3.57.75 \(\int \frac {1}{100} (25+50 x+(50 x-50 x^2-4 e^{10-2 x} x^2) \log (e^{\frac {1}{25} (e^{10-2 x}-25 x)} x)+50 x \log ^2(e^{\frac {1}{25} (e^{10-2 x}-25 x)} x)) \, dx\)

Optimal. Leaf size=32 \[ \frac {1}{4} x \left (1+x+x \log ^2\left (e^{\frac {1}{25} e^{10-2 x}-x} x\right )\right ) \]

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Rubi [F]  time = 0.81, 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 {1}{100} \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25 + 50*x + (50*x - 50*x^2 - 4*E^(10 - 2*x)*x^2)*Log[E^((E^(10 - 2*x) - 25*x)/25)*x] + 50*x*Log[E^((E^(10
- 2*x) - 25*x)/25)*x]^2)/100,x]

[Out]

(-7*E^(20 - 4*x))/20000 + x/4 - (3*E^(20 - 4*x)*x)/5000 - (E^(10 - 2*x)*x)/100 + x^2/8 - (E^(20 - 4*x)*x^2)/25
00 - (E^(10 - 2*x)*x^2)/100 + (5*x^3)/36 + (E^(10 - 2*x)*x^3)/150 - x^4/24 - (E^10*ExpIntegralEi[-2*x])/100 +
(E^(10 - 2*x)*Log[E^((E^(10 - 2*x) - 25*x)/25)*x])/100 + (E^(10 - 2*x)*x*Log[E^((E^(10 - 2*x) - 25*x)/25)*x])/
50 + (x^2*Log[E^((E^(10 - 2*x) - 25*x)/25)*x])/4 + (E^(10 - 2*x)*x^2*Log[E^((E^(10 - 2*x) - 25*x)/25)*x])/50 -
(x^3*Log[E^((E^(10 - 2*x) - 25*x)/25)*x])/6 + Defer[Int][x*Log[E^((E^(10 - 2*x) - 25*x)/25)*x]^2, x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{100} \int \left (25+50 x+\left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+50 x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}+\frac {1}{100} \int \left (50 x-50 x^2-4 e^{10-2 x} x^2\right ) \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{100} \int \frac {e^{-4 x} \left (25 e^{2 x} (-1+x)+2 e^{10} x\right ) \left (25 e^{2 x} x^2 (-3+2 x)-3 e^{10} \left (1+2 x+2 x^2\right )\right )}{75 x} \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {\int \frac {e^{-4 x} \left (25 e^{2 x} (-1+x)+2 e^{10} x\right ) \left (25 e^{2 x} x^2 (-3+2 x)-3 e^{10} \left (1+2 x+2 x^2\right )\right )}{x} \, dx}{7500}+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {\int \left (625 (-1+x) x (-3+2 x)-6 e^{20-4 x} \left (1+2 x+2 x^2\right )+\frac {25 e^{10-2 x} \left (3+3 x-12 x^3+4 x^4\right )}{x}\right ) \, dx}{7500}+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {\int e^{20-4 x} \left (1+2 x+2 x^2\right ) \, dx}{1250}-\frac {1}{300} \int \frac {e^{10-2 x} \left (3+3 x-12 x^3+4 x^4\right )}{x} \, dx-\frac {1}{12} \int (-1+x) x (-3+2 x) \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{4}+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {\int \left (e^{20-4 x}+2 e^{20-4 x} x+2 e^{20-4 x} x^2\right ) \, dx}{1250}-\frac {1}{300} \int \left (3 e^{10-2 x}+\frac {3 e^{10-2 x}}{x}-12 e^{10-2 x} x^2+4 e^{10-2 x} x^3\right ) \, dx-\frac {1}{12} \int \left (3 x-5 x^2+2 x^3\right ) \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=\frac {x}{4}+\frac {x^2}{8}+\frac {5 x^3}{36}-\frac {x^4}{24}+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {\int e^{20-4 x} \, dx}{1250}+\frac {1}{625} \int e^{20-4 x} x \, dx+\frac {1}{625} \int e^{20-4 x} x^2 \, dx-\frac {1}{100} \int e^{10-2 x} \, dx-\frac {1}{100} \int \frac {e^{10-2 x}}{x} \, dx-\frac {1}{75} \int e^{10-2 x} x^3 \, dx+\frac {1}{25} \int e^{10-2 