Optimal. Leaf size=30 \[ (15+x)^2+\frac {2}{\left (i \pi -\frac {1}{4} e^{3 x} x\right ) \log (x)} \]
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Rubi [F] time = 1.68, 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 {-32 i \pi +8 e^{3 x} x+e^{3 x} \left (8 x+24 x^2\right ) \log (x)+\left (-\pi ^2 \left (480 x+32 x^2\right )+i e^{3 x} \pi \left (-240 x^2-16 x^3\right )+e^{6 x} \left (30 x^3+2 x^4\right )\right ) \log ^2(x)}{\left (-16 \pi ^2 x-8 i e^{3 x} \pi x^2+e^{6 x} x^3\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} &=\int 2 \left (15+x+\frac {4}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)}+\frac {4 e^{3 x} (1+3 x)}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)}\right ) \, dx\\ &=2 \int \left (15+x+\frac {4}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)}+\frac {4 e^{3 x} (1+3 x)}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)}\right ) \, dx\\ &=30 x+x^2+8 \int \frac {1}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)} \, dx+8 \int \frac {e^{3 x} (1+3 x)}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)} \, dx\\ &=30 x+x^2+8 \int \left (-\frac {e^{3 x}}{\left (4 \pi +i e^{3 x} x\right )^2 \log (x)}+\frac {3 e^{3 x} x}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)}\right ) \, dx+8 \int \frac {1}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)} \, dx\\ &=30 x+x^2+8 \int \frac {1}{x \left (-4 i \pi +e^{3 x} x\right ) \log ^2(x)} \, dx-8 \int \frac {e^{3 x}}{\left (4 \pi +i e^{3 x} x\right )^2 \log (x)} \, dx+24 \int \frac {e^{3 x} x}{\left (-4 i \pi +e^{3 x} x\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.23, size = 37, normalized size = 1.23 \begin {gather*} 2 \left (15 x+\frac {x^2}{2}-\frac {4 i}{\left (4 \pi +i e^{3 x} x\right ) \log (x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.82, size = 47, normalized size = 1.57 \begin {gather*} \frac {{\left (-4 i \, \pi x^{2} - 120 i \, \pi x + {\left (x^{3} + 30 \, x^{2}\right )} e^{\left (3 \, x\right )}\right )} \log \relax (x) - 8}{{\left (-4 i \, \pi + x e^{\left (3 \, x\right )}\right )} \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.78, size = 57, normalized size = 1.90 \begin {gather*} -\frac {2 \, {\left (-i \, x^{3} e^{\left (3 \, x\right )} \log \relax (x) - 4 \, \pi x^{2} \log \relax (x) - 30 i \, x^{2} e^{\left (3 \, x\right )} \log \relax (x) - 120 \, \pi x \log \relax (x) + 8 i\right )}}{2 i \, x e^{\left (3 \, x\right )} \log \relax (x) + 8 \, \pi \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.23, size = 29, normalized size = 0.97
method | result | size |
risch | \(x^{2}+30 x -\frac {8 i}{\left (i x \,{\mathrm e}^{3 x}+4 \pi \right ) \ln \relax (x )}\) | \(29\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.50, size = 50, normalized size = 1.67 \begin {gather*} \frac {{\left (x^{3} + 30 \, x^{2}\right )} e^{\left (3 \, x\right )} \log \relax (x) - 4 \, {\left (i \, \pi x^{2} + 30 i \, \pi x\right )} \log \relax (x) - 8}{x e^{\left (3 \, x\right )} \log \relax (x) - 4 i \, \pi \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.30, size = 27, normalized size = 0.90 \begin {gather*} 30\,x+\frac {8}{\ln \relax (x)\,\left (-x\,{\mathrm {e}}^{3\,x}+\Pi \,4{}\mathrm {i}\right )}+x^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.50, size = 26, normalized size = 0.87 \begin {gather*} x^{2} + 30 x + \frac {8}{- x e^{3 x} \log {\relax (x )} + 4 i \pi \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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