Optimal. Leaf size=32 \[ 1+\frac {x}{e^3 \left (7+\frac {x-x^2+\log \left (5-x^2\right )}{5+x}\right )} \]
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Rubi [F]
time = 1.03, 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 {-875-350 x+100 x^2+68 x^3+13 x^4+\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (-6125-2800 x+1255 x^2+640 x^3-11 x^4-16 x^5+x^6\right )+e^3 \left (-350-80 x+80 x^2+16 x^3-2 x^4\right ) \log \left (5-x^2\right )+e^3 \left (-5+x^2\right ) \log ^2\left (5-x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {875+350 x-100 x^2-68 x^3-13 x^4-\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{e^3 \left (5-x^2\right ) \left (35+8 x-x^2+\log \left (5-x^2\right )\right )^2} \, dx\\ &=\frac {\int \frac {875+350 x-100 x^2-68 x^3-13 x^4-\left (-25-10 x+5 x^2+2 x^3\right ) \log \left (5-x^2\right )}{\left (5-x^2\right ) \left (35+8 x-x^2+\log \left (5-x^2\right )\right )^2} \, dx}{e^3}\\ &=\frac {\int \left (\frac {2 x \left (100-10 x-26 x^2+x^3+x^4\right )}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}+\frac {-5-2 x}{-35-8 x+x^2-\log \left (5-x^2\right )}\right ) \, dx}{e^3}\\ &=\frac {\int \frac {-5-2 x}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}+\frac {2 \int \frac {x \left (100-10 x-26 x^2+x^3+x^4\right )}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}\\ &=\frac {\int \left (-\frac {5}{-35-8 x+x^2-\log \left (5-x^2\right )}-\frac {2 x}{-35-8 x+x^2-\log \left (5-x^2\right )}\right ) \, dx}{e^3}+\frac {2 \int \left (-\frac {5}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}-\frac {21 x}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}+\frac {x^2}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}+\frac {x^3}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}-\frac {5 (5+x)}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}\right ) \, dx}{e^3}\\ &=\frac {2 \int \frac {x^2}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}+\frac {2 \int \frac {x^3}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {2 \int \frac {x}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {5 \int \frac {1}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {10 \int \frac {1}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {10 \int \frac {5+x}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {42 \int \frac {x}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}\\ &=\frac {2 \int \frac {x^2}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}+\frac {2 \int \frac {x^3}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {2 \int \frac {x}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {5 \int \frac {1}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {10 \int \left (\frac {5}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}+\frac {x}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}\right ) \, dx}{e^3}-\frac {10 \int \frac {1}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {42 \int \frac {x}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}\\ &=\frac {2 \int \frac {x^2}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}+\frac {2 \int \frac {x^3}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {2 \int \frac {x}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {5 \int \frac {1}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {10 \int \frac {1}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {10 \int \frac {x}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {42 \int \frac {x}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {50 \int \frac {1}{\left (-5+x^2\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}\\ &=\frac {2 \int \frac {x^2}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}+\frac {2 \int \frac {x^3}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {2 \int \frac {x}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {5 \int \frac {1}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {10 \int \left (-\frac {1}{2 \left (\sqrt {5}-x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}+\frac {1}{2 \left (\sqrt {5}+x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}\right ) \, dx}{e^3}-\frac {10 \int \frac {1}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {42 \int \frac {x}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {50 \int \left (-\frac {1}{2 \sqrt {5} \left (\sqrt {5}-x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}-\frac {1}{2 \sqrt {5} \left (\sqrt {5}+x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2}\right ) \, dx}{e^3}\\ &=\frac {2 \int \frac {x^2}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}+\frac {2 \int \frac {x^3}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {2 \int \frac {x}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}+\frac {5 \int \frac {1}{\left (\sqrt {5}-x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {5 \int \frac {1}{\left (\sqrt {5}+x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {5 \int \frac {1}{-35-8 x+x^2-\log \left (5-x^2\right )} \, dx}{e^3}-\frac {10 \int \frac {1}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}-\frac {42 \int \frac {x}{\left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}+\frac {\left (5 \sqrt {5}\right ) \int \frac {1}{\left (\sqrt {5}-x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}+\frac {\left (5 \sqrt {5}\right ) \int \frac {1}{\left (\sqrt {5}+x\right ) \left (-35-8 x+x^2-\log \left (5-x^2\right )\right )^2} \, dx}{e^3}\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.53, size = 28, normalized size = 0.88 \begin {gather*} \frac {x (5+x)}{e^3 \left (35+8 x-x^2+\log \left (5-x^2\right )\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.76, size = 29, normalized size = 0.91
method | result | size |
risch | \(-\frac {\left (5+x \right ) x \,{\mathrm e}^{-3}}{x^{2}-8 x -\ln \left (-x^{2}+5\right )-35}\) | \(29\) |
norman | \(\frac {-13 \,{\mathrm e}^{-3} x -{\mathrm e}^{-3} \ln \left (-x^{2}+5\right )-35 \,{\mathrm e}^{-3}}{x^{2}-8 x -\ln \left (-x^{2}+5\right )-35}\) | \(50\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, size = 39, normalized size = 1.22 \begin {gather*} -\frac {x^{2} + 5 \, x}{x^{2} e^{3} - 8 \, x e^{3} - e^{3} \log \left (-x^{2} + 5\right ) - 35 \, e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 35, normalized size = 1.09 \begin {gather*} -\frac {x^{2} + 5 \, x}{{\left (x^{2} - 8 \, x - 35\right )} e^{3} - e^{3} \log \left (-x^{2} + 5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.08, size = 34, normalized size = 1.06 \begin {gather*} \frac {x^{2} + 5 x}{- x^{2} e^{3} + 8 x e^{3} + e^{3} \log {\left (5 - x^{2} \right )} + 35 e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 0.43, size = 39, normalized size = 1.22 \begin {gather*} -\frac {x^{2} + 5 \, x}{x^{2} e^{3} - 8 \, x e^{3} - e^{3} \log \left (-x^{2} + 5\right ) - 35 \, e^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.65, size = 36, normalized size = 1.12 \begin {gather*} \frac {{\mathrm {e}}^{-3}\,\left (13\,x+\ln \left (5-x^2\right )+35\right )}{8\,x+\ln \left (5-x^2\right )-x^2+35} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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