3.28.25 \(\int \frac {-20 x-94 x^2-110 x^3+(-3 x^2-3 \log (2)) \log (x^2+\log (2))+(-20 x^2-100 x^3-125 x^4+(-20-100 x-125 x^2) \log (2)) \log ^2(x^2+\log (2))}{(20 x^2+100 x^3+125 x^4+(20+100 x+125 x^2) \log (2)) \log ^2(x^2+\log (2))} \, dx\)

Optimal. Leaf size=34 \[ -x+\frac {1+\frac {3}{2+\frac {x}{1+2 x}}}{5 \log \left (x^2+\log (2)\right )} \]

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Rubi [F]  time = 0.98, 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 {-20 x-94 x^2-110 x^3+\left (-3 x^2-3 \log (2)\right ) \log \left (x^2+\log (2)\right )+\left (-20 x^2-100 x^3-125 x^4+\left (-20-100 x-125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )}{\left (20 x^2+100 x^3+125 x^4+\left (20+100 x+125 x^2\right ) \log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-20*x - 94*x^2 - 110*x^3 + (-3*x^2 - 3*Log[2])*Log[x^2 + Log[2]] + (-20*x^2 - 100*x^3 - 125*x^4 + (-20 -
100*x - 125*x^2)*Log[2])*Log[x^2 + Log[2]]^2)/((20*x^2 + 100*x^3 + 125*x^4 + (20 + 100*x + 125*x^2)*Log[2])*Lo
g[x^2 + Log[2]]^2),x]

[Out]

-x + (2 + Log[2048])/((4 + 25*Log[2])*Log[x^2 + Log[2]]) + (12*Defer[Int][1/((2 + 5*x)*Log[x^2 + Log[2]]^2), x
])/(5*(4 + 25*Log[2])) - (2*Log[8]*Defer[Int][1/((x^2 + Log[2])*Log[x^2 + Log[2]]^2), x])/(5*(4 + 25*Log[2]))
- (3*Defer[Int][1/((2 + 5*x)^2*Log[x^2 + Log[2]]), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-1-\frac {2 x (5+11 x)}{5 (2+5 x) \left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )}-\frac {3}{5 (2+5 x)^2 \log \left (x^2+\log (2)\right )}\right ) \, dx\\ &=-x-\frac {2}{5} \int \frac {x (5+11 x)}{(2+5 x) \left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx\\ &=-x-\frac {2}{5} \int \left (-\frac {6}{(2+5 x) (4+25 \log (2)) \log ^2\left (x^2+\log (2)\right )}+\frac {\log (8)+5 x (2+\log (2048))}{\left (x^2+\log (2)\right ) (4+25 \log (2)) \log ^2\left (x^2+\log (2)\right )}\right ) \, dx-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx\\ &=-x-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx-\frac {2 \int \frac {\log (8)+5 x (2+\log (2048))}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}+\frac {12 \int \frac {1}{(2+5 x) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}\\ &=-x-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx-\frac {2 \int \left (\frac {\log (8)}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )}+\frac {5 x (2+\log (2048))}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )}\right ) \, dx}{5 (4+25 \log (2))}+\frac {12 \int \frac {1}{(2+5 x) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}\\ &=-x-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx+\frac {12 \int \frac {1}{(2+5 x) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2 \log (8)) \int \frac {1}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2 (2+\log (2048))) \int \frac {x}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx}{4+25 \log (2)}\\ &=-x-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx+\frac {12 \int \frac {1}{(2+5 x) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2 \log (8)) \int \frac {1}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2+\log (2048)) \operatorname {Subst}\left (\int \frac {1}{(x+\log (2)) \log ^2(x+\log (2))} \, dx,x,x^2\right )}{4+25 \log (2)}\\ &=-x-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx+\frac {12 \int \frac {1}{(2+5 x) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2 \log (8)) \int \frac {1}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2+\log (2048)) \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,x^2+\log (2)\right )}{4+25 \log (2)}\\ &=-x-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx+\frac {12 \int \frac {1}{(2+5 x) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2 \log (8)) \int \frac {1}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2+\log (2048)) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (x^2+\log (2)\right )\right )}{4+25 \log (2)}\\ &=-x+\frac {2+\log (2048)}{(4+25 \log (2)) \log \left (x^2+\log (2)\right )}-\frac {3}{5} \int \frac {1}{(2+5 x)^2 \log \left (x^2+\log (2)\right )} \, dx+\frac {12 \int \frac {1}{(2+5 x) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}-\frac {(2 \log (8)) \int \frac {1}{\left (x^2+\log (2)\right ) \log ^2\left (x^2+\log (2)\right )} \, dx}{5 (4+25 \log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 29, normalized size = 0.85 \begin {gather*} -x-\frac {-5-11 x}{5 (2+5 x) \log \left (x^2+\log (2)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20*x - 94*x^2 - 110*x^3 + (-3*x^2 - 3*Log[2])*Log[x^2 + Log[2]] + (-20*x^2 - 100*x^3 - 125*x^4 + (
-20 - 100*x - 125*x^2)*Log[2])*Log[x^2 + Log[2]]^2)/((20*x^2 + 100*x^3 + 125*x^4 + (20 + 100*x + 125*x^2)*Log[
2])*Log[x^2 + Log[2]]^2),x]

[Out]

-x - (-5 - 11*x)/(5*(2 + 5*x)*Log[x^2 + Log[2]])

