3.84.68 \(\int \frac {2-2 x+15 x^2+15 x^3+50 x^5+(-20 x^2-50 x^4) \log (2 x+5 x^3)+(-4 x^3-10 x^5+(4 x^2+10 x^4) \log (2 x+5 x^3)) \log (x-\log (2 x+5 x^3))}{10 x^2+25 x^4+(-10 x-25 x^3) \log (2 x+5 x^3)+(-2 x^2-5 x^4+(2 x+5 x^3) \log (2 x+5 x^3)) \log (x-\log (2 x+5 x^3))} \, dx\)

Optimal. Leaf size=24 \[ x^2+\log \left (-5+\log \left (x-\log \left (x+x \left (1+5 x^2\right )\right )\right )\right ) \]

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Rubi [F]  time = 7.48, 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 {2-2 x+15 x^2+15 x^3+50 x^5+\left (-20 x^2-50 x^4\right ) \log \left (2 x+5 x^3\right )+\left (-4 x^3-10 x^5+\left (4 x^2+10 x^4\right ) \log \left (2 x+5 x^3\right )\right ) \log \left (x-\log \left (2 x+5 x^3\right )\right )}{10 x^2+25 x^4+\left (-10 x-25 x^3\right ) \log \left (2 x+5 x^3\right )+\left (-2 x^2-5 x^4+\left (2 x+5 x^3\right ) \log \left (2 x+5 x^3\right )\right ) \log \left (x-\log \left (2 x+5 x^3\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 - 2*x + 15*x^2 + 15*x^3 + 50*x^5 + (-20*x^2 - 50*x^4)*Log[2*x + 5*x^3] + (-4*x^3 - 10*x^5 + (4*x^2 + 10
*x^4)*Log[2*x + 5*x^3])*Log[x - Log[2*x + 5*x^3]])/(10*x^2 + 25*x^4 + (-10*x - 25*x^3)*Log[2*x + 5*x^3] + (-2*
x^2 - 5*x^4 + (2*x + 5*x^3)*Log[2*x + 5*x^3])*Log[x - Log[2*x + 5*x^3]]),x]

[Out]

x^2 + 10*Defer[Int][x/(-5 + Log[x - Log[x*(2 + 5*x^2)]]), x] + Defer[Int][1/((x - Log[x*(2 + 5*x^2)])*(-5 + Lo
g[x - Log[x*(2 + 5*x^2)]])), x] - Defer[Int][1/(x*(x - Log[x*(2 + 5*x^2)])*(-5 + Log[x - Log[x*(2 + 5*x^2)]]))
, x] - 10*Defer[Int][x^2/((x - Log[x*(2 + 5*x^2)])*(-5 + Log[x - Log[x*(2 + 5*x^2)]])), x] + Sqrt[5]*Defer[Int
][1/((I*Sqrt[2] - Sqrt[5]*x)*(x - Log[x*(2 + 5*x^2)])*(-5 + Log[x - Log[x*(2 + 5*x^2)]])), x] - Sqrt[5]*Defer[
Int][1/((I*Sqrt[2] + Sqrt[5]*x)*(x - Log[x*(2 + 5*x^2)])*(-5 + Log[x - Log[x*(2 + 5*x^2)]])), x] + 10*Defer[In
t][(x*Log[x*(2 + 5*x^2)])/((x - Log[x*(2 + 5*x^2)])*(-5 + Log[x - Log[x*(2 + 5*x^2)]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2-2 x+15 x^2+15 x^3+50 x^5+\left (-20 x^2-50 x^4\right ) \log \left (2 x+5 x^3\right )+\left (-4 x^3-10 x^5+\left (4 x^2+10 x^4\right ) \log \left (2 x+5 x^3\right )\right ) \log \left (x-\log \left (2 x+5 x^3\right )\right )}{x \left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (5-\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )} \, dx\\ &=\int \left (\frac {2}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )}-\frac {2}{x \left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )}-\frac {15 x}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )}-\frac {15 x^2}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )}-\frac {50 x^4}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )}+\frac {10 x \log \left (x \left (2+5 x^2\right )\right )}{\left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )}+\frac {2 x \log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )}{-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )}\right ) \, dx\\ &=2 \int \frac {1}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )} \, dx-2 \int \frac {1}{x \left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )} \, dx+2 \int \frac {x \log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )}{-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )} \, dx+10 \int \frac {x \log \left (x \left (2+5 x^2\right )\right )}{\left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )} \, dx-15 \int \frac {x}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )} \, dx-15 \int \frac {x^2}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )} \, dx-50 \int \frac {x^4}{\left (2+5 x^2\right ) \left (x-\log \left (x \left (2+5 x^2\right )\right )\right ) \left (-5+\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 24, normalized size = 1.00 \begin {gather*} x^2+\log \left (5-\log \left (x-\log \left (x \left (2+5 x^2\right )\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 2*x + 15*x^2 + 15*x^3 + 50*x^5 + (-20*x^2 - 50*x^4)*Log[2*x + 5*x^3] + (-4*x^3 - 10*x^5 + (4*x^
2 + 10*x^4)*Log[2*x + 5*x^3])*Log[x - Log[2*x + 5*x^3]])/(10*x^2 + 25*x^4 + (-10*x - 25*x^3)*Log[2*x + 5*x^3]
+ (-2*x^2 - 5*x^4 + (2*x + 5*x^3)*Log[2*x + 5*x^3])*Log[x - Log[2*x + 5*x^3]]),x]

