3.77.91 \(\int \frac {3 x+3 \log (3)+e^x (x+\log (3))+(-6 x-3 \log (3)+e^x (-2 x+x^2+(-1+x) \log (3))) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+(3 x^2+3 x \log (3)+e^x (x^2+x \log (3))) \log (x)} \, dx\)

Optimal. Leaf size=20 \[ \log \left (-10+\frac {\left (3+e^x\right ) \log (x)}{x (x+\log (3))}\right ) \]

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Rubi [F]  time = 3.23, 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 {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3*x + 3*Log[3] + E^x*(x + Log[3]) + (-6*x - 3*Log[3] + E^x*(-2*x + x^2 + (-1 + x)*Log[3]))*Log[x])/(-10*x
^4 - 20*x^3*Log[3] - 10*x^2*Log[3]^2 + (3*x^2 + 3*x*Log[3] + E^x*(x^2 + x*Log[3]))*Log[x]),x]

[Out]

x - Log[x] - Log[x + Log[3]] + Log[Log[x]] + 10*Log[3]*Defer[Int][(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x
])^(-1), x] + 10*(2 - Log[3])*Defer[Int][x/(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x]), x] - 10*Defer[Int][
x^2/(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x]), x] - 10*Log[3]*Defer[Int][1/(Log[x]*(10*x^2 + 10*x*Log[3]
- 3*Log[x] - E^x*Log[x])), x] - 10*Defer[Int][x/(Log[x]*(10*x^2 + 10*x*Log[3] - 3*Log[x] - E^x*Log[x])), x] -
3*Defer[Int][Log[x]/(-10*x^2 - 10*x*Log[3] + 3*Log[x] + E^x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3 x-3 \log (3)-e^x (x+\log (3))-\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{x (x+\log (3)) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx\\ &=\int \left (\frac {x+\log (3)+x^2 \log (x)-2 x \left (1-\frac {\log (3)}{2}\right ) \log (x)-\log (3) \log (x)}{x (x+\log (3)) \log (x)}+\frac {-10 x-10 \log (3)-10 x^2 \log (x)+20 x \left (1-\frac {\log (3)}{2}\right ) \log (x)+10 \log (3) \log (x)+3 \log ^2(x)}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )}\right ) \, dx\\ &=\int \frac {x+\log (3)+x^2 \log (x)-2 x \left (1-\frac {\log (3)}{2}\right ) \log (x)-\log (3) \log (x)}{x (x+\log (3)) \log (x)} \, dx+\int \frac {-10 x-10 \log (3)-10 x^2 \log (x)+20 x \left (1-\frac {\log (3)}{2}\right ) \log (x)+10 \log (3) \log (x)+3 \log ^2(x)}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx\\ &=\int \frac {x+\log (3)+\left (x^2+x (-2+\log (3))-\log (3)\right ) \log (x)}{x (x+\log (3)) \log (x)} \, dx+\int \left (-\frac {10 x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)}-\frac {10 x (-2+\log (3))}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)}+\frac {10 \log (3)}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)}-\frac {10 x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )}-\frac {10 \log (3)}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )}-\frac {3 \log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)}\right ) \, dx\\ &=-\left (3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx\right )-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+\int \left (\frac {x^2-x (2-\log (3))-\log (3)}{x (x+\log (3))}+\frac {1}{x \log (x)}\right ) \, dx\\ &=-\left (3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx\right )-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+\int \frac {x^2-x (2-\log (3))-\log (3)}{x (x+\log (3))} \, dx+\int \frac {1}{x \log (x)} \, dx\\ &=-\left (3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx\right )-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+\int \left (1-\frac {1}{x}+\frac {1}{-x-\log (3)}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=x-\log (x)-\log (x+\log (3))+\log (\log (x))-3 \int \frac {\log (x)}{-10 x^2-10 x \log (3)+3 \log (x)+e^x \log (x)} \, dx-10 \int \frac {x^2}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-10 \int \frac {x}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx+(10 (2-\log (3))) \int \frac {x}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx+(10 \log (3)) \int \frac {1}{10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)} \, dx-(10 \log (3)) \int \frac {1}{\log (x) \left (10 x^2+10 x \log (3)-3 \log (x)-e^x \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.65, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {3 x+3 \log (3)+e^x (x+\log (3))+\left (-6 x-3 \log (3)+e^x \left (-2 x+x^2+(-1+x) \log (3)\right )\right ) \log (x)}{-10 x^4-20 x^3 \log (3)-10 x^2 \log ^2(3)+\left (3 x^2+3 x \log (3)+e^x \left (x^2+x \log (3)\right )\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(3*x + 3*Log[3] + E^x*(x + Log[3]) + (-6*x - 3*Log[3] + E^x*(-2*x + x^2 + (-1 + x)*Log[3]))*Log[x])/
(-10*x^4 - 20*x^3*Log[3] - 10*x^2*Log[3]^2 + (3*x^2 + 3*x*Log[3] + E^x*(x^2 + x*Log[3]))*Log[x]),x]

