3.71.6 \(\int \frac {(-20+100 x-180 x^4+100 x^5) \log (x^2)+(40+40 x^4) \log (\frac {e^{5 x}}{x+2 x^5+x^9})}{x+x^5} \, dx\)

Optimal. Leaf size=23 \[ 20 \log \left (x^2\right ) \log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right ) \]

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Rubi [F]  time = 0.40, 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 {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x+x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-20 + 100*x - 180*x^4 + 100*x^5)*Log[x^2] + (40 + 40*x^4)*Log[E^(5*x)/(x + 2*x^5 + x^9)])/(x + x^5),x]

[Out]

-200*x + 100*x*Log[x^2] - 5*Log[x^2]^2 - 40*Log[x^2]*Log[1 + x^4] - 20*PolyLog[2, -x^4] + 40*Defer[Int][Log[E^
(5*x)/(x*(1 + x^4)^2)]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-20+100 x-180 x^4+100 x^5\right ) \log \left (x^2\right )+\left (40+40 x^4\right ) \log \left (\frac {e^{5 x}}{x+2 x^5+x^9}\right )}{x \left (1+x^4\right )} \, dx\\ &=\int \left (\frac {20 (-1+x) \left (1-4 x-4 x^2-4 x^3+5 x^4\right ) \log \left (x^2\right )}{x \left (1+x^4\right )}+\frac {40 \log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )}{x}\right ) \, dx\\ &=20 \int \frac {(-1+x) \left (1-4 x-4 x^2-4 x^3+5 x^4\right ) \log \left (x^2\right )}{x \left (1+x^4\right )} \, dx+40 \int \frac {\log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )}{x} \, dx\\ &=20 \int \left (5 \log \left (x^2\right )-\frac {\log \left (x^2\right )}{x}-\frac {8 x^3 \log \left (x^2\right )}{1+x^4}\right ) \, dx+40 \int \frac {\log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )}{x} \, dx\\ &=-\left (20 \int \frac {\log \left (x^2\right )}{x} \, dx\right )+40 \int \frac {\log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )}{x} \, dx+100 \int \log \left (x^2\right ) \, dx-160 \int \frac {x^3 \log \left (x^2\right )}{1+x^4} \, dx\\ &=-200 x+100 x \log \left (x^2\right )-5 \log ^2\left (x^2\right )-40 \log \left (x^2\right ) \log \left (1+x^4\right )+40 \int \frac {\log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )}{x} \, dx+80 \int \frac {\log \left (1+x^4\right )}{x} \, dx\\ &=-200 x+100 x \log \left (x^2\right )-5 \log ^2\left (x^2\right )-40 \log \left (x^2\right ) \log \left (1+x^4\right )-20 \text {Li}_2\left (-x^4\right )+40 \int \frac {\log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.14, size = 63, normalized size = 2.74 \begin {gather*} 20 \left (\log \left (x^2\right ) \left (5 x+\log \left (\frac {1}{x \left (1+x^4\right )^2}\right )\right )-2 \log (x) \left (5 x+\log \left (\frac {1}{x \left (1+x^4\right )^2}\right )-\log \left (\frac {e^{5 x}}{x \left (1+x^4\right )^2}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-20 + 100*x - 180*x^4 + 100*x^5)*Log[x^2] + (40 + 40*x^4)*Log[E^(5*x)/(x + 2*x^5 + x^9)])/(x + x^5
),x]

[Out]

20*(Log[x^2]*(5*x + Log[1/(x*(1 + x^4)^2)]) - 2*Log[x]*(5*x + Log[1/(x*(1 + x^4)^2)] - Log[E^(5*x)/(x*(1 + x^4
)^2)]))

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fricas [A]  time = 0.56, size = 24, normalized size = 1.04 \begin {gather*} 20 \, \log \left (x^{2}\right ) \log \left (\frac {e^{\left (5 \, x\right )}}{x^{9} + 2 \, x^{5} + x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^4+40)*log(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x-20)*log(x^2))/(x^5+x),x, algorithm="
fricas")

[Out]

20*log(x^2)*log(e^(5*x)/(x^9 + 2*x^5 + x))

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giac [A]  time = 0.18, size = 27, normalized size = 1.17 \begin {gather*} 200 \, x \log \relax (x) - 40 \, \log \left (x^{8} + 2 \, x^{4} + 1\right ) \log \relax (x) - 40 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^4+40)*log(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x-20)*log(x^2))/(x^5+x),x, algorithm="
giac")

[Out]

