3.26.8 \(\int \frac {240 x^2+100 x^3+e^x (240 x+100 x^2)+(-120 x+50 x^2+e^x (-20 x-50 x^2)) \log (x)+(e^x (-120-50 x)-120 x-50 x^2+(e^x (-120-50 x)-120 x-50 x^2) \log (x)) \log (\frac {144+120 x+25 x^2}{e^x+x})}{(12 x^3+5 x^4+e^x (12 x^2+5 x^3)) \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {10 \left (-2 x+\log \left (\frac {(x+4 (3+x))^2}{e^x+x}\right )\right )}{x \log (x)} \]

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Rubi [F]  time = 5.62, 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 {240 x^2+100 x^3+e^x \left (240 x+100 x^2\right )+\left (-120 x+50 x^2+e^x \left (-20 x-50 x^2\right )\right ) \log (x)+\left (e^x (-120-50 x)-120 x-50 x^2+\left (e^x (-120-50 x)-120 x-50 x^2\right ) \log (x)\right ) \log \left (\frac {144+120 x+25 x^2}{e^x+x}\right )}{\left (12 x^3+5 x^4+e^x \left (12 x^2+5 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(240*x^2 + 100*x^3 + E^x*(240*x + 100*x^2) + (-120*x + 50*x^2 + E^x*(-20*x - 50*x^2))*Log[x] + (E^x*(-120
- 50*x) - 120*x - 50*x^2 + (E^x*(-120 - 50*x) - 120*x - 50*x^2)*Log[x])*Log[(144 + 120*x + 25*x^2)/(E^x + x)])
/((12*x^3 + 5*x^4 + E^x*(12*x^2 + 5*x^3))*Log[x]^2),x]

[Out]

-20/Log[x] + 10*Defer[Int][1/((E^x + x)*Log[x]), x] - 10*Defer[Int][1/(x*(E^x + x)*Log[x]), x] + 10*Defer[Int]
[(-2 - 5*x)/(x*(12 + 5*x)*Log[x]), x] - 10*Defer[Int][Log[(12 + 5*x)^2/(E^x + x)]/(x^2*Log[x]^2), x] - 10*Defe
r[Int][Log[(12 + 5*x)^2/(E^x + x)]/(x^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 \left (2 x+\log (x) \left (-\frac {x \left (12-5 x+e^x (2+5 x)\right )}{\left (e^x+x\right ) (12+5 x)}-\log \left (\frac {(12+5 x)^2}{e^x+x}\right )\right )-\log \left (\frac {(12+5 x)^2}{e^x+x}\right )\right )}{x^2 \log ^2(x)} \, dx\\ &=10 \int \frac {2 x+\log (x) \left (-\frac {x \left (12-5 x+e^x (2+5 x)\right )}{\left (e^x+x\right ) (12+5 x)}-\log \left (\frac {(12+5 x)^2}{e^x+x}\right )\right )-\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log ^2(x)} \, dx\\ &=10 \int \left (\frac {-1+x}{x \left (e^x+x\right ) \log (x)}+\frac {24 x+10 x^2-2 x \log (x)-5 x^2 \log (x)-12 \log \left (\frac {(12+5 x)^2}{e^x+x}\right )-5 x \log \left (\frac {(12+5 x)^2}{e^x+x}\right )-12 \log (x) \log \left (\frac {(12+5 x)^2}{e^x+x}\right )-5 x \log (x) \log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 (12+5 x) \log ^2(x)}\right ) \, dx\\ &=10 \int \frac {-1+x}{x \left (e^x+x\right ) \log (x)} \, dx+10 \int \frac {24 x+10 x^2-2 x \log (x)-5 x^2 \log (x)-12 \log \left (\frac {(12+5 x)^2}{e^x+x}\right )-5 x \log \left (\frac {(12+5 x)^2}{e^x+x}\right )-12 \log (x) \log \left (\frac {(12+5 x)^2}{e^x+x}\right )-5 x \log (x) \log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 (12+5 x) \log ^2(x)} \, dx\\ &=10 \int \left (\frac {1}{\left (e^x+x\right ) \log (x)}-\frac {1}{x \left (e^x+x\right ) \log (x)}\right ) \, dx+10 \int \frac {(12+5 x) \left (2 x-\log \left (\frac {(12+5 x)^2}{e^x+x}\right )\right )-\log (x) \left (x (2+5 x)+(12+5 x) \log \left (\frac {(12+5 x)^2}{e^x+x}\right )\right )}{x^2 (12+5 x) \log ^2(x)} \, dx\\ &=10 \int \frac {1}{\left (e^x+x\right ) \log (x)} \, dx-10 \int \frac {1}{x \left (e^x+x\right ) \log (x)} \, dx+10 \int \left (\frac {24+10 x-2 \log (x)-5 x \log (x)}{x (12+5 x) \log ^2(x)}-\frac {(1+\log (x)) \log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log ^2(x)}\right ) \, dx\\ &=10 \int \frac {1}{\left (e^x+x\right ) \log (x)} \, dx-10 \int \frac {1}{x \left (e^x+x\right ) \log (x)} \, dx+10 \int \frac {24+10 x-2 \log (x)-5 x \log (x)}{x (12+5 x) \log ^2(x)} \, dx-10 \int \frac {(1+\log (x)) \log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log ^2(x)} \, dx\\ &=10 \int \left (\frac {2}{x \log ^2(x)}+\frac {-2-5 x}{x (12+5 x) \log (x)}\right ) \, dx+10 \int \frac {1}{\left (e^x+x\right ) \log (x)} \, dx-10 \int \frac {1}{x \left (e^x+x\right ) \log (x)} \, dx-10 \int \left (\frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log ^2(x)}+\frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log (x)}\right ) \, dx\\ &=10 \int \frac {1}{\left (e^x+x\right ) \log (x)} \, dx-10 \int \frac {1}{x \left (e^x+x\right ) \log (x)} \, dx+10 \int \frac {-2-5 x}{x (12+5 x) \log (x)} \, dx-10 \int \frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log ^2(x)} \, dx-10 \int \frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log (x)} \, dx+20 \int \frac {1}{x \log ^2(x)} \, dx\\ &=10 \int \frac {1}{\left (e^x+x\right ) \log (x)} \, dx-10 \int \frac {1}{x \left (e^x+x\right ) \log (x)} \, dx+10 \int \frac {-2-5 x}{x (12+5 x) \log (x)} \, dx-10 \int \frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log ^2(x)} \, dx-10 \int \frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log (x)} \, dx+20 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )\\ &=-\frac {20}{\log (x)}+10 \int \frac {1}{\left (e^x+x\right ) \log (x)} \, dx-10 \int \frac {1}{x \left (e^x+x\right ) \log (x)} \, dx+10 \int \frac {-2-5 x}{x (12+5 x) \log (x)} \, dx-10 \int \frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log ^2(x)} \, dx-10 \int \frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 33, normalized size = 1.06 \begin {gather*} 10 \left (-\frac {2}{\log (x)}+\frac {\log \left (\frac {(12+5 x)^2}{e^x+x}\right )}{x \log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(240*x^2 + 100*x^3 + E^x*(240*x + 100*x^2) + (-120*x + 50*x^2 + E^x*(-20*x - 50*x^2))*Log[x] + (E^x*
(-120 - 50*x) - 120*x - 50*x^2 + (E^x*(-120 - 50*x) - 120*x - 50*x^2)*Log[x])*Log[(144 + 120*x + 25*x^2)/(E^x
+ x)])/((12*x^3 + 5*x^4 + E^x*(12*x^2 + 5*x^3))*Log[x]^2),x]

