3.93.94 \(\int \frac {160+e^{\frac {e^x}{8 x}+x} (5-5 x)-128 x^2-160 \log (x)}{32 x^2} \, dx\)

Optimal. Leaf size=29 \[ x-5 \left (\frac {1}{4} e^{\frac {e^x}{8 x}}+x-\frac {\log (x)}{x}\right ) \]

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Rubi [F]  time = 0.30, 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 {gather*} \int \frac {160+e^{\frac {e^x}{8 x}+x} (5-5 x)-128 x^2-160 \log (x)}{32 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(160 + E^(E^x/(8*x) + x)*(5 - 5*x) - 128*x^2 - 160*Log[x])/(32*x^2),x]

[Out]

-4*x + (5*Log[x])/x + (5*Defer[Int][E^(E^x/(8*x) + x)/x^2, x])/32 - (5*Defer[Int][E^(E^x/(8*x) + x)/x, x])/32

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{32} \int \frac {160+e^{\frac {e^x}{8 x}+x} (5-5 x)-128 x^2-160 \log (x)}{x^2} \, dx\\ &=\frac {1}{32} \int \left (-\frac {5 e^{\frac {e^x}{8 x}+x} (-1+x)}{x^2}-\frac {32 \left (-5+4 x^2+5 \log (x)\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {5}{32} \int \frac {e^{\frac {e^x}{8 x}+x} (-1+x)}{x^2} \, dx\right )-\int \frac {-5+4 x^2+5 \log (x)}{x^2} \, dx\\ &=-\left (\frac {5}{32} \int \left (-\frac {e^{\frac {e^x}{8 x}+x}}{x^2}+\frac {e^{\frac {e^x}{8 x}+x}}{x}\right ) \, dx\right )-\int \left (\frac {-5+4 x^2}{x^2}+\frac {5 \log (x)}{x^2}\right ) \, dx\\ &=\frac {5}{32} \int \frac {e^{\frac {e^x}{8 x}+x}}{x^2} \, dx-\frac {5}{32} \int \frac {e^{\frac {e^x}{8 x}+x}}{x} \, dx-5 \int \frac {\log (x)}{x^2} \, dx-\int \frac {-5+4 x^2}{x^2} \, dx\\ &=\frac {5}{x}+\frac {5 \log (x)}{x}+\frac {5}{32} \int \frac {e^{\frac {e^x}{8 x}+x}}{x^2} \, dx-\frac {5}{32} \int \frac {e^{\frac {e^x}{8 x}+x}}{x} \, dx-\int \left (4-\frac {5}{x^2}\right ) \, dx\\ &=-4 x+\frac {5 \log (x)}{x}+\frac {5}{32} \int \frac {e^{\frac {e^x}{8 x}+x}}{x^2} \, dx-\frac {5}{32} \int \frac {e^{\frac {e^x}{8 x}+x}}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 27, normalized size = 0.93 \begin {gather*} -\frac {5}{4} e^{\frac {e^x}{8 x}}-4 x+\frac {5 \log (x)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(160 + E^(E^x/(8*x) + x)*(5 - 5*x) - 128*x^2 - 160*Log[x])/(32*x^2),x]

[Out]

(-5*E^(E^x/(8*x)))/4 - 4*x + (5*Log[x])/x

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fricas [A]  time = 0.62, size = 40, normalized size = 1.38 \begin {gather*} -\frac {{\left (16 \, x^{2} e^{x} + 5 \, x e^{\left (\frac {8 \, x^{2} + e^{x}}{8 \, x}\right )} - 20 \, e^{x} \log \relax (x)\right )} e^{\left (-x\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*((-5*x+5)*exp(x)*exp(1/16*exp(x)/x)^2-160*log(x)-128*x^2+160)/x^2,x, algorithm="fricas")

[Out]

-1/4*(16*x^2*e^x + 5*x*e^(1/8*(8*x^2 + e^x)/x) - 20*e^x*log(x))*e^(-x)/x

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giac [A]  time = 0.13, size = 40, normalized size = 1.38 \begin {gather*} -\frac {{\left (16 \, x^{2} e^{x} + 5 \, x e^{\left (\frac {8 \, x^{2} + e^{x}}{8 \, x}\right )} - 20 \, e^{x} \log \relax (x)\right )} e^{\left (-x\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*((-5*x+5)*exp(x)*exp(1/16*exp(x)/x)^2-160*log(x)-128*x^2+160)/x^2,x, algorithm="giac")

[Out]

-1/4*(16*x^2*e^x + 5*x*e^(1/8*(8*x^2 + e^x)/x) - 20*e^x*log(x))*e^(-x)/x

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maple [A]  time = 0.08, size = 22, normalized size = 0.76

method	result	size
risch	\(-4 x -\frac {5 \,{\mathrm e}^{\frac {{\mathrm e}^{x}}{8 x}}}{4}+\frac {5 \ln \relax (x )}{x}\)	\(22\)
default	\(-4 x -\frac {5 \,{\mathrm e}^{\frac {{\mathrm e}^{x}}{8 x}}}{4}+\frac {5 \ln \relax (x )}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/32*((-5*x+5)*exp(x)*exp(1/16*exp(x)/x)^2-160*ln(x)-128*x^2+160)/x^2,x,method=_RETURNVERBOSE)

[Out]

-4*x-5/4*exp(1/8*exp(x)/x)+5*ln(x)/x

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maxima [A]  time = 0.42, size = 30, normalized size = 1.03 \begin {gather*} -4 \, x - \frac {5 \, {\left (x e^{\left (\frac {e^{x}}{8 \, x}\right )} - 4 \, \log \relax (x) - 4\right )}}{4 \, x} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*((-5*x+5)*exp(x)*exp(1/16*exp(x)/x)^2-160*log(x)-128*x^2+160)/x^2,x, algorithm="maxima")

[Out]

-4*x - 5/4*(x*e^(1/8*e^x/x) - 4*log(x) - 4)/x - 5/x

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mupad [B]  time = 7.77, size = 21, normalized size = 0.72 \begin {gather*} \frac {5\,\ln \relax (x)}{x}-\frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{8\,x}}}{4}-4\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*log(x) + 4*x^2 + (exp(exp(x)/(8*x))*exp(x)*(5*x - 5))/32 - 5)/x^2,x)

[Out]

(5*log(x))/x - (5*exp(exp(x)/(8*x)))/4 - 4*x

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sympy [A]  time = 0.41, size = 20, normalized size = 0.69 \begin {gather*} - 4 x - \frac {5 e^{\frac {e^{x}}{8 x}}}{4} + \frac {5 \log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*((-5*x+5)*exp(x)*exp(1/16*exp(x)/x)**2-160*ln(x)-128*x**2+160)/x**2,x)

[Out]

-4*x - 5*exp(exp(x)/(8*x))/4 + 5*log(x)/x

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