3.73.8 \(\int \frac {e^{x+\frac {x^2}{128}} (-1024 x-16 x^2)+128 e^{x+\frac {x^2}{128}} \log (x)+e^{x+\frac {x^2}{128}} (64 x+x^2) \log ^2(x)}{64 x} \, dx\)

Optimal. Leaf size=18 \[ e^{x+\frac {x^2}{128}} \left (-16+\log ^2(x)\right ) \]

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Rubi [F]  time = 0.29, 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 {e^{x+\frac {x^2}{128}} \left (-1024 x-16 x^2\right )+128 e^{x+\frac {x^2}{128}} \log (x)+e^{x+\frac {x^2}{128}} \left (64 x+x^2\right ) \log ^2(x)}{64 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x + x^2/128)*(-1024*x - 16*x^2) + 128*E^(x + x^2/128)*Log[x] + E^(x + x^2/128)*(64*x + x^2)*Log[x]^2)/
(64*x),x]

[Out]

-16*E^(x + x^2/128) + 2*Log[x]*Defer[Int][E^(x + x^2/128)/x, x] + Defer[Int][E^(x + x^2/128)*Log[x]^2, x] + De
fer[Int][E^(x + x^2/128)*x*Log[x]^2, x]/64 - 2*Defer[Int][Defer[Int][E^(x + x^2/128)/x, x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{64} \int \frac {e^{x+\frac {x^2}{128}} \left (-1024 x-16 x^2\right )+128 e^{x+\frac {x^2}{128}} \log (x)+e^{x+\frac {x^2}{128}} \left (64 x+x^2\right ) \log ^2(x)}{x} \, dx\\ &=\frac {1}{64} \int \left (-1024 e^{x+\frac {x^2}{128}}-16 e^{x+\frac {x^2}{128}} x+\frac {128 e^{x+\frac {x^2}{128}} \log (x)}{x}+64 e^{x+\frac {x^2}{128}} \log ^2(x)+e^{x+\frac {x^2}{128}} x \log ^2(x)\right ) \, dx\\ &=\frac {1}{64} \int e^{x+\frac {x^2}{128}} x \log ^2(x) \, dx-\frac {1}{4} \int e^{x+\frac {x^2}{128}} x \, dx+2 \int \frac {e^{x+\frac {x^2}{128}} \log (x)}{x} \, dx-16 \int e^{x+\frac {x^2}{128}} \, dx+\int e^{x+\frac {x^2}{128}} \log ^2(x) \, dx\\ &=-16 e^{x+\frac {x^2}{128}}+\frac {1}{64} \int e^{x+\frac {x^2}{128}} x \log ^2(x) \, dx-2 \int \frac {\int \frac {e^{x+\frac {x^2}{128}}}{x} \, dx}{x} \, dx+16 \int e^{x+\frac {x^2}{128}} \, dx-\frac {16 \int e^{32 \left (1+\frac {x}{64}\right )^2} \, dx}{e^{32}}+(2 \log (x)) \int \frac {e^{x+\frac {x^2}{128}}}{x} \, dx+\int e^{x+\frac {x^2}{128}} \log ^2(x) \, dx\\ &=-16 e^{x+\frac {x^2}{128}}-\frac {64 \sqrt {2 \pi } \text {erfi}\left (\frac {64+x}{8 \sqrt {2}}\right )}{e^{32}}+\frac {1}{64} \int e^{x+\frac {x^2}{128}} x \log ^2(x) \, dx-2 \int \frac {\int \frac {e^{x+\frac {x^2}{128}}}{x} \, dx}{x} \, dx+\frac {16 \int e^{32 \left (1+\frac {x}{64}\right )^2} \, dx}{e^{32}}+(2 \log (x)) \int \frac {e^{x+\frac {x^2}{128}}}{x} \, dx+\int e^{x+\frac {x^2}{128}} \log ^2(x) \, dx\\ &=-16 e^{x+\frac {x^2}{128}}+\frac {1}{64} \int e^{x+\frac {x^2}{128}} x \log ^2(x) \, dx-2 \int \frac {\int \frac {e^{x+\frac {x^2}{128}}}{x} \, dx}{x} \, dx+(2 \log (x)) \int \frac {e^{x+\frac {x^2}{128}}}{x} \, dx+\int e^{x+\frac {x^2}{128}} \log ^2(x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} e^{x+\frac {x^2}{128}} \left (-16+\log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + x^2/128)*(-1024*x - 16*x^2) + 128*E^(x + x^2/128)*Log[x] + E^(x + x^2/128)*(64*x + x^2)*Log[
x]^2)/(64*x),x]

[Out]

E^(x + x^2/128)*(-16 + Log[x]^2)

