Optimal. Leaf size=19 \[ -\frac {15}{-9+e^x-x-\frac {6}{1+x}} \]
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Rubi [F] time = 0.85, 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 {75-30 x-15 x^2+e^x \left (15+30 x+15 x^2\right )}{225+300 x+130 x^2+20 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (-30-50 x-22 x^2-2 x^3\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 {15 \left (5-2 x-x^2+e^x (1+x)^2\right )}{\left (15+10 x+x^2-e^x (1+x)\right )^2} \, dx\\ &=15 \int \frac {5-2 x-x^2+e^x (1+x)^2}{\left (15+10 x+x^2-e^x (1+x)\right )^2} \, dx\\ &=15 \int \left (-\frac {1+x}{15-e^x+10 x-e^x x+x^2}+\frac {20+23 x+10 x^2+x^3}{\left (-15+e^x-10 x+e^x x-x^2\right )^2}\right ) \, dx\\ &=-\left (15 \int \frac {1+x}{15-e^x+10 x-e^x x+x^2} \, dx\right )+15 \int \frac {20+23 x+10 x^2+x^3}{\left (-15+e^x-10 x+e^x x-x^2\right )^2} \, dx\\ &=15 \int \left (\frac {20}{\left (-15+e^x-10 x+e^x x-x^2\right )^2}+\frac {23 x}{\left (15-e^x+10 x-e^x x+x^2\right )^2}+\frac {10 x^2}{\left (15-e^x+10 x-e^x x+x^2\right )^2}+\frac {x^3}{\left (15-e^x+10 x-e^x x+x^2\right )^2}\right ) \, dx-15 \int \left (-\frac {1}{-15+e^x-10 x+e^x x-x^2}+\frac {x}{15-e^x+10 x-e^x x+x^2}\right ) \, dx\\ &=15 \int \frac {1}{-15+e^x-10 x+e^x x-x^2} \, dx+15 \int \frac {x^3}{\left (15-e^x+10 x-e^x x+x^2\right )^2} \, dx-15 \int \frac {x}{15-e^x+10 x-e^x x+x^2} \, dx+150 \int \frac {x^2}{\left (15-e^x+10 x-e^x x+x^2\right )^2} \, dx+300 \int \frac {1}{\left (-15+e^x-10 x+e^x x-x^2\right )^2} \, dx+345 \int \frac {x}{\left (15-e^x+10 x-e^x x+x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.28, size = 23, normalized size = 1.21 \begin {gather*} \frac {15 (1+x)}{15+10 x+x^2-e^x (1+x)} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.60, size = 22, normalized size = 1.16 \begin {gather*} \frac {15 \, {\left (x + 1\right )}}{x^{2} - {\left (x + 1\right )} e^{x} + 10 \, x + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 24, normalized size = 1.26 \begin {gather*} \frac {15 \, {\left (x + 1\right )}}{x^{2} - x e^{x} + 10 \, x - e^{x} + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.08, size = 25, normalized size = 1.32
method | result | size |
risch | \(\frac {15 x +15}{x^{2}-{\mathrm e}^{x} x +10 x -{\mathrm e}^{x}+15}\) | \(25\) |
norman | \(\frac {15 x +15}{x^{2}-{\mathrm e}^{x} x +10 x -{\mathrm e}^{x}+15}\) | \(26\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.40, size = 22, normalized size = 1.16 \begin {gather*} \frac {15 \, {\left (x + 1\right )}}{x^{2} - {\left (x + 1\right )} e^{x} + 10 \, x + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 7.12, size = 24, normalized size = 1.26 \begin {gather*} -\frac {15\,x+15}{{\mathrm {e}}^x+x\,\left ({\mathrm {e}}^x-10\right )-x^2-15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.23, size = 20, normalized size = 1.05 \begin {gather*} \frac {- 15 x - 15}{- x^{2} - 10 x + \left (x + 1\right ) e^{x} - 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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