3.25.25 \(\int \frac {e^x (-160 x-32 x^2)-512 \log (x)+(e^{2 x} (50 x^2+20 x^3+2 x^4)+e^x (160 x-160 x^2-32 x^3) \log (x)) \log (\log (x))+e^{2 x} (60 x^3+22 x^4+2 x^5) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx\)

Optimal. Leaf size=18 \[ \left (-\frac {16}{x}+e^x (5+x) \log (\log (x))\right )^2 \]

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Rubi [F]  time = 3.01, 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 \left (-160 x-32 x^2\right )-512 \log (x)+\left (e^{2 x} \left (50 x^2+20 x^3+2 x^4\right )+e^x \left (160 x-160 x^2-32 x^3\right ) \log (x)\right ) \log (\log (x))+e^{2 x} \left (60 x^3+22 x^4+2 x^5\right ) \log (x) \log ^2(\log (x))}{x^3 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-160*x - 32*x^2) - 512*Log[x] + (E^(2*x)*(50*x^2 + 20*x^3 + 2*x^4) + E^x*(160*x - 160*x^2 - 32*x^3)*
Log[x])*Log[Log[x]] + E^(2*x)*(60*x^3 + 22*x^4 + 2*x^5)*Log[x]*Log[Log[x]]^2)/(x^3*Log[x]),x]

[Out]

256/x^2 - (32*E^x*(5*x*Log[x]*Log[Log[x]] + x^2*Log[x]*Log[Log[x]]))/(x^2*Log[x]) + 20*Defer[Int][(E^(2*x)*Log
[Log[x]])/Log[x], x] + 50*Defer[Int][(E^(2*x)*Log[Log[x]])/(x*Log[x]), x] + 2*Defer[Int][(E^(2*x)*x*Log[Log[x]
])/Log[x], x] + 60*Defer[Int][E^(2*x)*Log[Log[x]]^2, x] + 22*Defer[Int][E^(2*x)*x*Log[Log[x]]^2, x] + 2*Defer[
Int][E^(2*x)*x^2*Log[Log[x]]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (16-e^x x (5+x) \log (\log (x))\right ) \left (-e^x x (5+x)-\log (x) \left (16+e^x x^2 (6+x) \log (\log (x))\right )\right )}{x^3 \log (x)} \, dx\\ &=2 \int \frac {\left (16-e^x x (5+x) \log (\log (x))\right ) \left (-e^x x (5+x)-\log (x) \left (16+e^x x^2 (6+x) \log (\log (x))\right )\right )}{x^3 \log (x)} \, dx\\ &=2 \int \left (-\frac {256}{x^3}-\frac {16 e^x \left (5+x-5 \log (x) \log (\log (x))+5 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}+\frac {e^{2 x} (5+x) \log (\log (x)) \left (5+x+6 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x \log (x)}\right ) \, dx\\ &=\frac {256}{x^2}+2 \int \frac {e^{2 x} (5+x) \log (\log (x)) \left (5+x+6 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x \log (x)} \, dx-32 \int \frac {e^x \left (5+x-5 \log (x) \log (\log (x))+5 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx\\ &=\frac {256}{x^2}-\frac {32 e^x \left (5 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}+2 \int \frac {e^{2 x} (5+x) \log (\log (x)) (5+x+x (6+x) \log (x) \log (\log (x)))}{x \log (x)} \, dx\\ &=\frac {256}{x^2}-\frac {32 e^x \left (5 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}+2 \int \left (\frac {e^{2 x} (5+x)^2 \log (\log (x))}{x \log (x)}+e^{2 x} (5+x) (6+x) \log ^2(\log (x))\right ) \, dx\\ &=\frac {256}{x^2}-\frac {32 e^x \left (5 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}+2 \int \frac {e^{2 x} (5+x)^2 \log (\log (x))}{x \log (x)} \, dx+2 \int e^{2 x} (5+x) (6+x) \log ^2(\log (x)) \, dx\\ &=\frac {256}{x^2}-\frac {32 e^x \left (5 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}+2 \int \left (\frac {10 e^{2 x} \log (\log (x))}{\log (x)}+\frac {25 e^{2 x} \log (\log (x))}{x \log (x)}+\frac {e^{2 x} x \log (\log (x))}{\log (x)}\right ) \, dx+2 \int \left (30 e^{2 x} \log ^2(\log (x))+11 e^{2 x} x \log ^2(\log (x))+e^{2 x} x^2 \log ^2(\log (x))\right ) \, dx\\ &=\frac {256}{x^2}-\frac {32 e^x \left (5 x \log (x) \log (\log (x))+x^2 \log (x) \log (\log (x))\right )}{x^2 \log (x)}+2 \int \frac {e^{2 x} x \log (\log (x))}{\log (x)} \, dx+2 \int e^{2 x} x^2 \log ^2(\log (x)) \, dx+20 \int \frac {e^{2 x} \log (\log (x))}{\log (x)} \, dx+22 \int e^{2 x} x \log ^2(\log (x)) \, dx+50 \int \frac {e^{2 x} \log (\log (x))}{x \log (x)} \, dx+60 \int e^{2 x} \log ^2(\log (x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 19, normalized size = 1.06 \begin {gather*} \frac {\left (-16+e^x x (5+x) \log (\log (x))\right )^2}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-160*x - 32*x^2) - 512*Log[x] + (E^(2*x)*(50*x^2 + 20*x^3 + 2*x^4) + E^x*(160*x - 160*x^2 - 32
*x^3)*Log[x])*Log[Log[x]] + E^(2*x)*(60*x^3 + 22*x^4 + 2*x^5)*Log[x]*Log[Log[x]]^2)/(x^3*Log[x]),x]

