3.66.35 \(\int \frac {e^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+(e^4 (18 x+60 x^2-12 x^3)+(-18 x-60 x^2+12 x^3) \log (2)) \log (x)+(e^8 (-6 x+2 x^3)+e^4 (12 x-4 x^3) \log (2)+(-6 x+2 x^3) \log ^2(2)) \log ^2(x)}{225 x-90 x^2+9 x^3+(e^4 (30 x^2-6 x^3)+(-30 x^2+6 x^3) \log (2)) \log (x)+(e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)) \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ 2 x+\frac {6}{x+\frac {3 (5-x)}{\left (e^4-\log (2)\right ) \log (x)}} \]

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Rubi [F]  time = 2.07, 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^4 (90-18 x)+450 x-180 x^2+18 x^3+(-90+18 x) \log (2)+\left (e^4 \left (18 x+60 x^2-12 x^3\right )+\left (-18 x-60 x^2+12 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 \left (-6 x+2 x^3\right )+e^4 \left (12 x-4 x^3\right ) \log (2)+\left (-6 x+2 x^3\right ) \log ^2(2)\right ) \log ^2(x)}{225 x-90 x^2+9 x^3+\left (e^4 \left (30 x^2-6 x^3\right )+\left (-30 x^2+6 x^3\right ) \log (2)\right ) \log (x)+\left (e^8 x^3-2 e^4 x^3 \log (2)+x^3 \log ^2(2)\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^4*(90 - 18*x) + 450*x - 180*x^2 + 18*x^3 + (-90 + 18*x)*Log[2] + (E^4*(18*x + 60*x^2 - 12*x^3) + (-18*x
- 60*x^2 + 12*x^3)*Log[2])*Log[x] + (E^8*(-6*x + 2*x^3) + E^4*(12*x - 4*x^3)*Log[2] + (-6*x + 2*x^3)*Log[2]^2
)*Log[x]^2)/(225*x - 90*x^2 + 9*x^3 + (E^4*(30*x^2 - 6*x^3) + (-30*x^2 + 6*x^3)*Log[2])*Log[x] + (E^8*x^3 - 2*
E^4*x^3*Log[2] + x^3*Log[2]^2)*Log[x]^2),x]

[Out]