[Out]

Integrate[(3*x + 3*Log[3] + E^x*(x + Log[3]) + (-6*x - 3*Log[3] + E^x*(-2*x + x^2 + (-1 + x)*Log[3]))*Log[x])/
(-10*x^4 - 20*x^3*Log[3] - 10*x^2*Log[3]^2 + (3*x^2 + 3*x*Log[3] + E^x*(x^2 + x*Log[3]))*Log[x]), x]

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fricas [B]  time = 1.10, size = 45, normalized size = 2.25 \begin {gather*} -\log \left (x^{2} + x \log \relax (3)\right ) + \log \left (-\frac {10 \, x^{2} + 10 \, x \log \relax (3) - {\left (e^{x} + 3\right )} \log \relax (x)}{e^{x} + 3}\right ) + \log \left (e^{x} + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-1)*log(3)+x^2-2*x)*exp(x)-3*log(3)-6*x)*log(x)+(log(3)+x)*exp(x)+3*log(3)+3*x)/(((x*log(3)+x^2
)*exp(x)+3*x*log(3)+3*x^2)*log(x)-10*x^2*log(3)^2-20*x^3*log(3)-10*x^4),x, algorithm="fricas")

[Out]

-log(x^2 + x*log(3)) + log(-(10*x^2 + 10*x*log(3) - (e^x + 3)*log(x))/(e^x + 3)) + log(e^x + 3)

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giac [A]  time = 0.23, size = 33, normalized size = 1.65 \begin {gather*} \log \left (-10 \, x^{2} - 10 \, x \log \relax (3) + e^{x} \log \relax (x) + 3 \, \log \relax (x)\right ) - \log \left (x + \log \relax (3)\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-1)*log(3)+x^2-2*x)*exp(x)-3*log(3)-6*x)*log(x)+(log(3)+x)*exp(x)+3*log(3)+3*x)/(((x*log(3)+x^2
)*exp(x)+3*x*log(3)+3*x^2)*log(x)-10*x^2*log(3)^2-20*x^3*log(3)-10*x^4),x, algorithm="giac")

[Out]

log(-10*x^2 - 10*x*log(3) + e^x*log(x) + 3*log(x)) - log(x + log(3)) - log(x)

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maple [A]  time = 0.32, size = 35, normalized size = 1.75

method	result	size
norman	\(-\ln \relax (x )-\ln \left (\ln \relax (3)+x \right )+\ln \left (10 x \ln \relax (3)+10 x^{2}-{\mathrm e}^{x} \ln \relax (x )-3 \ln \relax (x )\right )\)	\(35\)
risch	\(-\ln \left (x \ln \relax (3)+x^{2}\right )+\ln \left (3+{\mathrm e}^{x}\right )+\ln \left (\ln \relax (x )-\frac {10 x \left (\ln \relax (3)+x \right )}{3+{\mathrm e}^{x}}\right )\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x-1)*ln(3)+x^2-2*x)*exp(x)-3*ln(3)-6*x)*ln(x)+(ln(3)+x)*exp(x)+3*ln(3)+3*x)/(((x*ln(3)+x^2)*exp(x)+3*x
*ln(3)+3*x^2)*ln(x)-10*x^2*ln(3)^2-20*x^3*ln(3)-10*x^4),x,method=_RETURNVERBOSE)

