3.17.41 $\int \frac{x^2+e^{e^{e^{\frac{e^{-15-5x+25x^2}+x^2}{x}}}+e^{\frac{e^{-15-5x+25x^2}+x^2}{x}}+\frac{e^{-15-5x+25x^2}+x^2}{x}}(x^2+e^{-15-5x+25x^2}(-1-5x+50x^2))}{x^2}dx$

Optimal. Leaf size=28 $$e^{e^{e^{\frac{e^{5(-3-x+5x^2)}}{x}+x}}}+x$$

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Rubi [F]  time = 4.41, 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 = {} $$\begin{array}{c}\int \frac{x^2+\exp (e^{e^{\frac{e^{-15-5x+25x^2}+x^2}{x}}}+e^{\frac{e^{-15-5x+25x^2}+x^2}{x}}+\frac{e^{-15-5x+25x^2}+x^2}{x})(x^2+e^{-15-5x+25x^2}(-1-5x+50x^2))}{x^2}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(x^2 + E^(E^E^((E^(-15 - 5*x + 25*x^2) + x^2)/x) + E^((E^(-15 - 5*x + 25*x^2) + x^2)/x) + (E^(-15 - 5*x +
25*x^2) + x^2)/x)*(x^2 + E^(-15 - 5*x + 25*x^2)*(-1 - 5*x + 50*x^2)))/x^2,x]

[Out]

x + Defer[Int][E^(E^E^(E^(-15 - 5*x + 25*x^2)/x + x) + E^(E^(-15 - 5*x + 25*x^2)/x + x) + E^(-15 - 5*x + 25*x^
2)/x + x), x] + 50*Defer[Int][E^(-15 + E^E^(E^(-15 - 5*x + 25*x^2)/x + x) + E^(E^(-15 - 5*x + 25*x^2)/x + x) +
E^(-15 - 5*x + 25*x^2)/x - 4*x + 25*x^2), x] - Defer[Int][E^(-15 + E^E^(E^(-15 - 5*x + 25*x^2)/x + x) + E^(E^
(-15 - 5*x + 25*x^2)/x + x) + E^(-15 - 5*x + 25*x^2)/x - 4*x + 25*x^2)/x^2, x] - 5*Defer[Int][E^(-15 + E^E^(E^
(-15 - 5*x + 25*x^2)/x + x) + E^(E^(-15 - 5*x + 25*x^2)/x + x) + E^(-15 - 5*x + 25*x^2)/x - 4*x + 25*x^2)/x, x
]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (1+\exp (e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}+x)+\frac{\exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)(-1+5x)(1+10x)}{x^2})dx\\ & =x+\int \exp (e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}+x)dx+\int \frac{\exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)(-1+5x)(1+10x)}{x^2}dx\\ & =x+\int \exp (e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}+x)dx+\int (50\exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)-\frac{\exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)}{x^2}-\frac{5\exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)}{x})dx\\ & =x-5\int \frac{\exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)}{x}dx+50\int \exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)dx+\int \exp (e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}+x)dx-\int \frac{\exp (-15+e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}+e^{\frac{e^{-15-5x+25x^2}}{x}+x}+\frac{e^{-15-5x+25x^2}}{x}-4x+25x^2)}{x^2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.21, size = 26, normalized size = 0.93 $$\begin{array}{c}{e}^{e^{e^{\frac{e^{-15-5x+25x^2}}{x}+x}}}+x\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + E^(E^E^((E^(-15 - 5*x + 25*x^2) + x^2)/x) + E^((E^(-15 - 5*x + 25*x^2) + x^2)/x) + (E^(-15 -
5*x + 25*x^2) + x^2)/x)*(x^2 + E^(-15 - 5*x + 25*x^2)*(-1 - 5*x + 50*x^2)))/x^2,x]

[Out]

E^E^E^(E^(-15 - 5*x + 25*x^2)/x + x) + x

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fricas [B]  time = 0.70, size = 154, normalized size = 5.50 $$\begin{array}{c}(x{e}^{(\frac{x^2+e^{(25x^2-5x-15)}}{x}+e^{\left(\frac{x^2+e^{(25x^2-5x-15)}}{x}\right)})}+e^{\left(\frac{x^2+x{e}^{\left(\frac{x^2+e^{(25x^2-5x-15)}}{x}\right)}+x{e}^{\left(e^{\left(\frac{x^2+e^{(25x^2-5x-15)}}{x}\right)}\right)}+e^{(25x^2-5x-15)}}{x}\right)})e^{(-\frac{x^2+e^{(25x^2-5x-15)}}{x}-e^{\left(\frac{x^2+e^{(25x^2-5x-15)}}{x}\right)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^2-5*x-1)*exp(25*x^2-5*x-15)+x^2)*exp((exp(25*x^2-5*x-15)+x^2)/x)*exp(exp((exp(25*x^2-5*x-15)
+x^2)/x))*exp(exp(exp((exp(25*x^2-5*x-15)+x^2)/x)))+x^2)/x^2,x, algorithm="fricas")