200*x*log(x) - 40*log(x^8 + 2*x^4 + 1)*log(x) - 40*log(x)^2

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maple [C]  time = 15.83, size = 368, normalized size = 16.00

method	result	size
default	\(40 \ln \relax (x ) \ln \left (\frac {{\mathrm e}^{5 x}}{x^{9}+2 x^{5}+x}\right )+40 \ln \relax (x )^{2}-200 x \ln \relax (x )+80 \ln \relax (x ) \ln \left (\frac {\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}-x}{\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \dilog \left (\frac {\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}-x}{\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \ln \relax (x ) \ln \left (\frac {-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}-x}{-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \dilog \left (\frac {-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}-x}{-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}}\right )+80 \ln \relax (x ) \ln \left (\frac {-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}-x}{-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \dilog \left (\frac {-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}-x}{-\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \ln \relax (x ) \ln \left (\frac {\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}-x}{\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+80 \dilog \left (\frac {\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}-x}{\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}}\right )+100 x \ln \left (x^{2}\right )-20 \ln \relax (x ) \ln \left (x^{2}\right )+20 \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}+1\right )}{\sum }\left (-2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (x^{2}\right )+4 \dilog \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )+4 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x}{\underline {\hspace {1.25 ex}}\alpha }\right )\right )\right )\)	\(368\)
risch	\(\text {Expression too large to display}\)	\(2623\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((40*x^4+40)*ln(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x-20)*ln(x^2))/(x^5+x),x,method=_RETURNVERBOS
E)

[Out]

40*ln(x)*ln(exp(5*x)/(x^9+2*x^5+x))+40*ln(x)^2-200*x*ln(x)+80*ln(x)*ln((1/2*2^(1/2)+1/2*I*2^(1/2)-x)/(1/2*2^(1
/2)+1/2*I*2^(1/2)))+80*dilog((1/2*2^(1/2)+1/2*I*2^(1/2)-x)/(1/2*2^(1/2)+1/2*I*2^(1/2)))+80*ln(x)*ln((-1/2*2^(1
/2)+1/2*I*2^(1/2)-x)/(-1/2*2^(1/2)+1/2*I*2^(1/2)))+80*dilog((-1/2*2^(1/2)+1/2*I*2^(1/2)-x)/(-1/2*2^(1/2)+1/2*I
*2^(1/2)))+80*ln(x)*ln((-1/2*2^(1/2)-1/2*I*2^(1/2)-x)/(-1/2*2^(1/2)-1/2*I*2^(1/2)))+80*dilog((-1/2*2^(1/2)-1/2
*I*2^(1/2)-x)/(-1/2*2^(1/2)-1/2*I*2^(1/2)))+80*ln(x)*ln((1/2*2^(1/2)-1/2*I*2^(1/2)-x)/(1/2*2^(1/2)-1/2*I*2^(1/
2)))+80*dilog((1/2*2^(1/2)-1/2*I*2^(1/2)-x)/(1/2*2^(1/2)-1/2*I*2^(1/2)))+100*x*ln(x^2)-20*ln(x)*ln(x^2)+20*Sum
(-2*ln(x-_alpha)*ln(x^2)+4*dilog(x/_alpha)+4*ln(x-_alpha)*ln(x/_alpha),_alpha=RootOf(_Z^4+1))

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maxima [A]  time = 0.52, size = 22, normalized size = 0.96 \begin {gather*} 200 \, x \log \relax (x) - 80 \, \log \left (x^{4} + 1\right ) \log \relax (x) - 40 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^4+40)*log(exp(5*x)/(x^9+2*x^5+x))+(100*x^5-180*x^4+100*x-20)*log(x^2))/(x^5+x),x, algorithm="
maxima")

[Out]

200*x*log(x) - 80*log(x^4 + 1)*log(x) - 40*log(x)^2

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mupad [B]  time = 4.18, size = 23, normalized size = 1.00 \begin {gather*} 20\,\ln \left (x^2\right )\,\left (5\,x+\ln \left (\frac {1}{x^9+2\,x^5+x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(exp(5*x)/(x + 2*x^5 + x^9))*(40*x^4 + 40) + log(x^2)*(100*x - 180*x^4 + 100*x^5 - 20))/(x + x^5),x)

[Out]

20*log(x^2)*(5*x + log(1/(x + 2*x^5 + x^9)))

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sympy [A]  time = 0.50, size = 22, normalized size = 0.96 \begin {gather*} 20 \log {\left (x^{2} \right )} \log {\left (\frac {e^{5 x}}{x^{9} + 2 x^{5} + x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x**4+40)*ln(exp(5*x)/(x**9+2*x**5+x))+(100*x**5-180*x**4+100*x-20)*ln(x**2))/(x**5+x),x)

[Out]

20*log(x**2)*log(exp(5*x)/(x**9 + 2*x**5 + x))

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