Optimal. Leaf size=29 \[ 9-e^{\frac {e^{25-\frac {-3+x}{e^5 x}}}{3 x}}+x \]
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Rubi [F] time = 0.99, 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 {3 e^5 x^3+\exp \left (\frac {e^{\frac {3-x+25 e^5 x}{e^5 x}}}{3 x}+\frac {3-x+25 e^5 x}{e^5 x}\right ) \left (3+e^5 x\right )}{3 e^5 x^3} \, 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 {3 e^5 x^3+\exp \left (\frac {e^{\frac {3-x+25 e^5 x}{e^5 x}}}{3 x}+\frac {3-x+25 e^5 x}{e^5 x}\right ) \left (3+e^5 x\right )}{x^3} \, dx}{3 e^5}\\ &=\frac {\int \left (3 e^5+\frac {\exp \left (25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right ) \left (3+e^5 x\right )}{x^3}\right ) \, dx}{3 e^5}\\ &=x+\frac {\int \frac {\exp \left (25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right ) \left (3+e^5 x\right )}{x^3} \, dx}{3 e^5}\\ &=x+\frac {\int \left (\frac {3 \exp \left (25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right )}{x^3}+\frac {\exp \left (5+25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right )}{x^2}\right ) \, dx}{3 e^5}\\ &=x+\frac {\int \frac {\exp \left (5+25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right )}{x^2} \, dx}{3 e^5}+\frac {\int \frac {\exp \left (25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right )}{x^3} \, dx}{e^5}\\ &=x-\frac {\operatorname {Subst}\left (\int \exp \left (5+25 \left (1-\frac {1}{25 e^5}\right )+\frac {3 x}{e^5}+\frac {1}{3} e^{25-\frac {1}{e^5}+\frac {3 x}{e^5}} x\right ) \, dx,x,\frac {1}{x}\right )}{3 e^5}+\frac {\int \frac {\exp \left (25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right )}{x^3} \, dx}{e^5}\\ &=x-\frac {\operatorname {Subst}\left (\int \exp \left (5 \left (6-\frac {1}{5 e^5}\right )+\frac {3 x}{e^5}+\frac {1}{3} e^{25-\frac {1}{e^5}+\frac {3 x}{e^5}} x\right ) \, dx,x,\frac {1}{x}\right )}{3 e^5}+\frac {\int \frac {\exp \left (25 \left (1-\frac {1}{25 e^5}\right )+\frac {3}{e^5 x}+\frac {e^{25-\frac {1}{e^5}+\frac {3}{e^5 x}}}{3 x}\right )}{x^3} \, dx}{e^5}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.48, size = 28, normalized size = 0.97 \begin {gather*} -e^{\frac {e^{25+\frac {-1+\frac {3}{x}}{e^5}}}{3 x}}+x \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.95, size = 78, normalized size = 2.69 \begin {gather*} {\left (x e^{\left (\frac {{\left (25 \, x e^{5} - x + 3\right )} e^{\left (-5\right )}}{x}\right )} - e^{\left (\frac {{\left (75 \, x e^{5} - 3 \, x + e^{\left (\frac {{\left (25 \, x e^{5} - x + 3\right )} e^{\left (-5\right )}}{x} + 5\right )} + 9\right )} e^{\left (-5\right )}}{3 \, x}\right )}\right )} e^{\left (-\frac {{\left (25 \, x e^{5} - x + 3\right )} e^{\left (-5\right )}}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (3 \, x^{3} e^{5} + {\left (x e^{5} + 3\right )} e^{\left (\frac {{\left (25 \, x e^{5} - x + 3\right )} e^{\left (-5\right )}}{x} + \frac {e^{\left (\frac {{\left (25 \, x e^{5} - x + 3\right )} e^{\left (-5\right )}}{x}\right )}}{3 \, x}\right )}\right )} e^{\left (-5\right )}}{3 \, x^{3}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.19, size = 28, normalized size = 0.97
method | result | size |
risch | \(x -{\mathrm e}^{\frac {{\mathrm e}^{\frac {\left (25 x \,{\mathrm e}^{5}+3-x \right ) {\mathrm e}^{-5}}{x}}}{3 x}}\) | \(28\) |
norman | \(\frac {x^{3}-x^{2} {\mathrm e}^{\frac {{\mathrm e}^{\frac {\left (25 x \,{\mathrm e}^{5}+3-x \right ) {\mathrm e}^{-5}}{x}}}{3 x}}}{x^{2}}\) | \(39\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.86, size = 32, normalized size = 1.10 \begin {gather*} {\left (x e^{5} - e^{\left (\frac {e^{\left (\frac {3 \, e^{\left (-5\right )}}{x} - e^{\left (-5\right )} + 25\right )}}{3 \, x} + 5\right )}\right )} e^{\left (-5\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.52, size = 25, normalized size = 0.86 \begin {gather*} x-{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{-5}}{x}}\,{\mathrm {e}}^{-{\mathrm {e}}^{-5}}\,{\mathrm {e}}^{25}}{3\,x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.38, size = 22, normalized size = 0.76 \begin {gather*} x - e^{\frac {e^{\frac {- x + 25 x e^{5} + 3}{x e^{5}}}}{3 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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