3.62.61 \(\int \frac {(-20 x-10 x^2+10 x^3+e^4 (-20-10 x+10 x^2)+(-10 e^4-10 x) \log (4)) \log (\frac {5+x}{x})+(-10 x-2 x^2-5 x^3-x^4+e^4 (5 x-9 x^2-2 x^3)+(-5 x-x^2) \log (4)) \log ^2(\frac {5+x}{x})}{5 x^3+x^4+e^8 (5 x+x^2)+e^4 (10 x^2+2 x^3)} \, dx\)

Optimal. Leaf size=28 \[ \frac {\left (2+x-x^2+\log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{e^4+x} \]

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Rubi [F]  time = 5.26, 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 x-10 x^2+10 x^3+e^4 \left (-20-10 x+10 x^2\right )+\left (-10 e^4-10 x\right ) \log (4)\right ) \log \left (\frac {5+x}{x}\right )+\left (-10 x-2 x^2-5 x^3-x^4+e^4 \left (5 x-9 x^2-2 x^3\right )+\left (-5 x-x^2\right ) \log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{5 x^3+x^4+e^8 \left (5 x+x^2\right )+e^4 \left (10 x^2+2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-20*x - 10*x^2 + 10*x^3 + E^4*(-20 - 10*x + 10*x^2) + (-10*E^4 - 10*x)*Log[4])*Log[(5 + x)/x] + (-10*x -
2*x^2 - 5*x^3 - x^4 + E^4*(5*x - 9*x^2 - 2*x^3) + (-5*x - x^2)*Log[4])*Log[(5 + x)/x]^2)/(5*x^3 + x^4 + E^8*(
5*x + x^2) + E^4*(10*x^2 + 2*x^3)),x]

[Out]