6/x + 2*x + 18*Defer[Int][1/(x*(-15 + 3*x - E^4*x*(1 - Log[2]/E^4)*Log[x])), x] - 18*(E^4 - Log[2])*Defer[Int]
[(15 - 3*x + E^4*x*(1 - Log[2]/E^4)*Log[x])^(-2), x] - 1350*Defer[Int][1/(x^2*(15 - 3*x + E^4*x*(1 - Log[2]/E^
4)*Log[x])^2), x] + 90*(3 + E^4 - Log[2])*Defer[Int][1/(x*(15 - 3*x + E^4*x*(1 - Log[2]/E^4)*Log[x])^2), x] +
180*Defer[Int][1/(x^2*(15 - 3*x + E^4*x*(1 - Log[2]/E^4)*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (9 (-5+x) \left (-e^4-5 x+x^2+\log (2)\right )-3 x \left (-3-10 x+2 x^2\right ) \left (e^4-\log (2)\right ) \log (x)+x \left (-3+x^2\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)\right )}{x \left (3 (-5+x)-x \left (e^4-\log (2)\right ) \log (x)\right )^2} \, dx\\ &=2 \int \frac {9 (-5+x) \left (-e^4-5 x+x^2+\log (2)\right )-3 x \left (-3-10 x+2 x^2\right ) \left (e^4-\log (2)\right ) \log (x)+x \left (-3+x^2\right ) \left (e^4-\log (2)\right )^2 \log ^2(x)}{x \left (3 (-5+x)-x \left (e^4-\log (2)\right ) \log (x)\right )^2} \, dx\\ &=2 \int \left (\frac {-3+x^2}{x^2}+\frac {9 (5-x) \left (-15+x \left (e^4-\log (2)\right )\right )}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}+\frac {9 (10-x)}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}\right ) \, dx\\ &=2 \int \frac {-3+x^2}{x^2} \, dx+18 \int \frac {(5-x) \left (-15+x \left (e^4-\log (2)\right )\right )}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx+18 \int \frac {10-x}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx\\ &=2 \int \left (1-\frac {3}{x^2}\right ) \, dx+18 \int \left (-\frac {75}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}+\frac {5 \left (3+e^4-\log (2)\right )}{x \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}-\frac {e^4 \left (1-\frac {\log (2)}{e^4}\right )}{\left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2}\right ) \, dx+18 \int \left (\frac {1}{x \left (-15+3 x-e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}+\frac {10}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )}\right ) \, dx\\ &=\frac {6}{x}+2 x+18 \int \frac {1}{x \left (-15+3 x-e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx+180 \int \frac {1}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )} \, dx-1350 \int \frac {1}{x^2 \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx-\left (18 \left (e^4-\log (2)\right )\right ) \int \frac {1}{\left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx+\left (90 \left (3+e^4-\log (2)\right )\right ) \int \frac {1}{x \left (15-3 x+e^4 x \left (1-\frac {\log (2)}{e^4}\right ) \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 38, normalized size = 1.23 \begin {gather*} 2 \left (\frac {3}{x}+x+\frac {9 (-5+x)}{x \left (15-3 x+e^4 x \log (x)-x \log (2) \log (x)\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^4*(90 - 18*x) + 450*x - 180*x^2 + 18*x^3 + (-90 + 18*x)*Log[2] + (E^4*(18*x + 60*x^2 - 12*x^3) +
(-18*x - 60*x^2 + 12*x^3)*Log[2])*Log[x] + (E^8*(-6*x + 2*x^3) + E^4*(12*x - 4*x^3)*Log[2] + (-6*x + 2*x^3)*Lo
g[2]^2)*Log[x]^2)/(225*x - 90*x^2 + 9*x^3 + (E^4*(30*x^2 - 6*x^3) + (-30*x^2 + 6*x^3)*Log[2])*Log[x] + (E^8*x^
3 - 2*E^4*x^3*Log[2] + x^3*Log[2]^2)*Log[x]^2),x]

[Out]

2*(3/x + x + (9*(-5 + x))/(x*(15 - 3*x + E^4*x*Log[x] - x*Log[2]*Log[x])))

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fricas [A]  time = 0.76, size = 53, normalized size = 1.71 \begin {gather*} -\frac {2 \, {\left (3 \, x^{2} - {\left ({\left (x^{2} + 3\right )} e^{4} - {\left (x^{2} + 3\right )} \log \relax (2)\right )} \log \relax (x) - 15 \, x\right )}}{{\left (x e^{4} - x \log \relax (2)\right )} \log \relax (x) - 3 \, x + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-6*x)*log(2)^2+(-4*x^3+12*x)*exp(4)*log(2)+(2*x^3-6*x)*exp(4)^2)*log(x)^2+((12*x^3-60*x^2-18
*x)*log(2)+(-12*x^3+60*x^2+18*x)*exp(4))*log(x)+(18*x-90)*log(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3
*log(2)^2-2*x^3*exp(4)*log(2)+x^3*exp(4)^2)*log(x)^2+((6*x^3-30*x^2)*log(2)+(-6*x^3+30*x^2)*exp(4))*log(x)+9*x
^3-90*x^2+225*x),x, algorithm="fricas")

[Out]

-2*(3*x^2 - ((x^2 + 3)*e^4 - (x^2 + 3)*log(2))*log(x) - 15*x)/((x*e^4 - x*log(2))*log(x) - 3*x + 15)

