3.33.0 ∫ 4−x−x2+log(3)+ex(4−6x+x2+(1−x) log(3))+(−4+2x−log(3)) log(x) 32x2−16x3+2x4+(16x2−4x3) log(3)+2x2 log2(3) dx [3290]

Integrand size = 86, antiderivative size = 25 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {e^x+x-\log (x)}{2 x (-4+x-\log (3))} \]

Rubi [B] (veriﬁed)

Leaf count is larger than twice the leaf count of optimal. \(101\) vs. \(2(25)=50\).

Time = 0.94 (sec) , antiderivative size = 101, normalized size of antiderivative = 4.04, number of steps used = 23, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {6, 6873, 27, 12, 6874, 46, 2208, 2209, 2404, 2341, 2351, 31} \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {x \log (x)}{2 (4+\log (3))^2 (-x+4+\log (3))}+\frac {\log (x)}{2 x (4+\log (3))}+\frac {\log (x)}{2 (4+\log (3))^2}-\frac {e^x}{2 (4+\log (3)) (-x+4+\log (3))}-\frac {1}{2 (-x+4+\log (3))}-\frac {e^x}{2 x (4+\log (3))} \]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+x^2 \left (32+2 \log ^2(3)\right )} \, dx \\ & = \int \frac {-x-x^2+4 \left (1+\frac {\log (3)}{4}\right )+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{x^2 \left (2 x^2-4 x (4+\log (3))+2 (4+\log (3))^2\right )} \, dx \\ & = \int \frac {-x-x^2+4 \left (1+\frac {\log (3)}{4}\right )+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{2 x^2 (-4+x-\log (3))^2} \, dx \\ & = \frac {1}{2} \int \frac {-x-x^2+4 \left (1+\frac {\log (3)}{4}\right )+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{x^2 (-4+x-\log (3))^2} \, dx \\ & = \frac {1}{2} \int \left (-\frac {1}{(-4+x-\log (3))^2}-\frac {1}{x (-4+x-\log (3))^2}+\frac {4+\log (3)}{x^2 (-4+x-\log (3))^2}+\frac {e^x \left (4+x^2+\log (3)-x (6+\log (3))\right )}{x^2 (4-x+\log (3))^2}+\frac {(-4+2 x-\log (3)) \log (x)}{x^2 (-4+x-\log (3))^2}\right ) \, dx \\ & = -\frac {1}{2 (4-x+\log (3))}-\frac {1}{2} \int \frac {1}{x (-4+x-\log (3))^2} \, dx+\frac {1}{2} \int \frac {e^x \left (4+x^2+\log (3)-x (6+\log (3))\right )}{x^2 (4-x+\log (3))^2} \, dx+\frac {1}{2} \int \frac {(-4+2 x-\log (3)) \log (x)}{x^2 (-4+x-\log (3))^2} \, dx+\frac {1}{2} (4+\log (3)) \int \frac {1}{x^2 (-4+x-\log (3))^2} \, dx \\ & = -\frac {1}{2 (4-x+\log (3))}-\frac {1}{2} \int \left (\frac {1}{x (4+\log (3))^2}-\frac {1}{(-4+x-\log (3)) (4+\log (3))^2}+\frac {1}{(4+\log (3)) (4-x+\log (3))^2}\right ) \, dx+\frac {1}{2} \int \left (\frac {e^x}{x^2 (4+\log (3))}-\frac {e^x}{x (4+\log (3))}+\frac {e^x}{(-4+x-\log (3)) (4+\log (3))}-\frac {e^x}{(4+\log (3)) (4-x+\log (3))^2}\right ) \, dx+\frac {1}{2} \int \left (-\frac {\log (x)}{x^2 (4+\log (3))}+\frac {\log (x)}{(4+\log (3)) (4-x+\log (3))^2}\right ) \, dx+\frac {1}{2} (4+\log (3)) \int \left (\frac {2}{x (4+\log (3))^3}-\frac {2}{(-4+x-\log (3)) (4+\log (3))^3}+\frac {1}{x^2 (4+\log (3))^2}+\frac {1}{(4+\log (3))^2 (4-x+\log (3))^2}\right ) \, dx \\ & = -\frac {1}{2 x (4+\log (3))}-\frac {1}{2 (4-x+\log (3))}+\frac {\log (x)}{2 (4+\log (3))^2}-\frac {\log (4-x+\log (3))}{2 (4+\log (3))^2}+\frac {\int \frac {e^x}{x^2} \, dx}{2 (4+\log (3))}-\frac {\int \frac {e^x}{x} \, dx}{2 (4+\log (3))}+\frac {\int \frac {e^x}{-4+x-\log (3)} \, dx}{2 (4+\log (3))}-\frac {\int \frac {e^x}{(4-x+\log (3))^2} \, dx}{2 (4+\log (3))}-\frac {\int \frac {\log (x)}{x^2} \, dx}{2 (4+\log (3))}+\frac {\int \frac {\log (x)}{(4-x+\log (3))^2} \, dx}{2 (4+\log (3))} \\ & = -\frac {e^x}{2 x (4+\log (3))}-\frac {\text {Ei}(x)}{2 (4+\log (3))}+\frac {3 e^4 \text {Ei}(-4+x-\log (3))}{2 (4+\log (3))}-\frac {1}{2 (4-x+\log (3))}-\frac {e^x}{2 (4+\log (3)) (4-x+\log (3))}+\frac {\log (x)}{2 (4+\log (3))^2}+\frac {\log (x)}{2 x (4+\log (3))}+\frac {x \log (x)}{2 (4+\log (3))^2 (4-x+\log (3))}-\frac {\log (4-x+\log (3))}{2 (4+\log (3))^2}-\frac {\int \frac {1}{4-x+\log (3)} \, dx}{2 (4+\log (3))^2}+\frac {\int \frac {e^x}{x} \, dx}{2 (4+\log (3))}+\frac {\int \frac {e^x}{4-x+\log (3)} \, dx}{2 (4+\log (3))} \\ & = -\frac {e^x}{2 x (4+\log (3))}-\frac {1}{2 (4-x+\log (3))}-\frac {e^x}{2 (4+\log (3)) (4-x+\log (3))}+\frac {\log (x)}{2 (4+\log (3))^2}+\frac {\log (x)}{2 x (4+\log (3))}+\frac {x \log (x)}{2 (4+\log (3))^2 (4-x+\log (3))} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 0.74 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {e^x+x-\log (x)}{2 x (-4+x-\log (3))} \]

