3.27.3 \(\int \frac {-10 x-3 x^2+x^3+e^4 (5 x-x^2)+(-10+e^4 (5-x)-3 x+x^2) \log (3)+(15 x+14 x^2-2 x^3+e^4 (-10 x+x^2)+(20+e^4 (-10+x)+13 x-2 x^2) \log (3)) \log (x)+(5 x-x^2+(5-x) \log (3)+(-10 x+x^2+(-10+x) \log (3)) \log (x)) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 (-16 x^4-8 x^5)+(16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 (-16 x^3-8 x^4)) \log (3)+(-16 x^4+8 e^4 x^4-8 x^5+(-16 x^3+8 e^4 x^3-8 x^4) \log (3)) \log (x+\log (3))+(4 x^4+4 x^3 \log (3)) \log ^2(x+\log (3))} \, dx\)

Optimal. Leaf size=29 \[ \frac {(-5+x) \log (x)}{4 x^2 \left (2-e^4+x-\log (x+\log (3))\right )} \]

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Rubi [F]  time = 5.47, 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 {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{16 x^4+4 e^8 x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x - 3*x^2 + x^3 + E^4*(5*x - x^2) + (-10 + E^4*(5 - x) - 3*x + x^2)*Log[3] + (15*x + 14*x^2 - 2*x^3 +
E^4*(-10*x + x^2) + (20 + E^4*(-10 + x) + 13*x - 2*x^2)*Log[3])*Log[x] + (5*x - x^2 + (5 - x)*Log[3] + (-10*x
+ x^2 + (-10 + x)*Log[3])*Log[x])*Log[x + Log[3]])/(16*x^4 + 4*E^8*x^4 + 16*x^5 + 4*x^6 + E^4*(-16*x^4 - 8*x^
5) + (16*x^3 + 4*E^8*x^3 + 16*x^4 + 4*x^5 + E^4*(-16*x^3 - 8*x^4))*Log[3] + (-16*x^4 + 8*E^4*x^4 - 8*x^5 + (-1
6*x^3 + 8*E^4*x^3 - 8*x^4)*Log[3])*Log[x + Log[3]] + (4*x^4 + 4*x^3*Log[3])*Log[x + Log[3]]^2),x]

[Out]

