Integrand size = 60, antiderivative size = 32 \[ \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )\right )}{\left (9+x^2\right ) \log (4)} \, dx=-e^{\frac {1}{4} (2-\log (4))} x+\frac {x^2}{\log (4)}+2 \log \left (9+x^2\right ) \]
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )\right )}{\left (9+x^2\right ) \log (4)} \, dx=\frac {18+2 x^2-\sqrt {2 e} x \log (4)+\log (256) \log \left (9+x^2\right )}{\log (16)} \]
Integrate[(E^((2 - Log[4])/4)*((-9 - x^2)*Log[4] + E^((-2 + Log[4])/4)*(18 *x + 2*x^3 + 4*x*Log[4])))/((9 + x^2)*Log[4]),x]
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.66, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {27, 25, 2341, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (e^{\frac {1}{4} (\log (4)-2)} \left (2 x^3+18 x+4 x \log (4)\right )+\left (-x^2-9\right ) \log (4)\right )}{\left (x^2+9\right ) \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {\frac {e}{2}} \int -\frac {\left (x^2+9\right ) \log (4)-2 \sqrt {\frac {2}{e}} \left (x^3+2 \log (4) x+9 x\right )}{x^2+9}dx}{\log (4)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {\frac {e}{2}} \int \frac {\left (x^2+9\right ) \log (4)-2 \sqrt {\frac {2}{e}} \left (x^3+2 \log (4) x+9 x\right )}{x^2+9}dx}{\log (4)}\) |
\(\Big \downarrow \) 2341 |
\(\displaystyle -\frac {\sqrt {\frac {e}{2}} \int \left (-\frac {4 \sqrt {\frac {2}{e}} \log (4) x}{x^2+9}-2 \sqrt {\frac {2}{e}} x+\log (4)\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {\frac {e}{2}} \left (-\sqrt {\frac {2}{e}} x^2-2 \sqrt {\frac {2}{e}} \log (4) \log \left (x^2+9\right )+x \log (4)\right )}{\log (4)}\) |
Int[(E^((2 - Log[4])/4)*((-9 - x^2)*Log[4] + E^((-2 + Log[4])/4)*(18*x + 2 *x^3 + 4*x*Log[4])))/((9 + x^2)*Log[4]),x]
3.12.13.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Time = 2.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {x^{2}}{2 \ln \left (2\right )}-\frac {\sqrt {2}\, {\mathrm e}^{\frac {1}{2}} x}{2}+2 \ln \left (x^{2}+9\right )\) | \(27\) |
norman | \(\frac {x^{2}}{2 \ln \left (2\right )}-\frac {\sqrt {2}\, {\mathrm e}^{\frac {1}{2}} x}{2}+2 \ln \left (x^{2}+9\right )\) | \(29\) |
default | \(\frac {{\mathrm e}^{\frac {1}{2}-\frac {\ln \left (2\right )}{2}} \left ({\mathrm e}^{\frac {\ln \left (2\right )}{2}-\frac {1}{2}} x^{2}-2 x \ln \left (2\right )+4 \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (2\right )}{2}-\frac {1}{2}} \ln \left (x^{2}+9\right )\right )}{2 \ln \left (2\right )}\) | \(50\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {1}{2}-\frac {\ln \left (2\right )}{2}} \left ({\mathrm e}^{\frac {\ln \left (2\right )}{2}-\frac {1}{2}} x^{2}-2 x \ln \left (2\right )+4 \ln \left (2\right ) {\mathrm e}^{\frac {\ln \left (2\right )}{2}-\frac {1}{2}} \ln \left (x^{2}+9\right )\right )}{2 \ln \left (2\right )}\) | \(50\) |
meijerg | \(-3 \,{\mathrm e}^{\frac {1}{2}-\frac {\ln \left (2\right )}{2}} \arctan \left (\frac {x}{3}\right )+\frac {\frac {x^{2}}{2}-\frac {9 \ln \left (1+\frac {x^{2}}{9}\right )}{2}}{\ln \left (2\right )}+\frac {9 \left (\frac {4 \ln \left (2\right )}{9}+1\right ) \ln \left (1+\frac {x^{2}}{9}\right )}{2 \ln \left (2\right )}-\frac {3 \,{\mathrm e}^{\frac {1}{2}-\frac {\ln \left (2\right )}{2}} \left (\frac {2 x}{3}-2 \arctan \left (\frac {x}{3}\right )\right )}{2}\) | \(76\) |
