3.73.85 $\int \frac{e^{-x}(e^x+e^{e^{-x}(e^{15}{x}^2+e^{15+x}{x}^2)}(2e^{15+x}{x}^3+e^{15}(2x^3-x^4))-e^x\log (x))}{x^2}dx$

Optimal. Leaf size=24 $$-1+e^{e^{15}x(x+e^{-x}x)}+\frac{\log (x)}{x}$$

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Rubi [F]  time = 1.69, 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 = {} $$\begin{array}{c}\int \frac{e^{-x}(e^x+e^{e^{-x}(e^{15}{x}^2+e^{15+x}{x}^2)}(2e^{15+x}{x}^3+e^{15}(2x^3-x^4))-e^x\log (x))}{x^2}dx\end{array}$$

Verification is not applicable to the result.

[In]

Int[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E^15*(2*x^3 - x^4)) - E^x*Log[x])/(E^x*x^2)
,x]

[Out]

Log[x]/x + 2*Defer[Int][E^(15 + E^(15 - x)*(1 + E^x)*x^2)*x, x] + 2*Defer[Int][E^(15 - x + E^(15 - x)*(1 + E^x
)*x^2)*x, x] - Defer[Int][E^(15 - x + E^(15 - x)*(1 + E^x)*x^2)*x^2, x]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{1+e^{15-x+e^{15-x}(1+e^x)x^2}(2+2e^x-x)x^3-\log (x)}{x^2}dx\\ & =\int (e^{15-x+e^{15-x}(1+e^x)x^2}(2+2e^x-x)x+\frac{1-\log (x)}{x^2})dx\\ & =\int e^{15-x+e^{15-x}(1+e^x)x^2}(2+2e^x-x)xdx+\int \frac{1-\log (x)}{x^2}dx\\ & =\frac{\log (x)}{x}+\int (2e^{15+e^{15-x}(1+e^x)x^2}x-e^{15-x+e^{15-x}(1+e^x)x^2}(-2+x)x)dx\\ & =\frac{\log (x)}{x}+2\int e^{15+e^{15-x}(1+e^x)x^2}xdx-\int e^{15-x+e^{15-x}(1+e^x)x^2}(-2+x)xdx\\ & =\frac{\log (x)}{x}+2\int e^{15+e^{15-x}(1+e^x)x^2}xdx-\int (-2e^{15-x+e^{15-x}(1+e^x)x^2}x+e^{15-x+e^{15-x}(1+e^x)x^2}{x}^2)dx\\ & =\frac{\log (x)}{x}+2\int e^{15+e^{15-x}(1+e^x)x^2}xdx+2\int e^{15-x+e^{15-x}(1+e^x)x^2}xdx-\int e^{15-x+e^{15-x}(1+e^x)x^2}{x}^{2}dx\end{array}\end{array}$$

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Mathematica [A]  time = 0.43, size = 25, normalized size = 1.04 $$\begin{array}{c}{e}^{e^{15-x}(1+e^x)x^2}+\frac{\log (x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(E^x + E^((E^15*x^2 + E^(15 + x)*x^2)/E^x)*(2*E^(15 + x)*x^3 + E^15*(2*x^3 - x^4)) - E^x*Log[x])/(E^
x*x^2),x]

[Out]

E^(E^(15 - x)*(1 + E^x)*x^2) + Log[x]/x

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fricas [A]  time = 0.58, size = 32, normalized size = 1.33 $$\begin{array}{c}\frac{x{e}^{\left((x^2{e}^{30}+x^2{e}^{(x+30)})e^{(-x-15)}\right)}+\log (x)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="fricas")

[Out]

(x*e^((x^2*e^30 + x^2*e^(x + 30))*e^(-x - 15)) + log(x))/x

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giac [A]  time = 0.17, size = 25, normalized size = 1.04 $$\begin{array}{c}\frac{\log (x)}{x}+e^{(x^2{e}^{15}+x^2{e}^{(-x+15)})}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="giac")

[Out]

log(x)/x + e^(x^2*e^15 + x^2*e^(-x + 15))

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maple [A]  time = 0.15, size = 24, normalized size = 1.00

method	result	size
risch	$\frac{\ln (x)}{x}+{\mathrm{e}}^{x^2({\mathrm{e}}^{x+15}+{\mathrm{e}}^{15}){\mathrm{e}}^{-x}}$	$24$
default	${\mathrm{e}}^{x^2({\mathrm{e}}^{15}{\mathrm{e}}^x+{\mathrm{e}}^{15}){\mathrm{e}}^{-x}}+\frac{\ln (x)}{x}$	$29$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*ln(x)+exp
(x))/exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

ln(x)/x+exp(x^2*(exp(x+15)+exp(15))*exp(-x))

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maxima [A]  time = 0.49, size = 34, normalized size = 1.42 $$\begin{array}{c}\frac{x{e}^{(x^2{e}^{15}+x^2{e}^{(-x+15)})}+\log (x)+1}{x}-\frac{1}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3*exp(15)*exp(x)+(-x^4+2*x^3)*exp(15))*exp((x^2*exp(15)*exp(x)+x^2*exp(15))/exp(x))-exp(x)*log
(x)+exp(x))/exp(x)/x^2,x, algorithm="maxima")

[Out]

(x*e^(x^2*e^15 + x^2*e^(-x + 15)) + log(x) + 1)/x - 1/x

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mupad [B]  time = 4.46, size = 26, normalized size = 1.08 $$\begin{array}{c}\frac{\ln (x)}{x}+{\mathrm{e}}^{x^2{\mathrm{e}}^{15}}{\mathrm{e}}^{x^2{\mathrm{e}}^{-x}{\mathrm{e}}^{15}}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(x) - exp(x)*log(x) + exp(exp(-x)*(x^2*exp(15) + x^2*exp(15)*exp(x)))*(exp(15)*(2*x^3 - x^4)
+ 2*x^3*exp(15)*exp(x))))/x^2,x)

[Out]

log(x)/x + exp(x^2*exp(15))*exp(x^2*exp(-x)*exp(15))

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sympy [A]  time = 0.53, size = 26, normalized size = 1.08 $$\begin{array}{c}{e}^{(x^2{e}^{15}{e}^x+x^2{e}^{15})e^{-x}}+\frac{\log (x)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3*exp(15)*exp(x)+(-x**4+2*x**3)*exp(15))*exp((x**2*exp(15)*exp(x)+x**2*exp(15))/exp(x))-exp(x
)*ln(x)+exp(x))/exp(x)/x**2,x)

[Out]

exp((x**2*exp(15)*exp(x) + x**2*exp(15))*exp(-x)) + log(x)/x

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