3.48.97 \(\int \frac {3065 x^2+(-200-550 x) \log (4)+25 \log ^2(4)+(40 x^2+1100 x^3+(-80 x-210 x^2) \log (4)+10 x \log ^2(4)) \log (x)+(100 x^4+(-8 x^2-20 x^3) \log (4)+x^2 \log ^2(4)) \log ^2(x)}{3025 x^2-550 x \log (4)+25 \log ^2(4)+(1100 x^3-210 x^2 \log (4)+10 x \log ^2(4)) \log (x)+(100 x^4-20 x^3 \log (4)+x^2 \log ^2(4)) \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ x+\frac {x}{x+\frac {1}{8} \left (2 x-\log (4)+\frac {5 x}{5+x \log (x)}\right )} \]

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Rubi [F]  time = 2.13, 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 {3065 x^2+(-200-550 x) \log (4)+25 \log ^2(4)+\left (40 x^2+1100 x^3+\left (-80 x-210 x^2\right ) \log (4)+10 x \log ^2(4)\right ) \log (x)+\left (100 x^4+\left (-8 x^2-20 x^3\right ) \log (4)+x^2 \log ^2(4)\right ) \log ^2(x)}{3025 x^2-550 x \log (4)+25 \log ^2(4)+\left (1100 x^3-210 x^2 \log (4)+10 x \log ^2(4)\right ) \log (x)+\left (100 x^4-20 x^3 \log (4)+x^2 \log ^2(4)\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(3065*x^2 + (-200 - 550*x)*Log[4] + 25*Log[4]^2 + (40*x^2 + 1100*x^3 + (-80*x - 210*x^2)*Log[4] + 10*x*Log
[4]^2)*Log[x] + (100*x^4 + (-8*x^2 - 20*x^3)*Log[4] + x^2*Log[4]^2)*Log[x]^2)/(3025*x^2 - 550*x*Log[4] + 25*Lo
g[4]^2 + (1100*x^3 - 210*x^2*Log[4] + 10*x*Log[4]^2)*Log[x] + (100*x^4 - 20*x^3*Log[4] + x^2*Log[4]^2)*Log[x]^
2),x]

[Out]

