3.40.26 \(\int \frac {4 x^2+e^{4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)} (160+272 x-992 x^2-128 x^3+(-64-264 x-32 x^2) \log (x))}{x^2} \, dx\)

Optimal. Leaf size=23 \[ 4 \left (x-\frac {e^{(-2+4 x+\log (x))^2} (8+x)}{x}\right ) \]

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Rubi [F]  time = 3.28, 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 {gather*} \int \frac {4 x^2+\exp \left (4-16 x+16 x^2+(-4+8 x) \log (x)+\log ^2(x)\right ) \left (160+272 x-992 x^2-128 x^3+\left (-64-264 x-32 x^2\right ) \log (x)\right )}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4*x^2 + E^(4 - 16*x + 16*x^2 + (-4 + 8*x)*Log[x] + Log[x]^2)*(160 + 272*x - 992*x^2 - 128*x^3 + (-64 - 26
4*x - 32*x^2)*Log[x]))/x^2,x]

[Out]

4*x + 160*Defer[Int][E^(4*(1 - 2*x)^2 + Log[x]^2)*x^(-6 + 8*x), x] + 272*Defer[Int][E^(4*(1 - 2*x)^2 + Log[x]^
2)*x^(-5 + 8*x), x] - 992*Defer[Int][E^(4*(1 - 2*x)^2 + Log[x]^2)*x^(-4 + 8*x), x] - 128*Defer[Int][E^(4*(1 -
2*x)^2 + Log[x]^2)*x^(-3 + 8*x), x] - 64*Defer[Int][E^(4*(1 - 2*x)^2 + Log[x]^2)*x^(-6 + 8*x)*Log[x], x] - 264
*Defer[Int][E^(4*(1 - 2*x)^2 + Log[x]^2)*x^(-5 + 8*x)*Log[x], x] - 32*Defer[Int][E^(4*(1 - 2*x)^2 + Log[x]^2)*
x^(-4 + 8*x)*Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (4-8 e^{4 (1-2 x)^2+\log ^2(x)} x^{-6+8 x} \left (-20-34 x+124 x^2+16 x^3+8 \log (x)+33 x \log (x)+4 x^2 \log (x)\right )\right ) \, dx\\ &=4 x-8 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-6+8 x} \left (-20-34 x+124 x^2+16 x^3+8 \log (x)+33 x \log (x)+4 x^2 \log (x)\right ) \, dx\\ &=4 x-8 \int \left (-20 e^{4 (1-2 x)^2+\log ^2(x)} x^{-6+8 x}-34 e^{4 (1-2 x)^2+\log ^2(x)} x^{-5+8 x}+124 e^{4 (1-2 x)^2+\log ^2(x)} x^{-4+8 x}+16 e^{4 (1-2 x)^2+\log ^2(x)} x^{-3+8 x}+8 e^{4 (1-2 x)^2+\log ^2(x)} x^{-6+8 x} \log (x)+33 e^{4 (1-2 x)^2+\log ^2(x)} x^{-5+8 x} \log (x)+4 e^{4 (1-2 x)^2+\log ^2(x)} x^{-4+8 x} \log (x)\right ) \, dx\\ &=4 x-32 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-4+8 x} \log (x) \, dx-64 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-6+8 x} \log (x) \, dx-128 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-3+8 x} \, dx+160 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-6+8 x} \, dx-264 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-5+8 x} \log (x) \, dx+272 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-5+8 x} \, dx-992 \int e^{4 (1-2 x)^2+\log ^2(x)} x^{-4+8 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.48, size = 46, normalized size = 2.00 \begin {gather*} 4 x-\frac {4 e^{4-16 x+16 x^2+\log ^2(x)} x^{-5+8 x} \left (8+33 x+4 x^2\right )}{1+4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x^2 + E^(4 - 16*x + 16*x^2 + (-4 + 8*x)*Log[x] + Log[x]^2)*(160 + 272*x - 992*x^2 - 128*x^3 + (-6
4 - 264*x - 32*x^2)*Log[x]))/x^2,x]

[Out]

4*x - (4*E^(4 - 16*x + 16*x^2 + Log[x]^2)*x^(-5 + 8*x)*(8 + 33*x + 4*x^2))/(1 + 4*x)

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fricas [A]  time = 0.52, size = 38, normalized size = 1.65 \begin {gather*} \frac {4 \, {\left (x^{2} - {\left (x + 8\right )} e^{\left (16 \, x^{2} + 4 \, {\left (2 \, x - 1\right )} \log \relax (x) + \log \relax (x)^{2} - 16 \, x + 4\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-32*x^2-264*x-64)*log(x)-128*x^3-992*x^2+272*x+160)*exp(log(x)^2+(8*x-4)*log(x)+16*x^2-16*x+4)+4*
x^2)/x^2,x, algorithm="fricas")

[Out]