(-5*(1 - Log[3])*Defer[Int][Log[x]/(x^2*(2*(1 - E^4/2) + x - Log[x + Log[3]])^2), x])/(4*Log[3]) + ((5 + Log[3
] - Log[3]^2)*Defer[Int][Log[x]/(x*(2*(1 - E^4/2) + x - Log[x + Log[3]])^2), x])/(4*Log[3]^2) - ((5 + Log[3])*
Defer[Int][Log[x]/((x + Log[3])*(2*(1 - E^4/2) + x - Log[x + Log[3]])^2), x])/(4*Log[3]^2) + Defer[Int][1/(x^2
*(2*(1 - E^4/2) + x - Log[x + Log[3]])), x]/4 + (5*Defer[Int][Log[x]/(x^3*(2*(1 - E^4/2) + x - Log[x + Log[3]]
)), x])/2 + (5*Defer[Int][1/(x^3*(-2*(1 - E^4/2) - x + Log[x + Log[3]])), x])/4 + Defer[Int][Log[x]/(x^2*(-2*(
1 - E^4/2) - x + Log[x + Log[3]])), x]/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 x-3 x^2+x^3+e^4 \left (5 x-x^2\right )+\left (-10+e^4 (5-x)-3 x+x^2\right ) \log (3)+\left (15 x+14 x^2-2 x^3+e^4 \left (-10 x+x^2\right )+\left (20+e^4 (-10+x)+13 x-2 x^2\right ) \log (3)\right ) \log (x)+\left (5 x-x^2+(5-x) \log (3)+\left (-10 x+x^2+(-10+x) \log (3)\right ) \log (x)\right ) \log (x+\log (3))}{\left (16+4 e^8\right ) x^4+16 x^5+4 x^6+e^4 \left (-16 x^4-8 x^5\right )+\left (16 x^3+4 e^8 x^3+16 x^4+4 x^5+e^4 \left (-16 x^3-8 x^4\right )\right ) \log (3)+\left (-16 x^4+8 e^4 x^4-8 x^5+\left (-16 x^3+8 e^4 x^3-8 x^4\right ) \log (3)\right ) \log (x+\log (3))+\left (4 x^4+4 x^3 \log (3)\right ) \log ^2(x+\log (3))} \, dx\\ &=\int \frac {(-5+x) (x+\log (3)) \left (2-e^4+x-\log (x+\log (3))\right )+\log (x) \left (-2 x^3-10 \left (-2+e^4\right ) \log (3)+x \left (15+e^4 (-10+\log (3))+13 \log (3)\right )+x^2 \left (14+e^4-\log (9)\right )+(-10+x) (x+\log (3)) \log (x+\log (3))\right )}{4 x^3 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {(-5+x) (x+\log (3)) \left (2-e^4+x-\log (x+\log (3))\right )+\log (x) \left (-2 x^3-10 \left (-2+e^4\right ) \log (3)+x \left (15+e^4 (-10+\log (3))+13 \log (3)\right )+x^2 \left (14+e^4-\log (9)\right )+(-10+x) (x+\log (3)) \log (x+\log (3))\right )}{x^3 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {(5-x) (-1+x+\log (3)) \log (x)}{x^2 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}+\frac {-5+x+10 \log (x)-x \log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {(5-x) (-1+x+\log (3)) \log (x)}{x^2 (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx+\frac {1}{4} \int \frac {-5+x+10 \log (x)-x \log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )} \, dx\\ &=\frac {1}{4} \int \left (\frac {5 (-1+\log (3)) \log (x)}{x^2 \log (3) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}+\frac {(-5-\log (3)) \log (x)}{\log ^2(3) (x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}+\frac {\left (5+\log (3)-\log ^2(3)\right ) \log (x)}{x \log ^2(3) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2}\right ) \, dx+\frac {1}{4} \int \left (\frac {1}{x^2 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )}+\frac {10 \log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )}+\frac {5}{x^3 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )}+\frac {\log (x)}{x^2 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )}\right ) \, dx\\ &=\frac {1}{4} \int \frac {1}{x^2 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )} \, dx+\frac {1}{4} \int \frac {\log (x)}{x^2 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )} \, dx+\frac {5}{4} \int \frac {1}{x^3 \left (-2 \left (1-\frac {e^4}{2}\right )-x+\log (x+\log (3))\right )} \, dx+\frac {5}{2} \int \frac {\log (x)}{x^3 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )} \, dx-\frac {(5 (1-\log (3))) \int \frac {\log (x)}{x^2 \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx}{4 \log (3)}-\frac {(5+\log (3)) \int \frac {\log (x)}{(x+\log (3)) \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx}{4 \log ^2(3)}+\frac {\left (5+\log (3)-\log ^2(3)\right ) \int \frac {\log (x)}{x \left (2 \left (1-\frac {e^4}{2}\right )+x-\log (x+\log (3))\right )^2} \, dx}{4 \log ^2(3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 29, normalized size = 1.00 \begin {gather*} \frac {(-5+x) \log (x)}{4 x^2 \left (2-e^4+x-\log (x+\log (3))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x - 3*x^2 + x^3 + E^4*(5*x - x^2) + (-10 + E^4*(5 - x) - 3*x + x^2)*Log[3] + (15*x + 14*x^2 - 2
*x^3 + E^4*(-10*x + x^2) + (20 + E^4*(-10 + x) + 13*x - 2*x^2)*Log[3])*Log[x] + (5*x - x^2 + (5 - x)*Log[3] +
(-10*x + x^2 + (-10 + x)*Log[3])*Log[x])*Log[x + Log[3]])/(16*x^4 + 4*E^8*x^4 + 16*x^5 + 4*x^6 + E^4*(-16*x^4
- 8*x^5) + (16*x^3 + 4*E^8*x^3 + 16*x^4 + 4*x^5 + E^4*(-16*x^3 - 8*x^4))*Log[3] + (-16*x^4 + 8*E^4*x^4 - 8*x^5
+ (-16*x^3 + 8*E^4*x^3 - 8*x^4)*Log[3])*Log[x + Log[3]] + (4*x^4 + 4*x^3*Log[3])*Log[x + Log[3]]^2),x]