x + (4*Log[4])/(5*(10*x - Log[4])) - 220*Defer[Int][x/(55*x - 5*Log[4] + 10*x^2*Log[x] - x*Log[4]*Log[x])^2, x
] + 40*Defer[Int][x^2/(55*x - 5*Log[4] + 10*x^2*Log[x] - x*Log[4]*Log[x])^2, x] - 6*Log[4]^2*Defer[Int][1/((10
*x - Log[4])*(55*x - 5*Log[4] + 10*x^2*Log[x] - x*Log[4]*Log[x])^2), x] + 4*Defer[Int][(55*x - 5*Log[4] + 10*x
^2*Log[x] - x*Log[4]*Log[x])^(-1), x] + 4*Log[64]*Defer[Int][1/((10*x - Log[4])*(55*x - 5*Log[4] + 10*x^2*Log[
x] - x*Log[4]*Log[x])), x] - 4*Log[4]*Defer[Int][(-55*x + 5*Log[4] - 10*x^2*Log[x] + x*Log[4]*Log[x])^(-2), x]
- 2*Log[4]^3*Defer[Int][1/((-10*x + Log[4])^2*(-55*x + 5*Log[4] - 10*x^2*Log[x] + x*Log[4]*Log[x])^2), x] - 8
*Log[4]^2*Defer[Int][1/((-10*x + Log[4])^2*(-55*x + 5*Log[4] - 10*x^2*Log[x] + x*Log[4]*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 \left (613 x^2-110 x \log (4)+5 (-8+\log (4)) \log (4)\right )+10 x \left (110 x^2+x (4-21 \log (4))+(-8+\log (4)) \log (4)\right ) \log (x)+x^2 \left (100 x^2-20 x \log (4)+(-8+\log (4)) \log (4)\right ) \log ^2(x)}{(55 x-5 \log (4)+x (10 x-\log (4)) \log (x))^2} \, dx\\ &=\int \left (\frac {100 x^2-20 x \log (4)-(8-\log (4)) \log (4)}{(10 x-\log (4))^2}+\frac {40 x \left (100 x^3-5 \log ^2(4)+x \log (4) (100+\log (4))-10 x^2 (55+\log (16))\right )}{(10 x-\log (4))^2 \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2}+\frac {40 x (10 x+\log (4))}{(10 x-\log (4))^2 \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )}\right ) \, dx\\ &=40 \int \frac {x \left (100 x^3-5 \log ^2(4)+x \log (4) (100+\log (4))-10 x^2 (55+\log (16))\right )}{(10 x-\log (4))^2 \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2} \, dx+40 \int \frac {x (10 x+\log (4))}{(10 x-\log (4))^2 \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )} \, dx+\int \frac {100 x^2-20 x \log (4)-(8-\log (4)) \log (4)}{(10 x-\log (4))^2} \, dx\\ &=40 \int \left (-\frac {11 x}{2 \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2}+\frac {x^2}{\left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2}-\frac {3 \log ^2(4)}{20 (10 x-\log (4)) \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2}-\frac {\log (4)}{10 \left (-55 x+5 \log (4)-10 x^2 \log (x)+x \log (4) \log (x)\right )^2}-\frac {\log ^3(4)}{20 (-10 x+\log (4))^2 \left (-55 x+5 \log (4)-10 x^2 \log (x)+x \log (4) \log (x)\right )^2}\right ) \, dx+40 \int \left (\frac {1}{10 \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )}+\frac {\log (64)}{10 (10 x-\log (4)) \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )}-\frac {\log ^2(4)}{5 (-10 x+\log (4))^2 \left (-55 x+5 \log (4)-10 x^2 \log (x)+x \log (4) \log (x)\right )}\right ) \, dx+\int \left (1-\frac {8 \log (4)}{(-10 x+\log (4))^2}\right ) \, dx\\ &=x+\frac {4 \log (4)}{5 (10 x-\log (4))}+4 \int \frac {1}{55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)} \, dx+40 \int \frac {x^2}{\left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2} \, dx-220 \int \frac {x}{\left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2} \, dx-(4 \log (4)) \int \frac {1}{\left (-55 x+5 \log (4)-10 x^2 \log (x)+x \log (4) \log (x)\right )^2} \, dx-\left (6 \log ^2(4)\right ) \int \frac {1}{(10 x-\log (4)) \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )^2} \, dx-\left (8 \log ^2(4)\right ) \int \frac {1}{(-10 x+\log (4))^2 \left (-55 x+5 \log (4)-10 x^2 \log (x)+x \log (4) \log (x)\right )} \, dx-\left (2 \log ^3(4)\right ) \int \frac {1}{(-10 x+\log (4))^2 \left (-55 x+5 \log (4)-10 x^2 \log (x)+x \log (4) \log (x)\right )^2} \, dx+(4 \log (64)) \int \frac {1}{(10 x-\log (4)) \left (55 x-5 \log (4)+10 x^2 \log (x)-x \log (4) \log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 52, normalized size = 1.68 \begin {gather*} x+\frac {\log (256)}{50 x-5 \log (4)}-\frac {40 x^2}{(10 x-\log (4)) (55 x-5 \log (4)+x (10 x-\log (4)) \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3065*x^2 + (-200 - 550*x)*Log[4] + 25*Log[4]^2 + (40*x^2 + 1100*x^3 + (-80*x - 210*x^2)*Log[4] + 10
*x*Log[4]^2)*Log[x] + (100*x^4 + (-8*x^2 - 20*x^3)*Log[4] + x^2*Log[4]^2)*Log[x]^2)/(3025*x^2 - 550*x*Log[4] +
25*Log[4]^2 + (1100*x^3 - 210*x^2*Log[4] + 10*x*Log[4]^2)*Log[x] + (100*x^4 - 20*x^3*Log[4] + x^2*Log[4]^2)*L
og[x]^2),x]

[Out]

x + Log[256]/(50*x - 5*Log[4]) - (40*x^2)/((10*x - Log[4])*(55*x - 5*Log[4] + x*(10*x - Log[4])*Log[x]))

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fricas [B]  time = 0.63, size = 68, normalized size = 2.19 \begin {gather*} \frac {275 \, x^{2} - 10 \, {\left (5 \, x - 4\right )} \log \relax (2) + 2 \, {\left (25 \, x^{3} - {\left (5 \, x^{2} - 4 \, x\right )} \log \relax (2)\right )} \log \relax (x) - 20 \, x}{5 \, {\left (2 \, {\left (5 \, x^{2} - x \log \relax (2)\right )} \log \relax (x) + 55 \, x - 10 \, \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2*log(2)^2+2*(-20*x^3-8*x^2)*log(2)+100*x^4)*log(x)^2+(40*x*log(2)^2+2*(-210*x^2-80*x)*log(2)+
1100*x^3+40*x^2)*log(x)+100*log(2)^2+2*(-550*x-200)*log(2)+3065*x^2)/((4*x^2*log(2)^2-40*x^3*log(2)+100*x^4)*l
og(x)^2+(40*x*log(2)^2-420*x^2*log(2)+1100*x^3)*log(x)+100*log(2)^2-1100*x*log(2)+3025*x^2),x, algorithm="fric
as")

