Integrand size = 116, antiderivative size = 29 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=e^{1-x} \left (4+\frac {1}{x}+x \log (3)-\log ^2\left (4-9 x^2\right )\right ) \]
Time = 1.48 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=-e^{-x} \left (-4 e-\frac {e}{x}-e x \log (3)\right )-e^{1-x} \log ^2\left (4-9 x^2\right ) \]
Integrate[(E^(1 - x)*(4 + 4*x + 7*x^2 - 9*x^3 - 36*x^4 + (-4*x^2 + 4*x^3 + 9*x^4 - 9*x^5)*Log[3]) - 36*E^(1 - x)*x^3*Log[4 - 9*x^2] + E^(1 - x)*(-4* x^2 + 9*x^4)*Log[4 - 9*x^2]^2)/(-4*x^2 + 9*x^4),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.
\(\displaystyle \int \frac {e^{1-x} \left (9 x^4-4 x^2\right ) \log ^2\left (4-9 x^2\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-36 x^4-9 x^3+7 x^2+\left (-9 x^5+9 x^4+4 x^3-4 x^2\right ) \log (3)+4 x+4\right )}{9 x^4-4 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{1-x} \left (9 x^4-4 x^2\right ) \log ^2\left (4-9 x^2\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-36 x^4-9 x^3+7 x^2+\left (-9 x^5+9 x^4+4 x^3-4 x^2\right ) \log (3)+4 x+4\right )}{x^2 \left (9 x^2-4\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {9 e^{1-x} \left (9 x^5 \log (3)+36 x^4 \left (1-\frac {\log (3)}{4}\right )+9 x^3 \left (1-\frac {4 \log (3)}{9}\right )+4 x^2 \log ^2\left (4-9 x^2\right )-7 x^2 \left (1-\frac {4 \log (3)}{7}\right )-9 x^4 \log ^2\left (4-9 x^2\right )+36 x^3 \log \left (4-9 x^2\right )-4 x-4\right )}{4 \left (4-9 x^2\right )}+\frac {e^{1-x} \left (9 x^5 \log (3)+36 x^4 \left (1-\frac {\log (3)}{4}\right )+9 x^3 \left (1-\frac {4 \log (3)}{9}\right )+4 x^2 \log ^2\left (4-9 x^2\right )-7 x^2 \left (1-\frac {4 \log (3)}{7}\right )-9 x^4 \log ^2\left (4-9 x^2\right )+36 x^3 \log \left (4-9 x^2\right )-4 x-4\right )}{4 x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 6 e^{5/3} \int \frac {\operatorname {ExpIntegralEi}\left (-x-\frac {2}{3}\right )}{3 x-2}dx+6 e^{5/3} \int \frac {\operatorname {ExpIntegralEi}\left (-x-\frac {2}{3}\right )}{3 x+2}dx+6 \sqrt [3]{e} \int \frac {\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{3 x-2}dx+6 \sqrt [3]{e} \int \frac {\operatorname {ExpIntegralEi}\left (\frac {2}{3}-x\right )}{3 x+2}dx+\int e^{1-x} \log ^2\left (4-9 x^2\right )dx-2 e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-3 x-2)\right ) \log \left (4-9 x^2\right )-2 \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right ) \log \left (4-9 x^2\right )-\frac {1}{4} e^{5/3} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-3 x-2)\right )+\frac {5}{4} \sqrt [3]{e} \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right )-\frac {1}{18} e^{5/3} (9-\log (81)) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-3 x-2)\right )-\frac {1}{18} \sqrt [3]{e} (9-\log (81)) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right )-\frac {1}{12} e^{5/3} (7-\log (81)) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-3 x-2)\right )+\frac {1}{12} \sqrt [3]{e} (7-\log (81)) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right )-\frac {2}{9} e^{5/3} \log (3) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-3 x-2)\right )-\frac {2}{9} \sqrt [3]{e} \log (3) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right )+\frac {1}{3} e^{5/3} (4-\log (3)) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (-3 x-2)\right )-\frac {1}{3} \sqrt [3]{e} (4-\log (3)) \operatorname {ExpIntegralEi}\left (\frac {1}{3} (2-3 x)\right )+\frac {e^{1-x}}{x}+e^{1-x} x \log (3)-\frac {1}{4} e^{1-x} (7-\log (81))+e^{1-x} \log (3)+e^{1-x} (4-\log (3))+\frac {1}{4} e^{1-x} (7-4 \log (3))\) |