[Out]

((-5 + x)*Log[x])/(4*x^2*(2 - E^4 + x - Log[x + Log[3]]))

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fricas [A]  time = 0.87, size = 35, normalized size = 1.21 \begin {gather*} \frac {{\left (x - 5\right )} \log \relax (x)}{4 \, {\left (x^{3} - x^{2} e^{4} - x^{2} \log \left (x + \log \relax (3)\right ) + 2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-10)*log(3)+x^2-10*x)*log(x)+(5-x)*log(3)-x^2+5*x)*log(log(3)+x)+(((x-10)*exp(4)-2*x^2+13*x+20)
*log(3)+(x^2-10*x)*exp(4)-2*x^3+14*x^2+15*x)*log(x)+((5-x)*exp(4)+x^2-3*x-10)*log(3)+(-x^2+5*x)*exp(4)+x^3-3*x
^2-10*x)/((4*x^3*log(3)+4*x^4)*log(log(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)*log(3)+8*x^4*exp(4)-8*x^5-16*x^4)*
log(log(3)+x)+(4*x^3*exp(4)^2+(-8*x^4-16*x^3)*exp(4)+4*x^5+16*x^4+16*x^3)*log(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4
)*exp(4)+4*x^6+16*x^5+16*x^4),x, algorithm="fricas")

[Out]

1/4*(x - 5)*log(x)/(x^3 - x^2*e^4 - x^2*log(x + log(3)) + 2*x^2)

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giac [A]  time = 0.84, size = 39, normalized size = 1.34 \begin {gather*} \frac {x \log \relax (x) - 5 \, \log \relax (x)}{4 \, {\left (x^{3} - x^{2} e^{4} - x^{2} \log \left (x + \log \relax (3)\right ) + 2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-10)*log(3)+x^2-10*x)*log(x)+(5-x)*log(3)-x^2+5*x)*log(log(3)+x)+(((x-10)*exp(4)-2*x^2+13*x+20)
*log(3)+(x^2-10*x)*exp(4)-2*x^3+14*x^2+15*x)*log(x)+((5-x)*exp(4)+x^2-3*x-10)*log(3)+(-x^2+5*x)*exp(4)+x^3-3*x
^2-10*x)/((4*x^3*log(3)+4*x^4)*log(log(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)*log(3)+8*x^4*exp(4)-8*x^5-16*x^4)*
log(log(3)+x)+(4*x^3*exp(4)^2+(-8*x^4-16*x^3)*exp(4)+4*x^5+16*x^4+16*x^3)*log(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4
)*exp(4)+4*x^6+16*x^5+16*x^4),x, algorithm="giac")

[Out]

1/4*(x*log(x) - 5*log(x))/(x^3 - x^2*e^4 - x^2*log(x + log(3)) + 2*x^2)

