Optimal. Leaf size=32 \[ \frac {5 x}{3 \left (-e^x+\frac {-1+x}{x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}\right )} \]
________________________________________________________________________________________
Rubi [F] time = 24.20, 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 {540 x-315 x^2+50 x^3-25 x^4+e^x \left (-1125 x^2+1125 x^3-125 x^4+125 x^5\right )+\left (-90 x+45 x^2-10 x^3+5 x^4+e^x \left (450 x^2-450 x^3+50 x^4-50 x^5\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^x \left (-45 x^2+45 x^3-5 x^4+5 x^5\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )}{27-54 x+30 x^2-6 x^3+3 x^4+e^x \left (-270 x+270 x^2-30 x^3+30 x^4\right )+e^{2 x} \left (675 x^2+75 x^4\right )+\left (e^{2 x} \left (-270 x^2-30 x^4\right )+e^x \left (54 x-54 x^2+6 x^3-6 x^4\right )\right ) \log \left (\frac {9+x^2}{x^2}\right )+e^{2 x} \left (27 x^2+3 x^4\right ) \log ^2\left (\frac {9+x^2}{x^2}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5 x \left (108-9 \left (7+25 e^x\right ) x+5 \left (2+45 e^x\right ) x^2-5 \left (1+5 e^x\right ) x^3+25 e^x x^4-\left (9+x^2\right ) \left (2-\left (1+10 e^x\right ) x+10 e^x x^2\right ) \log \left (1+\frac {9}{x^2}\right )+e^x x \left (-9+9 x-x^2+x^3\right ) \log ^2\left (1+\frac {9}{x^2}\right )\right )}{3 \left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\frac {5}{3} \int \frac {x \left (108-9 \left (7+25 e^x\right ) x+5 \left (2+45 e^x\right ) x^2-5 \left (1+5 e^x\right ) x^3+25 e^x x^4-\left (9+x^2\right ) \left (2-\left (1+10 e^x\right ) x+10 e^x x^2\right ) \log \left (1+\frac {9}{x^2}\right )+e^x x \left (-9+9 x-x^2+x^3\right ) \log ^2\left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\frac {5}{3} \int \left (\frac {(-1+x) x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4-9 \log \left (1+\frac {9}{x^2}\right )-9 x \log \left (1+\frac {9}{x^2}\right )+8 x^2 \log \left (1+\frac {9}{x^2}\right )-x^3 \log \left (1+\frac {9}{x^2}\right )+x^4 \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}\right ) \, dx\\ &=\frac {5}{3} \int \frac {(-1+x) x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx+\frac {5}{3} \int \frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4-9 \log \left (1+\frac {9}{x^2}\right )-9 x \log \left (1+\frac {9}{x^2}\right )+8 x^2 \log \left (1+\frac {9}{x^2}\right )-x^3 \log \left (1+\frac {9}{x^2}\right )+x^4 \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\frac {5}{3} \int \frac {x \left (63+27 x-40 x^2+5 x^3-5 x^4+\left (-9-9 x+8 x^2-x^3+x^4\right ) \log \left (1+\frac {9}{x^2}\right )\right )}{\left (9+x^2\right ) \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+\frac {5}{3} \int \left (-\frac {x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx\\ &=-\left (\frac {5}{3} \int \frac {x \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx\right )+\frac {5}{3} \int \frac {x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )} \, dx+\frac {5}{3} \int \left (\frac {63 x}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {27 x^2}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {40 x^3}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {5 x^4}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {5 x^5}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {9 x \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {9 x^2 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {8 x^3 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}-\frac {x^4 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}+\frac {x^5 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2}\right ) \, dx\\ &=-\left (\frac {5}{3} \int \frac {x^4 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\right )+\frac {5}{3} \int \frac {x^5 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {5}{3} \int \left (-\frac {5 x}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x \log \left (1+\frac {9}{x^2}\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx+\frac {5}{3} \int \left (-\frac {5 x^2}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}+\frac {x^2 \log \left (1+\frac {9}{x^2}\right )}{1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )}\right ) \, dx+\frac {25}{3} \int \frac {x^4}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {25}{3} \int \frac {x^5}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+\frac {40}{3} \int \frac {x^3 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-15 \int \frac {x \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-15 \int \frac {x^2 \log \left (1+\frac {9}{x^2}\right )}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+45 \int \frac {x^2}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx-\frac {200}{3} \int \frac {x^3}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx+105 \int \frac {x}{\left (9+x^2\right ) \left (1-x-5 e^x x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.16, size = 44, normalized size = 1.38 \begin {gather*} -\frac {5 x^2 \left (-5+\log \left (1+\frac {9}{x^2}\right )\right )}{3 \left (1-\left (1+5 e^x\right ) x+e^x x \log \left (1+\frac {9}{x^2}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 1.00, size = 48, normalized size = 1.50 \begin {gather*} -\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.95, size = 48, normalized size = 1.50 \begin {gather*} -\frac {5 \, {\left (x^{2} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x^{2}\right )}}{3 \, {\left (x e^{x} \log \left (\frac {x^{2} + 9}{x^{2}}\right ) - 5 \, x e^{x} - x + 1\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [C] time = 0.20, size = 216, normalized size = 6.75
method | result | size |
risch | \(-\frac {5 x \,{\mathrm e}^{-x}}{3}+\frac {10 i x \left (x -1\right ) {\mathrm e}^{-x}}{3 \left (\pi x \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right ) {\mathrm e}^{x}-2 \pi x \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2} {\mathrm e}^{x}+\pi x \mathrm {csgn}\left (i x^{2}\right )^{3} {\mathrm e}^{x}-\pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \left (x^{2}+9\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right ) {\mathrm e}^{x}+\pi x \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )^{2} {\mathrm e}^{x}+\pi x \,\mathrm {csgn}\left (i \left (x^{2}+9\right )\right ) \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )^{2} {\mathrm e}^{x}-\pi x \mathrm {csgn}\left (\frac {i \left (x^{2}+9\right )}{x^{2}}\right )^{3} {\mathrm e}^{x}+4 i x \,{\mathrm e}^{x} \ln \relax (x )-2 i x \,{\mathrm e}^{x} \ln \left (x^{2}+9\right )+10 i x \,{\mathrm e}^{x}+2 i x -2 i\right )}\) | \(216\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.63, size = 69, normalized size = 2.16 \begin {gather*} -\frac {5 \, {\left (x^{2} e^{x} \log \left (x^{2} + 9\right ) - {\left (2 \, x^{2} \log \relax (x) + 5 \, x^{2}\right )} e^{x}\right )}}{3 \, {\left (x e^{\left (2 \, x\right )} \log \left (x^{2} + 9\right ) - {\left (2 \, x \log \relax (x) + 5 \, x\right )} e^{\left (2 \, x\right )} - {\left (x - 1\right )} e^{x}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.35, size = 44, normalized size = 1.38 \begin {gather*} \frac {5\,x^2\,\left (\ln \left (\frac {x^2+9}{x^2}\right )-5\right )}{3\,\left (x+5\,x\,{\mathrm {e}}^x-x\,{\mathrm {e}}^x\,\ln \left (\frac {x^2+9}{x^2}\right )-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.63, size = 44, normalized size = 1.38 \begin {gather*} \frac {- 5 x^{2} \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} + 25 x^{2}}{- 3 x + \left (3 x \log {\left (\frac {x^{2} + 9}{x^{2}} \right )} - 15 x\right ) e^{x} + 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________