3.73.59 $\int \frac{e^{\frac{2(2x^2-x^3-2x\log (\frac{x}{2})-{\log }^2(\frac{x}{2}))}{x^2}}(-4x^2-2x^4+e^4(-4x-x^2-2x^3)+(-4x+4x^2+e^4(-4+4x))\log (\frac{x}{2})+(4e^4+4x){\log }^2(\frac{x}{2}))}{x^4}dx$

Optimal. Leaf size=31 $$\frac{e^{6-2x-\frac{2{(x+\log \left(\frac{x}{2}\right))}^2}{x^2}}(e^4+x)}{x}$$

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Rubi [F]  time = 11.30, 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 = {} $$\begin{array}{c}\int \frac{\exp \left(\frac{2(2x^2-x^3-2x\log \left(\frac{x}{2}\right)-{\log }^2\left(\frac{x}{2}\right))}{x^2}\right)(-4x^2-2x^4+e^4(-4x-x^2-2x^3)+(-4x+4x^2+e^4(-4+4x))\log \left(\frac{x}{2}\right)+(4e^4+4x){\log }^2\left(\frac{x}{2}\right))}{x^4}dx\end{array}$$

Verification is not applicable to the result.

[In]

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

[Out]

-2*Defer[Int][E^((2*(2*x^2 - x^3 - 2*x*Log[x/2] - Log[x/2]^2))/x^2), x] - 4*Defer[Int][1/(E^((2*(-4*x^2 + x^3
+ 2*x*Log[x/2] + Log[x/2]^2))/x^2)*x^3), x] - (4 + E^4)*Defer[Int][E^((2*(2*x^2 - x^3 - 2*x*Log[x/2] - Log[x/2
]^2))/x^2)/x^2, x] - 2*Defer[Int][1/(E^((2*(-4*x^2 + x^3 + 2*x*Log[x/2] + Log[x/2]^2))/x^2)*x), x] - 4*Defer[I
nt][Log[x/2]/(E^((2*(-4*x^2 + x^3 + 2*x*Log[x/2] + Log[x/2]^2))/x^2)*x^4), x] - 4*(1 - E^4)*Defer[Int][(E^((2*
(2*x^2 - x^3 - 2*x*Log[x/2] - Log[x/2]^2))/x^2)*Log[x/2])/x^3, x] + 4*Defer[Int][(E^((2*(2*x^2 - x^3 - 2*x*Log
[x/2] - Log[x/2]^2))/x^2)*Log[x/2])/x^2, x] + 4*Defer[Int][Log[x/2]^2/(E^((2*(-4*x^2 + x^3 + 2*x*Log[x/2] + Lo
g[x/2]^2))/x^2)*x^4), x] + 4*Defer[Int][(E^((2*(2*x^2 - x^3 - 2*x*Log[x/2] - Log[x/2]^2))/x^2)*Log[x/2]^2)/x^3
, x]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 41, normalized size = 1.32 $$\begin{array}{c}{16}^{\frac{1}{x}}{e}^{4-2x-\frac{2{\log }^2\left(\frac{x}{2}\right)}{x^2}}{x}^{-\frac{4+x}{x}}(e^4+x)\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(16^x^(-1)*E^(4 - 2*x - (2*Log[x/2]^2)/x^2)*(E^4 + x))/x^((4 + x)/x)

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fricas [A]  time = 0.66, size = 36, normalized size = 1.16 $$\begin{array}{c}\frac{(x+e^4)e^{(-\frac{2(x^3-2x^2+2x\log \left(\frac{1}{2}x\right)+\log {\left(\frac{1}{2}x\right)}^2)}{x^2})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(4)+4*x)*log(1/2*x)^2+((4*x-4)*exp(4)+4*x^2-4*x)*log(1/2*x)+(-2*x^3-x^2-4*x)*exp(4)-2*x^4-4*x
^2)*exp((-log(1/2*x)^2-2*x*log(1/2*x)-x^3+2*x^2)/x^2)^2/x^4,x, algorithm="fricas")

[Out]