[Out]

10*(-2/Log[x] + Log[(12 + 5*x)^2/(E^x + x)]/(x*Log[x]))

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fricas [A]  time = 1.05, size = 33, normalized size = 1.06 \begin {gather*} -\frac {10 \, {\left (2 \, x - \log \left (\frac {25 \, x^{2} + 120 \, x + 144}{x + e^{x}}\right )\right )}}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-50*x-120)*exp(x)-50*x^2-120*x)*log(x)+(-50*x-120)*exp(x)-50*x^2-120*x)*log((25*x^2+120*x+144)/(
exp(x)+x))+((-50*x^2-20*x)*exp(x)+50*x^2-120*x)*log(x)+(100*x^2+240*x)*exp(x)+100*x^3+240*x^2)/((5*x^3+12*x^2)
*exp(x)+5*x^4+12*x^3)/log(x)^2,x, algorithm="fricas")

[Out]

-10*(2*x - log((25*x^2 + 120*x + 144)/(x + e^x)))/(x*log(x))

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giac [A]  time = 0.35, size = 31, normalized size = 1.00 \begin {gather*} -\frac {10 \, {\left (2 \, x - \log \left (25 \, x^{2} + 120 \, x + 144\right ) + \log \left (x + e^{x}\right )\right )}}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-50*x-120)*exp(x)-50*x^2-120*x)*log(x)+(-50*x-120)*exp(x)-50*x^2-120*x)*log((25*x^2+120*x+144)/(
exp(x)+x))+((-50*x^2-20*x)*exp(x)+50*x^2-120*x)*log(x)+(100*x^2+240*x)*exp(x)+100*x^3+240*x^2)/((5*x^3+12*x^2)
*exp(x)+5*x^4+12*x^3)/log(x)^2,x, algorithm="giac")

[Out]

-10*(2*x - log(25*x^2 + 120*x + 144) + log(x + e^x))/(x*log(x))

