3.33.14 $\int \frac{5+5e^{3}x+e^{36e^4-24e^{2}x+4x^2}(x+x^2-24e^2{x}^2+8x^3)+5x\log (x)}{5e^{3}x+e^{36e^4-24e^{2}x+4x^2}{x}^2+5x\log (x)}dx$

Optimal. Leaf size=29 $$x+\log (e^3+\frac{1}{5}{e}^{4{(3e^2-x)}^2}x+\log (x))$$

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Rubi [F]  time = 11.09, 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 = {} $$\begin{array}{c}\int \frac{5+5e^{3}x+e^{36e^4-24e^{2}x+4x^2}(x+x^2-24e^2{x}^2+8x^3)+5x\log (x)}{5e^{3}x+e^{36e^4-24e^{2}x+4x^2}{x}^2+5x\log (x)}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(5 + 5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*(x + x^2 - 24*E^2*x^2 + 8*x^3) + 5*x*Log[x])/(5*E^3*x + E^(36
*E^4 - 24*E^2*x + 4*x^2)*x^2 + 5*x*Log[x]),x]

[Out]

(1 - 24*E^2)*x + 4*x^2 + Log[x] + 120*Defer[Int][E^(5 + 24*E^2*x)/(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*x^2)*x +
5*E^(24*E^2*x)*Log[x]), x] + 5*(1 - E^3)*Defer[Int][E^(24*E^2*x)/(x*(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*x^2)*
x + 5*E^(24*E^2*x)*Log[x])), x] - 40*Defer[Int][(E^(3 + 24*E^2*x)*x)/(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*x^2)*
x + 5*E^(24*E^2*x)*Log[x]), x] + 120*Defer[Int][(E^(2 + 24*E^2*x)*Log[x])/(5*E^(3 + 24*E^2*x) + E^(36*E^4 + 4*
x^2)*x + 5*E^(24*E^2*x)*Log[x]), x] - 5*Defer[Int][(E^(24*E^2*x)*Log[x])/(x*(5*E^(3 + 24*E^2*x) + E^(36*E^4 +
4*x^2)*x + 5*E^(24*E^2*x)*Log[x])), x] - 40*Defer[Int][(E^(24*E^2*x)*x*Log[x])/(5*E^(3 + 24*E^2*x) + E^(36*E^4
+ 4*x^2)*x + 5*E^(24*E^2*x)*Log[x]), x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{1+(1-24e^2)x+8x^2}{x}-\frac{5e^{24e^{2}x}(-1+e^3-24e^{5}x+8e^3{x}^2+\log (x)-24e^{2}x\log (x)+8x^2\log (x))}{x(5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x))})dx\\ & =-(5\int \frac{e^{24e^{2}x}(-1+e^3-24e^{5}x+8e^3{x}^2+\log (x)-24e^{2}x\log (x)+8x^2\log (x))}{x(5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x))}dx)+\int \frac{1+(1-24e^2)x+8x^2}{x}dx\\ & =-(5\int (-\frac{24e^{5+24e^{2}x}}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)}-\frac{e^{24e^{2}x}(1-e^3)}{x(5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x))}+\frac{8e^{3+24e^{2}x}x}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)}-\frac{24e^{2+24e^{2}x}\log (x)}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)}+\frac{e^{24e^{2}x}\log (x)}{x(5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x))}+\frac{8e^{24e^{2}x}x\log (x)}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)})dx)+\int (1-24e^2+\frac{1}{x}+8x)dx\\ & =(1-24e^2)x+4x^2+\log (x)-5\int \frac{e^{24e^{2}x}\log (x)}{x(5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x))}dx-40\int \frac{e^{3+24e^{2}x}x}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)}dx-40\int \frac{e^{24e^{2}x}x\log (x)}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)}dx+120\int \frac{e^{5+24e^{2}x}}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)}dx+120\int \frac{e^{2+24e^{2}x}\log (x)}{5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x)}dx+\left(5(1-e^3)\right)\int \frac{e^{24e^{2}x}}{x(5e^{3+24e^{2}x}+e^{36e^4+4x^2}x+5e^{24e^{2}x}\log (x))}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.68, size = 34, normalized size = 1.17 $$\begin{array}{c}x+\log (5e^3+e^{36e^4-24e^{2}x+4x^2}x+5\log (x))\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(5 + 5*E^3*x + E^(36*E^4 - 24*E^2*x + 4*x^2)*(x + x^2 - 24*E^2*x^2 + 8*x^3) + 5*x*Log[x])/(5*E^3*x +
E^(36*E^4 - 24*E^2*x + 4*x^2)*x^2 + 5*x*Log[x]),x]

[Out]

x + Log[5*E^3 + E^(36*E^4 - 24*E^2*x + 4*x^2)*x + 5*Log[x]]