[Out]

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

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fricas [A]  time = 0.58, size = 22, normalized size = 0.92 \begin {gather*} x^{2} + \log \left (\log \left (x - \log \left (5 \, x^{3} + 2 \, x\right )\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^4+4*x^2)*log(5*x^3+2*x)-10*x^5-4*x^3)*log(-log(5*x^3+2*x)+x)+(-50*x^4-20*x^2)*log(5*x^3+2*x)
+50*x^5+15*x^3+15*x^2-2*x+2)/(((5*x^3+2*x)*log(5*x^3+2*x)-5*x^4-2*x^2)*log(-log(5*x^3+2*x)+x)+(-25*x^3-10*x)*l
og(5*x^3+2*x)+25*x^4+10*x^2),x, algorithm="fricas")

[Out]

x^2 + log(log(x - log(5*x^3 + 2*x)) - 5)

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giac [A]  time = 0.43, size = 22, normalized size = 0.92 \begin {gather*} x^{2} + \log \left (\log \left (x - \log \left (5 \, x^{3} + 2 \, x\right )\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^4+4*x^2)*log(5*x^3+2*x)-10*x^5-4*x^3)*log(-log(5*x^3+2*x)+x)+(-50*x^4-20*x^2)*log(5*x^3+2*x)
+50*x^5+15*x^3+15*x^2-2*x+2)/(((5*x^3+2*x)*log(5*x^3+2*x)-5*x^4-2*x^2)*log(-log(5*x^3+2*x)+x)+(-25*x^3-10*x)*l
og(5*x^3+2*x)+25*x^4+10*x^2),x, algorithm="giac")

[Out]

