3.101.81 \(\int \frac {e^x (1-2 x-4 x^2)+e^x (-4-5 x) \log (x)+(e^x (x^2+x^3)+e^x (x+x^2) \log (x)) \log (\frac {10 x^4+10 x^5}{x+\log (x)})}{(x^2+x^3+(x+x^2) \log (x)) \log ^2(\frac {10 x^4+10 x^5}{x+\log (x)})} \, dx\)

Optimal. Leaf size=23 \[ \frac {e^x}{\log \left (\frac {10 x^3 \left (x+x^2\right )}{x+\log (x)}\right )} \]

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Rubi [F]  time = 17.99, 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 {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{\left (x^2+x^3+\left (x+x^2\right ) \log (x)\right ) \log ^2\left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(1 - 2*x - 4*x^2) + E^x*(-4 - 5*x)*Log[x] + (E^x*(x^2 + x^3) + E^x*(x + x^2)*Log[x])*Log[(10*x^4 + 10
*x^5)/(x + Log[x])])/((x^2 + x^3 + (x + x^2)*Log[x])*Log[(10*x^4 + 10*x^5)/(x + Log[x])]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (1-2 x-4 x^2\right )+e^x (-4-5 x) \log (x)+\left (e^x \left (x^2+x^3\right )+e^x \left (x+x^2\right ) \log (x)\right ) \log \left (\frac {10 x^4+10 x^5}{x+\log (x)}\right )}{x (1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ &=\int \left (\frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)+x^2 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^3 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^2 \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )\right )}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)+x^2 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^3 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^2 \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )\right )}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx\\ &=\int \frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)+x^2 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^3 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^2 \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )\right )}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-\int \frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)+x^2 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^3 \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+x^2 \log (x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )\right )}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ &=\int \frac {e^x \left (1-2 x-4 x^2+x^2 (1+x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+\log (x) \left (-4-5 x+x (1+x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )\right )\right )}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-\int \frac {e^x \left (1-2 x-4 x^2+x^2 (1+x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )+\log (x) \left (-4-5 x+x (1+x) \log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )\right )\right )}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ &=-\int \left (\frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)\right )}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}+\frac {e^x x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx+\int \left (\frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)\right )}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}+\frac {e^x (1+x)}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx\\ &=\int \frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)\right )}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-\int \frac {e^x \left (1-2 x-4 x^2-4 \log (x)-5 x \log (x)\right )}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-\int \frac {e^x x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+\int \frac {e^x (1+x)}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ &=\int \left (-\frac {2 e^x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}+\frac {e^x}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {4 e^x x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {5 e^x \log (x)}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {4 e^x \log (x)}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx-\int \left (\frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {2 e^x x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {4 e^x x^2}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {4 e^x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {5 e^x x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx+\int \left (\frac {e^x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}+\frac {e^x x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx-\int \frac {e^x x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ &=-\left (2 \int \frac {e^x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\right )+2 \int \frac {e^x x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-4 \int \frac {e^x x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+4 \int \frac {e^x x^2}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-4 \int \frac {e^x \log (x)}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+4 \int \frac {e^x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-5 \int \frac {e^x \log (x)}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+5 \int \frac {e^x x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+\int \frac {e^x}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-\int \frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+\int \frac {e^x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ &=2 \int \left (\frac {e^x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx-2 \int \frac {e^x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+4 \int \left (-\frac {e^x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}+\frac {e^x x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}+\frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx-4 \int \frac {e^x x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-4 \int \frac {e^x \log (x)}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+4 \int \frac {e^x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+5 \int \left (\frac {e^x \log (x)}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}-\frac {e^x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )}\right ) \, dx-5 \int \frac {e^x \log (x)}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+\int \frac {e^x}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-\int \frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+\int \frac {e^x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ &=-\left (2 \int \frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\right )-4 \int \frac {e^x}{(x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+4 \int \frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-4 \int \frac {e^x \log (x)}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+4 \int \frac {e^x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-5 \int \frac {e^x \log (x)}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+\int \frac {e^x}{x (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx-\int \frac {e^x}{(1+x) (x+\log (x)) \log ^2\left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx+\int \frac {e^x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 21, normalized size = 0.91 \begin {gather*} \frac {e^x}{\log \left (\frac {10 x^4 (1+x)}{x+\log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 - 2*x - 4*x^2) + E^x*(-4 - 5*x)*Log[x] + (E^x*(x^2 + x^3) + E^x*(x + x^2)*Log[x])*Log[(10*x^
4 + 10*x^5)/(x + Log[x])])/((x^2 + x^3 + (x + x^2)*Log[x])*Log[(10*x^4 + 10*x^5)/(x + Log[x])]^2),x]

[Out]

