^2*log(x) + log(x)^3 + x^3)), x)

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sympy [B]  time = 0.70, size = 90, normalized size = 4.09 $$\begin{array}{c}\frac{x{e}^x}{x^2\log {(x)}^2+2x^2\log (3)\log (x)+x^2\log {(3)}^2+2x\log {(x)}^3+4x\log (3)\log {(x)}^2+2x\log {(3)}^2\log (x)+\log {(x)}^4+2\log (3)\log {(x)}^3+\log {(3)}^2\log {(x)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4*x+4)*exp(x)*ln(x)+(4*x**2-4*x-8)*exp(x))*ln(9*x**2)-16*exp(x)*ln(x)-16*exp(x)*x)/(ln(x)**3+3*x*
ln(x)**2+3*x**2*ln(x)+x**3)/ln(9*x**2)**3,x)

[Out]

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

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