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fricas [A]  time = 0.47, size = 41, normalized size = 1.21 \begin {gather*} -\frac {5 \, {\left (5 \, x^{2} + 2 \, x\right )} \log \left (x^{2} + \log \relax (2)\right ) - 11 \, x - 5}{5 \, {\left (5 \, x + 2\right )} \log \left (x^{2} + \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-125*x^2-100*x-20)*log(2)-125*x^4-100*x^3-20*x^2)*log(log(2)+x^2)^2+(-3*log(2)-3*x^2)*log(log(2)+
x^2)-110*x^3-94*x^2-20*x)/((125*x^2+100*x+20)*log(2)+125*x^4+100*x^3+20*x^2)/log(log(2)+x^2)^2,x, algorithm="f
ricas")

[Out]

-1/5*(5*(5*x^2 + 2*x)*log(x^2 + log(2)) - 11*x - 5)/((5*x + 2)*log(x^2 + log(2)))

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giac [A]  time = 0.25, size = 33, normalized size = 0.97 \begin {gather*} -x + \frac {11 \, x + 5}{5 \, {\left (5 \, x \log \left (x^{2} + \log \relax (2)\right ) + 2 \, \log \left (x^{2} + \log \relax (2)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-125*x^2-100*x-20)*log(2)-125*x^4-100*x^3-20*x^2)*log(log(2)+x^2)^2+(-3*log(2)-3*x^2)*log(log(2)+
x^2)-110*x^3-94*x^2-20*x)/((125*x^2+100*x+20)*log(2)+125*x^4+100*x^3+20*x^2)/log(log(2)+x^2)^2,x, algorithm="g
iac")

[Out]

-x + 1/5*(11*x + 5)/(5*x*log(x^2 + log(2)) + 2*log(x^2 + log(2)))

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maple [A]  time = 0.09, size = 28, normalized size = 0.82

method	result	size
risch	\(-x +\frac {11 x +5}{5 \left (5 x +2\right ) \ln \left (\ln \relax (2)+x^{2}\right )}\)	\(28\)
norman	\(\frac {1+\frac {4 \ln \left (\ln \relax (2)+x^{2}\right )}{5}+\frac {11 x}{5}-5 \ln \left (\ln \relax (2)+x^{2}\right ) x^{2}}{\left (5 x +2\right ) \ln \left (\ln \relax (2)+x^{2}\right )}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-125*x^2-100*x-20)*ln(2)-125*x^4-100*x^3-20*x^2)*ln(ln(2)+x^2)^2+(-3*ln(2)-3*x^2)*ln(ln(2)+x^2)-110*x^3
-94*x^2-20*x)/((125*x^2+100*x+20)*ln(2)+125*x^4+100*x^3+20*x^2)/ln(ln(2)+x^2)^2,x,method=_RETURNVERBOSE)

[Out]

-x+1/5*(11*x+5)/(5*x+2)/ln(ln(2)+x^2)

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maxima [A]  time = 0.75, size = 41, normalized size = 1.21 \begin {gather*} -\frac {5 \, {\left (5 \, x^{2} + 2 \, x\right )} \log \left (x^{2} + \log \relax (2)\right ) - 11 \, x - 5}{5 \, {\left (5 \, x + 2\right )} \log \left (x^{2} + \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-125*x^2-100*x-20)*log(2)-125*x^4-100*x^3-20*x^2)*log(log(2)+x^2)^2+(-3*log(2)-3*x^2)*log(log(2)+
x^2)-110*x^3-94*x^2-20*x)/((125*x^2+100*x+20)*log(2)+125*x^4+100*x^3+20*x^2)/log(log(2)+x^2)^2,x, algorithm="m
axima")

[Out]

-1/5*(5*(5*x^2 + 2*x)*log(x^2 + log(2)) - 11*x - 5)/((5*x + 2)*log(x^2 + log(2)))

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mupad [B]  time = 0.64, size = 88, normalized size = 2.59 \begin {gather*} \frac {\frac {11\,x+5}{5\,\left (5\,x+2\right )}+\frac {\ln \left (x^2+\ln \relax (2)\right )\,\left (3\,x^2+\ln \relax (8)\right )}{10\,x\,\left (25\,x^2+20\,x+4\right )}}{\ln \left (x^2+\ln \relax (2)\right )}-x-\frac {3\,x^2+3\,\ln \relax (2)}{250\,x^3+200\,x^2+40\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(20*x + log(log(2) + x^2)^2*(log(2)*(100*x + 125*x^2 + 20) + 20*x^2 + 100*x^3 + 125*x^4) + log(log(2) + x
^2)*(3*log(2) + 3*x^2) + 94*x^2 + 110*x^3)/(log(log(2) + x^2)^2*(log(2)*(100*x + 125*x^2 + 20) + 20*x^2 + 100*
x^3 + 125*x^4)),x)

[Out]

((11*x + 5)/(5*(5*x + 2)) + (log(log(2) + x^2)*(log(8) + 3*x^2))/(10*x*(20*x + 25*x^2 + 4)))/log(log(2) + x^2)
- x - (3*log(2) + 3*x^2)/(40*x + 200*x^2 + 250*x^3)

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sympy [A]  time = 0.27, size = 19, normalized size = 0.56 \begin {gather*} - x + \frac {11 x + 5}{\left (25 x + 10\right ) \log {\left (x^{2} + \log {\relax (2 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-125*x**2-100*x-20)*ln(2)-125*x**4-100*x**3-20*x**2)*ln(ln(2)+x**2)**2+(-3*ln(2)-3*x**2)*ln(ln(2)
+x**2)-110*x**3-94*x**2-20*x)/((125*x**2+100*x+20)*ln(2)+125*x**4+100*x**3+20*x**2)/ln(ln(2)+x**2)**2,x)

[Out]

-x + (11*x + 5)/((25*x + 10)*log(x**2 + log(2)))

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