int(1/2*((8*x*ln(2)+2*x^3+18*x)*exp(1/2*ln(2)-1/2)+2*(-x^2-9)*ln(2))/(x^2+ 9)/ln(2)/exp(1/2*ln(2)-1/2),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )\right )}{\left (9+x^2\right ) \log (4)} \, dx=\frac {{\left (x^{2} e^{\left (\frac {1}{2} \, \log \left (2\right ) - \frac {1}{2}\right )} + 4 \, e^{\left (\frac {1}{2} \, \log \left (2\right ) - \frac {1}{2}\right )} \log \left (2\right ) \log \left (x^{2} + 9\right ) - 2 \, x \log \left (2\right )\right )} e^{\left (-\frac {1}{2} \, \log \left (2\right ) + \frac {1}{2}\right )}}{2 \, \log \left (2\right )} \]
integrate(1/2*((8*x*log(2)+2*x^3+18*x)*exp(1/2*log(2)-1/2)+2*(-x^2-9)*log( 2))/(x^2+9)/log(2)/exp(1/2*log(2)-1/2),x, algorithm=\
1/2*(x^2*e^(1/2*log(2) - 1/2) + 4*e^(1/2*log(2) - 1/2)*log(2)*log(x^2 + 9) - 2*x*log(2))*e^(-1/2*log(2) + 1/2)/log(2)
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )\right )}{\left (9+x^2\right ) \log (4)} \, dx=\frac {x^{2}}{2 \log {\left (2 \right )}} - \frac {\sqrt {2} x e^{\frac {1}{2}}}{2} + 2 \log {\left (x^{2} + 9 \right )} \]
integrate(1/2*((8*x*ln(2)+2*x**3+18*x)*exp(1/2*ln(2)-1/2)+2*(-x**2-9)*ln(2 ))/(x**2+9)/ln(2)/exp(1/2*ln(2)-1/2),x)
Time = 0.18 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.59 \[ \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )\right )}{\left (9+x^2\right ) \log (4)} \, dx=\frac {{\left (4 \, e^{\left (\frac {1}{2} \, \log \left (2\right ) - \frac {1}{2}\right )} \log \left (2\right ) \log \left (x^{2} + 9\right ) - {\left (2 \, x e^{\frac {1}{2}} \log \left (2\right ) - \sqrt {2} x^{2}\right )} e^{\left (-\frac {1}{2}\right )}\right )} e^{\left (-\frac {1}{2} \, \log \left (2\right ) + \frac {1}{2}\right )}}{2 \, \log \left (2\right )} \]
integrate(1/2*((8*x*log(2)+2*x^3+18*x)*exp(1/2*log(2)-1/2)+2*(-x^2-9)*log( 2))/(x^2+9)/log(2)/exp(1/2*log(2)-1/2),x, algorithm=\
1/2*(4*e^(1/2*log(2) - 1/2)*log(2)*log(x^2 + 9) - (2*x*e^(1/2)*log(2) - sq rt(2)*x^2)*e^(-1/2))*e^(-1/2*log(2) + 1/2)/log(2)
Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )\right )}{\left (9+x^2\right ) \log (4)} \, dx=\frac {{\left (x^{2} e^{\left (\frac {1}{2} \, \log \left (2\right ) - \frac {1}{2}\right )} + 4 \, e^{\left (\frac {1}{2} \, \log \left (2\right ) - \frac {1}{2}\right )} \log \left (2\right ) \log \left (x^{2} + 9\right ) - 2 \, x \log \left (2\right )\right )} e^{\left (-\frac {1}{2} \, \log \left (2\right ) + \frac {1}{2}\right )}}{2 \, \log \left (2\right )} \]
integrate(1/2*((8*x*log(2)+2*x^3+18*x)*exp(1/2*log(2)-1/2)+2*(-x^2-9)*log( 2))/(x^2+9)/log(2)/exp(1/2*log(2)-1/2),x, algorithm=\
1/2*(x^2*e^(1/2*log(2) - 1/2) + 4*e^(1/2*log(2) - 1/2)*log(2)*log(x^2 + 9) - 2*x*log(2))*e^(-1/2*log(2) + 1/2)/log(2)
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {1}{4} (2-\log (4))} \left (\left (-9-x^2\right ) \log (4)+e^{\frac {1}{4} (-2+\log (4))} \left (18 x+2 x^3+4 x \log (4)\right )\right )}{\left (9+x^2\right ) \log (4)} \, dx=2\,\ln \left (x^2+9\right )+\frac {x^2}{2\,\ln \left (2\right )}-\frac {\sqrt {2}\,x\,\sqrt {\mathrm {e}}}{2} \]