[Out]

(-16 + E^x*x*(5 + x)*Log[Log[x]])^2/x^2

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fricas [B]  time = 0.63, size = 44, normalized size = 2.44 \begin {gather*} \frac {{\left (x^{4} + 10 \, x^{3} + 25 \, x^{2}\right )} e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )^{2} - 32 \, {\left (x^{2} + 5 \, x\right )} e^{x} \log \left (\log \relax (x)\right ) + 256}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+22*x^4+60*x^3)*exp(x)^2*log(x)*log(log(x))^2+((-32*x^3-160*x^2+160*x)*exp(x)*log(x)+(2*x^4+2
0*x^3+50*x^2)*exp(x)^2)*log(log(x))-512*log(x)+(-32*x^2-160*x)*exp(x))/x^3/log(x),x, algorithm="fricas")

[Out]

((x^4 + 10*x^3 + 25*x^2)*e^(2*x)*log(log(x))^2 - 32*(x^2 + 5*x)*e^x*log(log(x)) + 256)/x^2

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giac [B]  time = 1.82, size = 65, normalized size = 3.61 \begin {gather*} \frac {x^{4} e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )^{2} + 10 \, x^{3} e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )^{2} + 25 \, x^{2} e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )^{2} - 32 \, x^{2} e^{x} \log \left (\log \relax (x)\right ) - 160 \, x e^{x} \log \left (\log \relax (x)\right ) + 256}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+22*x^4+60*x^3)*exp(x)^2*log(x)*log(log(x))^2+((-32*x^3-160*x^2+160*x)*exp(x)*log(x)+(2*x^4+2
0*x^3+50*x^2)*exp(x)^2)*log(log(x))-512*log(x)+(-32*x^2-160*x)*exp(x))/x^3/log(x),x, algorithm="giac")

[Out]

(x^4*e^(2*x)*log(log(x))^2 + 10*x^3*e^(2*x)*log(log(x))^2 + 25*x^2*e^(2*x)*log(log(x))^2 - 32*x^2*e^x*log(log(
x)) - 160*x*e^x*log(log(x)) + 256)/x^2

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maple [B]  time = 0.05, size = 38, normalized size = 2.11

method	result	size
risch	\(\left (x^{2}+10 x +25\right ) {\mathrm e}^{2 x} \ln \left (\ln \relax (x )\right )^{2}-\frac {32 \left (5+x \right ) {\mathrm e}^{x} \ln \left (\ln \relax (x )\right )}{x}+\frac {256}{x^{2}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^5+22*x^4+60*x^3)*exp(x)^2*ln(x)*ln(ln(x))^2+((-32*x^3-160*x^2+160*x)*exp(x)*ln(x)+(2*x^4+20*x^3+50*x
^2)*exp(x)^2)*ln(ln(x))-512*ln(x)+(-32*x^2-160*x)*exp(x))/x^3/ln(x),x,method=_RETURNVERBOSE)