[Out]

1/5*(275*x^2 - 10*(5*x - 4)*log(2) + 2*(25*x^3 - (5*x^2 - 4*x)*log(2))*log(x) - 20*x)/(2*(5*x^2 - x*log(2))*lo
g(x) + 55*x - 10*log(2))

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giac [B]  time = 0.28, size = 65, normalized size = 2.10 \begin {gather*} x - \frac {20 \, x^{2}}{50 \, x^{3} \log \relax (x) - 20 \, x^{2} \log \relax (2) \log \relax (x) + 2 \, x \log \relax (2)^{2} \log \relax (x) + 275 \, x^{2} - 105 \, x \log \relax (2) + 10 \, \log \relax (2)^{2}} + \frac {4 \, \log \relax (2)}{5 \, {\left (5 \, x - \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2*log(2)^2+2*(-20*x^3-8*x^2)*log(2)+100*x^4)*log(x)^2+(40*x*log(2)^2+2*(-210*x^2-80*x)*log(2)+
1100*x^3+40*x^2)*log(x)+100*log(2)^2+2*(-550*x-200)*log(2)+3065*x^2)/((4*x^2*log(2)^2-40*x^3*log(2)+100*x^4)*l
og(x)^2+(40*x*log(2)^2-420*x^2*log(2)+1100*x^3)*log(x)+100*log(2)^2-1100*x*log(2)+3025*x^2),x, algorithm="giac
")

[Out]

x - 20*x^2/(50*x^3*log(x) - 20*x^2*log(2)*log(x) + 2*x*log(2)^2*log(x) + 275*x^2 - 105*x*log(2) + 10*log(2)^2)
+ 4/5*log(2)/(5*x - log(2))

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maple [B]  time = 0.18, size = 64, normalized size = 2.06

method	result	size
risch	\(\frac {5 x \ln \relax (2)-25 x^{2}-4 \ln \relax (2)}{5 \ln \relax (2)-25 x}-\frac {20 x^{2}}{\left (\ln \relax (2)-5 x \right ) \left (2 x \ln \relax (2) \ln \relax (x )-10 x^{2} \ln \relax (x )+10 \ln \relax (2)-55 x \right )}\)	\(64\)
norman	\(\frac {\left (4-\ln \relax (2)\right ) x +\left (\frac {2 \ln \relax (2)^{2}}{5}-\frac {8 \ln \relax (2)}{5}\right ) x \ln \relax (x )-55 x^{2}-10 x^{3} \ln \relax (x )+2 \ln \relax (2)^{2}-8 \ln \relax (2)}{2 x \ln \relax (2) \ln \relax (x )-10 x^{2} \ln \relax (x )+10 \ln \relax (2)-55 x}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2*ln(2)^2+2*(-20*x^3-8*x^2)*ln(2)+100*x^4)*ln(x)^2+(40*x*ln(2)^2+2*(-210*x^2-80*x)*ln(2)+1100*x^3+40
*x^2)*ln(x)+100*ln(2)^2+2*(-550*x-200)*ln(2)+3065*x^2)/((4*x^2*ln(2)^2-40*x^3*ln(2)+100*x^4)*ln(x)^2+(40*x*ln(
2)^2-420*x^2*ln(2)+1100*x^3)*ln(x)+100*ln(2)^2-1100*x*ln(2)+3025*x^2),x,method=_RETURNVERBOSE)

[Out]

1/5*(5*x*ln(2)-25*x^2-4*ln(2))/(ln(2)-5*x)-20*x^2/(ln(2)-5*x)/(2*x*ln(2)*ln(x)-10*x^2*ln(x)+10*ln(2)-55*x)

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maxima [B]  time = 0.49, size = 68, normalized size = 2.19 \begin {gather*} \frac {275 \, x^{2} - 10 \, x {\left (5 \, \log \relax (2) + 2\right )} + 2 \, {\left (25 \, x^{3} - 5 \, x^{2} \log \relax (2) + 4 \, x \log \relax (2)\right )} \log \relax (x) + 40 \, \log \relax (2)}{5 \, {\left (2 \, {\left (5 \, x^{2} - x \log \relax (2)\right )} \log \relax (x) + 55 \, x - 10 \, \log \relax (2)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2*log(2)^2+2*(-20*x^3-8*x^2)*log(2)+100*x^4)*log(x)^2+(40*x*log(2)^2+2*(-210*x^2-80*x)*log(2)+
1100*x^3+40*x^2)*log(x)+100*log(2)^2+2*(-550*x-200)*log(2)+3065*x^2)/((4*x^2*log(2)^2-40*x^3*log(2)+100*x^4)*l
og(x)^2+(40*x*log(2)^2-420*x^2*log(2)+1100*x^3)*log(x)+100*log(2)^2-1100*x*log(2)+3025*x^2),x, algorithm="maxi
ma")