4*(x^2 - (x + 8)*e^(16*x^2 + 4*(2*x - 1)*log(x) + log(x)^2 - 16*x + 4))/x

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giac [B]  time = 0.23, size = 62, normalized size = 2.70 \begin {gather*} \frac {4 \, {\left (x^{2} - x e^{\left (16 \, x^{2} + 8 \, x \log \relax (x) + \log \relax (x)^{2} - 16 \, x - 4 \, \log \relax (x) + 4\right )} - 8 \, e^{\left (16 \, x^{2} + 8 \, x \log \relax (x) + \log \relax (x)^{2} - 16 \, x - 4 \, \log \relax (x) + 4\right )}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-32*x^2-264*x-64)*log(x)-128*x^3-992*x^2+272*x+160)*exp(log(x)^2+(8*x-4)*log(x)+16*x^2-16*x+4)+4*
x^2)/x^2,x, algorithm="giac")

[Out]

4*(x^2 - x*e^(16*x^2 + 8*x*log(x) + log(x)^2 - 16*x - 4*log(x) + 4) - 8*e^(16*x^2 + 8*x*log(x) + log(x)^2 - 16
*x - 4*log(x) + 4))/x

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maple [A]  time = 0.05, size = 35, normalized size = 1.52

method	result	size
risch	\(4 x -\frac {4 \left (x +8\right ) x^{8 x -4} {\mathrm e}^{\ln \relax (x )^{2}+4+16 x^{2}-16 x}}{x}\)	\(35\)
default	\(4 x +\frac {-4 x \,{\mathrm e}^{\ln \relax (x )^{2}+\left (8 x -4\right ) \ln \relax (x )+16 x^{2}-16 x +4}-32 \,{\mathrm e}^{\ln \relax (x )^{2}+\left (8 x -4\right ) \ln \relax (x )+16 x^{2}-16 x +4}}{x}\)	\(61\)
norman	\(\frac {4 x^{2}-4 x \,{\mathrm e}^{\ln \relax (x )^{2}+\left (8 x -4\right ) \ln \relax (x )+16 x^{2}-16 x +4}-32 \,{\mathrm e}^{\ln \relax (x )^{2}+\left (8 x -4\right ) \ln \relax (x )+16 x^{2}-16 x +4}}{x}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-32*x^2-264*x-64)*ln(x)-128*x^3-992*x^2+272*x+160)*exp(ln(x)^2+(8*x-4)*ln(x)+16*x^2-16*x+4)+4*x^2)/x^2,
x,method=_RETURNVERBOSE)

[Out]

4*x-4*(x+8)/x*x^(8*x-4)*exp(ln(x)^2+4+16*x^2-16*x)

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maxima [A]  time = 0.43, size = 37, normalized size = 1.61 \begin {gather*} 4 \, x - \frac {4 \, {\left (x e^{4} + 8 \, e^{4}\right )} e^{\left (16 \, x^{2} + 8 \, x \log \relax (x) + \log \relax (x)^{2} - 16 \, x\right )}}{x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-32*x^2-264*x-64)*log(x)-128*x^3-992*x^2+272*x+160)*exp(log(x)^2+(8*x-4)*log(x)+16*x^2-16*x+4)+4*
x^2)/x^2,x, algorithm="maxima")

[Out]

4*x - 4*(x*e^4 + 8*e^4)*e^(16*x^2 + 8*x*log(x) + log(x)^2 - 16*x)/x^5

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mupad [B]  time = 2.31, size = 58, normalized size = 2.52 \begin {gather*} 4\,x-\frac {4\,x^{8\,x}\,{\mathrm {e}}^{-16\,x}\,{\mathrm {e}}^4\,{\mathrm {e}}^{16\,x^2}\,{\mathrm {e}}^{{\ln \relax (x)}^2}}{x^4}-\frac {32\,x^{8\,x}\,{\mathrm {e}}^{-16\,x}\,{\mathrm {e}}^4\,{\mathrm {e}}^{16\,x^2}\,{\mathrm {e}}^{{\ln \relax (x)}^2}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(x)^2 - 16*x + log(x)*(8*x - 4) + 16*x^2 + 4)*(log(x)*(264*x + 32*x^2 + 64) - 272*x + 992*x^2 + 1
28*x^3 - 160) - 4*x^2)/x^2,x)

[Out]

4*x - (4*x^(8*x)*exp(-16*x)*exp(4)*exp(16*x^2)*exp(log(x)^2))/x^4 - (32*x^(8*x)*exp(-16*x)*exp(4)*exp(16*x^2)*
exp(log(x)^2))/x^5

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sympy [A]  time = 0.34, size = 36, normalized size = 1.57 \begin {gather*} 4 x + \frac {\left (- 4 x - 32\right ) e^{16 x^{2} - 16 x + \left (8 x - 4\right ) \log {\relax (x )} + \log {\relax (x )}^{2} + 4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-32*x**2-264*x-64)*ln(x)-128*x**3-992*x**2+272*x+160)*exp(ln(x)**2+(8*x-4)*ln(x)+16*x**2-16*x+4)+
4*x**2)/x**2,x)

[Out]

4*x + (-4*x - 32)*exp(16*x**2 - 16*x + (8*x - 4)*log(x) + log(x)**2 + 4)/x

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