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maple [A]  time = 0.13, size = 25, normalized size = 0.86

method	result	size
risch	\(-\frac {\left (x -5\right ) \ln \relax (x )}{4 x^{2} \left ({\mathrm e}^{4}+\ln \left (\ln \relax (3)+x \right )-x -2\right )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x-10)*ln(3)+x^2-10*x)*ln(x)+(5-x)*ln(3)-x^2+5*x)*ln(ln(3)+x)+(((x-10)*exp(4)-2*x^2+13*x+20)*ln(3)+(x^2
-10*x)*exp(4)-2*x^3+14*x^2+15*x)*ln(x)+((5-x)*exp(4)+x^2-3*x-10)*ln(3)+(-x^2+5*x)*exp(4)+x^3-3*x^2-10*x)/((4*x
^3*ln(3)+4*x^4)*ln(ln(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)*ln(3)+8*x^4*exp(4)-8*x^5-16*x^4)*ln(ln(3)+x)+(4*x^3
*exp(4)^2+(-8*x^4-16*x^3)*exp(4)+4*x^5+16*x^4+16*x^3)*ln(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4)*exp(4)+4*x^6+16*x^5
+16*x^4),x,method=_RETURNVERBOSE)

[Out]

-1/4*(x-5)*ln(x)/x^2/(exp(4)+ln(ln(3)+x)-x-2)

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maxima [A]  time = 0.74, size = 32, normalized size = 1.10 \begin {gather*} \frac {{\left (x - 5\right )} \log \relax (x)}{4 \, {\left (x^{3} - x^{2} {\left (e^{4} - 2\right )} - x^{2} \log \left (x + \log \relax (3)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-10)*log(3)+x^2-10*x)*log(x)+(5-x)*log(3)-x^2+5*x)*log(log(3)+x)+(((x-10)*exp(4)-2*x^2+13*x+20)
*log(3)+(x^2-10*x)*exp(4)-2*x^3+14*x^2+15*x)*log(x)+((5-x)*exp(4)+x^2-3*x-10)*log(3)+(-x^2+5*x)*exp(4)+x^3-3*x
^2-10*x)/((4*x^3*log(3)+4*x^4)*log(log(3)+x)^2+((8*x^3*exp(4)-8*x^4-16*x^3)*log(3)+8*x^4*exp(4)-8*x^5-16*x^4)*
log(log(3)+x)+(4*x^3*exp(4)^2+(-8*x^4-16*x^3)*exp(4)+4*x^5+16*x^4+16*x^3)*log(3)+4*x^4*exp(4)^2+(-8*x^5-16*x^4
)*exp(4)+4*x^6+16*x^5+16*x^4),x, algorithm="maxima")

[Out]

1/4*(x - 5)*log(x)/(x^3 - x^2*(e^4 - 2) - x^2*log(x + log(3)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\ln \left (x+\ln \relax (3)\right )\,\left (5\,x-\ln \relax (3)\,\left (x-5\right )-x^2+\ln \relax (x)\,\left (\ln \relax (3)\,\left (x-10\right )-10\,x+x^2\right )\right )-10\,x+{\mathrm {e}}^4\,\left (5\,x-x^2\right )-\ln \relax (3)\,\left (3\,x+{\mathrm {e}}^4\,\left (x-5\right )-x^2+10\right )-3\,x^2+x^3+\ln \relax (x)\,\left (15\,x-{\mathrm {e}}^4\,\left (10\,x-x^2\right )+\ln \relax (3)\,\left (13\,x+{\mathrm {e}}^4\,\left (x-10\right )-2\,x^2+20\right )+14\,x^2-2\,x^3\right )}{4\,x^4\,{\mathrm {e}}^8-{\mathrm {e}}^4\,\left (8\,x^5+16\,x^4\right )+{\ln \left (x+\ln \relax (3)\right )}^2\,\left (4\,x^4+4\,\ln \relax (3)\,x^3\right )-\ln \left (x+\ln \relax (3)\right )\,\left (\ln \relax (3)\,\left (16\,x^3-8\,x^3\,{\mathrm {e}}^4+8\,x^4\right )-8\,x^4\,{\mathrm {e}}^4+16\,x^4+8\,x^5\right )+16\,x^4+16\,x^5+4\,x^6+\ln \relax (3)\,\left (4\,x^3\,{\mathrm {e}}^8-{\mathrm {e}}^4\,\left (8\,x^4+16\,x^3\right )+16\,x^3+16\,x^4+4\,x^5\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + log(3))*(5*x - log(3)*(x - 5) - x^2 + log(x)*(log(3)*(x - 10) - 10*x + x^2)) - 10*x + exp(4)*(5*x
- x^2) - log(3)*(3*x + exp(4)*(x - 5) - x^2 + 10) - 3*x^2 + x^3 + log(x)*(15*x - exp(4)*(10*x - x^2) + log(3)
*(13*x + exp(4)*(x - 10) - 2*x^2 + 20) + 14*x^2 - 2*x^3))/(4*x^4*exp(8) - exp(4)*(16*x^4 + 8*x^5) + log(x + lo
g(3))^2*(4*x^3*log(3) + 4*x^4) - log(x + log(3))*(log(3)*(16*x^3 - 8*x^3*exp(4) + 8*x^4) - 8*x^4*exp(4) + 16*x
^4 + 8*x^5) + 16*x^4 + 16*x^5 + 4*x^6 + log(3)*(4*x^3*exp(8) - exp(4)*(16*x^3 + 8*x^4) + 16*x^3 + 16*x^4 + 4*x
^5)),x)