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giac [B]  time = 0.25, size = 60, normalized size = 1.94 \begin {gather*} \frac {2 \, {\left (x^{2} e^{4} \log \relax (x) - x^{2} \log \relax (2) \log \relax (x) - 3 \, x^{2} + 3 \, e^{4} \log \relax (x) - 3 \, \log \relax (2) \log \relax (x) + 15 \, x\right )}}{x e^{4} \log \relax (x) - x \log \relax (2) \log \relax (x) - 3 \, x + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-6*x)*log(2)^2+(-4*x^3+12*x)*exp(4)*log(2)+(2*x^3-6*x)*exp(4)^2)*log(x)^2+((12*x^3-60*x^2-18
*x)*log(2)+(-12*x^3+60*x^2+18*x)*exp(4))*log(x)+(18*x-90)*log(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3
*log(2)^2-2*x^3*exp(4)*log(2)+x^3*exp(4)^2)*log(x)^2+((6*x^3-30*x^2)*log(2)+(-6*x^3+30*x^2)*exp(4))*log(x)+9*x
^3-90*x^2+225*x),x, algorithm="giac")

[Out]

2*(x^2*e^4*log(x) - x^2*log(2)*log(x) - 3*x^2 + 3*e^4*log(x) - 3*log(2)*log(x) + 15*x)/(x*e^4*log(x) - x*log(2
)*log(x) - 3*x + 15)

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maple [A]  time = 0.24, size = 40, normalized size = 1.29

method	result	size
risch	\(\frac {2 x^{2}+6}{x}+\frac {18 x -90}{x \left (x \,{\mathrm e}^{4} \ln \relax (x )-x \ln \relax (2) \ln \relax (x )-3 x +15\right )}\)	\(40\)
norman	\(\frac {\left (6 \,{\mathrm e}^{4}-6 \ln \relax (2)\right ) \ln \relax (x )+\left (2 \,{\mathrm e}^{4}-2 \ln \relax (2)\right ) x^{2} \ln \relax (x )+\left (10 \,{\mathrm e}^{4}-10 \ln \relax (2)\right ) x \ln \relax (x )-6 x^{2}+150}{x \,{\mathrm e}^{4} \ln \relax (x )-x \ln \relax (2) \ln \relax (x )-3 x +15}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3-6*x)*ln(2)^2+(-4*x^3+12*x)*exp(4)*ln(2)+(2*x^3-6*x)*exp(4)^2)*ln(x)^2+((12*x^3-60*x^2-18*x)*ln(2)
+(-12*x^3+60*x^2+18*x)*exp(4))*ln(x)+(18*x-90)*ln(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3*ln(2)^2-2*x
^3*exp(4)*ln(2)+x^3*exp(4)^2)*ln(x)^2+((6*x^3-30*x^2)*ln(2)+(-6*x^3+30*x^2)*exp(4))*ln(x)+9*x^3-90*x^2+225*x),
x,method=_RETURNVERBOSE)

[Out]

2*(x^2+3)/x+18*(x-5)/x/(x*exp(4)*ln(x)-x*ln(2)*ln(x)-3*x+15)

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maxima [A]  time = 1.15, size = 53, normalized size = 1.71 \begin {gather*} -\frac {2 \, {\left (3 \, x^{2} - {\left (x^{2} {\left (e^{4} - \log \relax (2)\right )} + 3 \, e^{4} - 3 \, \log \relax (2)\right )} \log \relax (x) - 15 \, x\right )}}{x {\left (e^{4} - \log \relax (2)\right )} \log \relax (x) - 3 \, x + 15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-6*x)*log(2)^2+(-4*x^3+12*x)*exp(4)*log(2)+(2*x^3-6*x)*exp(4)^2)*log(x)^2+((12*x^3-60*x^2-18
*x)*log(2)+(-12*x^3+60*x^2+18*x)*exp(4))*log(x)+(18*x-90)*log(2)+(-18*x+90)*exp(4)+18*x^3-180*x^2+450*x)/((x^3
*log(2)^2-2*x^3*exp(4)*log(2)+x^3*exp(4)^2)*log(x)^2+((6*x^3-30*x^2)*log(2)+(-6*x^3+30*x^2)*exp(4))*log(x)+9*x
^3-90*x^2+225*x),x, algorithm="maxima")

[Out]

