Optimal. Leaf size=29 \[ \log (x)+\frac {-\frac {2}{3}-x+\frac {-2+\frac {1}{5} \log (1+x)}{e^4}}{\log (x)} \]
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Rubi [F]
time = 1.31, 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 {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{e^4 \left (15 x+15 x^2\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{\left (15 x+15 x^2\right ) \log ^2(x)} \, dx}{e^4}\\ &=\frac {\int \frac {30+30 x+e^4 \left (10+25 x+15 x^2\right )+\left (3 x+e^4 \left (-15 x-15 x^2\right )\right ) \log (x)+e^4 (15+15 x) \log ^2(x)+(-3-3 x) \log (1+x)}{x (15+15 x) \log ^2(x)} \, dx}{e^4}\\ &=\frac {\int \left (\frac {30 \left (1+\frac {e^4}{3}\right )+30 \left (1+\frac {5 e^4}{6}\right ) x+15 e^4 x^2+3 \left (1-5 e^4\right ) x \log (x)-15 e^4 x^2 \log (x)+15 e^4 \log ^2(x)+15 e^4 x \log ^2(x)}{15 x (1+x) \log ^2(x)}-\frac {\log (1+x)}{5 x \log ^2(x)}\right ) \, dx}{e^4}\\ &=\frac {\int \frac {30 \left (1+\frac {e^4}{3}\right )+30 \left (1+\frac {5 e^4}{6}\right ) x+15 e^4 x^2+3 \left (1-5 e^4\right ) x \log (x)-15 e^4 x^2 \log (x)+15 e^4 \log ^2(x)+15 e^4 x \log ^2(x)}{x (1+x) \log ^2(x)} \, dx}{15 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ &=\frac {\int \frac {5 \left (6 (1+x)+e^4 \left (2+5 x+3 x^2\right )\right )-3 x \left (-1+5 e^4 (1+x)\right ) \log (x)+15 e^4 (1+x) \log ^2(x)}{x (1+x) \log ^2(x)} \, dx}{15 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ &=\frac {\int \left (\frac {15 e^4}{x}+\frac {5 \left (6+2 e^4+3 e^4 x\right )}{x \log ^2(x)}-\frac {3 \left (-1+5 e^4+5 e^4 x\right )}{(1+x) \log (x)}\right ) \, dx}{15 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ &=\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\int \frac {6+2 e^4+3 e^4 x}{x \log ^2(x)} \, dx}{3 e^4}\\ &=\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\int \left (\frac {3 e^4}{\log ^2(x)}+\frac {2 \left (3+e^4\right )}{x \log ^2(x)}\right ) \, dx}{3 e^4}\\ &=\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\left (2 \left (3+e^4\right )\right ) \int \frac {1}{x \log ^2(x)} \, dx}{3 e^4}+\int \frac {1}{\log ^2(x)} \, dx\\ &=-\frac {x}{\log (x)}+\log (x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}+\frac {\left (2 \left (3+e^4\right )\right ) \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (x)\right )}{3 e^4}+\int \frac {1}{\log (x)} \, dx\\ &=-\frac {2 \left (3+e^4\right )}{3 e^4 \log (x)}-\frac {x}{\log (x)}+\log (x)+\text {li}(x)-\frac {\int \frac {-1+5 e^4+5 e^4 x}{(1+x) \log (x)} \, dx}{5 e^4}-\frac {\int \frac {\log (1+x)}{x \log ^2(x)} \, dx}{5 e^4}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.44, size = 42, normalized size = 1.45 \begin {gather*} -\frac {6+2 e^4+3 e^4 x}{3 e^4 \log (x)}+\log (x)+\frac {\log (1+x)}{5 e^4 \log (x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 27.01, size = 41, normalized size = 1.41
method | result | size |
risch | \(\frac {{\mathrm e}^{-4} \ln \left (x +1\right )}{5 \ln \left (x \right )}+\frac {{\mathrm e}^{-4} \left (3 \,{\mathrm e}^{4} \ln \left (x \right )^{2}-3 x \,{\mathrm e}^{4}-2 \,{\mathrm e}^{4}-6\right )}{3 \ln \left (x \right )}\) | \(41\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.29, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left (15 \, e^{4} \log \left (x\right )^{2} - 15 \, x e^{4} - 10 \, e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.42, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left (15 \, e^{4} \log \left (x\right )^{2} - 5 \, {\left (3 \, x + 2\right )} e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 33, normalized size = 1.14 \begin {gather*} \frac {{\left (15 \, e^{4} \log \left (x\right )^{2} - 15 \, x e^{4} - 10 \, e^{4} + 3 \, \log \left (x + 1\right ) - 30\right )} e^{\left (-4\right )}}{15 \, \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 5.62, size = 28, normalized size = 0.97 \begin {gather*} \ln \left (x\right )-\frac {{\mathrm {e}}^{-4}\,\left (10\,{\mathrm {e}}^4-3\,\ln \left (x+1\right )+15\,x\,{\mathrm {e}}^4+30\right )}{15\,\ln \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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