x} x^2 \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=-\frac {e^{20-4 x}}{5000}+\frac {1}{200} e^{10-2 x}+\frac {x}{4}-\frac {e^{20-4 x} x}{2500}+\frac {x^2}{8}-\frac {e^{20-4 x} x^2}{2500}-\frac {1}{50} e^{10-2 x} x^2+\frac {5 x^3}{36}+\frac {1}{150} e^{10-2 x} x^3-\frac {x^4}{24}-\frac {1}{100} e^{10} \text {Ei}(-2 x)+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {\int e^{20-4 x} \, dx}{2500}+\frac {\int e^{20-4 x} x \, dx}{1250}-\frac {1}{50} \int e^{10-2 x} x^2 \, dx+\frac {1}{25} \int e^{10-2 x} x \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=-\frac {3 e^{20-4 x}}{10000}+\frac {1}{200} e^{10-2 x}+\frac {x}{4}-\frac {3 e^{20-4 x} x}{5000}-\frac {1}{50} e^{10-2 x} x+\frac {x^2}{8}-\frac {e^{20-4 x} x^2}{2500}-\frac {1}{100} e^{10-2 x} x^2+\frac {5 x^3}{36}+\frac {1}{150} e^{10-2 x} x^3-\frac {x^4}{24}-\frac {1}{100} e^{10} \text {Ei}(-2 x)+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {\int e^{20-4 x} \, dx}{5000}+\frac {1}{50} \int e^{10-2 x} \, dx-\frac {1}{50} \int e^{10-2 x} x \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=-\frac {7 e^{20-4 x}}{20000}-\frac {1}{200} e^{10-2 x}+\frac {x}{4}-\frac {3 e^{20-4 x} x}{5000}-\frac {1}{100} e^{10-2 x} x+\frac {x^2}{8}-\frac {e^{20-4 x} x^2}{2500}-\frac {1}{100} e^{10-2 x} x^2+\frac {5 x^3}{36}+\frac {1}{150} e^{10-2 x} x^3-\frac {x^4}{24}-\frac {1}{100} e^{10} \text {Ei}(-2 x)+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{100} \int e^{10-2 x} \, dx+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ &=-\frac {7 e^{20-4 x}}{20000}+\frac {x}{4}-\frac {3 e^{20-4 x} x}{5000}-\frac {1}{100} e^{10-2 x} x+\frac {x^2}{8}-\frac {e^{20-4 x} x^2}{2500}-\frac {1}{100} e^{10-2 x} x^2+\frac {5 x^3}{36}+\frac {1}{150} e^{10-2 x} x^3-\frac {x^4}{24}-\frac {1}{100} e^{10} \text {Ei}(-2 x)+\frac {1}{100} e^{10-2 x} \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{4} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{50} e^{10-2 x} x^2 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )-\frac {1}{6} x^3 \log \left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right )+\frac {1}{2} \int x \log ^2\left (e^{\frac {1}{25} \left (e^{10-2 x}-25 x\right )} x\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 32, normalized size = 1.00 \begin {gather*} \frac {1}{4} x \left (1+x+x \log ^2\left (e^{\frac {1}{25} e^{10-2 x}-x} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25 + 50*x + (50*x - 50*x^2 - 4*E^(10 - 2*x)*x^2)*Log[E^((E^(10 - 2*x) - 25*x)/25)*x] + 50*x*Log[E^(
(E^(10 - 2*x) - 25*x)/25)*x]^2)/100,x]

[Out]

(x*(1 + x + x*Log[E^(E^(10 - 2*x)/25 - x)*x]^2))/4

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fricas [A]  time = 0.74, size = 32, normalized size = 1.00 \begin {gather*} \frac {1}{4} \, x^{2} \log \left (x e^{\left (-x + \frac {1}{25} \, e^{\left (-2 \, x + 10\right )}\right )}\right )^{2} + \frac {1}{4} \, x^{2} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*x*log(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-50*x^2+50*x)*log(x*exp(1/25*exp(5-x)^
2-x))+1/2*x+1/4,x, algorithm="fricas")

