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

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

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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 = {} $$\begin{array}{c}\int \frac{e^{x+\frac{x^2}{128}}(-1024x-16x^2)+128e^{x+\frac{x^2}{128}}\log (x)+e^{x+\frac{x^2}{128}}(64x+x^2){\log }^2(x)}{64x}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{64}\int \frac{e^{x+\frac{x^2}{128}}(-1024x-16x^2)+128e^{x+\frac{x^2}{128}}\log (x)+e^{x+\frac{x^2}{128}}(64x+x^2){\log }^2(x)}{x}dx\\ & =\frac{1}{64}\int (-1024e^{x+\frac{x^2}{128}}-16e^{x+\frac{x^2}{128}}x+\frac{128e^{x+\frac{x^2}{128}}\log (x)}{x}+64e^{x+\frac{x^2}{128}}{\log }^2(x)+e^{x+\frac{x^2}{128}}x{\log }^2(x))dx\\ & =\frac{1}{64}\int e^{x+\frac{x^2}{128}}x{\log }^2(x)dx-\frac{1}{4}\int e^{x+\frac{x^2}{128}}xdx+2\int \frac{e^{x+\frac{x^2}{128}}\log (x)}{x}dx-16\int e^{x+\frac{x^2}{128}}dx+\int e^{x+\frac{x^2}{128}}{\log }^2(x)dx\\ & =-16e^{x+\frac{x^2}{128}}+\frac{1}{64}\int e^{x+\frac{x^2}{128}}x{\log }^2(x)dx-2\int \frac{\int \frac{e^{x+\frac{x^2}{128}}}{x}dx}{x}dx+16\int e^{x+\frac{x^2}{128}}dx-\frac{16\int e^{32{(1+\frac{x}{64})}^2}dx}{e^{32}}+(2\log (x))\int \frac{e^{x+\frac{x^2}{128}}}{x}dx+\int e^{x+\frac{x^2}{128}}{\log }^2(x)dx\\ & =-16e^{x+\frac{x^2}{128}}-\frac{64\sqrt{2\pi }\text{erfi}\left(\frac{64+x}{8\sqrt{2}}\right)}{e^{32}}+\frac{1}{64}\int e^{x+\frac{x^2}{128}}x{\log }^2(x)dx-2\int \frac{\int \frac{e^{x+\frac{x^2}{128}}}{x}dx}{x}dx+\frac{16\int e^{32{(1+\frac{x}{64})}^2}dx}{e^{32}}+(2\log (x))\int \frac{e^{x+\frac{x^2}{128}}}{x}dx+\int e^{x+\frac{x^2}{128}}{\log }^2(x)dx\\ & =-16e^{x+\frac{x^2}{128}}+\frac{1}{64}\int e^{x+\frac{x^2}{128}}x{\log }^2(x)dx-2\int \frac{\int \frac{e^{x+\frac{x^2}{128}}}{x}dx}{x}dx+(2\log (x))\int \frac{e^{x+\frac{x^2}{128}}}{x}dx+\int e^{x+\frac{x^2}{128}}{\log }^2(x)dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}{e}^{(\frac{1}{128}{x}^2+x)}\log (x{)}^2-16e^{(\frac{1}{128}{x}^2+x)}\end{array}$$

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 $$\begin{array}{c}{e}^{(\frac{1}{128}{x}^2+x)}\log (x{)}^2-16e^{(\frac{1}{128}{x}^2+x)}\end{array}$$

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	$\ln (x{)}^2{\mathrm{e}}^{\frac{x(128+x)}{128}}-16{\mathrm{e}}^{\frac{x(128+x)}{128}}$	$23$

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 $$\begin{array}{c}64i\sqrt{2}\sqrt{\pi }\mathrm{erf}(\frac{1}{16}i\sqrt{2}x+4i\sqrt{2})e^{(-32)}+e^{(\frac{1}{128}{x}^2+x)}\log (x{)}^2+8\sqrt{2}(\frac{8\sqrt{2}\sqrt{\frac{1}{2}}\sqrt{\pi }(x+64)(\mathrm{erf}\left(\frac{1}{8}\sqrt{\frac{1}{2}}\sqrt{-{(x+64)}^2}\right)-1)}{\sqrt{-{(x+64)}^2}}-\sqrt{2}{e}^{\left(\frac{1}{128}{(x+64)}^2\right)})e^{(-32)}\end{array}$$

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 $$\begin{array}{c}{\mathrm{e}}^{\frac{x^2}{128}+x}({\ln (x)}^2-16)\end{array}$$

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 $$\begin{array}{c}(e^x\log {(x)}^2-16e^x)e^{\frac{x^2}{128}}\end{array}$$

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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