Optimal. Leaf size=35 \[ 2 \left (e^{2 x \left (x+\log \left (\frac {4}{x}\right )\right )}+\frac {5 \left (-x+x^2\right )}{3-\log (2 x)}\right ) \]
[Out]
________________________________________________________________________________________
Rubi [F]
time = 1.36, 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 {-40+70 x+(10-20 x) \log (2 x)+e^{2 x^2+2 x \log \left (\frac {4}{x}\right )} \left (-36+72 x+36 \log \left (\frac {4}{x}\right )+\left (24-48 x-24 \log \left (\frac {4}{x}\right )\right ) \log (2 x)+\left (-4+8 x+4 \log \left (\frac {4}{x}\right )\right ) \log ^2(2 x)\right )}{9-6 \log (2 x)+\log ^2(2 x)} \, dx \end {gather*}
Verification is not applicable to the result.
[In]
[Out]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-40+70 x+4^{1+2 x} e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) (-3+\log (2 x))^2+(10-20 x) \log (2 x)}{(3-\log (2 x))^2} \, dx\\ &=\int \left (4^{1+2 x} e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right )-\frac {10 (4-7 x-\log (2 x)+2 x \log (2 x))}{(-3+\log (2 x))^2}\right ) \, dx\\ &=-\left (10 \int \frac {4-7 x-\log (2 x)+2 x \log (2 x)}{(-3+\log (2 x))^2} \, dx\right )+\int 4^{1+2 x} e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) \, dx\\ &=-\left (10 \int \left (\frac {1-x}{(-3+\log (2 x))^2}+\frac {-1+2 x}{-3+\log (2 x)}\right ) \, dx\right )+\int 4 e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) \, dx\\ &=4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \left (-1+2 x+\log \left (\frac {4}{x}\right )\right ) \, dx-10 \int \frac {1-x}{(-3+\log (2 x))^2} \, dx-10 \int \frac {-1+2 x}{-3+\log (2 x)} \, dx\\ &=-\frac {10 (1-x) x}{3-\log (2 x)}+4 \int \left (-e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x}+2 e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x}+e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \log \left (\frac {4}{x}\right )\right ) \, dx-10 \int \left (-\frac {1}{-3+\log (2 x)}+\frac {2 x}{-3+\log (2 x)}\right ) \, dx+10 \int \frac {1}{-3+\log (2 x)} \, dx-20 \int \frac {1-x}{-3+\log (2 x)} \, dx\\ &=-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \log \left (\frac {4}{x}\right ) \, dx+5 \text {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log (2 x)\right )+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx+10 \int \frac {1}{-3+\log (2 x)} \, dx-20 \int \left (\frac {1}{-3+\log (2 x)}-\frac {x}{-3+\log (2 x)}\right ) \, dx-20 \int \frac {x}{-3+\log (2 x)} \, dx\\ &=5 e^3 \text {Ei}(-3+\log (2 x))-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int \frac {\int 16^x e^{2 x^2} \left (\frac {1}{x}\right )^{2 x} \, dx}{x} \, dx+5 \text {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log (2 x)\right )-5 \text {Subst}\left (\int \frac {e^{2 x}}{-3+x} \, dx,x,\log (2 x)\right )+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx-20 \int \frac {1}{-3+\log (2 x)} \, dx+20 \int \frac {x}{-3+\log (2 x)} \, dx+\left (4 \log \left (\frac {4}{x}\right )\right ) \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx\\ &=-5 e^6 \text {Ei}(-2 (3-\log (2 x)))+10 e^3 \text {Ei}(-3+\log (2 x))-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int \frac {\int e^{2 x^2+x \log (16)} \left (\frac {1}{x}\right )^{2 x} \, dx}{x} \, dx+5 \text {Subst}\left (\int \frac {e^{2 x}}{-3+x} \, dx,x,\log (2 x)\right )+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx-10 \text {Subst}\left (\int \frac {e^x}{-3+x} \, dx,x,\log (2 x)\right )+\left (4 \log \left (\frac {4}{x}\right )\right ) \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx\\ &=-\frac {10 (1-x) x}{3-\log (2 x)}-4 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx+4 \int \frac {\int e^{2 x^2+x \log (16)} \left (\frac {1}{x}\right )^{2 x} \, dx}{x} \, dx+8 \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{-1+2 x} \, dx+\left (4 \log \left (\frac {4}{x}\right )\right ) \int e^{2 x^2+2 x \log (4)} \left (\frac {1}{x}\right )^{2 x} \, dx\\ \end {aligned} \end {gather*}
________________________________________________________________________________________
Mathematica [A]
time = 2.06, size = 35, normalized size = 1.00 \begin {gather*} 2 \left (16^x e^{2 x^2} \left (\frac {1}{x}\right )^{2 x}-\frac {5 (-1+x) x}{-3+\log (2 x)}\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [A]
time = 7.45, size = 71, normalized size = 2.03
method | result | size |
risch | \(-\frac {20 x \left (x -1\right )}{-6+2 \ln \left (2\right )+2 \ln \left (x \right )}+2 x^{-2 x} 4^{2 x} {\mathrm e}^{2 x^{2}}\) | \(40\) |
default | \(\frac {10 x}{\ln \left (2\right )+\ln \left (x \right )-3}-\frac {10 x^{2}}{\ln \left (2\right )+\ln \left (x \right )-3}+\frac {\left (2 \ln \left (2\right )-6\right ) {\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x}+2 \ln \left (x \right ) {\mathrm e}^{2 \left (\ln \left (\frac {4}{x}\right )+x \right ) x}}{\ln \left (2\right )+\ln \left (x \right )-3}\) | \(71\) |
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A]
time = 0.37, size = 59, normalized size = 1.69 \begin {gather*} -\frac {2 \, {\left (5 \, x^{2} - {\left (3 \, \log \left (2\right ) - \log \left (\frac {4}{x}\right ) - 3\right )} e^{\left (2 \, x^{2} + 2 \, x \log \left (\frac {4}{x}\right )\right )} - 5 \, x\right )}}{3 \, \log \left (2\right ) - \log \left (\frac {4}{x}\right ) - 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [A]
time = 0.22, size = 34, normalized size = 0.97 \begin {gather*} \frac {- 10 x^{2} + 10 x}{\log {\left (2 x \right )} - 3} + 2 e^{2 x^{2} + 2 x \left (- \log {\left (2 x \right )} + \log {\left (8 \right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A]
time = 0.62, size = 61, normalized size = 1.74 \begin {gather*} -\frac {2 \, {\left (5 \, x^{2} - e^{\left (2 \, x^{2} + 4 \, x \log \left (2\right ) - 2 \, x \log \left (x\right )\right )} \log \left (2 \, x\right ) - 5 \, x + 3 \, e^{\left (2 \, x^{2} + 4 \, x \log \left (2\right ) - 2 \, x \log \left (x\right )\right )}\right )}}{\log \left (2 \, x\right ) - 3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Mupad [B]
time = 7.67, size = 60, normalized size = 1.71 \begin {gather*} 10\,x-\frac {10\,x\,\left (7\,x-4\right )-10\,x\,\ln \left (2\,x\right )\,\left (2\,x-1\right )}{\ln \left (2\,x\right )-3}-20\,x^2+2\,2^{4\,x}\,{\mathrm {e}}^{2\,x^2}\,{\left (\frac {1}{x}\right )}^{2\,x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________