-((5 + x)*Log[1 + 5/x]^2) + (40*(2 - E^4 - E^8 + Log[4])*Log[x])/(E^4*(5 - E^4)) + (2*(28 - Log[4])*Log[5]*Log
[x])/(5 - E^4) + (2*(28 - Log[4])*Log[1 + 5/x]*Log[5 + x])/(5 - E^4) - ((28 - Log[4])*Log[5 + x]^2)/(5 - E^4)
+ (10*(2 - E^4 - E^8 + Log[4])*Log[1 + 5/x]*Log[E^4 + x])/(E^4*(5 - E^4)) - (10*(2 - E^4 - E^8 + Log[4])*Log[(
5 + x)/(5 - E^4)]*Log[E^4 + x])/(E^4*(5 - E^4)) - 10*PolyLog[2, -5/x] - (2*(28 - Log[4])*PolyLog[2, -1/5*x])/(
5 - E^4) - (10*(2 - E^4 - E^8 + Log[4])*PolyLog[2, -(x/E^4)])/(E^4*(5 - E^4)) - (10*(2 - E^4 - E^8 + Log[4])*P
olyLog[2, -((E^4 + x)/(5 - E^4))])/(E^4*(5 - E^4)) - (2*(2 + Log[4])*PolyLog[2, 1 - (5 + x)/x])/E^4 - (2 - E^4
- E^8 + Log[4])*Defer[Int][Log[1 + 5/x]^2/(E^4 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\log \left (1+\frac {5}{x}\right ) \left (\frac {10 \left (e^4+x\right ) \left (-2-x+x^2-\log (4)\right )}{x (5+x)}-\left (2+x^2+e^4 (-1+2 x)+\log (4)\right ) \log \left (\frac {5+x}{x}\right )\right )}{\left (e^4+x\right )^2} \, dx\\ &=\int \left (\frac {10 \left (-2-x+x^2-\log (4)\right ) \log \left (1+\frac {5}{x}\right )}{x (5+x) \left (e^4+x\right )}+\frac {\left (-2+e^4-2 e^4 x-x^2-\log (4)\right ) \log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2}\right ) \, dx\\ &=10 \int \frac {\left (-2-x+x^2-\log (4)\right ) \log \left (1+\frac {5}{x}\right )}{x (5+x) \left (e^4+x\right )} \, dx+\int \frac {\left (-2+e^4-2 e^4 x-x^2-\log (4)\right ) \log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx\\ &=10 \int \frac {\left (-2-x+x^2-\log (4)\right ) \log \left (\frac {5+x}{x}\right )}{x (5+x) \left (e^4+x\right )} \, dx+\int \frac {\left (-2+e^4-2 e^4 x-x^2-\log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{\left (e^4+x\right )^2} \, dx\\ &=10 \int \left (\frac {(-2-\log (4)) \log \left (\frac {5+x}{x}\right )}{5 e^4 x}+\frac {\left (-2+e^4+e^8-\log (4)\right ) \log \left (\frac {5+x}{x}\right )}{e^4 \left (-5+e^4\right ) \left (e^4+x\right )}+\frac {(-28+\log (4)) \log \left (\frac {5+x}{x}\right )}{5 \left (-5+e^4\right ) (5+x)}\right ) \, dx+\int \left (-\log ^2\left (\frac {5+x}{x}\right )+\frac {\left (-2+e^4+e^8-\log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{\left (e^4+x\right )^2}\right ) \, dx\\ &=\frac {(2 (28-\log (4))) \int \frac {\log \left (\frac {5+x}{x}\right )}{5+x} \, dx}{5-e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (\frac {5+x}{x}\right )}{\left (e^4+x\right )^2} \, dx-\frac {(2 (2+\log (4))) \int \frac {\log \left (\frac {5+x}{x}\right )}{x} \, dx}{e^4}+\frac {\left (10 \left (2-e^4-e^8+\log (4)\right )\right ) \int \frac {\log \left (\frac {5+x}{x}\right )}{e^4+x} \, dx}{e^4 \left (5-e^4\right )}-\int \log ^2\left (\frac {5+x}{x}\right ) \, dx\\ &=-\frac {2 (2+\log (4)) \text {Li}_2\left (1-\frac {5+x}{x}\right )}{e^4}+\frac {(2 (28-\log (4))) \int \frac {\log \left (1+\frac {5}{x}\right )}{5+x} \, dx}{5-e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx+\frac {\left (10 \left (2-e^4-e^8+\log (4)\right )\right ) \int \frac {\log \left (1+\frac {5}{x}\right )}{e^4+x} \, dx}{e^4 \left (5-e^4\right )}-\int \log ^2\left (1+\frac {5}{x}\right ) \, dx\\ &=-\left ((5+x) \log ^2\left (1+\frac {5}{x}\right )\right )+\frac {2 (28-\log (4)) \log \left (1+\frac {5}{x}\right ) \log (5+x)}{5-e^4}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (1+\frac {5}{x}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-\frac {2 (2+\log (4)) \text {Li}_2\left (1-\frac {5+x}{x}\right )}{e^4}-10 \int \frac {\log \left (1+\frac {5}{x}\right )}{x} \, dx+\frac {(10 (28-\log (4))) \int \frac {\log (5+x)}{\left (1+\frac {5}{x}\right ) x^2} \, dx}{5-e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx+\frac {\left (50 \left (2-e^4-e^8+\log (4)\right )\right ) \int \frac {\log \left (e^4+x\right )}{\left (1+\frac {5}{x}\right ) x^2} \, dx}{e^4 \left (5-e^4\right )}\\ &=-\left ((5+x) \log ^2\left (1+\frac {5}{x}\right )\right )+\frac {2 (28-\log (4)) \log \left (1+\frac {5}{x}\right ) \log (5+x)}{5-e^4}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (1+\frac {5}{x}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-10 \text {Li}_2\left (-\frac {5}{x}\right )-\frac {2 (2+\log (4)) \text {Li}_2\left (1-\frac {5+x}{x}\right )}{e^4}+\frac {(10 (28-\log (4))) \int \left (\frac {\log (5+x)}{5 x}-\frac {\log (5+x)}{5 (5+x)}\right ) \, dx}{5-e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx+\frac {\left (50 \left (2-e^4-e^8+\log (4)\right )\right ) \int \left (\frac {\log \left (e^4+x\right )}{5 x}-\frac {\log \left (e^4+x\right )}{5 (5+x)}\right ) \, dx}{e^4 \left (5-e^4\right )}\\ &=-\left ((5+x) \log ^2\left (1+\frac {5}{x}\right )\right )+\frac {2 (28-\log (4)) \log \left (1+\frac {5}{x}\right ) \log (5+x)}{5-e^4}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (1+\frac {5}{x}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-10 \text {Li}_2\left (-\frac {5}{x}\right )-\frac {2 (2+\log (4)) \text {Li}_2\left (1-\frac {5+x}{x}\right )}{e^4}+\frac {(2 (28-\log (4))) \int \frac {\log (5+x)}{x} \, dx}{5-e^4}-\frac {(2 (28-\log (4))) \int \frac {\log (5+x)}{5+x} \, dx}{5-e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx+\frac {\left (10 \left (2-e^4-e^8+\log (4)\right )\right ) \int \frac {\log \left (e^4+x\right )}{x} \, dx}{e^4 \left (5-e^4\right )}-\frac {\left (10 \left (2-e^4-e^8+\log (4)\right )\right ) \int \frac {\log \left (e^4+x\right )}{5+x} \, dx}{e^4 \left (5-e^4\right )}\\ &=-\left ((5+x) \log ^2\left (1+\frac {5}{x}\right )\right )+\frac {40 \left (2-e^4-e^8+\log (4)\right ) \log (x)}{e^4 \left (5-e^4\right )}+\frac {2 (28-\log (4)) \log (5) \log (x)}{5-e^4}+\frac {2 (28-\log (4)) \log \left (1+\frac {5}{x}\right ) \log (5+x)}{5-e^4}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (1+\frac {5}{x}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (\frac {5+x}{5-e^4}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-10 \text {Li}_2\left (-\frac {5}{x}\right )-\frac {2 (2+\log (4)) \text {Li}_2\left (1-\frac {5+x}{x}\right )}{e^4}+\frac {(2 (28-\log (4))) \int \frac {\log \left (1+\frac {x}{5}\right )}{x} \, dx}{5-e^4}-\frac {(2 (28-\log (4))) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x\right )}{5-e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx+\frac {\left (10 \left (2-e^4-e^8+\log (4)\right )\right ) \int \frac {\log \left (\frac {5+x}{5-e^4}\right )}{e^4+x} \, dx}{e^4 \left (5-e^4\right )}+\frac {\left (10 \left (2-e^4-e^8+\log (4)\right )\right ) \int \frac {\log \left (1+\frac {x}{e^4}\right )}{x} \, dx}{e^4 \left (5-e^4\right )}\\ &=-\left ((5+x) \log ^2\left (1+\frac {5}{x}\right )\right )+\frac {40 \left (2-e^4-e^8+\log (4)\right ) \log (x)}{e^4 \left (5-e^4\right )}+\frac {2 (28-\log (4)) \log (5) \log (x)}{5-e^4}+\frac {2 (28-\log (4)) \log \left (1+\frac {5}{x}\right ) \log (5+x)}{5-e^4}-\frac {(28-\log (4)) \log ^2(5+x)}{5-e^4}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (1+\frac {5}{x}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (\frac {5+x}{5-e^4}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-10 \text {Li}_2\left (-\frac {5}{x}\right )-\frac {2 (28-\log (4)) \text {Li}_2\left (-\frac {x}{5}\right )}{5-e^4}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \text {Li}_2\left (-\frac {x}{e^4}\right )}{e^4 \left (5-e^4\right )}-\frac {2 (2+\log (4)) \text {Li}_2\left (1-\frac {5+x}{x}\right )}{e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx+\frac {\left (10 \left (2-e^4-e^8+\log (4)\right )\right ) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5-e^4}\right )}{x} \, dx,x,e^4+x\right )}{e^4 \left (5-e^4\right )}\\ &=-\left ((5+x) \log ^2\left (1+\frac {5}{x}\right )\right )+\frac {40 \left (2-e^4-e^8+\log (4)\right ) \log (x)}{e^4 \left (5-e^4\right )}+\frac {2 (28-\log (4)) \log (5) \log (x)}{5-e^4}+\frac {2 (28-\log (4)) \log \left (1+\frac {5}{x}\right ) \log (5+x)}{5-e^4}-\frac {(28-\log (4)) \log ^2(5+x)}{5-e^4}+\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (1+\frac {5}{x}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \log \left (\frac {5+x}{5-e^4}\right ) \log \left (e^4+x\right )}{e^4 \left (5-e^4\right )}-10 \text {Li}_2\left (-\frac {5}{x}\right )-\frac {2 (28-\log (4)) \text {Li}_2\left (-\frac {x}{5}\right )}{5-e^4}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \text {Li}_2\left (-\frac {x}{e^4}\right )}{e^4 \left (5-e^4\right )}-\frac {10 \left (2-e^4-e^8+\log (4)\right ) \text {Li}_2\left (-\frac {e^4+x}{5-e^4}\right )}{e^4 \left (5-e^4\right )}-\frac {2 (2+\log (4)) \text {Li}_2\left (1-\frac {5+x}{x}\right )}{e^4}+\left (-2+e^4+e^8-\log (4)\right ) \int \frac {\log ^2\left (1+\frac {5}{x}\right )}{\left (e^4+x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.07, size = 28, normalized size = 1.00 \begin {gather*} \frac {\left (2+x-x^2+\log (4)\right ) \log ^2\left (\frac {5+x}{x}\right )}{e^4+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-20*x - 10*x^2 + 10*x^3 + E^4*(-20 - 10*x + 10*x^2) + (-10*E^4 - 10*x)*Log[4])*Log[(5 + x)/x] + (-
10*x - 2*x^2 - 5*x^3 - x^4 + E^4*(5*x - 9*x^2 - 2*x^3) + (-5*x - x^2)*Log[4])*Log[(5 + x)/x]^2)/(5*x^3 + x^4 +
E^8*(5*x + x^2) + E^4*(10*x^2 + 2*x^3)),x]

