3.71.83 \(\int \frac {e^{\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))} (288+68 x+5 \log (5)+20 x \log (x))}{48 x} \, dx\)

Optimal. Leaf size=21 \[ e^{x+\left (6+\frac {5}{12} \left (x+\frac {\log (5)}{4}\right )\right ) \log (x)} \]

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Rubi [F]  time = 1.23, 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 {\exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) (288+68 x+5 \log (5)+20 x \log (x))}{48 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((48*x + (288 + 20*x + 5*Log[5])*Log[x])/48)*(288 + 68*x + 5*Log[5] + 20*x*Log[x]))/(48*x),x]

[Out]

(17*Defer[Int][E^((48*x + (288 + 20*x + 5*Log[5])*Log[x])/48), x])/12 + ((288 + 5*Log[5])*Defer[Int][E^((48*x
+ (288 + 20*x + 5*Log[5])*Log[x])/48)/x, x])/48 + (5*Defer[Int][E^((48*x + (288 + 20*x + 5*Log[5])*Log[x])/48)
*Log[x], x])/12

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{48} \int \frac {\exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) (288+68 x+5 \log (5)+20 x \log (x))}{x} \, dx\\ &=\frac {1}{48} \int \frac {\exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) \left (68 x+288 \left (1+\frac {5 \log (5)}{288}\right )+20 x \log (x)\right )}{x} \, dx\\ &=\frac {1}{48} \int \left (\frac {\exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) (288+68 x+5 \log (5))}{x}+20 \exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) \log (x)\right ) \, dx\\ &=\frac {1}{48} \int \frac {\exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) (288+68 x+5 \log (5))}{x} \, dx+\frac {5}{12} \int \exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) \log (x) \, dx\\ &=\frac {1}{48} \int \left (68 \exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right )+\frac {\exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) (288+5 \log (5))}{x}\right ) \, dx+\frac {5}{12} \int \exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) \log (x) \, dx\\ &=\frac {5}{12} \int \exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) \log (x) \, dx+\frac {17}{12} \int \exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right ) \, dx+\frac {1}{48} (288+5 \log (5)) \int \frac {\exp \left (\frac {1}{48} (48 x+(288+20 x+5 \log (5)) \log (x))\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 19, normalized size = 0.90 \begin {gather*} e^x x^{1+\frac {5}{48} (48+4 x+\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((48*x + (288 + 20*x + 5*Log[5])*Log[x])/48)*(288 + 68*x + 5*Log[5] + 20*x*Log[x]))/(48*x),x]

[Out]

E^x*x^(1 + (5*(48 + 4*x + Log[5]))/48)

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fricas [A]  time = 1.59, size = 16, normalized size = 0.76 \begin {gather*} e^{\left (\frac {1}{48} \, {\left (20 \, x + 5 \, \log \relax (5) + 288\right )} \log \relax (x) + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(20*x*log(x)+5*log(5)+68*x+288)*exp(1/48*(5*log(5)+20*x+288)*log(x)+x)/x,x, algorithm="fricas")

[Out]

e^(1/48*(20*x + 5*log(5) + 288)*log(x) + x)

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giac [A]  time = 0.13, size = 18, normalized size = 0.86 \begin {gather*} e^{\left (\frac {5}{12} \, x \log \relax (x) + \frac {5}{48} \, \log \relax (5) \log \relax (x) + x + 6 \, \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(20*x*log(x)+5*log(5)+68*x+288)*exp(1/48*(5*log(5)+20*x+288)*log(x)+x)/x,x, algorithm="giac")

[Out]

e^(5/12*x*log(x) + 5/48*log(5)*log(x) + x + 6*log(x))

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maple [A]  time = 0.04, size = 15, normalized size = 0.71

method	result	size
risch	\(x^{6+\frac {5 x}{12}+\frac {5 \ln \relax (5)}{48}} {\mathrm e}^{x}\)	\(15\)
norman	\({\mathrm e}^{\frac {\left (5 \ln \relax (5)+20 x +288\right ) \ln \relax (x )}{48}+x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/48*(20*x*ln(x)+5*ln(5)+68*x+288)*exp(1/48*(5*ln(5)+20*x+288)*ln(x)+x)/x,x,method=_RETURNVERBOSE)

[Out]

x^(6+5/12*x+5/48*ln(5))*exp(x)

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maxima [A]  time = 0.54, size = 18, normalized size = 0.86 \begin {gather*} x^{6} e^{\left (\frac {5}{12} \, x \log \relax (x) + \frac {5}{48} \, \log \relax (5) \log \relax (x) + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(20*x*log(x)+5*log(5)+68*x+288)*exp(1/48*(5*log(5)+20*x+288)*log(x)+x)/x,x, algorithm="maxima")

[Out]

x^6*e^(5/12*x*log(x) + 5/48*log(5)*log(x) + x)

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mupad [B]  time = 4.30, size = 17, normalized size = 0.81 \begin {gather*} x^{\frac {5\,x}{12}+6}\,{\mathrm {e}}^{x+\frac {5\,\ln \relax (5)\,\ln \relax (x)}{48}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + (log(x)*(20*x + 5*log(5) + 288))/48)*(68*x + 5*log(5) + 20*x*log(x) + 288))/(48*x),x)

[Out]

x^((5*x)/12 + 6)*exp(x + (5*log(5)*log(x))/48)

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sympy [A]  time = 0.42, size = 19, normalized size = 0.90 \begin {gather*} e^{x + \left (\frac {5 x}{12} + \frac {5 \log {\relax (5 )}}{48} + 6\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/48*(20*x*ln(x)+5*ln(5)+68*x+288)*exp(1/48*(5*ln(5)+20*x+288)*ln(x)+x)/x,x)

[Out]

exp(x + (5*x/12 + 5*log(5)/48 + 6)*log(x))

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