x^2 + log(log(x - log(5*x^3 + 2*x)) - 5)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (10 x^{4}+4 x^{2}\right ) \ln \left (5 x^{3}+2 x \right )-10 x^{5}-4 x^{3}\right ) \ln \left (-\ln \left (5 x^{3}+2 x \right )+x \right )+\left (-50 x^{4}-20 x^{2}\right ) \ln \left (5 x^{3}+2 x \right )+50 x^{5}+15 x^{3}+15 x^{2}-2 x +2}{\left (\left (5 x^{3}+2 x \right ) \ln \left (5 x^{3}+2 x \right )-5 x^{4}-2 x^{2}\right ) \ln \left (-\ln \left (5 x^{3}+2 x \right )+x \right )+\left (-25 x^{3}-10 x \right ) \ln \left (5 x^{3}+2 x \right )+25 x^{4}+10 x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x^4+4*x^2)*ln(5*x^3+2*x)-10*x^5-4*x^3)*ln(-ln(5*x^3+2*x)+x)+(-50*x^4-20*x^2)*ln(5*x^3+2*x)+50*x^5+15
*x^3+15*x^2-2*x+2)/(((5*x^3+2*x)*ln(5*x^3+2*x)-5*x^4-2*x^2)*ln(-ln(5*x^3+2*x)+x)+(-25*x^3-10*x)*ln(5*x^3+2*x)+
25*x^4+10*x^2),x)

[Out]

int((((10*x^4+4*x^2)*ln(5*x^3+2*x)-10*x^5-4*x^3)*ln(-ln(5*x^3+2*x)+x)+(-50*x^4-20*x^2)*ln(5*x^3+2*x)+50*x^5+15
*x^3+15*x^2-2*x+2)/(((5*x^3+2*x)*ln(5*x^3+2*x)-5*x^4-2*x^2)*ln(-ln(5*x^3+2*x)+x)+(-25*x^3-10*x)*ln(5*x^3+2*x)+
25*x^4+10*x^2),x)

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maxima [A]  time = 0.41, size = 24, normalized size = 1.00 \begin {gather*} x^{2} + \log \left (\log \left (x - \log \left (5 \, x^{2} + 2\right ) - \log \relax (x)\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^4+4*x^2)*log(5*x^3+2*x)-10*x^5-4*x^3)*log(-log(5*x^3+2*x)+x)+(-50*x^4-20*x^2)*log(5*x^3+2*x)
+50*x^5+15*x^3+15*x^2-2*x+2)/(((5*x^3+2*x)*log(5*x^3+2*x)-5*x^4-2*x^2)*log(-log(5*x^3+2*x)+x)+(-25*x^3-10*x)*l
og(5*x^3+2*x)+25*x^4+10*x^2),x, algorithm="maxima")

[Out]

x^2 + log(log(x - log(5*x^2 + 2) - log(x)) - 5)

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mupad [B]  time = 5.51, size = 22, normalized size = 0.92 \begin {gather*} \ln \left (\ln \left (x-\ln \left (5\,x^3+2\,x\right )\right )-5\right )+x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*x^2 - log(2*x + 5*x^3)*(20*x^2 + 50*x^4) - log(x - log(2*x + 5*x^3))*(4*x^3 - log(2*x + 5*x^3)*(4*x^2
+ 10*x^4) + 10*x^5) - 2*x + 15*x^3 + 50*x^5 + 2)/(log(x - log(2*x + 5*x^3))*(2*x^2 - log(2*x + 5*x^3)*(2*x +
5*x^3) + 5*x^4) + log(2*x + 5*x^3)*(10*x + 25*x^3) - 10*x^2 - 25*x^4),x)

[Out]

log(log(x - log(2*x + 5*x^3)) - 5) + x^2

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sympy [A]  time = 0.78, size = 19, normalized size = 0.79 \begin {gather*} x^{2} + \log {\left (\log {\left (x - \log {\left (5 x^{3} + 2 x \right )} \right )} - 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x**4+4*x**2)*ln(5*x**3+2*x)-10*x**5-4*x**3)*ln(-ln(5*x**3+2*x)+x)+(-50*x**4-20*x**2)*ln(5*x**3
+2*x)+50*x**5+15*x**3+15*x**2-2*x+2)/(((5*x**3+2*x)*ln(5*x**3+2*x)-5*x**4-2*x**2)*ln(-ln(5*x**3+2*x)+x)+(-25*x
**3-10*x)*ln(5*x**3+2*x)+25*x**4+10*x**2),x)

[Out]

x**2 + log(log(x - log(5*x**3 + 2*x)) - 5)

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