Optimal. Leaf size=24 \[ \frac {125 e^8 x}{\left (2+\frac {1}{2+\frac {e^{-x} x}{2}}\right )^2} \]
________________________________________________________________________________________
Rubi [F] time = 1.24, 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 {10000 e^{8+3 x}+125 e^8 x^3+e^{8+2 x} \left (8000 x-1000 x^2\right )+e^{8+x} \left (1875 x^2-250 x^3\right )}{500 e^{3 x}+300 e^{2 x} x+60 e^x x^2+4 x^3} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {125 e^8 \left (80 e^{3 x}-8 e^{2 x} (-8+x) x+x^3-e^x x^2 (-15+2 x)\right )}{4 \left (5 e^x+x\right )^3} \, dx\\ &=\frac {1}{4} \left (125 e^8\right ) \int \frac {80 e^{3 x}-8 e^{2 x} (-8+x) x+x^3-e^x x^2 (-15+2 x)}{\left (5 e^x+x\right )^3} \, dx\\ &=\frac {1}{4} \left (125 e^8\right ) \int \left (\frac {16}{25}+\frac {2 (-1+x) x^3}{25 \left (5 e^x+x\right )^3}-\frac {8 (-2+x) x}{25 \left (5 e^x+x\right )}+\frac {x^2 (-5+6 x)}{25 \left (5 e^x+x\right )^2}\right ) \, dx\\ &=20 e^8 x+\frac {1}{4} \left (5 e^8\right ) \int \frac {x^2 (-5+6 x)}{\left (5 e^x+x\right )^2} \, dx+\frac {1}{2} \left (5 e^8\right ) \int \frac {(-1+x) x^3}{\left (5 e^x+x\right )^3} \, dx-\left (10 e^8\right ) \int \frac {(-2+x) x}{5 e^x+x} \, dx\\ &=20 e^8 x+\frac {1}{4} \left (5 e^8\right ) \int \left (-\frac {5 x^2}{\left (5 e^x+x\right )^2}+\frac {6 x^3}{\left (5 e^x+x\right )^2}\right ) \, dx+\frac {1}{2} \left (5 e^8\right ) \int \left (-\frac {x^3}{\left (5 e^x+x\right )^3}+\frac {x^4}{\left (5 e^x+x\right )^3}\right ) \, dx-\left (10 e^8\right ) \int \left (-\frac {2 x}{5 e^x+x}+\frac {x^2}{5 e^x+x}\right ) \, dx\\ &=20 e^8 x-\frac {1}{2} \left (5 e^8\right ) \int \frac {x^3}{\left (5 e^x+x\right )^3} \, dx+\frac {1}{2} \left (5 e^8\right ) \int \frac {x^4}{\left (5 e^x+x\right )^3} \, dx-\frac {1}{4} \left (25 e^8\right ) \int \frac {x^2}{\left (5 e^x+x\right )^2} \, dx+\frac {1}{2} \left (15 e^8\right ) \int \frac {x^3}{\left (5 e^x+x\right )^2} \, dx-\left (10 e^8\right ) \int \frac {x^2}{5 e^x+x} \, dx+\left (20 e^8\right ) \int \frac {x}{5 e^x+x} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.22, size = 26, normalized size = 1.08 \begin {gather*} \frac {125 e^8 x \left (4 e^x+x\right )^2}{4 \left (5 e^x+x\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.74, size = 51, normalized size = 2.12 \begin {gather*} \frac {125 \, {\left (x^{3} e^{24} + 8 \, x^{2} e^{\left (x + 24\right )} + 16 \, x e^{\left (2 \, x + 24\right )}\right )}}{4 \, {\left (x^{2} e^{16} + 10 \, x e^{\left (x + 16\right )} + 25 \, e^{\left (2 \, x + 16\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [B] time = 0.23, size = 44, normalized size = 1.83 \begin {gather*} \frac {125 \, {\left (x^{3} e^{8} + 8 \, x^{2} e^{\left (x + 8\right )} + 16 \, x e^{\left (2 \, x + 8\right )}\right )}}{4 \, {\left (x^{2} + 10 \, x e^{x} + 25 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.10, size = 30, normalized size = 1.25
method | result | size |
risch | \(20 x \,{\mathrm e}^{8}+\frac {5 \left (9 x +40 \,{\mathrm e}^{x}\right ) x^{2} {\mathrm e}^{8}}{4 \left (5 \,{\mathrm e}^{x}+x \right )^{2}}\) | \(30\) |
norman | \(\frac {\frac {125 x^{3} {\mathrm e}^{8}}{4}+500 x \,{\mathrm e}^{8} {\mathrm e}^{2 x}+250 \,{\mathrm e}^{8} {\mathrm e}^{x} x^{2}}{\left (5 \,{\mathrm e}^{x}+x \right )^{2}}\) | \(42\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.57, size = 44, normalized size = 1.83 \begin {gather*} \frac {125 \, {\left (x^{3} e^{8} + 8 \, x^{2} e^{\left (x + 8\right )} + 16 \, x e^{\left (2 \, x + 8\right )}\right )}}{4 \, {\left (x^{2} + 10 \, x e^{x} + 25 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 0.15, size = 44, normalized size = 1.83 \begin {gather*} \frac {125\,x\,\left (16\,{\mathrm {e}}^{2\,x+8}+8\,x\,{\mathrm {e}}^{x+8}+x^2\,{\mathrm {e}}^8\right )}{4\,\left (25\,{\mathrm {e}}^{2\,x}+10\,x\,{\mathrm {e}}^x+x^2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [B] time = 0.16, size = 44, normalized size = 1.83 \begin {gather*} 20 x e^{8} + \frac {45 x^{3} e^{8} + 200 x^{2} e^{8} e^{x}}{4 x^{2} + 40 x e^{x} + 100 e^{2 x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________