Integrand size = 137, antiderivative size = 32 \[ \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx=\frac {e^{-\left (1+\log (4)-\log \left (x^2\right )\right )^2}}{2 \left (-5+\frac {3}{x}+x+\log (x)\right )} \]
\[ \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx=\int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx \]
Integrate[(E^(-1 - Log[4]^2 - (-2 - 2*Log[4])*Log[x^2] - Log[x^2]^2)*(15 - 21*x + 3*x^2 + (12 - 20*x + 4*x^2)*Log[4] + (4*x + 4*x*Log[4])*Log[x] + ( -12 + 20*x - 4*x^2 - 4*x*Log[x])*Log[x^2]))/(16*(18 - 60*x + 62*x^2 - 20*x ^3 + 2*x^4 + (12*x - 20*x^2 + 4*x^3)*Log[x] + 2*x^2*Log[x]^2)),x]
Integrate[(E^(-1 - Log[4]^2 - (-2 - 2*Log[4])*Log[x^2] - Log[x^2]^2)*(15 - 21*x + 3*x^2 + (12 - 20*x + 4*x^2)*Log[4] + (4*x + 4*x*Log[4])*Log[x] + ( -12 + 20*x - 4*x^2 - 4*x*Log[x])*Log[x^2]))/(18 - 60*x + 62*x^2 - 20*x^3 + 2*x^4 + (12*x - 20*x^2 + 4*x^3)*Log[x] + 2*x^2*Log[x]^2), x]/16
Leaf count is larger than twice the leaf count of optimal. \(94\) vs. \(2(32)=64\).
Time = 0.73 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {27, 27, 34, 2726}
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.
| \(\displaystyle \int \frac {\left (3 x^2+\left (-4 x^2+20 x-4 x \log (x)-12\right ) \log \left (x^2\right )+\left (4 x^2-20 x+12\right ) \log (4)-21 x+(4 x+4 x \log (4)) \log (x)+15\right ) \exp \left (-\log ^2\left (x^2\right )-(-2-2 \log (4)) \log \left (x^2\right )-1-\log ^2(4)\right )}{16 \left (2 x^4-20 x^3+62 x^2+2 x^2 \log ^2(x)+\left (4 x^3-20 x^2+12 x\right ) \log (x)-60 x+18\right )} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{16} \int \frac {e^{-\log ^2\left (x^2\right )-\log ^2(4)-1} \left (x^2\right )^{2+2 \log (4)} \left (3 x^2+(4+\log (256)) \log (x) x-21 x-4 \left (x^2+\log (x) x-5 x+3\right ) \log \left (x^2\right )+4 \left (x^2-5 x+3\right ) \log (4)+15\right )}{2 \left (x^4-10 x^3+\log ^2(x) x^2+31 x^2-30 x+2 \left (x^3-5 x^2+3 x\right ) \log (x)+9\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{32} \int \frac {e^{-\log ^2\left (x^2\right )-\log ^2(4)-1} \left (x^2\right )^{2+\log (16)} \left (3 x^2+(4+\log (256)) \log (x) x-21 x-4 \left (x^2+\log (x) x-5 x+3\right ) \log \left (x^2\right )+4 \left (x^2-5 x+3\right ) \log (4)+15\right )}{x^4-10 x^3+\log ^2(x) x^2+31 x^2-30 x+2 \left (x^3-5 x^2+3 x\right ) \log (x)+9}dx\) |
| \(\Big \downarrow \) 34 |
| \(\displaystyle \frac {1}{32} x^{-2 \log (16)} \left (x^2\right )^{\log (16)} \int \frac {e^{-\log ^2\left (x^2\right )-\log ^2(4)-1} x^{4+\log (256)} \left (3 x^2+(4+\log (256)) \log (x) x-21 x-4 \left (x^2+\log (x) x-5 x+3\right ) \log \left (x^2\right )+4 \left (x^2-5 x+3\right ) \log (4)+15\right )}{x^4-10 x^3+\log ^2(x) x^2+31 x^2-30 x+2 \left (x^3-5 x^2+3 x\right ) \log (x)+9}dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle \frac {e^{-\log ^2\left (x^2\right )-1-\log ^2(4)} x^{5-2 \log (16)+\log (256)} \left (x^2\right )^{\log (16)} \left (x^2-5 x+x \log (x)+3\right )}{32 \left (x^4-10 x^3+31 x^2+x^2 \log ^2(x)+2 \left (x^3-5 x^2+3 x\right ) \log (x)-30 x+9\right )}\) |
Int[(E^(-1 - Log[4]^2 - (-2 - 2*Log[4])*Log[x^2] - Log[x^2]^2)*(15 - 21*x + 3*x^2 + (12 - 20*x + 4*x^2)*Log[4] + (4*x + 4*x*Log[4])*Log[x] + (-12 + 20*x - 4*x^2 - 4*x*Log[x])*Log[x^2]))/(16*(18 - 60*x + 62*x^2 - 20*x^3 + 2 *x^4 + (12*x - 20*x^2 + 4*x^3)*Log[x] + 2*x^2*Log[x]^2)),x]