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fricas [A]  time = 0.63, size = 24, normalized size = 1.33 \begin {gather*} e^{\left (\frac {1}{128} \, x^{2} + x\right )} \log \relax (x)^{2} - 16 \, e^{\left (\frac {1}{128} \, x^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((x^2+64*x)*exp(x)*exp(1/128*x^2)*log(x)^2+128*exp(x)*exp(1/128*x^2)*log(x)+(-16*x^2-1024*x)*ex
p(x)*exp(1/128*x^2))/x,x, algorithm="fricas")

[Out]

e^(1/128*x^2 + x)*log(x)^2 - 16*e^(1/128*x^2 + x)

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giac [A]  time = 1.80, size = 24, normalized size = 1.33 \begin {gather*} e^{\left (\frac {1}{128} \, x^{2} + x\right )} \log \relax (x)^{2} - 16 \, e^{\left (\frac {1}{128} \, x^{2} + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((x^2+64*x)*exp(x)*exp(1/128*x^2)*log(x)^2+128*exp(x)*exp(1/128*x^2)*log(x)+(-16*x^2-1024*x)*ex
p(x)*exp(1/128*x^2))/x,x, algorithm="giac")

[Out]

e^(1/128*x^2 + x)*log(x)^2 - 16*e^(1/128*x^2 + x)

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maple [A]  time = 0.03, size = 23, normalized size = 1.28

method	result	size
risch	\(\ln \relax (x )^{2} {\mathrm e}^{\frac {x \left (128+x \right )}{128}}-16 \,{\mathrm e}^{\frac {x \left (128+x \right )}{128}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/64*((x^2+64*x)*exp(x)*exp(1/128*x^2)*ln(x)^2+128*exp(x)*exp(1/128*x^2)*ln(x)+(-16*x^2-1024*x)*exp(x)*exp
(1/128*x^2))/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)^2*exp(1/128*x*(128+x))-16*exp(1/128*x*(128+x))

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maxima [C]  time = 0.56, size = 98, normalized size = 5.44 \begin {gather*} 64 i \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{16} i \, \sqrt {2} x + 4 i \, \sqrt {2}\right ) e^{\left (-32\right )} + e^{\left (\frac {1}{128} \, x^{2} + x\right )} \log \relax (x)^{2} + 8 \, \sqrt {2} {\left (\frac {8 \, \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {\pi } {\left (x + 64\right )} {\left (\operatorname {erf}\left (\frac {1}{8} \, \sqrt {\frac {1}{2}} \sqrt {-{\left (x + 64\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (x + 64\right )}^{2}}} - \sqrt {2} e^{\left (\frac {1}{128} \, {\left (x + 64\right )}^{2}\right )}\right )} e^{\left (-32\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((x^2+64*x)*exp(x)*exp(1/128*x^2)*log(x)^2+128*exp(x)*exp(1/128*x^2)*log(x)+(-16*x^2-1024*x)*ex
p(x)*exp(1/128*x^2))/x,x, algorithm="maxima")

[Out]

64*I*sqrt(2)*sqrt(pi)*erf(1/16*I*sqrt(2)*x + 4*I*sqrt(2))*e^(-32) + e^(1/128*x^2 + x)*log(x)^2 + 8*sqrt(2)*(8*
sqrt(2)*sqrt(1/2)*sqrt(pi)*(x + 64)*(erf(1/8*sqrt(1/2)*sqrt(-(x + 64)^2)) - 1)/sqrt(-(x + 64)^2) - sqrt(2)*e^(
1/128*(x + 64)^2))*e^(-32)

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mupad [B]  time = 4.48, size = 15, normalized size = 0.83 \begin {gather*} {\mathrm {e}}^{\frac {x^2}{128}+x}\,\left ({\ln \relax (x)}^2-16\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(x^2/128)*exp(x)*log(x) - (exp(x^2/128)*exp(x)*(1024*x + 16*x^2))/64 + (exp(x^2/128)*exp(x)*log(x)^2
*(64*x + x^2))/64)/x,x)

[Out]

exp(x + x^2/128)*(log(x)^2 - 16)

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sympy [A]  time = 28.91, size = 19, normalized size = 1.06 \begin {gather*} \left (e^{x} \log {\relax (x )}^{2} - 16 e^{x}\right ) e^{\frac {x^{2}}{128}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*((x**2+64*x)*exp(x)*exp(1/128*x**2)*ln(x)**2+128*exp(x)*exp(1/128*x**2)*ln(x)+(-16*x**2-1024*x)
*exp(x)*exp(1/128*x**2))/x,x)

[Out]

(exp(x)*log(x)**2 - 16*exp(x))*exp(x**2/128)

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