Optimal. Leaf size=30 \[ \frac {x}{5}+\frac {1}{1+x}-\frac {3^x x}{-e^{1+x}+5 x} \]
________________________________________________________________________________________
Rubi [F] time = 2.76, 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 {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{125 x^2+250 x^3+125 x^4+e^{2+2 x} \left (5+10 x+5 x^2\right )+e^{1+x} \left (-50 x-100 x^2-50 x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{5 \left (e^{1+x}-5 x\right )^2 (1+x)^2} \, dx\\ &=\frac {1}{5} \int \frac {-100 x^2+50 x^3+25 x^4+e^{2+2 x} \left (-4+2 x+x^2\right )+e^{1+x} \left (40 x-20 x^2-10 x^3\right )+3^x \left (\left (-25 x^2-50 x^3-25 x^4\right ) \log (3)+e^{1+x} \left (5+5 x-5 x^2-5 x^3+\left (5 x+10 x^2+5 x^3\right ) \log (3)\right )\right )}{\left (e^{1+x}-5 x\right )^2 (1+x)^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {25\ 3^x (-1+x) x}{\left (e^{1+x}-5 x\right )^2}+\frac {-4+2 x+x^2}{(1+x)^2}+\frac {5\ 3^x (1-x (1-\log (3)))}{e^{1+x}-5 x}\right ) \, dx\\ &=\frac {1}{5} \int \frac {-4+2 x+x^2}{(1+x)^2} \, dx-5 \int \frac {3^x (-1+x) x}{\left (e^{1+x}-5 x\right )^2} \, dx+\int \frac {3^x (1-x (1-\log (3)))}{e^{1+x}-5 x} \, dx\\ &=\frac {1}{5} \int \left (1-\frac {5}{(1+x)^2}\right ) \, dx-5 \int \left (-\frac {3^x x}{\left (e^{1+x}-5 x\right )^2}+\frac {3^x x^2}{\left (e^{1+x}-5 x\right )^2}\right ) \, dx+\int \frac {3^x (1+x (-1+\log (3)))}{e^{1+x}-5 x} \, dx\\ &=\frac {x}{5}+\frac {1}{1+x}+5 \int \frac {3^x x}{\left (e^{1+x}-5 x\right )^2} \, dx-5 \int \frac {3^x x^2}{\left (e^{1+x}-5 x\right )^2} \, dx+\int \left (\frac {3^x}{e^{1+x}-5 x}+\frac {3^x x (-1+\log (3))}{e^{1+x}-5 x}\right ) \, dx\\ &=\frac {x}{5}+\frac {1}{1+x}+5 \int \frac {3^x x}{\left (e^{1+x}-5 x\right )^2} \, dx-5 \int \frac {3^x x^2}{\left (e^{1+x}-5 x\right )^2} \, dx+(-1+\log (3)) \int \frac {3^x x}{e^{1+x}-5 x} \, dx+\int \frac {3^x}{e^{1+x}-5 x} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A] time = 0.16, size = 30, normalized size = 1.00 \begin {gather*} \frac {1}{5} \left (x+\frac {5\ 3^x x}{e^{1+x}-5 x}+\frac {5}{1+x}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [B] time = 0.89, size = 58, normalized size = 1.93 \begin {gather*} \frac {5 \, x^{3} - 5 \, {\left (x^{2} + x\right )} 3^{x} + 5 \, x^{2} - {\left (x^{2} + x + 5\right )} e^{\left (x + 1\right )} + 25 \, x}{5 \, {\left (5 \, x^{2} - {\left (x + 1\right )} e^{\left (x + 1\right )} + 5 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {25 \, x^{4} + 50 \, x^{3} - 5 \, {\left ({\left (x^{3} + x^{2} - {\left (x^{3} + 2 \, x^{2} + x\right )} \log \relax (3) - x - 1\right )} e^{\left (x + 1\right )} + 5 \, {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} \log \relax (3)\right )} 3^{x} - 100 \, x^{2} + {\left (x^{2} + 2 \, x - 4\right )} e^{\left (2 \, x + 2\right )} - 10 \, {\left (x^{3} + 2 \, x^{2} - 4 \, x\right )} e^{\left (x + 1\right )}}{5 \, {\left (25 \, x^{4} + 50 \, x^{3} + 25 \, x^{2} + {\left (x^{2} + 2 \, x + 1\right )} e^{\left (2 \, x + 2\right )} - 10 \, {\left (x^{3} + 2 \, x^{2} + x\right )} e^{\left (x + 1\right )}\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [B] time = 0.45, size = 66, normalized size = 2.20
method | result | size |
norman | \(\frac {x^{3}+4 x -\frac {4 \,{\mathrm e}^{x +1}}{5}-x \,{\mathrm e}^{x \ln \relax (3)}-x^{2} {\mathrm e}^{x \ln \relax (3)}-\frac {x^{2} {\mathrm e}^{x +1}}{5}}{5 x^{2}-x \,{\mathrm e}^{x +1}+5 x -{\mathrm e}^{x +1}}\) | \(66\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [B] time = 0.52, size = 67, normalized size = 2.23 \begin {gather*} \frac {5 \, x^{3} - 5 \, {\left (x^{2} + x\right )} 3^{x} + 5 \, x^{2} - {\left (x^{2} e + x e + 5 \, e\right )} e^{x} + 25 \, x}{5 \, {\left (5 \, x^{2} - {\left (x e + e\right )} e^{x} + 5 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 5.48, size = 27, normalized size = 0.90 \begin {gather*} \frac {x}{5}+\frac {1}{x+1}-\frac {3^x\,x}{5\,\left (x-\frac {{\mathrm {e}}^{x+1}}{5}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________