-2*(3*x^2 - (x^2*(e^4 - log(2)) + 3*e^4 - 3*log(2))*log(x) - 15*x)/(x*(e^4 - log(2))*log(x) - 3*x + 15)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {450\,x+\ln \relax (2)\,\left (18\,x-90\right )-{\ln \relax (x)}^2\,\left ({\mathrm {e}}^8\,\left (6\,x-2\,x^3\right )+{\ln \relax (2)}^2\,\left (6\,x-2\,x^3\right )-{\mathrm {e}}^4\,\ln \relax (2)\,\left (12\,x-4\,x^3\right )\right )+\ln \relax (x)\,\left ({\mathrm {e}}^4\,\left (-12\,x^3+60\,x^2+18\,x\right )-\ln \relax (2)\,\left (-12\,x^3+60\,x^2+18\,x\right )\right )-180\,x^2+18\,x^3-{\mathrm {e}}^4\,\left (18\,x-90\right )}{225\,x+{\ln \relax (x)}^2\,\left (x^3\,{\ln \relax (2)}^2+x^3\,{\mathrm {e}}^8-2\,x^3\,{\mathrm {e}}^4\,\ln \relax (2)\right )+\ln \relax (x)\,\left ({\mathrm {e}}^4\,\left (30\,x^2-6\,x^3\right )-\ln \relax (2)\,\left (30\,x^2-6\,x^3\right )\right )-90\,x^2+9\,x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((450*x + log(2)*(18*x - 90) - log(x)^2*(exp(8)*(6*x - 2*x^3) + log(2)^2*(6*x - 2*x^3) - exp(4)*log(2)*(12*
x - 4*x^3)) + log(x)*(exp(4)*(18*x + 60*x^2 - 12*x^3) - log(2)*(18*x + 60*x^2 - 12*x^3)) - 180*x^2 + 18*x^3 -
exp(4)*(18*x - 90))/(225*x + log(x)^2*(x^3*log(2)^2 + x^3*exp(8) - 2*x^3*exp(4)*log(2)) + log(x)*(exp(4)*(30*x
^2 - 6*x^3) - log(2)*(30*x^2 - 6*x^3)) - 90*x^2 + 9*x^3),x)

[Out]

int((450*x + log(2)*(18*x - 90) - log(x)^2*(exp(8)*(6*x - 2*x^3) + log(2)^2*(6*x - 2*x^3) - exp(4)*log(2)*(12*
x - 4*x^3)) + log(x)*(exp(4)*(18*x + 60*x^2 - 12*x^3) - log(2)*(18*x + 60*x^2 - 12*x^3)) - 180*x^2 + 18*x^3 -
exp(4)*(18*x - 90))/(225*x + log(x)^2*(x^3*log(2)^2 + x^3*exp(8) - 2*x^3*exp(4)*log(2)) + log(x)*(exp(4)*(30*x
^2 - 6*x^3) - log(2)*(30*x^2 - 6*x^3)) - 90*x^2 + 9*x^3), x)

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sympy [A]  time = 0.36, size = 36, normalized size = 1.16 \begin {gather*} 2 x + \frac {18 x - 90}{- 3 x^{2} + 15 x + \left (- x^{2} \log {\relax (2 )} + x^{2} e^{4}\right ) \log {\relax (x )}} + \frac {6}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3-6*x)*ln(2)**2+(-4*x**3+12*x)*exp(4)*ln(2)+(2*x**3-6*x)*exp(4)**2)*ln(x)**2+((12*x**3-60*x*
*2-18*x)*ln(2)+(-12*x**3+60*x**2+18*x)*exp(4))*ln(x)+(18*x-90)*ln(2)+(-18*x+90)*exp(4)+18*x**3-180*x**2+450*x)
/((x**3*ln(2)**2-2*x**3*exp(4)*ln(2)+x**3*exp(4)**2)*ln(x)**2+((6*x**3-30*x**2)*ln(2)+(-6*x**3+30*x**2)*exp(4)
)*ln(x)+9*x**3-90*x**2+225*x),x)

[Out]

2*x + (18*x - 90)/(-3*x**2 + 15*x + (-x**2*log(2) + x**2*exp(4))*log(x)) + 6/x

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