(x + e^4)*e^(-2*(x^3 - 2*x^2 + 2*x*log(1/2*x) + log(1/2*x)^2)/x^2)/x

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giac [B]  time = 0.58, size = 63, normalized size = 2.03 $$\begin{array}{c}\frac{x{e}^{(-\frac{2(x^3-2x^2+2x\log \left(\frac{1}{2}x\right)+\log {\left(\frac{1}{2}x\right)}^2)}{x^2})}+e^{(-\frac{2(x^3-4x^2+2x\log \left(\frac{1}{2}x\right)+\log {\left(\frac{1}{2}x\right)}^2)}{x^2})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(4)+4*x)*log(1/2*x)^2+((4*x-4)*exp(4)+4*x^2-4*x)*log(1/2*x)+(-2*x^3-x^2-4*x)*exp(4)-2*x^4-4*x
^2)*exp((-log(1/2*x)^2-2*x*log(1/2*x)-x^3+2*x^2)/x^2)^2/x^4,x, algorithm="giac")

[Out]

(x*e^(-2*(x^3 - 2*x^2 + 2*x*log(1/2*x) + log(1/2*x)^2)/x^2) + e^(-2*(x^3 - 4*x^2 + 2*x*log(1/2*x) + log(1/2*x)
^2)/x^2))/x

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maple [A]  time = 0.05, size = 37, normalized size = 1.19

method	result	size
risch	$\frac{(x+{\mathrm{e}}^4){\mathrm{e}}^{-\frac{2(x^3+\ln {\left(\frac{x}{2}\right)}^2+2x\ln \left(\frac{x}{2}\right)-2x^2)}{x^2}}}{x}$	$37$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*exp(4)+4*x)*ln(1/2*x)^2+((4*x-4)*exp(4)+4*x^2-4*x)*ln(1/2*x)+(-2*x^3-x^2-4*x)*exp(4)-2*x^4-4*x^2)*exp(
(-ln(1/2*x)^2-2*x*ln(1/2*x)-x^3+2*x^2)/x^2)^2/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.56, size = 57, normalized size = 1.84 $$\begin{array}{c}\frac{(x{e}^4+e^8)e^{(-2x+\frac{4\log (2)}{x}-\frac{2\log (2{)}^2}{x^2}-\frac{4\log (x)}{x}+\frac{4\log (2)\log (x)}{x^2}-\frac{2\log (x{)}^2}{x^2})}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(4)+4*x)*log(1/2*x)^2+((4*x-4)*exp(4)+4*x^2-4*x)*log(1/2*x)+(-2*x^3-x^2-4*x)*exp(4)-2*x^4-4*x
^2)*exp((-log(1/2*x)^2-2*x*log(1/2*x)-x^3+2*x^2)/x^2)^2/x^4,x, algorithm="maxima")

[Out]

(x*e^4 + e^8)*e^(-2*x + 4*log(2)/x - 2*log(2)^2/x^2 - 4*log(x)/x + 4*log(2)*log(x)/x^2 - 2*log(x)^2/x^2)/x

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mupad [B]  time = 4.64, size = 57, normalized size = 1.84 $$\begin{array}{c}\frac{2^{4/x}{x}^{\frac{4\ln (2)}{x^2}}{\mathrm{e}}^{4-\frac{2{\ln (2)}^2}{x^2}-\frac{2{\ln (x)}^2}{x^2}-2x}(x+{\mathrm{e}}^4)}{x^{4/x}x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(4/x)*x^((4*log(2))/x^2)*exp(4 - (2*log(2)^2)/x^2 - (2*log(x)^2)/x^2 - 2*x)*(x + exp(4)))/(x^(4/x)*x)

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sympy [A]  time = 0.49, size = 37, normalized size = 1.19 $$\begin{array}{c}\frac{(x+e^4)e^{\frac{2(-x^3+2x^2-2x\log \left(\frac{x}{2}\right)-\log {\left(\frac{x}{2}\right)}^2)}{x^2}}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*exp(4)+4*x)*ln(1/2*x)**2+((4*x-4)*exp(4)+4*x**2-4*x)*ln(1/2*x)+(-2*x**3-x**2-4*x)*exp(4)-2*x**4-
4*x**2)*exp((-ln(1/2*x)**2-2*x*ln(1/2*x)-x**3+2*x**2)/x**2)**2/x**4,x)

[Out]

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

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