[Out]

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

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fricas [A]  time = 0.56, size = 30, normalized size = 1.07 \begin {gather*} -\frac {{\left (x^{2} - x - 2 \, \log \relax (2) - 2\right )} \log \left (\frac {x + 5}{x}\right )^{2}}{x + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x^2-5*x)*log(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2-10*x)*log(1/x*(5+x))^2+(2*(-10*exp(4
)-10*x)*log(2)+(10*x^2-10*x-20)*exp(4)+10*x^3-10*x^2-20*x)*log(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*
exp(4)+x^4+5*x^3),x, algorithm="fricas")

[Out]

-(x^2 - x - 2*log(2) - 2)*log((x + 5)/x)^2/(x + e^4)

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giac [B]  time = 0.23, size = 142, normalized size = 5.07 \begin {gather*} \frac {\frac {2 \, {\left (x + 5\right )}^{2} \log \relax (2) \log \left (\frac {x + 5}{x}\right )^{2}}{x^{2}} - \frac {4 \, {\left (x + 5\right )} \log \relax (2) \log \left (\frac {x + 5}{x}\right )^{2}}{x} + 2 \, \log \relax (2) \log \left (\frac {x + 5}{x}\right )^{2} + \frac {2 \, {\left (x + 5\right )}^{2} \log \left (\frac {x + 5}{x}\right )^{2}}{x^{2}} + \frac {{\left (x + 5\right )} \log \left (\frac {x + 5}{x}\right )^{2}}{x} - 28 \, \log \left (\frac {x + 5}{x}\right )^{2}}{\frac {{\left (x + 5\right )}^{2} e^{4}}{x^{2}} - \frac {2 \, {\left (x + 5\right )} e^{4}}{x} + \frac {5 \, {\left (x + 5\right )}}{x} + e^{4} - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x^2-5*x)*log(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2-10*x)*log(1/x*(5+x))^2+(2*(-10*exp(4
)-10*x)*log(2)+(10*x^2-10*x-20)*exp(4)+10*x^3-10*x^2-20*x)*log(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*
exp(4)+x^4+5*x^3),x, algorithm="giac")

[Out]

(2*(x + 5)^2*log(2)*log((x + 5)/x)^2/x^2 - 4*(x + 5)*log(2)*log((x + 5)/x)^2/x + 2*log(2)*log((x + 5)/x)^2 + 2
*(x + 5)^2*log((x + 5)/x)^2/x^2 + (x + 5)*log((x + 5)/x)^2/x - 28*log((x + 5)/x)^2)/((x + 5)^2*e^4/x^2 - 2*(x
+ 5)*e^4/x + 5*(x + 5)/x + e^4 - 5)

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maple [A]  time = 5.47, size = 53, normalized size = 1.89

method	result	size
norman	\(\frac {x \ln \left (\frac {5+x}{x}\right )^{2}+\left (2+2 \ln \relax (2)\right ) \ln \left (\frac {5+x}{x}\right )^{2}-\ln \left (\frac {5+x}{x}\right )^{2} x^{2}}{x +{\mathrm e}^{4}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(-x^2-5*x)*ln(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2-10*x)*ln(1/x*(5+x))^2+(2*(-10*exp(4)-10*x)*
ln(2)+(10*x^2-10*x-20)*exp(4)+10*x^3-10*x^2-20*x)*ln(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*exp(4)+x^4
+5*x^3),x,method=_RETURNVERBOSE)