Maple [A] (veriﬁed)

Time = 1.77 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00


method	result	size

parallelrisch	\(\frac {-{\mathrm e}^{x}-x +\ln \left (x \right )}{2 x \left (4+\ln \left (3\right )-x \right )}\)	\(25\)
norman	\(\frac {-\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {\ln \left (x \right )}{2}}{x \left (4+\ln \left (3\right )-x \right )}\)	\(26\)
default	\(\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {1}{2 x -8-2 \ln \left (3\right )}-\frac {\ln \left (x -4-\ln \left (3\right )\right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right ) x}+\frac {\ln \left (4+\ln \left (3\right )-x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (x \right ) x}{2 \left (\ln \left (3\right )+4\right )^{2} \left (4+\ln \left (3\right )-x \right )}-\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{2 \ln \left (3\right )+8}-\frac {{\mathrm e}^{x}}{2 \left (x -4-\ln \left (3\right )\right )}-\frac {{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{2}-\frac {4 \,\operatorname {Ei}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} x}+\frac {\operatorname {Ei}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}+\frac {4 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}-\frac {5 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}+\frac {3 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {3 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\ln \left (3\right )+4}-\frac {\ln \left (3\right ) \operatorname {Ei}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} x}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}\)	\(402\)
parts	\(\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {1}{2 x -8-2 \ln \left (3\right )}-\frac {\ln \left (x -4-\ln \left (3\right )\right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right ) x}+\frac {\ln \left (4+\ln \left (3\right )-x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (x \right ) x}{2 \left (\ln \left (3\right )+4\right )^{2} \left (4+\ln \left (3\right )-x \right )}-\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{2 \ln \left (3\right )+8}-\frac {{\mathrm e}^{x}}{2 \left (x -4-\ln \left (3\right )\right )}-\frac {{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{2}-\frac {4 \,\operatorname {Ei}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} x}+\frac {\operatorname {Ei}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}+\frac {4 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}-\frac {5 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}+\frac {3 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {3 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\ln \left (3\right )+4}-\frac {\ln \left (3\right ) \operatorname {Ei}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} x}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {Ei}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}\)	\(402\)

Fricas [A] (veriﬁcation not implemented)

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {x + e^{x} - \log \left (x\right )}{2 \, {\left (x^{2} - x \log \left (3\right ) - 4 \, x\right )}} \]

Sympy [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).

Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {e^{x}}{2 x^{2} - 8 x - 2 x \log {\left (3 \right )}} - \frac {\log {\left (x \right )}}{2 x^{2} - 8 x - 2 x \log {\left (3 \right )}} + \frac {1}{2 x - 8 - 2 \log {\left (3 \right )}} \]

Maxima [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (22) = 44\).

Time = 0.31 (sec) , antiderivative size = 362, normalized size of antiderivative = 14.48 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=-\frac {1}{2} \, {\left (\frac {2 \, x - \log \left (3\right ) - 4}{{\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} x^{2} - {\left (\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64\right )} x} + \frac {2 \, \log \left (x - \log \left (3\right ) - 4\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64} - \frac {2 \, \log \left (x\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64}\right )} \log \left (3\right ) + \frac {x {\left (\log \left (3\right ) + 4\right )} + {\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} e^{x} - \log \left (3\right )^{2} - {\left (x^{2} - x {\left (\log \left (3\right ) + 4\right )} + \log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} \log \left (x\right ) - 8 \, \log \left (3\right ) - 16}{2 \, {\left ({\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} x^{2} - {\left (\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64\right )} x\right )}} - \frac {2 \, {\left (2 \, x - \log \left (3\right ) - 4\right )}}{{\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} x^{2} - {\left (\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64\right )} x} - \frac {4 \, \log \left (x - \log \left (3\right ) - 4\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64} + \frac {\log \left (x - \log \left (3\right ) - 4\right )}{\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16} + \frac {4 \, \log \left (x\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64} - \frac {\log \left (x\right )}{2 \, {\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )}} + \frac {1}{2 \, {\left (x {\left (\log \left (3\right ) + 4\right )} - \log \left (3\right )^{2} - 8 \, \log \left (3\right ) - 16\right )}} + \frac {1}{2 \, {\left (x - \log \left (3\right ) - 4\right )}} \]

Giac [A] (veriﬁcation not implemented)

Mupad [B] (veriﬁcation not implemented)

Time = 14.82 (sec) , antiderivative size = 7592, normalized size of antiderivative = 303.68 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\text {Too large to display} \]

\(\int \frac {4-x-x^2+\log (3)+e^x (4-6 x+x^2+(1-x) \log (3))+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+(16 x^2-4 x^3) \log (3)+2 x^2 \log ^2(3)} \, dx\) [3290]

Optimal result