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

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fricas [A]  time = 0.48, size = 21, normalized size = 0.91 \begin {gather*} \frac {e^{x}}{\log \left (\frac {10 \, {\left (x^{5} + x^{4}\right )}}{x + \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x/log(10*(x^5 + x^4)/(x + log(x)))

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giac [A]  time = 0.40, size = 23, normalized size = 1.00 \begin {gather*} \frac {e^{x}}{\log \left (10 \, x + 10\right ) - \log \left (x + \log \relax (x)\right ) + 4 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.16, size = 441, normalized size = 19.17

method	result	size
risch	\(\frac {2 i {\mathrm e}^{x}}{\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{x +\ln \relax (x )}\right )+\pi \,\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x^{4} \left (x +1\right )}{x +\ln \relax (x )}\right )+2 i \ln \relax (2)+2 i \ln \relax (5)+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-\pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i x^{4} \left (x +1\right )}{x +\ln \relax (x )}\right )^{3}+\pi \mathrm {csgn}\left (i x^{4}\right )^{3}+\pi \mathrm {csgn}\left (\frac {i \left (x +1\right )}{x +\ln \relax (x )}\right )^{3}-\pi \,\mathrm {csgn}\left (\frac {i}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{x +\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (x +1\right )\right ) \mathrm {csgn}\left (\frac {i \left (x +1\right )}{x +\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-\pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-\pi \,\mathrm {csgn}\left (i x^{4}\right ) \mathrm {csgn}\left (\frac {i x^{4} \left (x +1\right )}{x +\ln \relax (x )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i \left (x +1\right )}{x +\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x^{4} \left (x +1\right )}{x +\ln \relax (x )}\right )^{2}+\pi \mathrm {csgn}\left (i x^{3}\right )^{3}-2 i \ln \left (x +\ln \relax (x )\right )+2 i \ln \left (x +1\right )+8 i \ln \relax (x )}\)	\(441\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*exp(x)/(Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)+Pi*csgn(I*x^2)^3+Pi*csgn(I*x)*csgn(I*x^3)*csgn(I*x^4)+Pi*csgn
(I/(x+ln(x)))*csgn(I*(x+1))*csgn(I*(x+1)/(x+ln(x)))+Pi*csgn(I*x^4)*csgn(I*(x+1)/(x+ln(x)))*csgn(I*x^4/(x+ln(x)
)*(x+1))+2*I*ln(2)+2*I*ln(5)+Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2-Pi*csgn(I*x)*csgn(I*x^3)^
2-Pi*csgn(I*x^2)*csgn(I*x^3)^2+Pi*csgn(I*x^4/(x+ln(x))*(x+1))^3+Pi*csgn(I*x^4)^3+Pi*csgn(I*(x+1)/(x+ln(x)))^3-
Pi*csgn(I/(x+ln(x)))*csgn(I*(x+1)/(x+ln(x)))^2-Pi*csgn(I*(x+1))*csgn(I*(x+1)/(x+ln(x)))^2-Pi*csgn(I*x)*csgn(I*
x^4)^2-Pi*csgn(I*x^3)*csgn(I*x^4)^2-Pi*csgn(I*x^4)*csgn(I*x^4/(x+ln(x))*(x+1))^2-Pi*csgn(I*(x+1)/(x+ln(x)))*cs
gn(I*x^4/(x+ln(x))*(x+1))^2+Pi*csgn(I*x^3)^3-2*I*ln(x+ln(x))+2*I*ln(x+1)+8*I*ln(x))

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maxima [A]  time = 0.50, size = 25, normalized size = 1.09 \begin {gather*} \frac {e^{x}}{\log \relax (5) + \log \relax (2) - \log \left (x + \log \relax (x)\right ) + \log \left (x + 1\right ) + 4 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^x/(log(5) + log(2) - log(x + log(x)) + log(x + 1) + 4*log(x))

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mupad [B]  time = 7.25, size = 24, normalized size = 1.04 \begin {gather*} \frac {{\mathrm {e}}^x}{\ln \left (\frac {10\,x^5+10\,x^4}{x+\ln \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(2*x + 4*x^2 - 1) - log((10*x^4 + 10*x^5)/(x + log(x)))*(exp(x)*(x^2 + x^3) + exp(x)*log(x)*(x +
x^2)) + exp(x)*log(x)*(5*x + 4))/(log((10*x^4 + 10*x^5)/(x + log(x)))^2*(x^2 + x^3 + log(x)*(x + x^2))),x)

[Out]

exp(x)/log((10*x^4 + 10*x^5)/(x + log(x)))

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sympy [A]  time = 0.72, size = 19, normalized size = 0.83 \begin {gather*} \frac {e^{x}}{\log {\left (\frac {10 x^{5} + 10 x^{4}}{x + \log {\relax (x )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(x)/log((10*x**5 + 10*x**4)/(x + log(x)))

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