Optimal. Leaf size=34 \[ -e^3+\frac {1}{4} e^4 \left (5+\frac {e^x x}{2-e^{3-x}+x}\right ) \]
________________________________________________________________________________________
Rubi [F] time = 2.20, 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 {e^x \left (e^{7-x} (-1-2 x)+e^4 \left (2+2 x+x^2\right )\right )}{16+4 e^{6-2 x}+e^{3-x} (-16-8 x)+16 x+4 x^2} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{3 x} \left (e^{7-x} (-1-2 x)+e^4 \left (2+2 x+x^2\right )\right )}{4 \left (e^3-2 e^x-e^x x\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {e^{3 x} \left (e^{7-x} (-1-2 x)+e^4 \left (2+2 x+x^2\right )\right )}{\left (e^3-2 e^x-e^x x\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-e^{1+2 x} (1+2 x)-\frac {e^{4+3 x} x (3+x)}{\left (e^3-2 e^x-e^x x\right )^2}-\frac {e^{1+3 x} \left (2+5 x+2 x^2\right )}{e^3-2 e^x-e^x x}\right ) \, dx\\ &=-\left (\frac {1}{4} \int e^{1+2 x} (1+2 x) \, dx\right )-\frac {1}{4} \int \frac {e^{4+3 x} x (3+x)}{\left (e^3-2 e^x-e^x x\right )^2} \, dx-\frac {1}{4} \int \frac {e^{1+3 x} \left (2+5 x+2 x^2\right )}{e^3-2 e^x-e^x x} \, dx\\ &=-\frac {1}{8} e^{1+2 x} (1+2 x)+\frac {1}{4} \int e^{1+2 x} \, dx-\frac {1}{4} \int \left (\frac {3 e^{4+3 x} x}{\left (-e^3+2 e^x+e^x x\right )^2}+\frac {e^{4+3 x} x^2}{\left (-e^3+2 e^x+e^x x\right )^2}\right ) \, dx-\frac {1}{4} \int \left (-\frac {2 e^{1+3 x}}{-e^3+2 e^x+e^x x}-\frac {5 e^{1+3 x} x}{-e^3+2 e^x+e^x x}-\frac {2 e^{1+3 x} x^2}{-e^3+2 e^x+e^x x}\right ) \, dx\\ &=\frac {1}{8} e^{1+2 x}-\frac {1}{8} e^{1+2 x} (1+2 x)-\frac {1}{4} \int \frac {e^{4+3 x} x^2}{\left (-e^3+2 e^x+e^x x\right )^2} \, dx+\frac {1}{2} \int \frac {e^{1+3 x}}{-e^3+2 e^x+e^x x} \, dx+\frac {1}{2} \int \frac {e^{1+3 x} x^2}{-e^3+2 e^x+e^x x} \, dx-\frac {3}{4} \int \frac {e^{4+3 x} x}{\left (-e^3+2 e^x+e^x x\right )^2} \, dx+\frac {5}{4} \int \frac {e^{1+3 x} x}{-e^3+2 e^x+e^x x} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.53, size = 27, normalized size = 0.79 \begin {gather*} \frac {e^{4+2 x} x}{4 \left (-e^3+e^x (2+x)\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 1.40, size = 22, normalized size = 0.65 \begin {gather*} \frac {x e^{\left (2 \, x + 4\right )}}{4 \, {\left ({\left (x + 2\right )} e^{x} - e^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.22, size = 28, normalized size = 0.82 \begin {gather*} \frac {x e^{4}}{4 \, {\left (x e^{\left (-x\right )} + 2 \, e^{\left (-x\right )} - e^{\left (-2 \, x + 3\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.20, size = 24, normalized size = 0.71
method | result | size |
norman | \(-\frac {x \,{\mathrm e}^{4} {\mathrm e}^{2 x}}{4 \left (-{\mathrm e}^{x} x +{\mathrm e}^{3}-2 \,{\mathrm e}^{x}\right )}\) | \(24\) |
risch | \(\frac {x \,{\mathrm e}^{7}}{4 x^{2}+16 x +16}+\frac {x \,{\mathrm e}^{4+x}}{4 x +8}-\frac {{\mathrm e}^{10} x}{4 \left (x^{2}+4 x +4\right ) \left (-{\mathrm e}^{x} x +{\mathrm e}^{3}-2 \,{\mathrm e}^{x}\right )}\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.49, size = 22, normalized size = 0.65 \begin {gather*} \frac {x e^{\left (2 \, x + 4\right )}}{4 \, {\left ({\left (x + 2\right )} e^{x} - e^{3}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 4.49, size = 53, normalized size = 1.56 \begin {gather*} \frac {x\,\left (2\,{\mathrm {e}}^{x+4}\,{\mathrm {e}}^x-{\mathrm {e}}^{x+4}\,{\mathrm {e}}^3+{\mathrm {e}}^7\,{\mathrm {e}}^x+x\,{\mathrm {e}}^{x+4}\,{\mathrm {e}}^x\right )}{4\,\left (x+2\right )\,\left (2\,{\mathrm {e}}^x-{\mathrm {e}}^3+x\,{\mathrm {e}}^x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 0.33, size = 71, normalized size = 2.09 \begin {gather*} \frac {x e^{10}}{- 4 x^{2} e^{3} - 16 x e^{3} + \left (4 x^{3} + 24 x^{2} + 48 x + 32\right ) e^{x} - 16 e^{3}} + \frac {x e^{7}}{4 x^{2} + 16 x + 16} + \frac {x e^{4} e^{x}}{4 x + 8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________