Int[(E^(1 - x)*(4 + 4*x + 7*x^2 - 9*x^3 - 36*x^4 + (-4*x^2 + 4*x^3 + 9*x^4 - 9*x^5)*Log[3]) - 36*E^(1 - x)*x^3*Log[4 - 9*x^2] + E^(1 - x)*(-4*x^2 + 9*x^4)*Log[4 - 9*x^2]^2)/(-4*x^2 + 9*x^4),x]
3.13.75.3.1 Defintions of rubi rules used
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_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 2.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41
method | result | size |
risch | \(-{\mathrm e}^{1-x} \ln \left (-9 x^{2}+4\right )^{2}+\frac {\left (x^{2} \ln \left (3\right )+4 x +1\right ) {\mathrm e}^{1-x}}{x}\) | \(41\) |
parallelrisch | \(\frac {81 \ln \left (3\right ) {\mathrm e}^{1-x} x^{2}-81 \ln \left (-9 x^{2}+4\right )^{2} x \,{\mathrm e}^{1-x}+324 x \,{\mathrm e}^{1-x}+81 \,{\mathrm e}^{1-x}}{81 x}\) | \(56\) |
int(((9*x^4-4*x^2)*exp(1-x)*ln(-9*x^2+4)^2-36*x^3*exp(1-x)*ln(-9*x^2+4)+(( -9*x^5+9*x^4+4*x^3-4*x^2)*ln(3)-36*x^4-9*x^3+7*x^2+4*x+4)*exp(1-x))/(9*x^4 -4*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=-\frac {x e^{\left (-x + 1\right )} \log \left (-9 \, x^{2} + 4\right )^{2} - {\left (x^{2} \log \left (3\right ) + 4 \, x + 1\right )} e^{\left (-x + 1\right )}}{x} \]
integrate(((9*x^4-4*x^2)*exp(1-x)*log(-9*x^2+4)^2-36*x^3*exp(1-x)*log(-9*x ^2+4)+((-9*x^5+9*x^4+4*x^3-4*x^2)*log(3)-36*x^4-9*x^3+7*x^2+4*x+4)*exp(1-x ))/(9*x^4-4*x^2),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {\left (x^{2} \log {\left (3 \right )} - x \log {\left (4 - 9 x^{2} \right )}^{2} + 4 x + 1\right ) e^{1 - x}}{x} \]
integrate(((9*x**4-4*x**2)*exp(1-x)*ln(-9*x**2+4)**2-36*x**3*exp(1-x)*ln(- 9*x**2+4)+((-9*x**5+9*x**4+4*x**3-4*x**2)*ln(3)-36*x**4-9*x**3+7*x**2+4*x+ 4)*exp(1-x))/(9*x**4-4*x**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (28) = 56\).
Time = 0.35 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.31 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {{\left (x^{2} e \log \left (3\right ) - x e \log \left (3 \, x + 2\right )^{2} - 2 \, x e \log \left (3 \, x + 2\right ) \log \left (-3 \, x + 2\right ) - x e \log \left (-3 \, x + 2\right )^{2} + 4 \, x e + e\right )} e^{\left (-x\right )}}{x} \]
integrate(((9*x^4-4*x^2)*exp(1-x)*log(-9*x^2+4)^2-36*x^3*exp(1-x)*log(-9*x ^2+4)+((-9*x^5+9*x^4+4*x^3-4*x^2)*log(3)-36*x^4-9*x^3+7*x^2+4*x+4)*exp(1-x ))/(9*x^4-4*x^2),x, algorithm=\
(x^2*e*log(3) - x*e*log(3*x + 2)^2 - 2*x*e*log(3*x + 2)*log(-3*x + 2) - x* e*log(-3*x + 2)^2 + 4*x*e + e)*e^(-x)/x
Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.76 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {x^{2} e^{\left (-x + 1\right )} \log \left (3\right ) - x e^{\left (-x + 1\right )} \log \left (-9 \, x^{2} + 4\right )^{2} + 4 \, x e^{\left (-x + 1\right )} + e^{\left (-x + 1\right )}}{x} \]
integrate(((9*x^4-4*x^2)*exp(1-x)*log(-9*x^2+4)^2-36*x^3*exp(1-x)*log(-9*x ^2+4)+((-9*x^5+9*x^4+4*x^3-4*x^2)*log(3)-36*x^4-9*x^3+7*x^2+4*x+4)*exp(1-x ))/(9*x^4-4*x^2),x, algorithm=\
Time = 11.60 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {e^{1-x} \left (4+4 x+7 x^2-9 x^3-36 x^4+\left (-4 x^2+4 x^3+9 x^4-9 x^5\right ) \log (3)\right )-36 e^{1-x} x^3 \log \left (4-9 x^2\right )+e^{1-x} \left (-4 x^2+9 x^4\right ) \log ^2\left (4-9 x^2\right )}{-4 x^2+9 x^4} \, dx=\frac {{\mathrm {e}}^{1-x}\,\left (\ln \left (3\right )\,x^2+4\,x+1\right )}{x}-{\ln \left (4-9\,x^2\right )}^2\,{\mathrm {e}}^{1-x} \]