[Out]

1/4*x^2*log(x*e^(-x + 1/25*e^(-2*x + 10)))^2 + 1/4*x^2 + 1/4*x

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giac [B]  time = 0.22, size = 65, normalized size = 2.03 \begin {gather*} \frac {1}{4} \, x^{4} - \frac {1}{50} \, x^{3} e^{\left (-2 \, x + 10\right )} - \frac {1}{2} \, x^{3} \log \relax (x) + \frac {1}{50} \, x^{2} e^{\left (-2 \, x + 10\right )} \log \relax (x) + \frac {1}{4} \, x^{2} \log \relax (x)^{2} + \frac {1}{2500} \, x^{2} e^{\left (-4 \, x + 20\right )} + \frac {1}{4} \, x^{2} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*x*log(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-50*x^2+50*x)*log(x*exp(1/25*exp(5-x)^
2-x))+1/2*x+1/4,x, algorithm="giac")

[Out]

1/4*x^4 - 1/50*x^3*e^(-2*x + 10) - 1/2*x^3*log(x) + 1/50*x^2*e^(-2*x + 10)*log(x) + 1/4*x^2*log(x)^2 + 1/2500*
x^2*e^(-4*x + 20) + 1/4*x^2 + 1/4*x

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maple [C]  time = 2.04, size = 768, normalized size = 24.00

method	result	size
risch	\(\frac {x}{4}+\frac {x^{2}}{4}+\frac {x^{2} \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2}}{4}+\frac {\left (50 x^{2} \ln \relax (x )-25 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )+25 i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2}+25 i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2}-25 i \pi \,x^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{3}\right ) \ln \left ({\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )}{100}+\frac {x^{2} \ln \relax (x )^{2}}{4}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \ln \relax (x )}{4}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{3} \ln \relax (x )}{4}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2} \ln \relax (x )}{4}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2} \ln \relax (x )}{4}-\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2}}{16}+\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{3}}{8}-\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{4}}{16}+\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{3}}{8}-\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{4}}{4}+\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{5}}{8}-\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{4}}{16}+\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i {\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{5}}{8}-\frac {\pi ^{2} x^{2} \mathrm {csgn}\left (i x \,{\mathrm e}^{\frac {{\mathrm e}^{-2 x +10}}{25}-x}\right )^{6}}{16}\)	\(768\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*x*ln(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-50*x^2+50*x)*ln(x*exp(1/25*exp(5-x)^2-x))+1/
2*x+1/4,x,method=_RETURNVERBOSE)

[Out]

1/4*x+1/4*x^2+1/4*x^2*ln(exp(1/25*exp(-2*x+10)-x))^2+1/100*(50*x^2*ln(x)-25*I*Pi*x^2*csgn(I*x)*csgn(I*exp(1/25
*exp(-2*x+10)-x))*csgn(I*x*exp(1/25*exp(-2*x+10)-x))+25*I*Pi*x^2*csgn(I*x)*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^
2+25*I*Pi*x^2*csgn(I*exp(1/25*exp(-2*x+10)-x))*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^2-25*I*Pi*x^2*csgn(I*x*exp(1
/25*exp(-2*x+10)-x))^3)*ln(exp(1/25*exp(-2*x+10)-x))+1/4*x^2*ln(x)^2-1/4*I*Pi*x^2*csgn(I*x)*csgn(I*exp(1/25*ex
p(-2*x+10)-x))*csgn(I*x*exp(1/25*exp(-2*x+10)-x))*ln(x)-1/4*I*Pi*x^2*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^3*ln(x
)+1/4*I*Pi*x^2*csgn(I*x)*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^2*ln(x)+1/4*I*Pi*x^2*csgn(I*exp(1/25*exp(-2*x+10)-
x))*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^2*ln(x)-1/16*Pi^2*x^2*csgn(I*x)^2*csgn(I*exp(1/25*exp(-2*x+10)-x))^2*cs
gn(I*x*exp(1/25*exp(-2*x+10)-x))^2+1/8*Pi^2*x^2*csgn(I*x)^2*csgn(I*exp(1/25*exp(-2*x+10)-x))*csgn(I*x*exp(1/25
*exp(-2*x+10)-x))^3-1/16*Pi^2*x^2*csgn(I*x)^2*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^4+1/8*Pi^2*x^2*csgn(I*x)*csgn
(I*exp(1/25*exp(-2*x+10)-x))^2*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^3-1/4*Pi^2*x^2*csgn(I*x)*csgn(I*exp(1/25*exp
(-2*x+10)-x))*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^4+1/8*Pi^2*x^2*csgn(I*x)*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^5
-1/16*Pi^2*x^2*csgn(I*exp(1/25*exp(-2*x+10)-x))^2*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^4+1/8*Pi^2*x^2*csgn(I*exp
(1/25*exp(-2*x+10)-x))*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^5-1/16*Pi^2*x^2*csgn(I*x*exp(1/25*exp(-2*x+10)-x))^6

