Integrand size = 75, antiderivative size = 24 \[ \int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx=\frac {2}{\log \left (e^{-\frac {5 (3-x)}{-7+x+x^3}} x\right )} \]
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx=\frac {2}{\log \left (e^{\frac {5 (-3+x)}{-7+x+x^3}} x\right )} \]
Integrate[(-98 + 68*x - 2*x^2 - 62*x^3 + 16*x^4 - 2*x^6)/((49*x - 14*x^2 + x^3 - 14*x^4 + 2*x^5 + x^7)*Log[x/E^((15 - 5*x)/(-7 + x + x^3))]^2),x]
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.
\(\displaystyle \int \frac {-2 x^6+16 x^4-62 x^3-2 x^2+68 x-98}{\left (x^7+2 x^5-14 x^4+x^3-14 x^2+49 x\right ) \log ^2\left (e^{-\frac {15-5 x}{x^3+x-7}} x\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-2 x^6+16 x^4-62 x^3-2 x^2+68 x-98}{x \left (x^6+2 x^4-14 x^3+x^2-14 x+49\right ) \log ^2\left (e^{-\frac {15-5 x}{x^3+x-7}} x\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \frac {-2 x^6+16 x^4-62 x^3-2 x^2+68 x-98}{x \left (x^3+x-7\right )^2 \log ^2\left (e^{-\frac {15-5 x}{x^3+x-7}} x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {2 \left (-x^6+8 x^4-31 x^3-x^2+34 x-49\right )}{x \left (-x^3-x+7\right )^2 \log ^2\left (e^{-\frac {15-5 x}{x^3+x-7}} x\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {x^6-8 x^4+31 x^3+x^2-34 x+49}{x \left (-x^3-x+7\right )^2 \log ^2\left (e^{\frac {5 (3-x)}{-x^3-x+7}} x\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {x^6-8 x^4+31 x^3+x^2-34 x+49}{x \left (-x^3-x+7\right )^2 \log ^2\left (e^{\frac {5 (3-x)}{-x^3-x+7}} x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {5 \left (9 x^2+2 x-18\right )}{\left (x^3+x-7\right )^2 \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}+\frac {1}{x \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}-\frac {10}{\left (x^3+x-7\right ) \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (\int \frac {1}{x \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}dx-90 \int \frac {1}{\left (x^3+x-7\right )^2 \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}dx+10 \int \frac {x}{\left (x^3+x-7\right )^2 \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}dx-10 \int \frac {1}{\left (x^3+x-7\right ) \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}dx+45 \int \frac {x^2}{\left (x^3+x-7\right )^2 \log ^2\left (e^{\frac {5 (x-3)}{x^3+x-7}} x\right )}dx\right )\) |
Int[(-98 + 68*x - 2*x^2 - 62*x^3 + 16*x^4 - 2*x^6)/((49*x - 14*x^2 + x^3 - 14*x^4 + 2*x^5 + x^7)*Log[x/E^((15 - 5*x)/(-7 + x + x^3))]^2),x]
3.17.93.3.1 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[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 8.09 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75
method | result | size |
default | \(-\frac {-2940 x^{3}-2940 x +20580}{1470 \ln \left (x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right ) \left (x^{3}+x -7\right )}\) | \(42\) |
parallelrisch | \(-\frac {-5880 x^{3}-5880 x +41160}{2940 \ln \left (x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right ) \left (x^{3}+x -7\right )}\) | \(42\) |
risch | \(\frac {4 i}{\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )^{2}+\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{\frac {5 x -15}{x^{3}+x -7}}\right )^{3}+2 i \ln \left (x \right )-2 i \ln \left ({\mathrm e}^{-\frac {5 \left (-3+x \right )}{x^{3}+x -7}}\right )}\) | \(168\) |
int((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^2+49*x )/ln(x/exp((15-5*x)/(x^3+x-7)))^2,x,method=_RETURNVERBOSE)
Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx=\frac {2}{\log \left (x e^{\left (\frac {5 \, {\left (x - 3\right )}}{x^{3} + x - 7}\right )}\right )} \]
integrate((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^ 2+49*x)/log(x/exp((15-5*x)/(x^3+x-7)))^2,x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx=\frac {2}{\log {\left (x e^{- \frac {15 - 5 x}{x^{3} + x - 7}} \right )}} \]
integrate((-2*x**6+16*x**4-62*x**3-2*x**2+68*x-98)/(x**7+2*x**5-14*x**4+x* *3-14*x**2+49*x)/ln(x/exp((15-5*x)/(x**3+x-7)))**2,x)
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx=\frac {2 \, {\left (x^{3} + x - 7\right )}}{{\left (x^{3} + x - 7\right )} \log \left (x\right ) + 5 \, x - 15} \]
integrate((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^ 2+49*x)/log(x/exp((15-5*x)/(x^3+x-7)))^2,x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx=\frac {2 \, {\left (x^{3} + x - 7\right )}}{x^{3} \log \left (x\right ) + x \log \left (x\right ) + 5 \, x - 7 \, \log \left (x\right ) - 15} \]
integrate((-2*x^6+16*x^4-62*x^3-2*x^2+68*x-98)/(x^7+2*x^5-14*x^4+x^3-14*x^ 2+49*x)/log(x/exp((15-5*x)/(x^3+x-7)))^2,x, algorithm=\
Time = 11.18 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {-98+68 x-2 x^2-62 x^3+16 x^4-2 x^6}{\left (49 x-14 x^2+x^3-14 x^4+2 x^5+x^7\right ) \log ^2\left (e^{-\frac {15-5 x}{-7+x+x^3}} x\right )} \, dx=\frac {2\,x^3+2\,x-14}{5\,x-7\,\ln \left (x\right )+x^3\,\ln \left (x\right )+x\,\ln \left (x\right )-15} \]