[Out]

-ln(x)-ln(ln(3)+x)+ln(10*x*ln(3)+10*x^2-exp(x)*ln(x)-3*ln(x))

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maxima [B]  time = 0.51, size = 43, normalized size = 2.15 \begin {gather*} -\log \left (x + \log \relax (3)\right ) - \log \relax (x) + \log \left (-\frac {10 \, x^{2} + 10 \, x \log \relax (3) - e^{x} \log \relax (x) - 3 \, \log \relax (x)}{\log \relax (x)}\right ) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-1)*log(3)+x^2-2*x)*exp(x)-3*log(3)-6*x)*log(x)+(log(3)+x)*exp(x)+3*log(3)+3*x)/(((x*log(3)+x^2
)*exp(x)+3*x*log(3)+3*x^2)*log(x)-10*x^2*log(3)^2-20*x^3*log(3)-10*x^4),x, algorithm="maxima")

[Out]

-log(x + log(3)) - log(x) + log(-(10*x^2 + 10*x*log(3) - e^x*log(x) - 3*log(x))/log(x)) + log(log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int \frac {3\,x+3\,\ln \relax (3)-\ln \relax (x)\,\left (6\,x+3\,\ln \relax (3)-{\mathrm {e}}^x\,\left (\ln \relax (3)\,\left (x-1\right )-2\,x+x^2\right )\right )+{\mathrm {e}}^x\,\left (x+\ln \relax (3)\right )}{10\,x^2\,{\ln \relax (3)}^2+20\,x^3\,\ln \relax (3)-\ln \relax (x)\,\left (3\,x\,\ln \relax (3)+{\mathrm {e}}^x\,\left (x^2+\ln \relax (3)\,x\right )+3\,x^2\right )+10\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x + 3*log(3) - log(x)*(6*x + 3*log(3) - exp(x)*(log(3)*(x - 1) - 2*x + x^2)) + exp(x)*(x + log(3)))/(1
0*x^2*log(3)^2 + 20*x^3*log(3) - log(x)*(3*x*log(3) + exp(x)*(x*log(3) + x^2) + 3*x^2) + 10*x^4),x)

[Out]

-int((3*x + 3*log(3) - log(x)*(6*x + 3*log(3) - exp(x)*(log(3)*(x - 1) - 2*x + x^2)) + exp(x)*(x + log(3)))/(1
0*x^2*log(3)^2 + 20*x^3*log(3) - log(x)*(3*x*log(3) + exp(x)*(x*log(3) + x^2) + 3*x^2) + 10*x^4), x)

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sympy [B]  time = 0.57, size = 39, normalized size = 1.95 \begin {gather*} - \log {\left (x^{2} + x \log {\relax (3 )} \right )} + \log {\left (\frac {- 10 x^{2} - 10 x \log {\relax (3 )} + 3 \log {\relax (x )}}{\log {\relax (x )}} + e^{x} \right )} + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-1)*ln(3)+x**2-2*x)*exp(x)-3*ln(3)-6*x)*ln(x)+(ln(3)+x)*exp(x)+3*ln(3)+3*x)/(((x*ln(3)+x**2)*ex
p(x)+3*x*ln(3)+3*x**2)*ln(x)-10*x**2*ln(3)**2-20*x**3*ln(3)-10*x**4),x)

[Out]

-log(x**2 + x*log(3)) + log((-10*x**2 - 10*x*log(3) + 3*log(x))/log(x) + exp(x)) + log(log(x))

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