Optimal. Leaf size=25 \[ \frac {e^x}{8+5 \log (3)-4 \left (1+x^2-4 (-9+\log (5))\right )} \]
________________________________________________________________________________________
Rubi [B] time = 0.40, antiderivative size = 73, normalized size of antiderivative = 2.92, number of steps used = 1, number of rules used = 1, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2288} \begin {gather*} -\frac {e^x \left (4 x^2+140-\log (243)-16 \log (5)\right )}{16 x^4+1120 x^2-32 \log (5) \left (4 x^2+5 (28-\log (3))\right )-40 \left (x^2+35\right ) \log (3)+19600+256 \log ^2(5)+25 \log ^2(3)} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 2288
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=-\frac {e^x \left (140+4 x^2-16 \log (5)-\log (243)\right )}{19600+1120 x^2+16 x^4-40 \left (35+x^2\right ) \log (3)+25 \log ^2(3)-32 \left (4 x^2+5 (28-\log (3))\right ) \log (5)+256 \log ^2(5)}\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.20, size = 34, normalized size = 1.36 \begin {gather*} \frac {e^x \left (-140-4 x^2+16 \log (5)+\log (243)\right )}{\left (140+4 x^2-5 \log (3)-16 \log (5)\right )^2} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 1.04, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^{x}}{4 \, x^{2} - 16 \, \log \relax (5) - 5 \, \log \relax (3) + 140} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.20, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^{x}}{4 \, x^{2} - 16 \, \log \relax (5) - 5 \, \log \relax (3) + 140} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.16, size = 21, normalized size = 0.84
method | result | size |
gosper | \(\frac {{\mathrm e}^{x}}{-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)}\) | \(21\) |
norman | \(\frac {{\mathrm e}^{x}}{-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)}\) | \(21\) |
default | \(-\frac {70 \,{\mathrm e}^{x} x}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}+\frac {35 \,{\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{4 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}+\frac {35 \,{\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{4 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}+\frac {5 i \ln \relax (3) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {2 i \ln \relax (5) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}+\frac {{\mathrm e}^{x}}{-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)}-\frac {5 i \ln \relax (3) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {i {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {{\mathrm e}^{x} x}{2 \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}+\frac {{\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16}+\frac {{\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16}+\frac {5 \ln \relax (3) {\mathrm e}^{x} x}{2 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}-\frac {5 \ln \relax (3) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}-\frac {5 \ln \relax (3) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right )}+\frac {2 i \ln \relax (5) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}-\frac {35 i {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{2 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}+\frac {8 \ln \relax (5) {\mathrm e}^{x} x}{\left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \left (-140-4 x^{2}+16 \ln \relax (5)+5 \ln \relax (3)\right )}-\frac {\ln \relax (5) {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \ln \relax (5)+5 \ln \relax (3)-140}-\frac {\ln \relax (5) {\mathrm e}^{-\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x -\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{16 \ln \relax (5)+5 \ln \relax (3)-140}+\frac {35 i {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{2 \left (16 \ln \relax (5)+5 \ln \relax (3)-140\right ) \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}+\frac {i {\mathrm e}^{\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}} \expIntegralEi \left (1, -x +\frac {i \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}{2}\right )}{8 \sqrt {-16 \ln \relax (5)-5 \ln \relax (3)+140}}\) | \(1038\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.49, size = 21, normalized size = 0.84 \begin {gather*} -\frac {e^{x}}{4 \, x^{2} - 16 \, \log \relax (5) - 5 \, \log \relax (3) + 140} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (-4\,x^2+8\,x+5\,\ln \relax (3)+16\,\ln \relax (5)-140\right )}{25\,{\ln \relax (3)}^2-\ln \relax (3)\,\left (40\,x^2+1400\right )-\ln \relax (5)\,\left (128\,x^2-160\,\ln \relax (3)+4480\right )+256\,{\ln \relax (5)}^2+1120\,x^2+16\,x^4+19600} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.17, size = 20, normalized size = 0.80 \begin {gather*} - \frac {e^{x}}{4 x^{2} - 16 \log {\relax (5 )} - 5 \log {\relax (3 )} + 140} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________