Optimal. Leaf size=22 \[ \frac {\left (-4+e^x\right )^2}{\left (1-x^2+(4+x)^2\right )^2} \]
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Rubi [A] time = 0.33, antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 6, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6688, 12, 6742, 2197} \begin {gather*} -\frac {8 e^x}{(8 x+17)^2}+\frac {e^{2 x}}{(8 x+17)^2}+\frac {16}{(8 x+17)^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2197
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (4-e^x\right ) \left (-32-e^x (9+8 x)\right )}{(17+8 x)^3} \, dx\\ &=2 \int \frac {\left (4-e^x\right ) \left (-32-e^x (9+8 x)\right )}{(17+8 x)^3} \, dx\\ &=2 \int \left (-\frac {128}{(17+8 x)^3}-\frac {4 e^x (1+8 x)}{(17+8 x)^3}+\frac {e^{2 x} (9+8 x)}{(17+8 x)^3}\right ) \, dx\\ &=\frac {16}{(17+8 x)^2}+2 \int \frac {e^{2 x} (9+8 x)}{(17+8 x)^3} \, dx-8 \int \frac {e^x (1+8 x)}{(17+8 x)^3} \, dx\\ &=\frac {16}{(17+8 x)^2}-\frac {8 e^x}{(17+8 x)^2}+\frac {e^{2 x}}{(17+8 x)^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.14, size = 15, normalized size = 0.68 \begin {gather*} \frac {\left (-4+e^x\right )^2}{(17+8 x)^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.57, size = 23, normalized size = 1.05 \begin {gather*} \frac {e^{\left (2 \, x\right )} - 8 \, e^{x} + 16}{64 \, x^{2} + 272 \, x + 289} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.22, size = 23, normalized size = 1.05 \begin {gather*} \frac {e^{\left (2 \, x\right )} - 8 \, e^{x} + 16}{64 \, x^{2} + 272 \, x + 289} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.04, size = 19, normalized size = 0.86
method | result | size |
norman | \(\frac {{\mathrm e}^{2 x}+16-8 \,{\mathrm e}^{x}}{\left (8 x +17\right )^{2}}\) | \(19\) |
default | \(\frac {16}{\left (8 x +17\right )^{2}}-\frac {{\mathrm e}^{x}}{8 \left (x +\frac {17}{8}\right )^{2}}+\frac {{\mathrm e}^{2 x}}{64 \left (x +\frac {17}{8}\right )^{2}}\) | \(31\) |
risch | \(\frac {1}{4 x^{2}+17 x +\frac {289}{16}}+\frac {{\mathrm e}^{2 x}}{\left (8 x +17\right )^{2}}-\frac {8 \,{\mathrm e}^{x}}{\left (8 x +17\right )^{2}}\) | \(37\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {{\left (8 \, x + 17\right )} e^{\left (2 \, x\right )} - 64 \, x e^{x}}{512 \, x^{3} + 3264 \, x^{2} + 6936 \, x + 4913} + \frac {e^{\left (-\frac {17}{8}\right )} E_{3}\left (-x - \frac {17}{8}\right )}{{\left (8 \, x + 17\right )}^{2}} + \frac {16}{64 \, x^{2} + 272 \, x + 289} - 2 \, \int \frac {32 \, {\left (16 \, x - 17\right )} e^{x}}{4096 \, x^{4} + 34816 \, x^{3} + 110976 \, x^{2} + 157216 \, x + 83521}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.12, size = 14, normalized size = 0.64 \begin {gather*} \frac {{\left ({\mathrm {e}}^x-4\right )}^2}{{\left (8\,x+17\right )}^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.15, size = 61, normalized size = 2.77 \begin {gather*} \frac {\left (- 512 x^{2} - 2176 x - 2312\right ) e^{x} + \left (64 x^{2} + 272 x + 289\right ) e^{2 x}}{4096 x^{4} + 34816 x^{3} + 110976 x^{2} + 157216 x + 83521} + \frac {256}{1024 x^{2} + 4352 x + 4624} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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