[Out]

1/5*(275*x^2 - 10*x*(5*log(2) + 2) + 2*(25*x^3 - 5*x^2*log(2) + 4*x*log(2))*log(x) + 40*log(2))/(2*(5*x^2 - x*
log(2))*log(x) + 55*x - 10*log(2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {100\,{\ln \relax (2)}^2-2\,\ln \relax (2)\,\left (550\,x+200\right )+{\ln \relax (x)}^2\,\left (4\,x^2\,{\ln \relax (2)}^2-2\,\ln \relax (2)\,\left (20\,x^3+8\,x^2\right )+100\,x^4\right )+3065\,x^2+\ln \relax (x)\,\left (40\,x\,{\ln \relax (2)}^2-2\,\ln \relax (2)\,\left (210\,x^2+80\,x\right )+40\,x^2+1100\,x^3\right )}{\ln \relax (x)\,\left (1100\,x^3-420\,\ln \relax (2)\,x^2+40\,{\ln \relax (2)}^2\,x\right )-1100\,x\,\ln \relax (2)+100\,{\ln \relax (2)}^2+3025\,x^2+{\ln \relax (x)}^2\,\left (100\,x^4-40\,\ln \relax (2)\,x^3+4\,{\ln \relax (2)}^2\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*log(2)^2 - 2*log(2)*(550*x + 200) + log(x)^2*(4*x^2*log(2)^2 - 2*log(2)*(8*x^2 + 20*x^3) + 100*x^4) +
3065*x^2 + log(x)*(40*x*log(2)^2 - 2*log(2)*(80*x + 210*x^2) + 40*x^2 + 1100*x^3))/(log(x)*(40*x*log(2)^2 - 4
20*x^2*log(2) + 1100*x^3) - 1100*x*log(2) + 100*log(2)^2 + 3025*x^2 + log(x)^2*(4*x^2*log(2)^2 - 40*x^3*log(2)
+ 100*x^4)),x)

[Out]

int((100*log(2)^2 - 2*log(2)*(550*x + 200) + log(x)^2*(4*x^2*log(2)^2 - 2*log(2)*(8*x^2 + 20*x^3) + 100*x^4) +
3065*x^2 + log(x)*(40*x*log(2)^2 - 2*log(2)*(80*x + 210*x^2) + 40*x^2 + 1100*x^3))/(log(x)*(40*x*log(2)^2 - 4
20*x^2*log(2) + 1100*x^3) - 1100*x*log(2) + 100*log(2)^2 + 3025*x^2 + log(x)^2*(4*x^2*log(2)^2 - 40*x^3*log(2)
+ 100*x^4)), x)

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sympy [B]  time = 0.39, size = 63, normalized size = 2.03 \begin {gather*} - \frac {20 x^{2}}{275 x^{2} - 105 x \log {\relax (2 )} + \left (50 x^{3} - 20 x^{2} \log {\relax (2 )} + 2 x \log {\relax (2 )}^{2}\right ) \log {\relax (x )} + 10 \log {\relax (2 )}^{2}} + x + \frac {4 \log {\relax (2 )}}{25 x - 5 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2*ln(2)**2+2*(-20*x**3-8*x**2)*ln(2)+100*x**4)*ln(x)**2+(40*x*ln(2)**2+2*(-210*x**2-80*x)*ln(
2)+1100*x**3+40*x**2)*ln(x)+100*ln(2)**2+2*(-550*x-200)*ln(2)+3065*x**2)/((4*x**2*ln(2)**2-40*x**3*ln(2)+100*x
**4)*ln(x)**2+(40*x*ln(2)**2-420*x**2*ln(2)+1100*x**3)*ln(x)+100*ln(2)**2-1100*x*ln(2)+3025*x**2),x)

[Out]

-20*x**2/(275*x**2 - 105*x*log(2) + (50*x**3 - 20*x**2*log(2) + 2*x*log(2)**2)*log(x) + 10*log(2)**2) + x + 4*
log(2)/(25*x - 5*log(2))

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