[Out]

(x*e^((x^2 + e^(25*x^2 - 5*x - 15))/x + e^((x^2 + e^(25*x^2 - 5*x - 15))/x)) + e^((x^2 + x*e^((x^2 + e^(25*x^2
- 5*x - 15))/x) + x*e^(e^((x^2 + e^(25*x^2 - 5*x - 15))/x)) + e^(25*x^2 - 5*x - 15))/x))*e^(-(x^2 + e^(25*x^2
- 5*x - 15))/x - e^((x^2 + e^(25*x^2 - 5*x - 15))/x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 $$\begin{array}{c}\int \frac{x^2+(x^2+(50x^2-5x-1)e^{(25x^2-5x-15)})e^{(\frac{x^2+e^{(25x^2-5x-15)}}{x}+e^{\left(\frac{x^2+e^{(25x^2-5x-15)}}{x}\right)}+e^{\left(e^{\left(\frac{x^2+e^{(25x^2-5x-15)}}{x}\right)}\right)})}}{x^2}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^2-5*x-1)*exp(25*x^2-5*x-15)+x^2)*exp((exp(25*x^2-5*x-15)+x^2)/x)*exp(exp((exp(25*x^2-5*x-15)
+x^2)/x))*exp(exp(exp((exp(25*x^2-5*x-15)+x^2)/x)))+x^2)/x^2,x, algorithm="giac")

[Out]

integrate((x^2 + (x^2 + (50*x^2 - 5*x - 1)*e^(25*x^2 - 5*x - 15))*e^((x^2 + e^(25*x^2 - 5*x - 15))/x + e^((x^2
+ e^(25*x^2 - 5*x - 15))/x) + e^(e^((x^2 + e^(25*x^2 - 5*x - 15))/x))))/x^2, x)

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maple [A]  time = 0.09, size = 25, normalized size = 0.89

method	result	size
risch	$x+{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{\frac{{\mathrm{e}}^{25x^2-5x-15}+x^2}{x}}}}$	$25$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((50*x^2-5*x-1)*exp(25*x^2-5*x-15)+x^2)*exp((exp(25*x^2-5*x-15)+x^2)/x)*exp(exp((exp(25*x^2-5*x-15)+x^2)/
x))*exp(exp(exp((exp(25*x^2-5*x-15)+x^2)/x)))+x^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x+exp(exp(exp((exp(25*x^2-5*x-15)+x^2)/x)))

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maxima [A]  time = 0.82, size = 22, normalized size = 0.79 $$\begin{array}{c}x+e^{\left(e^{\left(e^{(x+\frac{e^{(25x^2-5x-15)}}{x})}\right)}\right)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^2-5*x-1)*exp(25*x^2-5*x-15)+x^2)*exp((exp(25*x^2-5*x-15)+x^2)/x)*exp(exp((exp(25*x^2-5*x-15)
+x^2)/x))*exp(exp(exp((exp(25*x^2-5*x-15)+x^2)/x)))+x^2)/x^2,x, algorithm="maxima")

[Out]

x + e^(e^(e^(x + e^(25*x^2 - 5*x - 15)/x)))

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mupad [B]  time = 1.24, size = 24, normalized size = 0.86 $$\begin{array}{c}x+{\mathrm{e}}^{{\mathrm{e}}^{{\mathrm{e}}^{\frac{{\mathrm{e}}^{-5x}{\mathrm{e}}^{-15}{\mathrm{e}}^{25x^2}}{x}}{\mathrm{e}}^x}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - exp(exp((exp(25*x^2 - 5*x - 15) + x^2)/x))*exp(exp(exp((exp(25*x^2 - 5*x - 15) + x^2)/x)))*exp((exp
(25*x^2 - 5*x - 15) + x^2)/x)*(exp(25*x^2 - 5*x - 15)*(5*x - 50*x^2 + 1) - x^2))/x^2,x)

[Out]

x + exp(exp(exp((exp(-5*x)*exp(-15)*exp(25*x^2))/x)*exp(x)))

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sympy [A]  time = 14.72, size = 22, normalized size = 0.79 $$\begin{array}{c}x+e^{e^{e^{\frac{x^2+e^{25x^2-5x-15}}{x}}}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x**2-5*x-1)*exp(25*x**2-5*x-15)+x**2)*exp((exp(25*x**2-5*x-15)+x**2)/x)*exp(exp((exp(25*x**2-5
*x-15)+x**2)/x))*exp(exp(exp((exp(25*x**2-5*x-15)+x**2)/x)))+x**2)/x**2,x)

[Out]

x + exp(exp(exp((x**2 + exp(25*x**2 - 5*x - 15))/x)))

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