[Out]

(x^2+10*x+25)*exp(2*x)*ln(ln(x))^2-32/x*(5+x)*exp(x)*ln(ln(x))+256/x^2

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maxima [B]  time = 0.49, size = 43, normalized size = 2.39 \begin {gather*} \frac {{\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{\left (2 \, x\right )} \log \left (\log \relax (x)\right )^{2} - 32 \, {\left (x + 5\right )} e^{x} \log \left (\log \relax (x)\right )}{x} + \frac {256}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+22*x^4+60*x^3)*exp(x)^2*log(x)*log(log(x))^2+((-32*x^3-160*x^2+160*x)*exp(x)*log(x)+(2*x^4+2
0*x^3+50*x^2)*exp(x)^2)*log(log(x))-512*log(x)+(-32*x^2-160*x)*exp(x))/x^3/log(x),x, algorithm="maxima")

[Out]

((x^3 + 10*x^2 + 25*x)*e^(2*x)*log(log(x))^2 - 32*(x + 5)*e^x*log(log(x)))/x + 256/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \begin {gather*} -\int \frac {-{\mathrm {e}}^{2\,x}\,\ln \relax (x)\,\left (2\,x^5+22\,x^4+60\,x^3\right )\,{\ln \left (\ln \relax (x)\right )}^2+\left ({\mathrm {e}}^x\,\ln \relax (x)\,\left (32\,x^3+160\,x^2-160\,x\right )-{\mathrm {e}}^{2\,x}\,\left (2\,x^4+20\,x^3+50\,x^2\right )\right )\,\ln \left (\ln \relax (x)\right )+512\,\ln \relax (x)+{\mathrm {e}}^x\,\left (32\,x^2+160\,x\right )}{x^3\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(512*log(x) - log(log(x))*(exp(2*x)*(50*x^2 + 20*x^3 + 2*x^4) - exp(x)*log(x)*(160*x^2 - 160*x + 32*x^3))
+ exp(x)*(160*x + 32*x^2) - log(log(x))^2*exp(2*x)*log(x)*(60*x^3 + 22*x^4 + 2*x^5))/(x^3*log(x)),x)

[Out]

-int((512*log(x) - log(log(x))*(exp(2*x)*(50*x^2 + 20*x^3 + 2*x^4) - exp(x)*log(x)*(160*x^2 - 160*x + 32*x^3))
+ exp(x)*(160*x + 32*x^2) - log(log(x))^2*exp(2*x)*log(x)*(60*x^3 + 22*x^4 + 2*x^5))/(x^3*log(x)), x)

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sympy [B]  time = 0.50, size = 63, normalized size = 3.50 \begin {gather*} \frac {\left (- 32 x \log {\left (\log {\relax (x )} \right )} - 160 \log {\left (\log {\relax (x )} \right )}\right ) e^{x} + \left (x^{3} \log {\left (\log {\relax (x )} \right )}^{2} + 10 x^{2} \log {\left (\log {\relax (x )} \right )}^{2} + 25 x \log {\left (\log {\relax (x )} \right )}^{2}\right ) e^{2 x}}{x} + \frac {256}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**5+22*x**4+60*x**3)*exp(x)**2*ln(x)*ln(ln(x))**2+((-32*x**3-160*x**2+160*x)*exp(x)*ln(x)+(2*x*
*4+20*x**3+50*x**2)*exp(x)**2)*ln(ln(x))-512*ln(x)+(-32*x**2-160*x)*exp(x))/x**3/ln(x),x)

[Out]

((-32*x*log(log(x)) - 160*log(log(x)))*exp(x) + (x**3*log(log(x))**2 + 10*x**2*log(log(x))**2 + 25*x*log(log(x
))**2)*exp(2*x))/x + 256/x**2

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