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maxima [B]  time = 0.50, size = 63, normalized size = 1.97 \begin {gather*} \frac {1}{4} \, x^{4} - \frac {1}{2} \, x^{3} \log \relax (x) + \frac {1}{4} \, x^{2} \log \relax (x)^{2} + \frac {1}{2500} \, x^{2} e^{\left (-4 \, x + 20\right )} + \frac {1}{4} \, x^{2} - \frac {1}{50} \, {\left (x^{3} e^{10} - x^{2} e^{10} \log \relax (x)\right )} e^{\left (-2 \, x\right )} + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*x*log(x*exp(1/25*exp(5-x)^2-x))^2+1/100*(-4*x^2*exp(5-x)^2-50*x^2+50*x)*log(x*exp(1/25*exp(5-x)^
2-x))+1/2*x+1/4,x, algorithm="maxima")

[Out]

1/4*x^4 - 1/2*x^3*log(x) + 1/4*x^2*log(x)^2 + 1/2500*x^2*e^(-4*x + 20) + 1/4*x^2 - 1/50*(x^3*e^10 - x^2*e^10*l
og(x))*e^(-2*x) + 1/4*x

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mupad [B]  time = 3.85, size = 65, normalized size = 2.03 \begin {gather*} \frac {x}{4}-\frac {x^3\,\ln \relax (x)}{2}+\frac {x^2\,{\ln \relax (x)}^2}{4}-\frac {x^3\,{\mathrm {e}}^{10-2\,x}}{50}+\frac {x^2\,{\mathrm {e}}^{20-4\,x}}{2500}+\frac {x^2}{4}+\frac {x^4}{4}+\frac {x^2\,{\mathrm {e}}^{10-2\,x}\,\ln \relax (x)}{50} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/2 + (x*log(x*exp(exp(10 - 2*x)/25 - x))^2)/2 - (log(x*exp(exp(10 - 2*x)/25 - x))*(4*x^2*exp(10 - 2*x) -
50*x + 50*x^2))/100 + 1/4,x)

[Out]

x/4 - (x^3*log(x))/2 + (x^2*log(x)^2)/4 - (x^3*exp(10 - 2*x))/50 + (x^2*exp(20 - 4*x))/2500 + x^2/4 + x^4/4 +
(x^2*exp(10 - 2*x)*log(x))/50

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sympy [A]  time = 0.37, size = 29, normalized size = 0.91 \begin {gather*} \frac {x^{2} \log {\left (x e^{- x + \frac {e^{10 - 2 x}}{25}} \right )}^{2}}{4} + \frac {x^{2}}{4} + \frac {x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*x*ln(x*exp(1/25*exp(5-x)**2-x))**2+1/100*(-4*x**2*exp(5-x)**2-50*x**2+50*x)*ln(x*exp(1/25*exp(5-
x)**2-x))+1/2*x+1/4,x)

[Out]

x**2*log(x*exp(-x + exp(10 - 2*x)/25))**2/4 + x**2/4 + x/4

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