[Out]

int((log(x + log(3))*(5*x - log(3)*(x - 5) - x^2 + log(x)*(log(3)*(x - 10) - 10*x + x^2)) - 10*x + exp(4)*(5*x
- x^2) - log(3)*(3*x + exp(4)*(x - 5) - x^2 + 10) - 3*x^2 + x^3 + log(x)*(15*x - exp(4)*(10*x - x^2) + log(3)
*(13*x + exp(4)*(x - 10) - 2*x^2 + 20) + 14*x^2 - 2*x^3))/(4*x^4*exp(8) - exp(4)*(16*x^4 + 8*x^5) + log(x + lo
g(3))^2*(4*x^3*log(3) + 4*x^4) - log(x + log(3))*(log(3)*(16*x^3 - 8*x^3*exp(4) + 8*x^4) - 8*x^4*exp(4) + 16*x
^4 + 8*x^5) + 16*x^4 + 16*x^5 + 4*x^6 + log(3)*(4*x^3*exp(8) - exp(4)*(16*x^3 + 8*x^4) + 16*x^3 + 16*x^4 + 4*x
^5)), x)

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sympy [A]  time = 0.44, size = 39, normalized size = 1.34 \begin {gather*} \frac {- x \log {\relax (x )} + 5 \log {\relax (x )}}{- 4 x^{3} + 4 x^{2} \log {\left (x + \log {\relax (3 )} \right )} - 8 x^{2} + 4 x^{2} e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x-10)*ln(3)+x**2-10*x)*ln(x)+(5-x)*ln(3)-x**2+5*x)*ln(ln(3)+x)+(((x-10)*exp(4)-2*x**2+13*x+20)*l
n(3)+(x**2-10*x)*exp(4)-2*x**3+14*x**2+15*x)*ln(x)+((5-x)*exp(4)+x**2-3*x-10)*ln(3)+(-x**2+5*x)*exp(4)+x**3-3*
x**2-10*x)/((4*x**3*ln(3)+4*x**4)*ln(ln(3)+x)**2+((8*x**3*exp(4)-8*x**4-16*x**3)*ln(3)+8*x**4*exp(4)-8*x**5-16
*x**4)*ln(ln(3)+x)+(4*x**3*exp(4)**2+(-8*x**4-16*x**3)*exp(4)+4*x**5+16*x**4+16*x**3)*ln(3)+4*x**4*exp(4)**2+(
-8*x**5-16*x**4)*exp(4)+4*x**6+16*x**5+16*x**4),x)

[Out]

(-x*log(x) + 5*log(x))/(-4*x**3 + 4*x**2*log(x + log(3)) - 8*x**2 + 4*x**2*exp(4))

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