(E^(-1 - Log[4]^2 - Log[x^2]^2)*x^(5 - 2*Log[16] + Log[256])*(x^2)^Log[16] *(3 - 5*x + x^2 + x*Log[x]))/(32*(9 - 30*x + 31*x^2 - 10*x^3 + x^4 + 2*(3* x - 5*x^2 + x^3)*Log[x] + x^2*Log[x]^2))
3.7.59.3.1 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[(u_.)*((a_.)*(x_)^(m_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*x^m)^F racPart[p]/x^(m*FracPart[p])) Int[u*x^(m*p), x], x] /; FreeQ[{a, m, p}, x ] && !IntegerQ[p]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 3.36 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.56
| method | result | size |
| parallelrisch | \(\frac {x \,{\mathrm e}^{-\ln \left (x^{2}\right )^{2}+\left (4 \ln \left (2\right )+2\right ) \ln \left (x^{2}\right )-4 \ln \left (2\right )^{2}-1}}{32 x \ln \left (x \right )+32 x^{2}-160 x +96}\) | \(50\) |
| risch | \(\frac {x^{5} x^{8 \ln \left (2\right )} x^{-4 i \pi \,\operatorname {csgn}\left (i x \right )} 2^{4 i \pi \,\operatorname {csgn}\left (i x \right )} x^{4 i \pi \,\operatorname {csgn}\left (i x^{2}\right )} 2^{-4 i \pi \,\operatorname {csgn}\left (i x^{2}\right )} {\mathrm e}^{-4 \ln \left (x \right )^{2}-1-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+\frac {\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}}{4}-\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{5} \operatorname {csgn}\left (i x \right )+\frac {3 \pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{4} \operatorname {csgn}\left (i x \right )^{2}}{2}-\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{3} \operatorname {csgn}\left (i x \right )^{3}+\frac {\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )^{4}}{4}+2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-4 \ln \left (2\right )^{2}}}{32 x \ln \left (x \right )+32 x^{2}-160 x +96}\) | \(244\) |
int(((-4*x*ln(x)-4*x^2+20*x-12)*ln(x^2)+(8*x*ln(2)+4*x)*ln(x)+2*(4*x^2-20* x+12)*ln(2)+3*x^2-21*x+15)/(2*x^2*ln(x)^2+(4*x^3-20*x^2+12*x)*ln(x)+2*x^4- 20*x^3+62*x^2-60*x+18)/exp(ln(x^2)^2+(-4*ln(2)-2)*ln(x^2)+4*ln(2)^2+4*ln(2 )+1),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx=\frac {x e^{\left (-4 \, \log \left (2\right )^{2} + 4 \, {\left (2 \, \log \left (2\right ) + 1\right )} \log \left (x\right ) - 4 \, \log \left (x\right )^{2} - 4 \, \log \left (2\right ) - 1\right )}}{2 \, {\left (x^{2} + x \log \left (x\right ) - 5 \, x + 3\right )}} \]
integrate(((-4*x*log(x)-4*x^2+20*x-12)*log(x^2)+(8*x*log(2)+4*x)*log(x)+2* (4*x^2-20*x+12)*log(2)+3*x^2-21*x+15)/(2*x^2*log(x)^2+(4*x^3-20*x^2+12*x)* log(x)+2*x^4-20*x^3+62*x^2-60*x+18)/exp(log(x^2)^2+(-4*log(2)-2)*log(x^2)+ 4*log(2)^2+4*log(2)+1),x, algorithm=\
1/2*x*e^(-4*log(2)^2 + 4*(2*log(2) + 1)*log(x) - 4*log(x)^2 - 4*log(2) - 1 )/(x^2 + x*log(x) - 5*x + 3)
Timed out. \[ \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx=\text {Timed out} \]
integrate(((-4*x*ln(x)-4*x**2+20*x-12)*ln(x**2)+(8*x*ln(2)+4*x)*ln(x)+2*(4 *x**2-20*x+12)*ln(2)+3*x**2-21*x+15)/(2*x**2*ln(x)**2+(4*x**3-20*x**2+12*x )*ln(x)+2*x**4-20*x**3+62*x**2-60*x+18)/exp(ln(x**2)**2+(-4*ln(2)-2)*ln(x* *2)+4*ln(2)**2+4*ln(2)+1),x)
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (31) = 62\).