[Out]

(x*ln(1/x*(5+x))^2+(2+2*ln(2))*ln(1/x*(5+x))^2-ln(1/x*(5+x))^2*x^2)/(x+exp(4))

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maxima [B]  time = 0.50, size = 65, normalized size = 2.32 \begin {gather*} -\frac {{\left (x^{2} - x - 2 \, \log \relax (2) - 2\right )} \log \left (x + 5\right )^{2} - 2 \, {\left (x^{2} - x - 2 \, \log \relax (2) - 2\right )} \log \left (x + 5\right ) \log \relax (x) + {\left (x^{2} - x - 2 \, \log \relax (2) - 2\right )} \log \relax (x)^{2}}{x + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x^2-5*x)*log(2)+(-2*x^3-9*x^2+5*x)*exp(4)-x^4-5*x^3-2*x^2-10*x)*log(1/x*(5+x))^2+(2*(-10*exp(4
)-10*x)*log(2)+(10*x^2-10*x-20)*exp(4)+10*x^3-10*x^2-20*x)*log(1/x*(5+x)))/((x^2+5*x)*exp(4)^2+(2*x^3+10*x^2)*
exp(4)+x^4+5*x^3),x, algorithm="maxima")

[Out]

-((x^2 - x - 2*log(2) - 2)*log(x + 5)^2 - 2*(x^2 - x - 2*log(2) - 2)*log(x + 5)*log(x) + (x^2 - x - 2*log(2) -
2)*log(x)^2)/(x + e^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\left (10\,x+{\mathrm {e}}^4\,\left (2\,x^3+9\,x^2-5\,x\right )+2\,x^2+5\,x^3+x^4+2\,\ln \relax (2)\,\left (x^2+5\,x\right )\right )\,{\ln \left (\frac {x+5}{x}\right )}^2+\left (20\,x+{\mathrm {e}}^4\,\left (-10\,x^2+10\,x+20\right )+2\,\ln \relax (2)\,\left (10\,x+10\,{\mathrm {e}}^4\right )+10\,x^2-10\,x^3\right )\,\ln \left (\frac {x+5}{x}\right )}{{\mathrm {e}}^4\,\left (2\,x^3+10\,x^2\right )+{\mathrm {e}}^8\,\left (x^2+5\,x\right )+5\,x^3+x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log((x + 5)/x)*(20*x + exp(4)*(10*x - 10*x^2 + 20) + 2*log(2)*(10*x + 10*exp(4)) + 10*x^2 - 10*x^3) + lo
g((x + 5)/x)^2*(10*x + exp(4)*(9*x^2 - 5*x + 2*x^3) + 2*x^2 + 5*x^3 + x^4 + 2*log(2)*(5*x + x^2)))/(exp(4)*(10
*x^2 + 2*x^3) + exp(8)*(5*x + x^2) + 5*x^3 + x^4),x)

[Out]

int(-(log((x + 5)/x)*(20*x + exp(4)*(10*x - 10*x^2 + 20) + 2*log(2)*(10*x + 10*exp(4)) + 10*x^2 - 10*x^3) + lo
g((x + 5)/x)^2*(10*x + exp(4)*(9*x^2 - 5*x + 2*x^3) + 2*x^2 + 5*x^3 + x^4 + 2*log(2)*(5*x + x^2)))/(exp(4)*(10
*x^2 + 2*x^3) + exp(8)*(5*x + x^2) + 5*x^3 + x^4), x)

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sympy [A]  time = 0.36, size = 24, normalized size = 0.86 \begin {gather*} \frac {\left (- x^{2} + x + 2 \log {\relax (2 )} + 2\right ) \log {\left (\frac {x + 5}{x} \right )}^{2}}{x + e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(-x**2-5*x)*ln(2)+(-2*x**3-9*x**2+5*x)*exp(4)-x**4-5*x**3-2*x**2-10*x)*ln(1/x*(5+x))**2+(2*(-10*
exp(4)-10*x)*ln(2)+(10*x**2-10*x-20)*exp(4)+10*x**3-10*x**2-20*x)*ln(1/x*(5+x)))/((x**2+5*x)*exp(4)**2+(2*x**3
+10*x**2)*exp(4)+x**4+5*x**3),x)

[Out]

(-x**2 + x + 2*log(2) + 2)*log((x + 5)/x)**2/(x + exp(4))

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