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maple [C]  time = 0.18, size = 218, normalized size = 7.03

method	result	size
risch	\(-\frac {10 \ln \left ({\mathrm e}^{x}+x \right )}{x \ln \relax (x )}+\frac {-5 i \pi \mathrm {csgn}\left (i \left (x +\frac {12}{5}\right )\right )^{2} \mathrm {csgn}\left (i \left (x +\frac {12}{5}\right )^{2}\right )+10 i \pi \,\mathrm {csgn}\left (i \left (x +\frac {12}{5}\right )\right ) \mathrm {csgn}\left (i \left (x +\frac {12}{5}\right )^{2}\right )^{2}-5 i \pi \mathrm {csgn}\left (i \left (x +\frac {12}{5}\right )^{2}\right )^{3}-5 i \pi \,\mathrm {csgn}\left (i \left (x +\frac {12}{5}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {12}{5}\right )^{2}}{{\mathrm e}^{x}+x}\right )+5 i \pi \,\mathrm {csgn}\left (i \left (x +\frac {12}{5}\right )^{2}\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {12}{5}\right )^{2}}{{\mathrm e}^{x}+x}\right )^{2}+5 i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{x}+x}\right ) \mathrm {csgn}\left (\frac {i \left (x +\frac {12}{5}\right )^{2}}{{\mathrm e}^{x}+x}\right )^{2}-5 i \pi \mathrm {csgn}\left (\frac {i \left (x +\frac {12}{5}\right )^{2}}{{\mathrm e}^{x}+x}\right )^{3}+20 \ln \relax (5)-20 x +20 \ln \left (x +\frac {12}{5}\right )}{x \ln \relax (x )}\)	\(218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-50*x-120)*exp(x)-50*x^2-120*x)*ln(x)+(-50*x-120)*exp(x)-50*x^2-120*x)*ln((25*x^2+120*x+144)/(exp(x)+x
))+((-50*x^2-20*x)*exp(x)+50*x^2-120*x)*ln(x)+(100*x^2+240*x)*exp(x)+100*x^3+240*x^2)/((5*x^3+12*x^2)*exp(x)+5
*x^4+12*x^3)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-10/x/ln(x)*ln(exp(x)+x)+5*(-I*Pi*csgn(I*(x+12/5))^2*csgn(I*(x+12/5)^2)+2*I*Pi*csgn(I*(x+12/5))*csgn(I*(x+12/5
)^2)^2-I*Pi*csgn(I*(x+12/5)^2)^3-I*Pi*csgn(I*(x+12/5)^2)*csgn(I/(exp(x)+x))*csgn(I*(x+12/5)^2/(exp(x)+x))+I*Pi
*csgn(I*(x+12/5)^2)*csgn(I*(x+12/5)^2/(exp(x)+x))^2+I*Pi*csgn(I/(exp(x)+x))*csgn(I*(x+12/5)^2/(exp(x)+x))^2-I*
Pi*csgn(I*(x+12/5)^2/(exp(x)+x))^3+4*ln(5)-4*x+4*ln(x+12/5))/x/ln(x)

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maxima [A]  time = 0.40, size = 26, normalized size = 0.84 \begin {gather*} -\frac {10 \, {\left (2 \, x - 2 \, \log \left (5 \, x + 12\right ) + \log \left (x + e^{x}\right )\right )}}{x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-50*x-120)*exp(x)-50*x^2-120*x)*log(x)+(-50*x-120)*exp(x)-50*x^2-120*x)*log((25*x^2+120*x+144)/(
exp(x)+x))+((-50*x^2-20*x)*exp(x)+50*x^2-120*x)*log(x)+(100*x^2+240*x)*exp(x)+100*x^3+240*x^2)/((5*x^3+12*x^2)
*exp(x)+5*x^4+12*x^3)/log(x)^2,x, algorithm="maxima")

[Out]

-10*(2*x - 2*log(5*x + 12) + log(x + e^x))/(x*log(x))

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mupad [B]  time = 1.75, size = 33, normalized size = 1.06 \begin {gather*} -\frac {10\,\left (2\,x-\ln \left (\frac {25\,x^2+120\,x+144}{x+{\mathrm {e}}^x}\right )\right )}{x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(240*x + 100*x^2) - log(x)*(120*x + exp(x)*(20*x + 50*x^2) - 50*x^2) - log((120*x + 25*x^2 + 144)/
(x + exp(x)))*(120*x + log(x)*(120*x + exp(x)*(50*x + 120) + 50*x^2) + exp(x)*(50*x + 120) + 50*x^2) + 240*x^2
+ 100*x^3)/(log(x)^2*(exp(x)*(12*x^2 + 5*x^3) + 12*x^3 + 5*x^4)),x)

[Out]

-(10*(2*x - log((120*x + 25*x^2 + 144)/(x + exp(x)))))/(x*log(x))

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sympy [A]  time = 0.78, size = 27, normalized size = 0.87 \begin {gather*} - \frac {20}{\log {\relax (x )}} + \frac {10 \log {\left (\frac {25 x^{2} + 120 x + 144}{x + e^{x}} \right )}}{x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-50*x-120)*exp(x)-50*x**2-120*x)*ln(x)+(-50*x-120)*exp(x)-50*x**2-120*x)*ln((25*x**2+120*x+144)/
(exp(x)+x))+((-50*x**2-20*x)*exp(x)+50*x**2-120*x)*ln(x)+(100*x**2+240*x)*exp(x)+100*x**3+240*x**2)/((5*x**3+1
2*x**2)*exp(x)+5*x**4+12*x**3)/ln(x)**2,x)

[Out]

-20/log(x) + 10*log((25*x**2 + 120*x + 144)/(x + exp(x)))/(x*log(x))

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