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fricas [A]  time = 0.52, size = 30, normalized size = 1.03 $$\begin{array}{c}x+\log (x{e}^{(4x^2-24x{e}^2+36e^4)}+5e^3+5\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x
)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x, algorithm="fricas")

[Out]

x + log(x*e^(4*x^2 - 24*x*e^2 + 36*e^4) + 5*e^3 + 5*log(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{5x{e}^3+(8x^3-24x^2{e}^2+x^2+x)e^{(4x^2-24x{e}^2+36e^4)}+5x\log (x)+5}{x^2{e}^{(4x^2-24x{e}^2+36e^4)}+5x{e}^3+5x\log (x)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x
)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x, algorithm="giac")

[Out]

integrate((5*x*e^3 + (8*x^3 - 24*x^2*e^2 + x^2 + x)*e^(4*x^2 - 24*x*e^2 + 36*e^4) + 5*x*log(x) + 5)/(x^2*e^(4*
x^2 - 24*x*e^2 + 36*e^4) + 5*x*e^3 + 5*x*log(x)), x)

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maple [A]  time = 0.09, size = 28, normalized size = 0.97

method	result	size
risch	$x+\ln ({\mathrm{e}}^3+\frac{{\mathrm{e}}^{36{\mathrm{e}}^4-24{\mathrm{e}}^{2}x+4x^2}x}{5}+\ln (x))$	$28$
norman	$x+\ln ({\mathrm{e}}^{36{\mathrm{e}}^4-24{\mathrm{e}}^{2}x+4x^2}x+5\ln (x)+5{\mathrm{e}}^3)$	$33$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*ln(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*ln(x)+x^2*ex
p(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x,method=_RETURNVERBOSE)

[Out]

x+ln(exp(3)+1/5*exp(36*exp(4)-24*exp(2)*x+4*x^2)*x+ln(x))

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maxima [B]  time = 0.47, size = 53, normalized size = 1.83 $$\begin{array}{c}-x(24e^2-1)+\log \left(\frac{x{e}^{(4x^2+36e^4)}+5(e^3+\log (x))e^{\left(24x{e}^2\right)}}{5(e^3+\log (x))}\right)+\log (e^3+\log (x))\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*log(x)+(-24*x^2*exp(2)+8*x^3+x^2+x)*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)+5)/(5*x*log(x
)+x^2*exp(36*exp(2)^2-24*exp(2)*x+4*x^2)+5*x*exp(3)),x, algorithm="maxima")

[Out]

-x*(24*e^2 - 1) + log(1/5*(x*e^(4*x^2 + 36*e^4) + 5*(e^3 + log(x))*e^(24*x*e^2))/(e^3 + log(x))) + log(e^3 + l
og(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}\int \frac{5x{\mathrm{e}}^3+5x\ln (x)+{\mathrm{e}}^{4x^2-24{\mathrm{e}}^{2}x+36{\mathrm{e}}^4}(x-24x^2{\mathrm{e}}^2+x^2+8x^3)+5}{x^2{\mathrm{e}}^{4x^2-24{\mathrm{e}}^{2}x+36{\mathrm{e}}^4}+5x{\mathrm{e}}^3+5x\ln (x)}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*exp(3) + 5*x*log(x) + exp(36*exp(4) - 24*x*exp(2) + 4*x^2)*(x - 24*x^2*exp(2) + x^2 + 8*x^3) + 5)/(x^
2*exp(36*exp(4) - 24*x*exp(2) + 4*x^2) + 5*x*exp(3) + 5*x*log(x)),x)

[Out]

int((5*x*exp(3) + 5*x*log(x) + exp(36*exp(4) - 24*x*exp(2) + 4*x^2)*(x - 24*x^2*exp(2) + x^2 + 8*x^3) + 5)/(x^
2*exp(36*exp(4) - 24*x*exp(2) + 4*x^2) + 5*x*exp(3) + 5*x*log(x)), x)

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sympy [A]  time = 0.55, size = 36, normalized size = 1.24 $$\begin{array}{c}x+\log (x)+\log (e^{4x^2-24x{e}^2+36e^4}+\frac{5\log (x)+5e^3}{x})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*ln(x)+(-24*x**2*exp(2)+8*x**3+x**2+x)*exp(36*exp(2)**2-24*exp(2)*x+4*x**2)+5*x*exp(3)+5)/(5*x*l
n(x)+x**2*exp(36*exp(2)**2-24*exp(2)*x+4*x**2)+5*x*exp(3)),x)

[Out]

x + log(x) + log(exp(4*x**2 - 24*x*exp(2) + 36*exp(4)) + (5*log(x) + 5*exp(3))/x)

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