Time = 0.39 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.22 \[ \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx=\frac {x^{5} e^{\left (8 \, \log \left (2\right ) \log \left (x\right ) - 4 \, \log \left (x\right )^{2}\right )}}{32 \, {\left (x^{2} e^{\left (4 \, \log \left (2\right )^{2} + 1\right )} + x e^{\left (4 \, \log \left (2\right )^{2} + 1\right )} \log \left (x\right ) - 5 \, x e^{\left (4 \, \log \left (2\right )^{2} + 1\right )} + 3 \, e^{\left (4 \, \log \left (2\right )^{2} + 1\right )}\right )}} \]
integrate(((-4*x*log(x)-4*x^2+20*x-12)*log(x^2)+(8*x*log(2)+4*x)*log(x)+2* (4*x^2-20*x+12)*log(2)+3*x^2-21*x+15)/(2*x^2*log(x)^2+(4*x^3-20*x^2+12*x)* log(x)+2*x^4-20*x^3+62*x^2-60*x+18)/exp(log(x^2)^2+(-4*log(2)-2)*log(x^2)+ 4*log(2)^2+4*log(2)+1),x, algorithm=\
1/32*x^5*e^(8*log(2)*log(x) - 4*log(x)^2)/(x^2*e^(4*log(2)^2 + 1) + x*e^(4 *log(2)^2 + 1)*log(x) - 5*x*e^(4*log(2)^2 + 1) + 3*e^(4*log(2)^2 + 1))
Time = 0.59 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx=\frac {x e^{\left (-4 \, \log \left (2\right )^{2} + 8 \, \log \left (2\right ) \log \left (x\right ) - 4 \, \log \left (x\right )^{2} + 4 \, \log \left (x\right ) - 1\right )}}{32 \, {\left (x^{2} + x \log \left (x\right ) - 5 \, x + 3\right )}} \]
integrate(((-4*x*log(x)-4*x^2+20*x-12)*log(x^2)+(8*x*log(2)+4*x)*log(x)+2* (4*x^2-20*x+12)*log(2)+3*x^2-21*x+15)/(2*x^2*log(x)^2+(4*x^3-20*x^2+12*x)* log(x)+2*x^4-20*x^3+62*x^2-60*x+18)/exp(log(x^2)^2+(-4*log(2)-2)*log(x^2)+ 4*log(2)^2+4*log(2)+1),x, algorithm=\
1/32*x*e^(-4*log(2)^2 + 8*log(2)*log(x) - 4*log(x)^2 + 4*log(x) - 1)/(x^2 + x*log(x) - 5*x + 3)
Time = 8.70 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {e^{-1-\log ^2(4)-(-2-2 \log (4)) \log \left (x^2\right )-\log ^2\left (x^2\right )} \left (15-21 x+3 x^2+\left (12-20 x+4 x^2\right ) \log (4)+(4 x+4 x \log (4)) \log (x)+\left (-12+20 x-4 x^2-4 x \log (x)\right ) \log \left (x^2\right )\right )}{16 \left (18-60 x+62 x^2-20 x^3+2 x^4+\left (12 x-20 x^2+4 x^3\right ) \log (x)+2 x^2 \log ^2(x)\right )} \, dx=\frac {x^5\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-{\ln \left (x^2\right )}^2}\,{\mathrm {e}}^{-4\,{\ln \left (2\right )}^2}\,{\left (x^2\right )}^{4\,\ln \left (2\right )}}{32\,\left (x\,\ln \left (x\right )-5\,x+x^2+3\right )} \]
int((exp(log(x^2)*(4*log(2) + 2) - log(x^2)^2 - 4*log(2)^2 - 4*log(2) - 1) *(2*log(2)*(4*x^2 - 20*x + 12) - 21*x + log(x)*(4*x + 8*x*log(2)) - log(x^ 2)*(4*x*log(x) - 20*x + 4*x^2 + 12) + 3*x^2 + 15))/(2*x^2*log(x)^2 - 60*x + 62*x^2 - 20*x^3 + 2*x^4 + log(x)*(12*x - 20*x^2 + 4*x^3) + 18),x)